Opamps Requirements

ee101b: Op-amps 22 – 1

Lecture 22

ee101b: Op-amps

EE101BDepartment of Electrical Engineering Prof. M. HershensonStanford University Prof. K.V. Shenoy

• motivation

• ideal op-amp circuits

– the golden rules

– the inverting amplifier

– the non-inverting amplifier

– a unity-gain buffer

• real op-amp circuits

– the inverting amplifier

– the non-inverting amplifier

• final remarks


Reading for this lecture

Reading

• Sedra & Smith. Sections 2.1, 2.2, 2.3, 2.4, 2.5

• Sedra & Smith. Section 2.7

Practice problems

• Sedra & Smith. Example 2.1




Optional (advanced) reading

• Sedra & Smith. Section 8.4 (ideal case)

• Sedra & Smith. Section 8.6 (shunt-shunt)


Motivation

Why do we care about op-amps?

• They are a fundamental block of analog circuits.

• Op-amps are widely used in electronics. They areused to build amplifiers, filters, rectifiers,data-acquisition systems, oscillators, . . .

• Also an important block of mixed-mode IC’s(switched capacitor filters, comparators . . . )

• There are many types of op-amps: instrumentation,high-speed, high-voltage, power . . .


The ideal op-amp

-

+v+

v−

vout

Av(v+ − v−)

(a)

+-

+

-

vid

vout

Avvid

(b)

They are very high gain DC-coupled differentialamplifiers

vout = Av(v+ − v−)

• Infinite differential voltage gain (Av = ∞)

• Zero common-mode voltage gain (Acm = 0)

• Infinite input resistance (Rin = ∞)

• Zero output resistance (Rout = 0)

• Infinite output current capability (can sink or sourceinfinite current)

• Infinite output voltage range (output voltage−∞ ≤ vout ≤ ∞)

• Infinite open-loop bandwidth (the op-amp works upto ∞ frequency)


Negative feedback aroundop-amps

• We assume that there is some sort of negativefeedback applied to the op-amp.

• By negative feedback we mean that the outputsignal is fed-back to the input in such a way that itcancels some of the input signal. Typically this willmean that there is some connection from theoutput node to the negative input. This connectioncannot involve any inversions and must work at DC.


Golden rules

Since the voltage gain is very high (can considerinfinite)and assuming the op-amp has some sort ofnegative feedback around it,

• GR 1: Both input terminals are at the samepotential (v+ = v−)

Since the input resistance is infinite,

• GR 2:Input currents are zero (i+ = i− = 0)

Some comments

• You only need these two golden rules plusKirchoff’s laws to analyze any ideal op-amp circuit.

• In this lecture we will analyze two op-amp circuits(although there are many more!). We will find thevoltage gain (Av), the input resistance (Rin) andthe output resistance (Rout).


The inverting amplifier: gain

+

R1

R2

i−i1

i2

vinvout

Voltage Gain

• GR 1: v+ = v−.

v+ = 0 ⇒ v− = 0.

We say that v− is at virtual ground because it is atzero potential. But v− is not really connected toground since there is no direct DC path from v− toground.

• GR 2: i+ = i− = 0. KCL at v−,

i1 = i2 + i− ⇒ i1 = i2

• Ohm’s law across R1

i1 =vin

R1


−vout = i2R2 =vin

R1R2

so the voltage gain Av = vout

vinis

Av = −R2

R1


The inverting amplifier: Rout

+

R1

R2

i−i1

i2

itest

vout

Output resistance

• GR 1: v+ = v−.

v+ = 0 ⇒ v− = 0.


v− = i1R1 = 0 ⇒ i1 = 0

• GR 2: i+ = i− = 0. KCL at v−,

i1 = i2 + i− ⇒ i2 = 0


−vout = i2R2 = 0

so the output resistance Rout = vout

itestis

Rout = 0


The inverting amplifier: Rin

+

R1

R2

i−i1

i2

vinvout

Input resistanceThe input resistance is the resistance seen by thevoltage source vin,

Rin =vin

i1= R1

Summary inverting amplifier

• Gain is Av = −R2

R1. Gain is negative (that is why it

is called inverting amplifier).

• Rin = R1. In general, it is much smaller than theinput resistance of a real op-amp.

• Rout = 0.

ee101b: Op-amps 22 – 10

The inverting amplifier

(Real) design problem: Choosing resistor values tobuild an inverting amplifier with gain Av = −10.

1. One option is to choose R1 = 1MΩ andR2 = 10MΩ

+

1MΩ

10MΩ

i−i1

i2

vinvout

This choice of resistors would provide a gain of -10.However, the resistors are very large. We generallytry to stay away from using very large resistors:they are noisy (noise proportional to R), in an ICenvironment they occupy a lot of area (area meansmoney) and they limit the circuit frequencyresponse (we will see this later).

2. Another option is to choose R1 = 10Ω andR2 = 100Ω

+

10Ω

100Ω

i−i1

i2

vinvout

This translates into an input resistance of 10Ω.What this means is that this amplifier will besinking a lot of current from its input.

ee101b: Op-amps 22 – 11

Conclusion: You want resistors that are large enough(so not a lot of current is consumed, so impedancelevels are high enough) but you don’t want them toolarge (noisy).

ee101b: Op-amps 22 – 12

The non-inverting amplifier:gain

+

R1

R2

i−

i+

i1

i2

vinvout

Voltage Gain

• GR 1: v+ = v−.

v+ = vin ⇒ v− = vin.

• GR 2: i+ = i− = 0. KCL at v−,

i2 = i1 + i− ⇒ i1 = i2


i1 =vin

R1

• KVL from output node to ground

vout = i2R2 + v− =vin

R1R2 + vin

so the voltage gain Av = vout

vinis

Av = 1 +R2

R1

ee101b: Op-amps 22 – 13

The non-inverting amplifier:Rin and Rout

Input resistanceThe resistance seen by the voltage source vin,

Rin =vin

i+=

vin

0= ∞

Output resistance

+

R1

R2i−

i+

i1i2 itest

vout

• GR 1 v+ = v− = 0.

• Ohm’s law across R1, i1 = v−R1

= 0

• GR 2 i+ = i− = 0. KCL at v−, i2 = i1 + i− = 0

• KVL from output node to ground,vout = i2R2 + v− = 0

so the the output resistance Rout = vout

itestis

Rout = 0

ee101b: Op-amps 22 – 14

The non-inverting amplifier

Summary

• Gain is Av = 1 + R2

R1. Gain is positive (that is why it

is called non-inverting amplifier).

• Rin = ∞. (as large as the input resistance of anideal op-amp).

• Rout = 0.

Note that the non-inverting amplifier has a much higherinput resistance than the inverting amplifier.

ee101b: Op-amps 22 – 15

Voltage follower (unity-gainbuffer)

(A special case of the non-inverting amplifier withR1 = ∞ and R2 = 0)

+

vinvout

The voltage gain can be directly found from the firstgolden rule

v− = v+ ⇒ vout = vin Av = 1

Why is a circuit with unity gain useful?

• This op-amp circuit has Av = 1, Rin = ∞ andRout = 0. Therefore it maintains the signal level(i.e., the input and output signal are at the samevalue) and it serves as an impedancetransformation circuit.

ee101b: Op-amps 22 – 16

Voltage follower (unity-gainbuffer)

vin

vout

10kΩ

1kΩ

Assume we have a voltage source with source resistanceRs = 10kΩ and we want to drive a load of valueRL = 1kΩ. The output signal will be

vout =RL

Rs + RLvin = 0.09vin

This means we will loose most of the signal at thesource resistance.

Let’s now use a buffer and see what happens

+

vinvout

10kΩ

1kΩ

• Since i+ = 0, we have v+ = vin

• Since v+ = v=, we have v− = vout = vin

Note that all the signal is delivered to the load. (Inpractice, because of the nonidealities of the op-ampsome signal is lost)

ee101b: Op-amps 22 – 17

A more real operationalamplifier

We now consider the op-amp that has a finitedifferential gain that is also frequency dependent.The transfer function is given by

A =vout

v+ − v−=

Av

1 + sω3dB

(this is a good model, most compensated op-amps looklike one pole systems for frequencies bellow unity-gainbandwidth)

• it is a DC-coupled amplifier

• differential gain at midband is Av

• high 3dB frequency is ωH = Avω3dB

• unity gain bandwidth is Avω3dB

• In this lecture we will analyze the op-amp in afeedback configuration.

• We still assume a zero common-mode gain, aninfinite op-amp input impedance and zero outputimpedance.

• Due to the lack of time, we do not considerfrequency dependency of the gain

ee101b: Op-amps 22 – 18

Non-inverting amplifier

Now consider the same amplifier with some negativefeedback around it connected in the non-invertingamplifier configuration that we studied in the previousslides.

+

R1

R2

vinvoutve

vf

• The error signal

ve = vin − vf =vout

Av(1)

• If we assume that the input resistance is infinite, wehave

vf = voutR1

R1 + R2(2)

• substituting (2) in (1), we have

G =vout

vin=

Av

1 + AvR1

R1+R2

ee101b: Op-amps 22 – 19

• note that if A → ∞G ≈ R1 + R2

R1= Gideal

And a block diagram is

A

R1

R1+R2

voutvin

Note that we can write the closed-loop gain as,

G =R1

R1 + R2

T

1 + T

where T , the loop gain is

T = af = AvR1 + R2

R1

ee101b: Op-amps 22 – 20

Inverting amplifier

+

R1

R2

i−i1

i2

vinvout

For an ideal amplifier, the gain is given by

Gideal = −R2

R1

A first idea, could be to represent the inverting amplifierwith the following block diagram,

A

R2

R1

voutvin

This block diagram is wrong. To show why we computethe actual closed-loop gain from the nodal equations.

ee101b: Op-amps 22 – 21

+

R1

R2

vinve

vfvout

• We havevout = −Avvf (1)

• If we assume that the input resistance is infinite, wehave

vin − vf

R1=

vf − vout

R2(2)

• substituting (1) in (2), we have

Avvin + vout

R1= −vout(1 + Av)

R2which can be written as

G =vout

vin= − R2

R1 + R2

Av

1 + AvR1

R1+R2

= −R2

R1

T

1 + T

where tT , the loop gain is given by

T = AvR1

R1 + R2

• Note that the loop-gain has the same expression asfor the non-inverting amplifier

A possible block diagram for the inverting amplifier is

ee101b: Op-amps 22 – 22

A R2R1+R2

R1

R2

−R1

R2voutvin

Another possible block diagram is also

A R2R1R1+R2

1R2

− 1R1

voutvin

Note that even though many possible block diagramsare possible, obtaining a correct block diagram is nottrivial. Even the most intuitive choices can be wrong.One thing is important, the loop gain of whateverrepresentation we choose must be correct.

ee101b: Op-amps 22 – 23

Summary

Open-loop Non-inverting InvertingGain Av

R1+R2R1

−R2R1

Loop Gain - AvR1

R1+R2Av

R1R1+R2

3dB bandwidth ω3dB

(1 + Av

R1R1+R2

)ω3dB

(1 + Av

R1R1+R2

ω3dB

)

Unity-gain bandwidth Avω3dB Avω3dB Avω3dB

The loop gain T is a very important measure of thefeedback circuit,

• Closed-loop gain is reduced by a factor of 1 + T

• 3dB closed-loop bandwidth is extended by a factorof 1 + T

• By analyzing T we can determine (in most cases)whether the op-amp will be stable

ee101b: Op-amps 22 – 24

Final remarks

• there is lots more to know about op-amps

• many interesting circuits can be build usingop-amps, e.g., precision differential amplifiers,filters, buffers (Sedra & Smith has many examples)

• real op-amps have many issues we have notdiscussed (e.g., offset, finite swing, finite inputresistance, . . . ) but we can assume they are idealfor first order calculations

• what is inside an op-amp? many of the circuits wehave studied, e.g., differential amplifiers,common-source stages, source follower stages, . . .

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Opamps Requirements