Chapter 3(Analog).doc

Chapter 3 Amplifiers with Active Loads – CMOS Amplifiers

Section 3.1 Amplifiers with Active Loads

In the last chapter, we noticed that the load must be large. There are two problems here: (1) For IC design, this is not desirable because it is not cost effective to fabricate a desired resistor, not mentioning the fact that a large resistor will require a rather large space in the IC. (2) A large resistor may easily drive the transistor out of saturation as shown in Fig. 3.1-1.

Fig. 3.1-1 A large driving a transistor out of saturation

It will be desirable if we have a load curve, instead of a load line as shown in Fig. 3.1-2 below:

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Fig. 3.1-2 A desirable load curve

To achieve this desirable load curve, we may use an active load, instead of a passive load, such as a resistor. Let us consider the following PMOS and its I-V curve as shown in Fig. 3.1-3. Its vs relationship is shown in Fig. 3.1-4.

Fig. 3.1-3 A PMOS transistor and its I-V curve

Fig. 3.1-4 A PMOS transistor circuit with its I-V diagrams

From Fig. 3.1-4, we can see that a PMOS circuit can be used as a load for an NMOS amplifier, as shown in Fig. 3.1-5.

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Fig. 3.1-5 A CMOS transistor circuit with its I-V curves

It should be noted that both and have to be proper. In Fig. 3.1-6, we show improper ’s and in Fig. 3.1-7, we show improper ’s.

Fig. 3.1-6 Different ’s for a fixed

Fig. 3.1-7 Different ’s for a fixed

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Note that so far as Q1 is concerned, Q2 is its load and vice versa, as shown in the above figures. Since NMOS and PMOS are complementary to each other, we call this kind of circuits CMOS circuits.

For the CMOS amplifier shown in Fig. 3.1-5, let us assume that the circuit is properly biased. Fig. 3.1-8 shows the diagram of the I-V curves of Q1 and its load curve, which is the I-V curve of Q2.

. Fig. 31-8 A CMOS transistor circuit with its I-V curves of Q1 and a load curve for Q1

Let us imagine that increases. Initially, decreases rather slowly. After it reaches , it starts to drop quickly to . As can be seen, an ideal operating point

should be around . Fig. 3.1-9 shows the DC input-output diagram and

why it behaves as an amplifier..

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Fig. 3.1-9 The amplification of input signal

The small signal equivalent circuit of the CMOS amplifier is shown in Fig. 3.1-10. The impedances and are the output impedances of and respectively. For and , refer to Section 2.4.

Fig. 3.1-10 A CMOS transistor circuit and its small signal equivalent circuit

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As can be seen,

(3.1-1)

If , which is often the case, we have

(3.1-2)

If a passive load is used, . Since is much larger than which can be used, we have obtained a larger gain. By passive loads, we mean loads such as resistors, inductors and capacitors which do not require power supplies.Section 3.2 Some Experiments about CMOS Amplifiers

The following circuit shown in Fig. 3.2-1 will be used in our SPICE simulation experiments.

Fig. 3.2-1 The CMOS amplifier circuit for the Experiments in Section 3.2

Experiment 3.2-1. The I-V Curve of Q1 and the its Load Curve.

In Table 3.2-1, we display the SPICE simulation program of the experiment and in Fig. 3.2-2, we show the I-V curve of Q1 and its load curve. Note that the load curve of Q1 is the I-V curve of Q2.

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Table 3.2-1 Program of Experiment 3.2-1

simple.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod postVDD 11 0 3.3vR1 11 1 0kVSG2 11 2 0.9vV3 3 0 0v.param W1=5uM1 3 4 0 0+nch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'M2 3 2 1 1+pch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'VGS1 4 0 0.65v.DC V3 0 3.3v 0.1v.PROBE I(M2) I(M1) I(R1).end

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Fig. 3.2-2 The operating points of the circuit in Fig 3.2-1

Experiment 3.2-2 The Operating Point with the Same VGS1 and a Smaller VSG2.

In this experiment, we lowered from 0.9V to 0.8V. The program is shown in Table 3.2-2 and the resulting operating point can be seen in Fig. 3.2-3. In fact, this operating point is close to the ohmic region, which is undesirable.


simple.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod postVDD 11 0 3.3vR1 11 1 0kVSG2 11 2 0.8vV3 3 0 0v.param W1=5uM1 3 4 0 0+nch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)'

IDS

Operating point

I-V curve of Q2(load curve of Q1)

I-V curve of Q1 VDD

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Vout=VDS1

+AS='0.95u*W1' PS='2*(0.95u+W1)'M2 3 2 1 1+pch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'VGS1 4 0 0.65v.DC V3 0 3.3v 0.1v.PROBE I(M2) I(M1) I(R1).end

Fig. 3.2-3 The operating points of the amplifier circuit in Fig 3.2-1 with a smaller

Experiment 3.2-3 The Operating Point with the Same VGS1 and a Higher VSG2

In this experiment, we increased from 0.9V to 1.0V. The program is displayed in Table 3.2-3 and the result is in Fig. 3.2-4. Again, as can be seen, this new operating point is not ideal either.


simple

Vout=VDS1

IDS

I-V curve of Q2(load curve of Q1)

I-V curve of Q1

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.protect

.lib 'c:\mm0355v.l' TT

.unprotect

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.options nomod postVDD 11 0 3.3vR1 11 1 0kVSG2 11 2 1vV3 3 0 0v.param W1=5uM1 3 4 0 0+nch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'M2 3 2 1 1+pch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'VGS1 4 0 0.65v.DC V3 0 3.3v 0.1v.PROBE I(M2) I(M1) I(R1).end

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Fig. 3.2-4 The operating points of the amplifier circuit in Fig 3.2-1 with a higher

From the above experiments, we first conclude that to achieve an appropriate operating point, we must be careful in setting and . We also note that the I-V curves are not so flat as we wished. Therefore, we cannot expect a very high gain with this kind of simple CMOS circuits. As we shall learn in later chapters, the gain can be higher if we use a cascode design.

Experiment 3.2-4 The Gain with the Same Bias Voltages in Experiment 3.2-1

We used a signal with magnitude 0.001V and frequency 500kHz. The gain was found to be 30. The program is shown in Table 3.2-4 and the result is shown in Fig. 3.2-5.


simple.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod postVDD 11 0 3.3v

Vout=VDS1

IDSI-V curve of Q2(load curve of Q1)

I-V curve of Q1

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R1 11 1 0kVSG 11 2 0.9v

.param W1=5uM1 3 4 0 0+nch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'M2 3 2 1 1+pch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'VGS 4 5 0.65vVin 5 0 sin(0 0.001v 500k)

.tran 0.001us 15us

.end

Fig. 3.2-5 The gain of the CMOS amplifier for input signal with 500KHz

Vin

Vout

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Experiment 3.2-5 The Gain with the Bias Voltages of Experiment 3.2-2

In this experiment, we used the bias voltages in Experiment 3.2-2. That is, and . The program is in Table 3.2-5 and the result is in

Fig. 3.2-6. As can be seen, the gain was reduced to 20.

Table 3.2-5 The program for Experiment 3.2-5 with

3.2-5.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod post

VDD 11 0 3.3vR1 11 1 0kVSG2 11 2 0.8v.param W1=5uM1 3 4 0 0+nch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'M2 3 2 1 1+pch L=0.35u W='W1' m=1+AD='0.95u*W1' PD='2*(0.95u+W1)' +AS='0.95u*W1' PS='2*(0.95u+W1)'VGS1 4 5 0.65vVin 5 0 sin(0 0.001v 500k).tran 0.001us 15us .end

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Fig. 3.2-6 The result of the CMOS amplifier with

Section 3.3 A Desired Current Source

In a CMOS circuit, a has to be used, as shown in Fig. 3.3-1. In practice, it is not desirable to have many such power supplies all over the integrated circuit. In this section, we shall see how this can be replaced by a desired current source and a current mirror.

Fig. 3.3-1 A CMOS amplifier with

The purpose of is to produce a desired load curve of Q1 as shown in Fig. 3.3-2.

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Fig. 3.3-2 A CMOS amplifier, its I-V curves and load lines

The load curve of Q1, which corresponds to a particular I-V curve of Q2, is shown in Fig. 3.3-3. This load curve is determined by .

Fig. 3.3-3 A CMOS amplifier with a fixed and its I-V curves

It is natural for us to think that a proper is the only way to produce the desired load curve for Q1. Actually, there is another way. Note each load curve almost corresponds to a desired , as shown in Fig. 3.3-4. In other words, we may think of a way to produce a desired current in Q2, which of course is also the current in Q1.

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Fig. 3.3-4 An illustration of how a desired current determines the I-V curve

There are two problems here: (1) How can we generate a desired current? (2) How can we force Q2 to have the desired current?

To answer the first question, let us consider a typical NMOS circuit with a resistive load as shown in Fig. 3.3-5.

Fig. 3.3-5 An NMOS circuit with a resistive load

In the ohmic region, the relationship between the current and different voltages is expressed as below:

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(3.3-1)

(3.3-2)

Suppose we want to have a desired current . We may think that is a constant. But, from the above equations, we still have three variables, namely

and Since there are only two equations, we cannot find these three variables for a given desired .

In the boundary between ohmic and saturation regions where , the two equations governing current and voltages in the transistor are as follows:

(3.3-3)

and (3.3-4)

As can be seen, there are still three variables and only two equations.

There is a trick to solve the above problem. We may connect the drain to gate as shown in Fig. 3.3-6.

Fig. 3.3-6 The connection of the drain and the gate

After this is done, we have

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(3.3-5)

We have successfully eliminated one variable. Besides,

(3.3-6)

From Equation (3.3-6), we have

(3.3-7)

Thus, this connection makes sure that the transistor is in saturation region. Since it is in the saturation region, we have

(3.3-8)

and (3.3-9)

Although we often say that a transistor is in saturation if its drain is connected to its gate, we must understand it is in a very peculiar situation. Traditionally, a transistor has a family of -curves, each of which corresponds to a specified gate bias voltage and besides, the can be any value as illustrated in Fig. 3.3-2. Once the drain is connected to the gate, we note the following:

(1) We have lost because it is always equal to . Therefore, we do not have the traditional -curves any more.

(2) For each , since , we have . This transistor is in saturation. But it is rather close to the boundary between the ohmic region and the saturation region.

(3) Because of the above point, the relationship between current and voltage is the dotted line illustrated in Fig. 3.3-7.

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Fig. 3.3-7 The current in an NMOS transistor

(4) We may safely say that the transistor is no longer a transistor. It can be now viewed as a diode with only two terminals. The relationship between current and voltage is hyperbolic expressed in Equation 3.3-8.

(5) For a traditional transistor, is supplied by a bias voltage. Since there is no bias voltage, how do we determine ? Note that the desired current is related to

. This will be discussed in below.

Given a certain desired , can be determined by using Equations (3.3-8). Thus can be found by using Equation (3.3-9). We can also determine and graphically as shown in Fig. 3.3-8. This means that we can design a desired current source by using the circuit shown in Fig. 3.3-6. By adjusting the value of , we can get the desired current.

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Fig. 3.3-8 The generation of a desired current

Let us examine Fig. 3.3-6 again. We do not have to provide a bias voltage any more. This is a very desirable property which will become clear as we introduce current mirror. But, the reader should note that a does exist and it is produced.

In this section, we have discussed how to generate a desired current. In the next section, we shall show how we can force Q2 to have this desired current. This is done by the current mirror.

Section 3.4 The Current Mirror

Let us consider the circuit in Fig. 3.4-1.

Fig. 3.4-1 A current mirror

Assume Q1 and Q2 have the same Note that Q1 is in the saturation region and has a desired current in it. Assume Q2 is also in the saturation region. Since

, by using Equation (3.3-3), we have

(3.4-1)

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If and , from Equation (3.4-1), we have . Q1 is called a current mirror for Q2.

As indicated before, Q2 must be in the saturation region. So our question is: Under what condition would Q2 be out of saturation . Note that Q2 must be connected to a load. If the load is too high, this will cause it to be out of saturation as illustrated in Fig. 3.4-2. Of course, if an improper bias voltage is used, out of saturation may also occur.

Fig. 3.4-2 The out of saturation of Q2

The reader may be puzzled about one thing. We know that if an NMOS transistor is in the saturation region, its current is determined by . Is this still true in this case? Our answer is “Yes”. That is, for the circuit in Fig. 3.4-1,

is still determined by . This will be shown below.

Note that , and . Thus,

(3.4-2)

From Equation (3.4-2), we conclude that is determined by .

The advantage of using the current mirror is that no biasing voltage is needed to give a proper . There is still a . But this is equal to which is in

3-21

turn determined by . is determined by selecting a proper , as illustrated in Fig. 3.4-3.

Fig. 3.4-3 The determination of the biasing voltage in a current mirror

A current mirror can be based upon a PMOS transistor as in the CMOS amplifier case. Fig. 3.4-4 shows a CMOS amplifier with a current mirror.

Fig. 3.4-4 A PMOS current mirror

We must remember that the purpose of using a current mirror is to generate a proper I-V curve of Q2. This I-V curve serves as a load curve for Q1 as shown in Fig. 3.4-5. From Equation (3.3-8) and (3.3-9), we know that by adjusting the value of , we can obtain different current values in Q3, which mean different I-V curves in Q2. In other words, if we want a different load curve of Q1, we may

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simply change the value of .Although the above discussed circuit is usually called current mirror, it is

meaningful to call this as a gate voltage mirror. If the current mirror consists of two NMOS transistors, then because the source terminals are connected together to the ground and the gates are connected to each other. If the current mirror consists of two PMOS transistors, then because the drain terminals are connected together to and the gates are connected to each other.

Fig. 3.4-5 The obtaining of different I-V curves for an NMOS transistor through a current mirror

Section 3.5 Experiments for the CMOS Amplifiers with Current Mirrors

In this set of experiments, we used the circuit shown in Fig. 3.5-1.

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Fig. 3.5-1 The current mirror used in the experiments of Section 3.5

Experiment 3.5-1 The Operating Points of M1 and M3.

In this experiment, we like to find out whether I(M1) is equal to I(M3) or not. We first try to find the characteristics of M1. The program is shown in Table 3.5-1. We then do the same thing to M3. The program is shown in Table 3.5-2. The curves related to M1 are shown in Fig. 3.5-2. The curves related to M3 are shown in Fig. 3.5-3. Note the I-V curve of M3 is not a typical one for a transistor because the gate of M3 is connected to the drain of M3.

Table 3.5-1 Program for Experiment 3.5-1

Ex3.5-11.protect.lib 'C:\model\tsmc\MIXED035\mm0355v.l' TT.unprotect.op.options nomod post

VDD 1 0 3.3vR4 4 0 30kRdm1 1_1 0.param W1=10u W2=10u W3=10u W4=10u

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M1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1_1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

V2 2 0 0vVGS1 3 5 0.7vVin 5 0 0v

.DC V2 0 3.3v 0.1v

.PROBE I(M1) I(Rdm)

.end

Table 3.5-2 Another program for Experiment 3.5-1

Ex3.5-12.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod post

VDD 1 0 3.3vR4 4 0 30kRdm1 1_1 0

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'

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M2 2 4 1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1_1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

V3 4 0 0vVGS1 3 5 0.7vVin 5 0 0v

.DC V3 0 3.3v 0.1v

.PROBE I(R4) I(Rdm)

.end

Fig. 3.5-2 I-V curve and load curve for M1

Load Curve of M1(I-V Curve of M2)

I-V Curve of M1

Vout

IDS1

7.9x10-5

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Fig. 3.5-3 I-V curve and load line for M3 of the circuit in Fig 3.5-1

From this experiment, we conclude that I(M3)=I(M1) as expected.

Experiment 3.5-2 The Operating Point of M2

The I-V curve of M2 is the load curve of M1. The I-V curve of M2 is determined by the current mirror mechanism. We were told that the current mirror works only when M2 is in the saturation region. In this experiment, we first show the characteristics of M1. The program is shown in Table 3.5-3. The I-V curve of M2 and its load curve (M1 is the load of M2) are shown in Fig. 3.5-4.


Ex3.5-2.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod post

VDD 1 0 3.3vR4 4 0 30k

Load Line of R4

I-V Curve of M3

7.6x10-5

VDS

3-27

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1_1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

V2 2 0 0vVGS1 3 5 0.7vVin 5 0 0v

.DC V2 0 3.3v 0.1v

.PROBE I(M1) I(Rdm)Rdm1 1_1 0

.end

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Fig. 3.5-4 Operating points of M2 of the circuit in Fig 3.5-1

As shown in Fig. 3.5-4, M2 is in the saturation region.

To drive M2 out of the saturation region, we lowered from 0.7V to 0.6V. The program is shown in Table 3.5-4 and the curves are shown in Fig. 3.5-5.

Table 3.5-4 The program to drive M2 out of saturation

Ex3.5-2b.protect.lib 'c:\mm0355v.l' TT.unprotect.op.options nomod post

VDD 1 0 3.3vR4 4 0 30k

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1'

Vout

IDS2

I-V Curve of M2


3-29

+PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1_1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

V2 2 0 0vVGS1 3 5 0.6vVin 5 0 0v

.DC V2 0 3.3v 0.1v

.PROBE I(M1) I(Rdm)Rdm1 1_1 0

.end

Vout

IDS2

I-V Curve of M2


A smaller VGS1.

3-30

Fig. 3.5-5 The out of saturation of M2

From Fig. 3.5-5, we can see that M2 is now out of saturation. We then printed the essential data by using the SPICE simulation program in Table 3.5-5. We can see that I(M2) is quite different from I(M3) now. This is due to the fact that M2 is out of saturation.

Table 3.5-5 Experimental data for Experiment 3.5-2

subckt element 0:m1 0:m2 0:m3 model 0:nch.3 0:pch.3 0:pch.3 region Saturati Linear Saturati id 25.4028u -26.2672u -75.8881u ibs -3.880e-17 6.373e-18 1.832e-17 ibd -864.4522n 1.0916f 1.1504f vgs 600.0000m -1.0234 -1.0234 vds 3.2166 -83.3759m -1.0234 vbs 0. 0. 0. vth 545.8793m -719.8174m -688.8560m vdsat 85.3814m -293.2779m -318.3382m beta 6.7003m 1.3100m 1.3166m gam eff 591.1171m 485.8319m 485.8388m gm 441.4749u 97.6517u 411.0201u gds 8.7037u 268.5247u 18.5395u gmb 111.6472u 22.6467u 87.7975u cdtot 11.3338f 28.6456f 14.4251f cgtot 10.2012f 16.2693f 12.7341f cstot 21.1660f 31.1152f 30.5144f cbtot 27.1028f 38.9009f 33.8541f cgs 5.6263f 8.8368f 9.9096f cgd 2.0774f 7.2706f 1.8392f

The reader must be aware of one important thing. Even a current mirror pair transistors are both in saturation, they may have different currents. Suppose two transistors form a current mirror pair. Their currents will be equal when they have roughly the same load. To put it in another way, we may say that they will have the same current when their ’s are roughly the same. Consider the circuit in Fig. 3.5-1. Transistors M2 and M3 are a current mirror pair. We do the following two experiments :

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Experiment 3.5-3 The Current Mirror with Roughly the Same Load

We set . The program is in Table 3.5-6 and the result is in Table 3.5-7. From Table 3.5-7, we can see that . Thus the currents of M2 and M3 are roughly the same.

Table 3.5-6 Program for Experiment 3.5-3 with

.protect

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.op

.options nomod post

VDD 1 0 3.3vR4 4 0 30k.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

VGS1 3 5 0.7vVin 5 0 0v

.end

Table 3.5-7 The result of Experiment 3.5-3 with

subckt element 0:m1 0:m2 0:m3 model 0:nch.3 0:pch.3 0:pch.3

3-32

region Saturati Saturati Saturati id 79.5086u -79.5373u -75.8881u ibs -1.211e-16 1.919e-17 1.832e-17 ibd -28.6617n 5.6104f 1.1504f vgs 700.0000m -1.0234 -1.0234 vds 2.0764 -1.2236 -1.0234 vbs 0. 0. 0. vth 555.9314m -682.2505m -688.8560m vdsat 143.4528m -323.6750m -318.3382m beta 6.6571m 1.3180m 1.3166m gam eff 591.1204m 485.8393m 485.8388m gm 894.0008u 422.7631u 411.0201u gds 14.3304u 18.0766u 18.5395u gmb 229.7740u 90.3057u 87.7975u cdtot 12.1394f 13.8074f 14.4251f cgtot 13.6902f 12.7341f 12.7341f cstot 27.3788f 30.5142f 30.5144f cbtot 28.2726f 33.2362f 33.8541f cgs 10.1826f 9.9096f 9.9096f cgd 2.0774f 1.8392f 1.8392f

Experiment 3.5-4 The Current Mirror with Different Loads

In this experiment, we set . The program is the same as in Experiment 3.5-3. The result is Table 3.5-8. This higher makes sure that M2 still in saturation. But now, . A higher of a PMOS transistor represents a lower load of it Thus M2 has a smaller load than M3. Thus M2 has a larger current as can be seen In Table 3.5-8.

Table 3.5-8 Program for Experiment 3.5-4 with

subckt element 0:m1 0:m2 0:m3 model 0:nch.3 0:pch.3 0:pch.3 region Saturati Saturati Saturati id 104.3858u -104.3836u -75.8881u ibs -1.587e-16 2.512e-17 1.832e-17 ibd -13.2955f 2.2633n 1.1504f vgs 750.0000m -1.0234 -1.0234 vds 688.8019m -2.6112 -1.0234

3-33

vbs 0. 0. 0. vth 568.1324m -636.4667m -688.8560m vdsat 169.9490m -360.2912m -318.3382m beta 6.6283m 1.3277m 1.3166m gam eff 591.1219m 485.8428m 485.8388m gm 991.7799u 487.5038u 411.0201u gds 19.7567u 18.3672u 18.5395u gmb 262.4467u 104.1232u 87.7975u cdtot 14.0246f 11.0626f 14.4251f cgtot 14.0367f 12.7340f 12.7341f cstot 27.9954f 30.5126f 30.5144f cbtot 30.1844f 30.4899f 33.8541f cgs 10.6443f 9.9096f 9.9096f cgd 2.0774f 1.8392f 1.8392f

When two identical transistors form a current mirror pair are both in saturation region, according to Equation (3.4-1), they should have the same current. But, from the above experiment, we know that they may have different currents. What is wrong? Is Equation (3.4-1) wrong?

We must understand that Equation (3.4-1) is based upon one assumption: The IV curves of the transistors are absolutely flat as illustrated in Fig. 3.5-6(a). If this is indeed the case, then we can see that different loads will produce the same current. But in reality, most transistors do not have such kind of flat I-V curves. Instead, their I-V curves are as illustrated in Fig. 3.5-6(b). In such a situation, different loads will produce different currents.

3-34

Fig. 3.5-6 Two I-V curves

A better equation for the currents in a current mirror is as follows:

(3.5-1)

In the above equation, is related to the slope of the I-V curve. If the I-V curve is flat, is small and cannot be ignored if otherwise.

From Equation (3.5-1), we can see that for a current mirror pair, if is small or . Note that means that the two transistors have the same load.

In the later chapters, we shall introduce the cascoded transistors. They have very flat I-V curves.

3-35

Experiment 3.5-5 The DC Input-Output Relationship of M1.

In this experiment, we plotted versus . The program is in Table 3.5-9 and the DC input-output relationship is shown in Fig. 3.5-7.


.protect


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VDD 1 0 3.3vR4 4 0 30k

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'

VGS1 3 0 0v

.DC VGS1 0 3.3v 0.1v

.PROBE I(M1)

.end

3-36

Fig. 3.5-7 vs

Section 3.6 The Current Mirror with an Active Load

In the above sections, the current mirror has a resistive load. As we indicated before, a resistive load is not practical in VLSI design. Therefore, it can be replaced by an active load, namely a transistor. Fig. 3.6-1 shows a typical CMOS amplifier whose current mirror has an active load.

Fig. 3.6-1 A current mirror with an active load

VGS1

VDS1

3-37

In the above circuit, Q3 is a current mirror while Q4 is its load. Note that the main purpose of having a current mirror is to produce a desired basing current in Q2

which is equal to the current in Q3. To generate such a desired current, we use the I-V curve of Q3 and its load curve, which is the I-V curve of Q4. These curves are shown in Fig. 3.6-2.

Fig. 3.6-2 The determination of operating point for M4 in Fig. 3.6-1

Note that we have a desired current in our mind. So we just have to adjust such that its corresponding I-V curve intersects the I-V curve of Q3 at the proper place which gives us the desired current in Q4, which is also the current in Q3.

We indicated before that we like to use current mirrors because we do not like to have two biases as required in a CMOS circuit shown in Fig. 3.1-5. One may wonder at this point that we need two power supplies (constant voltage sources) for this current mirror circuit in the circuit shown in Fig. 3.6-1. Note that in this circuit, although there are two biases, they can be designed to be the same. Thus, actually, we only need one bias.

Besides, it will be shown in the next chapter that the current mirror actually has an entirely different function. That is, it provides a feedback in the differential amplifier which gives us a high gain.

Section 3.7 Experiments with the Current Mirror with an Active Load

In the following experiments, we used the amplifier circuit shown in Fig. 3.7-1.

3-38

Fig. 3.7-1 The current mirror circuit for experiments in Section 3.7

Experiment 3.7-1 The Operating Point of M4.

The program for this experiment is shown in Table 3.7-1. The curves are shown in Fig. 3.7-2. We like to point out again that the load curve of M4 is the I-V curve of M3. This I-V curve of M3 is hyperbola because the gate of M3 is connected to the drain of M3. The result shows that the current is 100u, a quite small value.


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.options nomod post

VDD 1 0 3.3v

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1_1 1+pch L=0.35u +W='W2' m=1 AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 3_1 1+pch L=0.35u

3-39

+W='W3' m=1 AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'M4 4 5 0 0+nch L=0.35u +W='W4' m=1 AD='0.95u*W4' PD='2*(0.95u+W4)' +AS='0.95u*W4' PS='2*(0.95u+W4)'

V4 4 0 0vVGS1 3 6 0.7vVGS4 5 0 0.7vVin 6 0 0v.DC V4 0 3.3v 0.1v.PROBE I(M4) I(Rm3)Rdm 1 1_1 0Rm3 1 3_1 0

.end

Fig. 3.7-2 Operating point of M4

The gain of this amplifier was found to be 30. The program for this testing is in

3-40

IDS4

VDS4

Table 3.7-2 and the signals are shown in Fig. 3.7-3.

Table 3.7-2 Program of the gain in Experiment 3.7-1

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.op

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VDD 1 0 3.3v

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'M2 2 4 1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'M4 4 5 0 0+nch L=0.35u +W='W4' m=1AD='0.95u*W4' PD='2*(0.95u+W4)' +AS='0.95u*W4' PS='2*(0.95u+W4)'

VGS1 3 6 0.7vVGS4 5 0 0.7vVin 6 0 sin(0v 0.01v 10Meg)

.tran 0.1ns 600ns

.end

3-41

Fig. 3.7-3 The gain of the amplifier in Experiment 3.7-1

Experiment 3.7-2 The Influence of VGS4

In this experiment, we increased from 0.7V to 0.75V. This will raise the current in M3 and consequently that of M2. The program to test the gain is shown in Table 3.7-3 and the signals are shown in Fig. 3.7-4. The gain was reduced to 6.


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.options nomod post

VDD 1 0 3.3v

.param W1=10u W2=10u W3=10u W4=10uM1 2 3 0 0+nch L=0.35u W='W1' m=1 AD='0.95u*W1' +PD='2*(0.95u+W1)' AS='0.95u*W1' PS='2*(0.95u+W1)'

3-42

M2 2 4 1 1+pch L=0.35u +W='W2' m=1AD='0.95u*W2' PD='2*(0.95u+W2)' +AS='0.95u*W2' PS='2*(0.95u+W2)'M3 4 4 1 1+pch L=0.35u +W='W3' m=1AD='0.95u*W3' PD='2*(0.95u+W3)' +AS='0.95u*W3' PS='2*(0.95u+W3)'M4 4 5 0 0+nch L=0.35u +W='W4' m=1AD='0.95u*W4' PD='2*(0.95u+W4)' +AS='0.95u*W4' PS='2*(0.95u+W4)'

VGS1 3 6 0.7vVGS4 5 0 0.75vVin 6 0 sin(0v 0.01v 10Meg)

.tran 0.1ns 600ns

.end

3-43

Fig. 3.7-4 The gain in Experiment 3.7-2

Section 3.8 A Summary of the Current Mirror Technology

To make the idea of the current mirror clear, let us make a summary of it as follows. Consider the CMOS transistor circuit as shown in Fig. 3.8-1.

Fig. 3.8-1 A CMOS transistor circuit

Suppose we have already selected a certain for Q1. This will produce an - curve as shown in Fig. 3.8-2. This can be so selected that it will produce a desired current in Q1.

Fig. 3.8-2 The curve of the NMOS transistor in Fig. 3.8-1 under the assumption that is specified

We now need to select an appropriate for Q2. This needs to produce an - curve for Q2 as shown in Fig. 3.8-3. In other words, these two - curves must match.

3-44

Fig. 3.8-3 The matching of the curves of the two transistors

A well-experienced reader will understand that is usually not equal to . Therefore, we have to use two different power supplies to bias our transistors. The current mirror technology tries to avoid the necessity of having two power supplies. Instead of thinking giving Q2 an appropriate bias voltage, we shall give it an appropriate current because we all know that a bias voltage will correspond to a current if the transistor is in the saturation region.. This appropriate current must be the desired current for Q1. Consider Fig. 3.8-4.

Fig. 3.8-4 A CMOS transistor where the gate and the drain of the PMOS transistor are connected together

Suppose Q3 is identical to Q1 in Fig. 3.8-1 and we have . Then the - curve for Q3 will be identical to that for Q1. Furthermore, since Q4 is in the

saturation region because of the connection of its drain to gate, the - curve for will be a hyperbola curve. Fig. 3.8-5 shows that we will have the desired current in

and .

3-45

Fig. 3.8-5 The current in Q3

We now connect these two circuits together to construct a current mirror as shown in Fig. 3.8-6.

Fig. 3.8-6 A complete current mirror circuit

Assuming that and are identical, we shall have a desired current in which is also the desired current for . Thus we have avoided having two different power supplies. But, we must realize that we still have given an appropriate . Note that . is created, not supplied. This can be understood by considering the - curves of shown in Fig. 3.8-7. We can now see that once we have the desired current for , we have also got the desired bias voltage for . Thus we may really call the current mirror circuit a “bias voltage mirror”.

3-46

Fig. 3.8-7 The current and in Q4

Section 3.9 A Further Discussion of Current Mirror

In the previous sections, we assumed that for a current mirror pair with identical transistors, if they are both in the saturation region, the currents will be the same. This was actually under the assumption that the curves are absolutely flat. Let us consider the current mirror pair as shown in Fig. 3.9-1. In the figure, the determination of the current in Q1 is also shown.

Fig. 3.9-1 A current mirror pair and the determination of current in

The determination of current in is shown in Fig. 3.9-2. In Fig. 3.9-2(a), we assume that the curve of is absolutely flat. Under such condition, different of loads will produce the same current. In Fig. 3.9-2(b), we assume that the curve of is not absolutely flat. As can be seen, different loads will produce different currents.

3-47

Fig. 3.9-2 The determination of current in

Different load lines determine different ’s. We therefore may say that for a current pair with identical transistors, the currents will be the same if and only if the two transistors are both in saturation region and have the same , Thus a more refined equation for the currents of a current mirror pair is as follows:

(3.9-1)

The parameter in Equation (3.9-1) reflects the slope of the curve of the transistor. If the curve is very flat, will be very small and the two currents will be almost the same.

Let us consider Fig. 3.9-3. This is a rather unusual case because the two transistors in the current mirror pair are fed with constant current sources. But it is used in the rail to rail comparator circuit introduced when we discuss operational

3-48

amplifier. In this circuit, if , by using Equation (3.9-1), we will have and vice versa. To put it in another way, if is large(small), will be

high(low).

Fig. 3.9-3 A current mirror fed with different current sources

Experiment 3.9-1 An Experiment to Confirm Equation (3.9-1)

In this experiment, we set and . The program is in Table 3.9-1 and the result is in Table 3.9-2. As shown in Table 3.9-2,

which confirms Equation (3.9-1)

3-49

Table 3.9-1 The program for Experiment 3.9-1Exp0325.protect.lib ‘c:\mm0355v.l’ TT.unprotect.op.options nomod post

VDD VDD! 0 3.3v

M1 1 1 0 0 NCH L=0.35U W=10U M=2M2 2 1 0 0 NCH L=0.35U W=10U M=2

i1 VDD! 1 1000ui2 VDD! 2 950u

.end

Table 3.9-2 The result of Experiment 3.9-1

subckt element 0:m1 0:m2 model 0:nch.3 0:nch.3 region Saturati Saturati id 1.0000m 950.0000u ibs -1.4627f -1.3916f ibd -26.6812f -26.1682f vgs 1.0310 1.0310 vds 1.0310 578.8480m vbs 0. 0. vth 565.4705m 569.4205m vdsat 343.8641m 341.7802m beta 12.9697m 12.9654m gam eff 591.1455m 591.1440m gm 3.2988m 3.1402m gds 84.2507u 172.9488u gmb 859.3150u 828.1345u

3-50

cdtot 23.7543f 25.2259f cgtot 28.4759f 28.4760f cstot 52.6396f 52.6405f cbtot 51.9305f 53.4031f cgs 21.9311f 21.9311f cgd 4.1549f 4.1549f

Section 3.10 A Rail to Rail Comparator Which Is Suitable for High and Low Inputs

In this section, we will introduce a comparator. The schematic diagram of a comparator is shown in Fig. 3.10-1

Fig. 3.10-1 The schematic diagram of a comparator

The comparator compares the inputs INP and INN. If , is high and if , is low. For a rail to rail comparator, we insist that the input voltages can be of any value. In other words, they can be very high or very low. Fig. 3.10-2 shows such a comparator.

Fig. 3.10-2 A rail to rail comparator

3-51

The basic ideas of the above circuit are as follows:

1. M8 and M12 are constant current sources while M7 and M11 determine the currents.

2. Because of the constant current sources, if I(M2) is large (small), I(M1) will be small (large). Similar argument applies to I(M10) and I(M9).

3. The following pairs of transistors are current mirrors: (M3,M4), (M5,M6), (M13,M17) and (M14,M18).

4. M17 and M18 constitute a pair of PMOS’s to have a Vout either high or low. We shall show that if , and will be high and if

, and will be low.

5. The inputs and are gate voltages of and respectively. There are the following two cases:

Case 1: At least one of and is turned on. This happens when at least one of them is high enough. In this case, among and , one of them will have larger current. Suppose , we can see that

. Because of the current pair effect, we can see that in this case, . Using the same argument, we can see if . .

Let us redraw the last stage of the circuit involving to as shown in Fig. 3.10-3. The currents of and are the same as the currents of and

respectively. We may consider and as constant current sources

feeding constant currents into the current mirror pair and . As discussed above, if , . In other words, will be high. Using the same argument, we can see that if , will be low.

3-52

Fig. 3.10-3 The last stage of the rail to rail comparator

There is an important point to be noted here. As experiments will show later, the currents in our circuits are all very small, around . As shown in Fig. 3.10-4, if

, will be very high and if , will be very low. If the current in the constant current source is not that small, we will not have this effect.

3-53

Fig. 10.3-4 The of the rail to rail comparator

In conclusion, for Case 1 where at least one of and is turned on, if , will be very high and if , will be very low.

Case 2: Both and are cutoff. This happens when both input voltages are very low. Note that the two input voltages are also input gate voltages of and

. Since both inputs are of low voltages, at least one of and will be

turned on.

Let us suppose . In this case, . Thus . But and due to the current

mirror effect. Therefore, . But is supplied by and is supplied by . Therefore, . Due to the current pair

effect, and will be high which is correct.

The same argument will give us the following conclusion: When both and are cutoff and , will be high which is correct.

Experiment 3.10-1 The Testing of a Rail to Rail Comparator

In this experiment, we tested the rail to rail comparator in Fig. 3.10-2. The program is in Table 3.10-1 and the result is in Fig. 3.10-5. As can be seen, it is indeed a rail to rail comparator. It works when both inputs are high and when both inputs are low.

Table 3.10-1 The program for the rail to rail comparator testingRail_to_Rail_Comparator.PROTECT.OPTION POST.lib "d:\model\tsmc\MIXED035\mm0355v.l" TT.UNPROTECT.op

M6 N1N322 N1N322 VSS VSS NCH L=2U W=3U M=2M3 N1N320 N1N320 VSS VSS NCH L=2U W=3U M=2M8 VVG I2U_S VDD VDD PCH L=5U W=6U M=2

3-54

M1 N1N320 INN VVG VDD PCH L=2U W=6U M=2M2 N1N322 INP VVG VDD PCH L=2U W=6U M=2M7 I2U_S I2U_S VDD VDD PCH L=5U W=6U M=2M5 N1N365 N1N322 VSS VSS NCH L=2U W=3U M=2M4 N1N362 N1N320 VSS VSS NCH L=2U W=3U M=2

M10 N1N362 INP N1N372 VSS NCH L=2U W=3U M=2M11 I2U I2U VSS VSS NCH L=5U W=5U M=2M12 N1N372 I2U VSS VSS NCH L=5U W=5U M=2M9 N1N365 INN N1N372 VSS NCH L=2U W=3U M=2M14 N1N362 N1N362 VDD VDD PCH L=2U W=6U M=2M13 N1N365 N1N365 VDD VDD PCH L=2U W=6U M=2

M17 N1N431 N1N365 VDD VDD PCH L=2U W=6U M=2M18 VOUT N1N362 VDD VDD PCH L=2U W=6U M=2M15 N1N431 N1N431 VSS VSS NCH L=2U W=3U M=2M16 VOUT N1N431 VSS VSS NCH L=2U W=3U M=2

vdd vdd gnd 1.5vss vss gnd -1.5v01 inn gnd -1.2v02 inp gnd pwl(0 -1.5 0.5m 1.5 1m -1.5)i01 i2u_s vss 2ui02 vdd i2u 2u

.tran 0.1u 1m

.probe v(VOUT)

.END

3-55

Fig. 3.10-5 The performance of the rail to rail comparator

Experiment 3.10-2 The First Analysis of the New Comparator

In this experiment, we set and . This case belongs to Case 1. Note that is turned off by which is too low. But is turned on by which is high enough Since we have , we should have to be low. The experimental result shows that it is indeed the case.

The program and the results are all in Table 3.10-2.

Table 3.10-2 Program and Results for and .****** Star-HSPICE -- 2001.4 (20011215) 16:53:22 02/01/2014 pcnt

3-56

INN

INPVOUT

INN

INP

VOUT

Copyright (C) 1985-2001 by Avant! Corporation. Unpublished-rights reserved under US copyright laws. This program is protected by law and is subject to the terms and conditions of the license agreement found in: C:\avanti\Hspice2001.4\license.txt Use of this program is your acceptance to be bound by this license agreement. Star-HSPICE is the trademark of Avant! Corporation. Input File: c:\work\test.sp lic: lic: FLEXlm:v7.2 USER:cwlu HOSTNAME:cwlu-PC HOSTID: PID:1023756 lic: Using FLEXlm license file: lic: c:\flexlm\license.dat lic: Checkout hspicewin; Encryption code: 26F03DEB9F8B lic: License/Maintenance for hspicewin will expire on 1-jan-0/2020.0 lic: NODE LOCKED DEMO license on host cwlu-PC lic: Init: read install configuration file: C:\avanti\Hspice2001.4\meta.cfg Init: hspice initialization file: C:\avanti\Hspice2001.4\hspice.ini * hspice.ini * * use ascii only for initial pc star-hspice release * .option post = 2 .op m6 n1n322 n1n322 vss vss nch l=2u w=3u m=2 m3 n1n320 n1n320 vss vss nch l=2u w=3u m=2 m8 vvg i2u_s vdd vdd pch l=5u w=6u m=2 m1 n1n320 inn vvg vdd pch l=2u w=6u m=2 m2 n1n322 inp vvg vdd pch l=2u w=6u m=2 m7 i2u_s i2u_s vdd vdd pch l=5u w=6u m=2 m5 n1n365 n1n322 vss vss nch l=2u w=3u m=2 m4 n1n362 n1n320 vss vss nch l=2u w=3u m=2 m10 n1n362 inp n1n372 vss nch l=2u w=3u m=2 m11 i2u i2u vss vss nch l=5u w=5u m=2

3-57

m12 n1n372 i2u vss vss nch l=5u w=5u m=2 m9 n1n365 inn n1n372 vss nch l=2u w=3u m=2 m14 n1n362 n1n362 vdd vdd pch l=2u w=6u m=2 m13 n1n365 n1n365 vdd vdd pch l=2u w=6u m=2 m17 n1n431 n1n365 vdd vdd pch l=2u w=6u m=2 m18 vout n1n362 vdd vdd pch l=2u w=6u m=2 m15 n1n431 n1n431 vss vss nch l=2u w=3u m=2 m16 vout n1n431 vss vss nch l=2u w=3u m=2 vdd vdd gnd 1.5 vss vss gnd -1.5 v01 inn gnd 1.2 v02 inp gnd pwl(0 -1.2 0.5m -1.2 1m -1.2) i01 i2u_s vss 2u i02 vdd i2u 2u .tran 0.1u 1m .probe v(vout) .end 1 ****** Star-HSPICE -- 2001.4 (20011215) 16:53:22 02/01/2014 pcnt ****** rail_to_rail_comparator ****** operating point information tnom= 25.000 temp= 25.000 ****** ***** operating point status is all simulation time is 0. node =voltage node =voltage node =voltage

+0:i2u =-891.2839m 0:i2u_s = 616.5880m 0:inn = 1.2000 +0:inp = -1.2000 0:n1n320 = -1.3953 0:n1n322 =-906.5721m +0:n1n362 = 1.2007 0:n1n365 = 626.2582m 0:n1n372 = 219.5969m +0:n1n431 =-850.6398m 0:vdd = 1.5000 0:vout = -1.5000 +0:vss = -1.5000 0:vvg =-113.5773m

3-58

**** voltage sources

subckt element 0:vdd 0:vss 0:v01 0:v02 volts 1.5000 -1.5000 1.2000 -1.2000 current -14.8383u 14.8383u 0. 0. power 22.2575u 22.2575u 0. 0.

total voltage source power dissipation= 44.5150u watts

***** current sources

subckt element 0:i01 0:i02 volts 2.1166 2.3913 current 2.0000u 2.0000u power -4.2332u -4.7826u

total current source power dissipation= -9.0157u watts

**** mosfets

subckt

element 0:m6 0:m3 0:m8 0:m1 0:m2 0:m7

model 0:nch.1 0:nch.1 0:pch.1 0:pch.1 0:pch.1 0:pch.1

region Saturati Cutoff Saturati Cutoff Saturati Saturati

id 2.0365u 7.9269p -2.0365u 0. -2.0365u -2.0000u

ibs -3.505e-18 -1.401e-23 4.073e-19 1.1318f 1.1318f 4.000e-19

ibd -9.2878f -9.1303f 18.3741f 1.1318f 1.1318f 1.1318f

vgs 593.4279m 104.7496m -883.4120m 1.3136 -1.0864 -883.4120m

vds 593.4279m 104.7496m -1.6136 -1.2817 -792.9949m -883.4120m

vbs 0. 0. 0. 1.6136 1.6136 0.

vth 543.2795m 543.9419m -748.1164m -1.0135 -1.0153 -749.3364m

3-59

vdsat 96.6985m 41.6315m -177.7796m -50.1839m -133.9215m -176.8155m

beta 606.5518u 609.4420u 155.7470u 349.2394u 344.2959u 155.7131u

gam eff 454.9560m 454.9563m 412.0847m 407.8679m 407.8679m 412.0847m

gm 32.9696u 233.1712p 21.0827u 0. 29.4172u 20.7946u

gds 93.0728n 3.9594p 46.4605n 0. 155.3928n 55.4900n

gmb 9.6691u 78.2604p 4.9809u 0. 3.8780u 4.9129u

cdtot 8.3255f 9.1574f 13.7487f 11.5579f 12.2445f 15.9078f

cgtot 33.0035f 18.1864f 187.9780f 30.4049f 62.5419f 187.9781f

cstot 38.7243f 9.4240f 224.9786f 13.7487f 71.2622f 224.9787f

cbtot 29.9577f 31.7010f 92.8041f 46.8543f 36.1089f 94.9632f

cgs 23.2469f 1.2667f 161.4899f 2.2143f 52.3653f 161.4900f

cgd 1.2667f 1.2667f 2.2143f 2.2143f 2.2143f 2.2143f

subckt

element 0:m5 0:m4 0:m10 0:m11 0:m12 0:m9

model 0:nch.1 0:nch.1 0:nch.1 0:nch.1 0:nch.1 0:nch.1

region Saturati Cutoff Cutoff Saturati Saturati Saturati

id 2.1777u 10.2317p 0. 2.0000u 2.0438u 2.0439u

ibs -3.748e-18 -2.691e-23 -9.2878f -3.199e-18 -3.269e-18 -9.2878f

ibd -209.3898p -23.3231f -9.2878f -14.1098f -4.1553p -9.2878f

vgs 593.4279m 104.7496m -1.4196 608.7161m 608.7161m 980.4031m

vds 2.1263 2.7007 981.0581m 608.7161m 1.7196 406.6613m

vbs 0. 0. -1.7196 0. 0. -1.7196

vth 541.1929m 540.4078m 929.3768m 537.8483m 536.8917m 930.1489m

vdsat 97.8961m 41.6315m 44.7769m 114.9326m 115.5472m 103.5167m

beta 606.6705u 609.7366u 580.1696u 402.3561u 402.3913u 575.8428u

gam eff 454.9560m 454.9563m 419.4380m 462.0946m 462.0946m 419.4378m

gm 34.6402u 300.9120p 0. 28.8390u 29.2914u 33.6203u

gds 118.4587n 1.8632p 0. 42.2651n 39.8855n 102.3707n

gmb 10.1640u 100.9692p 0. 8.6615u 8.7988u 6.1839u

cdtot 7.1737f 6.9351f 6.9351f 13.1056f 11.6350f 7.1737f

cgtot 33.0033f 18.1864f 16.9216f 144.5955f 144.5954f 37.9447f

cstot 38.7240f 9.4240f 7.3834f 163.3203f 163.3201f 43.4977f

cbtot 28.8059f 29.4787f 26.1734f 85.4179f 83.9473f 21.6511f

cgs 23.2466f 1.2667f 1.2667f 112.9988f 112.9987f 31.9434f

cgd 1.2667f 1.2667f 1.2667f 2.0919f 2.0919f 1.2667f

3-60

subckt

element 0:m14 0:m13 0:m17 0:m18 0:m15 0:m16

model 0:pch.1 0:pch.1 0:pch.1 0:pch.1 0:nch.1 0:nch.1

region Cutoff Saturati Saturati Cutoff Saturati Linear

id -27.2318p -4.2217u -4.5800u -38.1010p 4.5800u 50.1032p

ibs 5.567e-24 8.441e-19 9.157e-19 8.821e-24 -7.882e-18 -8.626e-23

ibd 1.1318f 1.1319f 10.6560p 2.1810f -9.2878f -2.894e-19

vgs -299.3450m -873.7418m -873.7418m -299.3450m 649.3602m 649.3602m

vds -299.3450m -873.7418m -2.3506 -3.0000 649.3602m 800.8438n

vbs 0. 0. 0. 0. 0. 0.

vth -755.5812m -753.3571m -747.6275m -745.1034m 543.2070m 544.0845m

vdsat -47.7563m -168.6889m -173.2276m -47.7570m 132.4373m 131.8442m

beta 379.0230u 371.9200u 372.2863u 380.2017u 604.5423u 604.4957u

gam eff 405.5279m 405.5279m 405.5279m 405.5279m 454.9557m 454.9563m

gm 757.9553p 46.7282u 49.6449u 1.0583n 58.5379u 398.8819p

gds 4.4157p 277.6103n 227.4293n 4.4269p 166.2485n 62.6001u

gmb 212.0479p 10.6467u 11.3100u 294.2883p 17.0758u 119.8347p

cdtot 18.8069f 15.9447f 12.3327f 11.4275f 8.2566f 51.5858f

cgtot 30.4154f 74.7540f 74.7538f 30.4154f 39.9154f 53.8147f

cstot 21.1724f 98.3694f 98.3690f 21.1724f 50.4498f 54.2498f

cbtot 61.5375f 56.4639f 52.8519f 54.1581f 30.4686f 32.3148f

cgs 2.2143f 62.8709f 62.8706f 2.2143f 32.0807f 25.4730f

cgd 2.2143f 2.2143f 2.2143f 2.2143f 1.2667f 25.4728f

Opening plot unit= 79

file=c:\work\test.tr0

***** job concluded

1 ****** Star-HSPICE -- 2001.4 (20011215) 16:53:22 02/01/2014 pcnt

******

rail_to_rail_comparator

****** job statistics summary tnom= 25.000 temp= 25.000

3-61

******

total memory used 475 kbytes

# nodes = 51 # elements= 24 # diodes= 0 # bjts = 0 # jfets = 0 # mosfets = 18

analysis time # points tot. iter conv.iter

op point 0.02 1 29 transient 0.18 10001 4006 2003 rev= 0 readin 0.48 errchk 0.03 setup 0.01 output 0.00 total cpu time 0.75 seconds job started at 16:53:22 02/01/2014 job ended at 16:53:23 02/01/2014

lic: Release hspicewin token(s)

For the convenience of discussion, we displayed the new comparator circuit below.

From Table 3.10-2, we can see that is turned off and is turned on. We can also note that is turned off and is turned on. . That is turned on

3-62

will cause to be on and to have a reasonable gate voltage equal to 0.649V. But, since is turned off and is also turned off, will be very low as illustrated in Fig. 3.10-4. This is proved to be true. From Table 3.10-2, we can see that which is very low.

Experiment 3.10-3 The Second Analysis of the New Comparator with M9 and M10 both on.

In the above experiment, M9 is on and M10 is cut off. There is current in M15 and no current in M16. In this experiment, both M9 and M10 are turned on. Besides, there will be currents in both M15 and M16. But they are not equal. Note that M15 and M16 form a current mirror pair. So, we have to figure out the value of the output voltage.

In this experiment, we set and We can see that both M9 and M10 will be turned on. Since , we expect that to be high. The program is in Table 3.10-3 and the result is in Table 3.10-4.

Table 3.10-3 Program for Experiment 3.10-3Rail_to_Rail_Comparator

.PROTECT

.OPTION POST

.lib "d:\model\tsmc\MIXED035\mm0355v.l" TT

.UNPROTECT

.op

M6 N1N322 N1N322 VSS VSS NCH L=2U W=3U M=2


M8 VVG I2U_S VDD VDD PCH L=5U W=6U M=2

M1 N1N320 INN VVG VDD PCH L=2U W=6U M=2

M2 N1N322 INP VVG VDD PCH L=2U W=6U M=2

M7 I2U_S I2U_S VDD VDD PCH L=5U W=6U M=2



M10 N1N362 INP N1N372 VSS NCH L=2U W=3U M=2

M11 I2U I2U VSS VSS NCH L=5U W=5U M=2

M12 N1N372 I2U VSS VSS NCH L=5U W=5U M=2

M9 N1N365 INN N1N372 VSS NCH L=2U W=3U M=2

3-63

M14 N1N362 N1N362 VDD VDD PCH L=2U W=6U M=2



M18 VOUT N1N362 VDD VDD PCH L=2U W=6U M=2


M16 VOUT N1N431 VSS VSS NCH L=2U W=3U M=2

vdd vdd gnd 1.5

vss vss gnd -1.5

v01 inn gnd 1.23

v02 inp gnd pwl(0 1.25 0.5m 1.25 1m 1.25)

i01 i2u_s vss 2u

i02 vdd i2u 2u

.tran 0.1u 1m

.probe v(VOUT)

.END

Table 3.10-4 Result of Experiment 3.10-3subckt

element 0:m6 0:m3 0:m8 0:m1 0:m2 0:m7


region Cutoff Cutoff Linear Cutoff Cutoff Saturati

id 20.3665p 27.3597p -37.4396p -16.5393p -9.4663p -2.0000u

ibs -3.553e-23 -4.761e-23 7.488e-24 7.885e-20 7.885e-20 4.000e-19

ibd -9.2420f -9.2568f 7.884e-20 1.4467f 1.3168f 1.1318f

vgs 136.4823m 146.4637m -883.4120m -269.9982m -249.9982m -883.4120m

vds 136.4823m 146.4637m -1.7898u -2.8535 -2.8635 -883.4120m

vbs 0. 0. 0. 1.7898u 1.7898u 0.

vth 543.8987m 543.8851m -750.8104m -745.6720m -745.6333m -749.3364m

vdsat 41.6323m 41.6328m -175.6604m -47.7549m -47.7542m -176.8155m

beta 609.4455u 609.4467u 155.6720u 380.1378u 380.1422u 155.7131u

gam eff 454.9563m 454.9563m 412.0847m 405.5279m 405.5279m 412.0847m

gm 597.8749p 802.4829p 232.4841p 461.8119p 264.9798p 20.7946u

gds 5.0595p 5.8219p 20.9219u 1.9351p 1.1099p 55.4900n

3-64

gmb 198.2757p 265.1281p 57.3248p 131.2395p 76.4010p 4.9129u

cdtot 9.0853f 9.0633f 232.7812f 11.6113f 11.5984f 15.9078f

cgtot 17.9555f 17.8851f 261.0278f 30.9014f 31.2499f 187.9781f

cstot 9.4240f 9.4240f 244.5662f 21.1723f 21.1723f 224.9787f

cbtot 31.3980f 31.3057f 102.9165f 54.8279f 55.1635f 94.9632f

cgs 1.2667f 1.2667f 126.5434f 2.2143f 2.2143f 161.4900f

cgd 1.2667f 1.2667f 126.5421f 2.2143f 2.2143f 2.2143f

subckt

element 0:m5 0:m4 0:m10 0:m11 0:m12 0:m9


region Cutoff Cutoff Saturati Saturati Saturati Saturati

id 24.4179p 32.6435p 1.2150u 2.0000u 2.0466u 831.6196n

ibs -4.977e-23 -6.385e-23 -9.2878f -3.199e-18 -3.274e-18 -9.2878f

ibd -14.9904f -16.0858f -9.2878f -14.1098f -7.2861p -9.2878f

vgs 136.4823m 146.4637m 962.8506m 608.7161m 608.7161m 942.8506m

vds 2.2448 2.2207 433.5150m 608.7161m 1.7871 457.6929m

vbs 0. 0. -1.7871 0. 0. -1.7871

vth 541.0283m 541.0612m 942.3986m 537.8483m 536.8336m 942.3659m

vdsat 41.6324m 41.6329m 86.5622m 114.9326m 115.5846m 76.9069m

beta 609.6850u 609.6822u 576.0724u 402.3561u 402.3935u 576.7799u

gam eff 454.9563m 454.9563m 418.3443m 462.0946m 462.0946m 418.3443m

gm 716.6385p 957.2058p 22.3626u 28.8390u 29.3188u 16.3132u

gds 2.5381p 3.3164p 62.7341n 42.2651n 40.5671n 44.3415n

gmb 237.6224p 316.1995p 4.0704u 8.6615u 8.8071u 2.9704u

cdtot 7.1196f 7.1304f 7.1304f 13.1056f 11.5748f 7.1196f

cgtot 17.9555f 17.8851f 34.3782f 144.5955f 144.5954f 30.3227f

cstot 9.4240f 9.4240f 38.5898f 163.3203f 163.3201f 33.0862f

cbtot 29.4324f 29.3728f 21.3791f 85.4179f 83.8871f 21.2625f

cgs 1.2667f 1.2667f 27.8610f 112.9988f 112.9987f 23.1740f

cgd 1.2667f 1.2667f 1.2667f 2.0919f 2.0919f 1.2667f

subckt

element 0:m14 0:m13 0:m17 0:m18 0:m15 0:m16


3-65

region Saturati Saturati Saturati Saturati Saturati Saturati

id -1.2150u -831.6554n -944.0701n -1.1144u 944.0821n 1.1122u

ibs 2.430e-19 1.663e-19 1.888e-19 2.229e-19 -1.625e-18 -1.915e-18

ibd 1.1318f 1.1318f 4.4359p 1.1271f -9.2878f -2.2270n

vgs -779.3356m -755.1576m -755.1576m -779.3356m 549.7193m 549.7193m

vds -779.3356m -755.1576m -2.4503 -140.7207m 549.7193m 2.8593

vbs 0. 0. 0. 0. 0. 0.

vth -753.7202m -753.8136m -747.2371m -756.1978m 543.3375m 540.1934m

vdsat -103.8531m -91.3506m -94.5463m -102.4894m 74.6185m 75.9963m

beta 375.8266u 376.5760u 377.1212u 375.6311u 607.7514u 607.9492u

gam eff 405.5279m 405.5279m 405.5279m 405.5279m 454.9562m 454.9561m

gm 18.3972u 13.3282u 14.9116u 16.6440u 17.7910u 20.4097u

gds 100.5750n 72.0932n 65.2785n 822.9689n 51.2357n 209.3923n

gmb 4.2936u 3.1335u 3.5027u 3.8885u 5.2414u 6.0184u

cdtot 16.3192f 16.4198f 12.1772f 19.9484f 8.3821f 6.8781f

cgtot 45.1908f 25.3103f 25.3103f 45.1912f 18.6641f 18.6635f

cstot 52.2239f 21.1723f 21.1723f 52.2245f 14.3984f 14.3975f

cbtot 55.1472f 54.0453f 49.8026f 58.7765f 28.7367f 27.2327f

cgs 26.5665f 2.2143f 2.2143f 26.5670f 4.9948f 4.9941f

cgd 2.2143f 2.2143f 2.2143f 2.2143f 1.2667f 1.2667f

Again we display the comparator circuit in below for the convenience of discussion.

Since both and are quite high, they turn on both M9 and M10. As seen in Table 3.10-4, both M9 and M10 are in saturation condition. This makes both M13

3-66

and M14 in saturation condition. Since and , . Therefore, . Because of the current pair effect, . This makes . But, M15 and M16 are a current mirror pair. . This further means that M16 will have a large . From Table 3.10-4. we can see that . Thus which is high and correct because .

Experiment 3.10-4 The Third Experiment of the New Comparator with Both M9 and M10 both off.

In this experiment, and . As can be seen, both input voltages are low and both and are turned off. Yet . Therefore, we expect to be low.

The program is in Table 3.10-5 and the result is in Table 3.10-6.

Table 3.10-5 Program for Experiment 3.10-4 Rail_to_Rail_Comparator

.PROTECT

.OPTION POST

.lib “d:\model\tsmc\MIXED035\mm0355v.l” TT

.UNPROTECT

.op



M8 VVG I2U_S VDD VDD PCH L=5U W=6U M=2

M1 N1N320 INN VVG VDD PCH L=2U W=6U M=2

M2 N1N322 INP VVG VDD PCH L=2U W=6U M=2

M7 I2U_S I2U_S VDD VDD PCH L=5U W=6U M=2



M10 N1N362 INP N1N372 VSS NCH L=2U W=3U M=2

M11 I2U I2U VSS VSS NCH L=5U W=5U M=2

M12 N1N372 I2U VSS VSS NCH L=5U W=5U M=2

M9 N1N365 INN N1N372 VSS NCH L=2U W=3U M=2

3-67




M18 VOUT N1N362 VDD VDD PCH L=2U W=6U M=2


M16 VOUT N1N431 VSS VSS NCH L=2U W=3U M=2

vdd vdd gnd 1.5

vss vss gnd -1.5

v01 inn gnd -1.2

v02 inp gnd pwl(0 -1.25 0.5m -1.25 1m -1.25)

i01 i2u_s vss 2u

i02 vdd i2u 2u

.tran 0.1u 1m

.probe v(VOUT)

.END

Table 3.10-6 The Result of Experiment 3.10-4subckt

element 0:m6 0:m3 0:m8 0:m1 0:m2 0:m7


region Saturati Cutoff Saturati Cutoff Saturati Saturati

id 1.4269u 612.5813n -2.0395u -612.5769n -1.4269u -2.0000u

ibs -2.456e-18 -1.055e-18 4.079e-19 1.1318f 1.1318f 4.000e-19

ibd -9.2878f -9.2878f 32.0797f 1.1318f 1.1318f 1.1318f

vgs 572.3731m 527.5418m -883.4120m -1.0216 -1.0716 -883.4120m

vds 572.3731m 527.5418m -1.6784 -794.0722m -749.2409m -883.4120m

vbs 0. 0. 0. 1.6784 1.6784 0.

vth 543.3073m 543.3672m -748.0081m -1.0235 -1.0237 -749.3364m

vdsat 85.3051m 65.8550m -177.8653m -88.3247m -117.3678m -176.8155m

beta 607.1749u 608.2183u 155.7500u 345.9420u 344.2817u 155.7131u

gam eff 454.9561m 454.9562m 412.0847m 407.9415m 407.9415m 412.0847m

gm 24.9772u 12.2686u 21.1053u 11.0512u 22.5051u 20.7946u

gds 71.0035n 35.9353n 46.0825n 54.0013n 116.7956n 55.4900n

3-68

gmb 7.3410u 3.6233u 4.9862u 1.4621u 2.9397u 4.9129u

cdtot 8.3525f 8.4118f 13.6010f 12.1435f 12.2119f 15.9078f

cgtot 27.6282f 15.8023f 187.9780f 18.0203f 52.7873f 187.9781f

cstot 29.6055f 9.4240f 224.9786f 13.6010f 58.5044f 224.9787f

cbtot 29.5114f 28.5713f 92.6564f 34.9076f 35.5964f 94.9632f

cgs 16.3993f 1.2667f 161.4899f 2.2143f 41.4400f 161.4900f

cgd 1.2667f 1.2667f 2.2143f 2.2143f 2.2143f 2.2143f

subckt

element 0:m5 0:m4 0:m10 0:m11 0:m12 0:m9


region Saturati Cutoff Cutoff Saturati Linear Cutoff

id 1.5441u 676.7954n 662.1709p 2.0000u 3.3861n 2.7150n

ibs -2.658e-18 -1.165e-18 -3.861e-17 -3.199e-18 -5.417e-21 -3.861e-17

ibd -238.2998p -149.0853p -173.1472f -14.1098f -5.864e-17 -531.4258f

vgs 572.3731m 527.5418m 249.8930m 608.7161m 608.7161m 299.8930m

vds 2.2044 2.2574 2.2573 608.7161m 107.0099u 2.2043

vbs 0. 0. -107.0099u 0. 0. -107.0099u

vth 541.0855m 541.0122m 541.0432m 537.8483m 538.3706m 541.1153m

vdsat 86.4417m 66.7030m 41.6697m 114.9326m 114.6016m 41.7883m

beta 607.3074u 608.3750u 609.6824u 402.3561u 402.3366u 609.6703u

gam eff 454.9561m 454.9562m 454.9533m 462.0946m 462.0947m 454.9533m

gm 26.5428u 13.3367u 18.9907n 28.8390u 31.5634n 75.7517n

gds 99.3345n 53.4127n 69.0381p 42.2651n 31.6327u 266.2652p

gmb 7.8053u 3.9412u 6.0328n 8.6615u 9.7016n 23.6264n

cdtot 7.1377f 7.1141f 7.1141f 13.1056f 167.1387f 7.1377f

cgtot 27.6280f 15.8023f 17.2131f 144.5955f 194.5720f 16.9216f

cstot 29.6051f 9.4240f 9.4237f 163.3203f 176.9682f 9.4237f

cbtot 28.2966f 27.2736f 28.6841f 85.4179f 89.6365f 28.4163f

cgs 16.3990f 1.2667f 1.2667f 112.9988f 89.3943f 1.2667f

cgd 1.2667f 1.2667f 1.2667f 2.0919f 89.2912f 1.2667f

subckt

element 0:m14 0:m13 0:m17 0:m18 0:m15 0:m16


region Cutoff Saturati Saturati Cutoff Saturati Linear

id -677.6219n -1.5471u -1.7287u -800.9205n 1.7287u 800.9508n

3-69

ibs 1.355e-19 3.094e-19 3.457e-19 1.602e-19 -2.976e-18 -1.379e-18

ibd 1.1318f 1.1318f 6.5751p 18.5384p -9.2878f -6.2942f

vgs -742.6043m -795.5597m -795.5597m -742.6043m 583.5252m 583.5252m

vds -742.6043m -795.5597m -2.4165 -2.9709 583.5252m 29.0923m

vbs 0. 0. 0. 0. 0. 0.

vth -753.8622m -753.6576m -747.3691m -745.2171m 543.2925m 544.0460m

vdsat -85.6282m -113.3005m -117.1631m -89.4961m 91.1744m 90.7688m

beta 376.9182u 375.2592u 375.7309u 377.6553u 606.8550u 606.8108u

gam eff 405.5279m 405.5279m 405.5279m 405.5279m 454.9561m 454.9562m

gm 11.1316u 22.4021u 24.5703u 12.9311u 29.0643u 10.2253u

gds 59.9847n 123.6645n 108.5950n 55.6181n 82.2642n 22.0000u

gmb 2.6268u 5.2030u 5.7022u 3.0483u 8.5324u 3.0453u

cdtot 16.4729f 16.2527f 12.2292f 11.4632f 8.3381f 17.5648f

cgtot 25.4104f 54.8434f 54.8429f 25.4104f 30.7551f 34.5483f

cstot 21.1723f 67.2908f 67.2900f 21.1723f 34.9100f 34.8575f

cbtot 54.1985f 55.6586f 51.6350f 49.1887f 29.7736f 30.9459f

cgs 2.2143f 38.3946f 38.3940f 2.2143f 20.3813f 20.4770f

cgd 2.2143f 2.2143f 2.2143f 2.2143f 1.2667f 5.9859f

Again we display the comparator circuit in below for the convenience of discussion.

Since and , they are both low and thus both M9 and M10 are turned off. Note that and both M1 and M2 are PMOS transistors, one of them will be turned on. From Table 3.10-6, we can see that M2 is

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turned on and M1 is turned off. Since M2 is on, M6 has current which makes M5 have current. The current of M5 is supplied by M13. Therefore, M13 has current while M14 is cutoff. This makes M17 have current and M18 cutoff. Thus, M15 has current and M16 has no current. Since M15 and M16 are a current mirror pair, M16 is forced to be in linear mode and which is very small. Finally,

which is low as expected because .

Constant Current Source

In the comparator, we need two constant current sources, one from a PMOS transistor and one from an NMOS transistor. Fig. 3.10-7 presents such a circuit. The currents can be identical if the loads are identical or the IV-curves are absolutely flat.

Fig.3.10-7 A constant current source which produces currents flowing out from a PMOS transistor an current flowing out from an NMOS transistor

The circuit in Fig. 3.10-7 is easy to understand. We have to understand that MP1 is load of MN2. The circuit works only when the IV-curve of MN2 matches with the IV-curve of MP1 as shown in Fig. 3.10-8. If the IV-curve of MP1 is too low or too high, the circuit will not work as I1 will not be equal to I2.

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Fig. 3.10-8 A good matching of the IV curves of MN2 and MP1

Note that there is no bias voltage in the circuit. The only parameter which we can use to adjust the location of an IV curve is the parameter W/L of the gate. A smaller W/L of MP1 will lower its IV-curve. It is easy to see that if the IV-curve of MP1 is too low, MN2 will be in the triode region. The circuit will not work in this case. If W/L of the gate of MP1 is large, it will not be ideal. But the situation will not be too bad.

Experiment 3-10-5 A Testing of of the gate of MP1 in Fig. 3.10-7.

In the first test, for NMOs, W=5u and L=0.35u. For PMOS, W=17.4u and L=0.35u. The program is in Table 3.10-7.

.Table 3.10-7 The program for Experiment 3.10-5protect.lib 'mm0355v.l' TT.unprotect.op.options nomod post

**source**VDD VDD 0 1.8R VDD N1 10kMN1 N1 N1 0 0 nch W=5u L=0.35uMN2 P1 N1 0 0 nch W=5u L=0.35uMP1 P1 P1 VDD VDD pch W=17.4u L=0.35uMNO N N1 0 0 nch W=5u L=0.35uMPO P P1 VDD VDD pch W=17.4u L=0.35uVP P 0 829.5532mVN N 0 829.5643m.print I(MNO)

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.end

Table 3.10-8 shows the result. As seen in Table 3.10-8, the circuit works perfectly.

Table 3.10-8 The result of W equal to 17.4u for PM1

We then change W=0.5u for PM1. The result is shown in Table 3.10-9. We can that the circuit is not working at all. Ip is not equal to IN

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Table 3.10-9 The result of W equal to 0.5u for PM1

We then enlarge W to 100u. The result is shown in Table 3.10-10. The circuit is not working. But the situation is not that bad.

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Table 3.10-10 The result of W equal to 100u for PM1

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Experiment 3.10-11 An Experiment to Show the Flatness of the IV-curve of M16

As we indicated before, the rail to rail comparator works because of one important characteristics of the circuit: The current must be small and as indicated in Fig. 3.10-4, the IV-curve of M16 must be quite flat. This experiment confirms this. The program is shown in Table.3.10-11 and the result is in Fig. 3.10-9. We can see that the IV-curve is indeed very flat.. It changes only for which is very small. The current is so small because the length of the transistor is very long, .

Table 3.10-11 The program for Experiment 3.10-60505.protect.lib ‘C:\mm0355v.l’ TT.unprotect.op.options nomod post

VDD 1 0 3.3vR1 1 2 280k

M1 2 3 0 0 NCH L=2U W=3U M=2

V2 2 0 0vVGS1 3 0 0.549v

.DC V2 0 3.3v 0.1v

.PROBE I(R1) I(M1)

.end

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Fig. 3.10-9 An IV-curve of M16

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Chapter 3(Analog).doc