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1. Introduction 8307011 Basic Analog Circuits Spring 2005 1
1. INTRODUCTION
1. Introduction 8307011 Basic Analog Circuits Spring 2005 2
1.1 Electronic Systems
Electronic-System Block Diagrams
Figure 1.1 Block diagram of a simple electronic system: an AM radio.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 3
Analog Versus Digital Systems
Figure 1.2 Analog signals take a continuum of amplitude values. Digital signals take a few discrete amplitudes.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 4
Conversion of Signals from Analog to Digital Form
Figure 1.4 Quantization error occurs when an analog signal is reconstructed from its digital form.
Figure 1.3 An analog signal is converted to an approximate digital equivalent by sampling. Each sample value is represented by a 3-bit code word. (Practical converters use longer code words.)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 5
Relative Advantages of Analog and Digital Systems
Figure 1.5 After noise is added, the original amplitudes of a digital signal can be determined. This is not true for an analog signal.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 6
1.2 The Design ProcessSystem Design Circuit Design
Figure 1.7 Flowchart of the circuit-design process.Figure 1.6 Typical flowchart for design of electronic
systems.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 7
Basic Lows from Circuit Theory (Reminder)
Ohm Low
i
R+ -
vA resistance R, the current i through it and the voltage
v across it.
If the Ohm low is valid, the resistor is linear.
Nonlinear resistor: the voltage is not proportional to the current. For it
( )ifv =
Rvi
Riv
=
=
1. Introduction 8307011 Basic Analog Circuits Spring 2005 8
Kirchhoff Current Low
The algebraic sum of all currents, flowing through the branches, connected to the same node, is 0.
i1
R1
i2
R2
i5
R5
i3R3 i4 i6
R6Vs is
ab
cFor node a: is – i1 – i5 = 0
For node b: i1 – i2 – i3 + i4 = 0
For node c: i2 + i5 – i6 = 0
An example of a circuit with all currents in it.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 9
Kirchhoff Voltage Low
The algebraic sum of all voltages in a loop is 0.
An example of a circuit with four loops marked in it.
R1R2
R5
R3 i4 R6Vs
+
-
++
+
+
-
- -
-
-+
v1 v2
v5
v3 v6
AB
C
D For loop A: –Vs + v1 + v3 = 0
For loop B: v2 + v6 – v3 = 0
For loop C: v5 – v2 – v1 = 0
For loop D: v5 + v6 – Vs = 0
1. Introduction 8307011 Basic Analog Circuits Spring 2005 10
1.4 Basic Amplifier Concepts
Ideally an amplifier produces an output signal with the same wave shape as the input signal, but with a larger amplitude.
( ) )(tvAtv ivo = (1.2)
Av is the voltage gain of the amplifier.
Figure 1.15 Electronic amplifier.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 11
Non-inverting amplifier: Av > 0.
Inverting amplifier: Av < 0.
Figure 1.16 Input waveform and corresponding output waveforms.
The Common Ground Node
Often, one of the amplifier input terminals and one of the output terminals are connected to a common ground.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 12
Exercise 1.1.A certain noninverting amplifier has a voltage gain magnitude of 50. The input voltage isvi(t)=0.1sin(2000πt) (a) Find a expression for the output voltage vo(t) . (b) Repeat Part (a) for an inverting amplifier.
Solution:
a)
( ) ( ) [ ] ( )[ ] ( )tttvAtvA
ivo
v
ππ 2000sin52000sin1.05050
=×===
b)
( ) ( ) [ ] ( )[ ] ( )tttvAtvA
ivo
v
ππ 2000sin52000sin1.05050
−=×−==−=
1. Introduction 8307011 Basic Analog Circuits Spring 2005 13
The Voltage-Amplifier Model
Input resistance: equivalent resistance when looking into the input terminals
i
ii i
vR =
Input impedance: a combination of input resistance, input capacitance and input inductance.
Output resistance: equivalent resistance when looking into the output terminals.
Figure 1.17 Model of an electronic amplifier, including input resistance Ri and output resistance Ro.
Open-circuit voltage gain Avo: the voltage gain when the output is an open circuit (RL = ∞). Then
ivoo vAv =
1. Introduction 8307011 Basic Analog Circuits Spring 2005 14
Power GainCurrent Gain
io
PPG =
io
i iiA = (1.5)(1.3)
Li
viiLo
io
i RRA
R/vR/v
iiA ===
Li
viviioo
io
RR)A(AA
IVIV
PPG 2==== (1.6)(1.4)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 15
Example 1.1 Using the Voltage Amplifier ModelA source with internal voltage Vs = 1mV rms and internal resistance of Rs = 1MΩ is connected to the input terminal of an amplifier having an open-circuit voltage gain of Avo=104 , and input resistance of Ri=2MΩ , and an output resistance of Ro= 2Ω. The load resistance is RL = 8Ω. Find the voltage gainsAvs=Vo/Vs and Av = Vo/Vi . Also, find the current gain and power gain.
Solution: We can apply the voltage - divider principle to the input circuit to write
rmsmV667.0=+
= ssi
ii V
RRRV
8000
Vrms336.5
=+
==
=+
=
oL
Lvo
i
ov
oL
Livoo
RRRA
VVA
RRRVAV
i
siis
si
isi R
RRVVRR
RVV +=⇒
+=
Figure 1.18 Source, amplifier model, and load for Example 1.1.
5333=++
==oL
L
si
ivo
s
ovs RR
RRR
RAVVA
9102×==L
ivi R
RAAThe voltage - controlled source gives
rmsV67.6104 == iivo VVA 121016×== iv AAG
1. Introduction 8307011 Basic Analog Circuits Spring 2005 16
The Voltage Divider Loading Effects
R1
R2Vs
-
+
-
Vo
+
iLoading effects: Reduction of the gain of the amplifier due to the voltage dividers formed at the input from the source resistance and amplifier input resistance; and at the output from the amplifier output resistance and load resistance.
The voltage divider circuit.
Assuming open circuit at the output:
21
2
21
22
21
RRRA
RRRViRV
RRVi
v
so
s
+=
+==
+=
1. Introduction 8307011 Basic Analog Circuits Spring 2005 17
1.5 Cascaded Amplifiers
Figure 1.19 Cascade connection of two amplifiers.
1
2
i
ov v
vA = 12 oi vv =
2
2
1
1
i
o
i
ov v
vvvA ×=
12
11
oo
io
v vv
vvA ×=
21 vvv AAA = (1.7)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 18
Example 1.2 Analysis of a cascaded amplifier
Consider the cascade connection of two amplifiers shown in Figure 1.20. Find the current gain, voltage gain, and power gain of each stage and for the overall cascade connection.
Figure 1.20 Cascaded amplifiers of Example 1.2.
Solution: Considering loading by the input resistance of the second stage that the voltage gain of the first stage is
150012
211 =
+=
RRRAA
ii
vov
5002
22 =+
=RR
RAAiL
iLvov
For the second stage
5
21
11 10==ii
vi RRAA
750222 ==
Li
vi RRAA
621 1075×== iii AAA
7111 1051 ×== .AAG iv
4222 10753 ×== .AAG ivThe overall voltage gain is
1121 106255 ×== .GGG750021 == vvv AAA
1. Introduction 8307011 Basic Analog Circuits Spring 2005 19
Simplified Models for Cascaded Amplifier Stages• Input impedance of the model is the input impedance of the first stage
•Output impedance of the model is the output impedance of the last stage
•Open-circuit voltage gain is determined when the output of the last stage is an open circuit.
ΩMRR ii 11 ==
Ω1002 == oo RR
Example 1.3 Determining the Overall Model for a Cascaded Amplifier
Determine the overall simplified model for the cascaded amplifier in Figure 1.20 (Example 1.2).
Solution:
15012
211 =
+=
oii
vov RRRAA
Figure 1.21 Simplified model for the cascaded amplifiers of Figure 1.20. See Example 1.2.
1002 == vov AA
321 1015×== vvvo AAA
1. Introduction 8307011 Basic Analog Circuits Spring 2005 20
1.6 Power Supplies and Efficiency
Figure 1.22 The power supply delivers power to the amplifier from several constant voltage sources.
BBBBAAAAs IVIVP +=Total power, delivered from the power supplies in Figure 1.22: (1.8)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 21
In fact the amplifier converts the dc power, taken from the power supplies into ac power of the output signal in the load.
(1.9)dosi PPPP +=+
Usually Pi << PS.
Figure 1.23 Illustration of power flow.
Power EfficiencyPower efficiency = (output signal power to load) / (power from the power supply)
so
PP
=η (1.10)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 22
1.7 Decibel NotationVoltage and Current Gains Expressed in Decibels
(1.11)GGdB 10log10=
(1.15)||log20 10 vdBv AA =(1.12)21GGG =
(1.16)||log20 10 idBi AA =
Voltages, Currents and Other Quantities Expressed in Decibels
( ) )(log10log10 211010 GGGGdB ==
volt1log20 10
voltsdBV
VV =)(log10)(log10 210110 GGGdB +=
millivolt1log20 10
voltsdBmV
VV =dBdBdB GGG 21 += (1.12)
watt1log10 10
wattsdBW
PP =
milliwatt1log10 10
wattsdBmW
PP =
1. Introduction 8307011 Basic Analog Circuits Spring 2005 23
Exercise 1.7
An amplifier has an input resistance of 2kΩ, an output resistance of 100Ω, and an open-circuit voltage gain of Avo = 2000. If this amplifier is operated with a load of RL = 300Ω, find the power gain in decibels and the voltage gain Av=vo/vi in decibels.
Solution:
1500300100
3002000 =+
=+
=Lo
Lvov RR
RAA
dB5.63log20 10dB == vv AA
1000030020001500 ===
L
ivi R
RAA
. 7105.1150010000 ×=×== vi AAG
dB8.71log10 10 == GGdB
1. Introduction 8307011 Basic Analog Circuits Spring 2005 24
1.8 Amplifier Models
The Current-Amplifier Model
Aisc – short circuit current gain.
i
oscisc i
iA =
iisc – output current when the output is short circuited.Figure 1.25 Current-amplifier model.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 25
Example 1.5 Converting a Voltage Amplifier to a Current Amplifier
A certain amplifier is modeled by the voltage- amplifier model shown in Figure 1.26. Find the current amplifier model
3101001000100 ====
o
ivo
i
oscisc R
RAi
iA
Figure 1.26 Voltage amplifier of Example 1.5.
Solution In order to find the short-circuit current gain, we connect a short circuit to the output terminals of the amplifier as shown in Figure 1.26.
Figure 1.27 Current-amplifier model equivalent to the voltage-amplifier model of Figure 1.26.
o
ivoosc
i
ii
RvAi
Rvi
=
=
1. Introduction 8307011 Basic Analog Circuits Spring 2005 26
The Transresistance-Amplifier ModelThe Transconductance-Amplifier Model
Figure 1.28 Transconductance-amplifier model. Figure 1.30 Transresistance-amplifier model.
Gmsc - short-circuit transconductance gain. Rmoc - open-circuit transresistance gain.
iosc
msc viG =
iooc
moc ivR =
1. Introduction 8307011 Basic Analog Circuits Spring 2005 27
1.9 The Ideal Amplifiers
Classifying Real Amplifiers• Ideal Voltage Amplifier
•Ideal Current Amplifier
• Ideal Transconductance Amplifier
• Ideal Transresistance Amplifier
0 ; =∞= oi RR
∞== oi RR ;0
∞=∞= oi RR ;
0 ;0 == oi RR
1. Introduction 8307011 Basic Analog Circuits Spring 2005 28
1.10 Amplifier Frequency ResponseSignal as a Sum Sinusoids of Various Frequencies
Every periodic signal v(t) with period T can be represented as an infinite sum of the sinusoids (Fourierseries):
( ) ( ) ( ) ...3sin2sinsin
...23sin22sin2sin)(
3032021010
3322110
+++++++=
+
++
++
++=
ϕωϕωϕω
ϕπϕπϕπ
tatataa
tT
atT
atT
aatv
Tf
Tf ππω 22 frequency; lfundamenta - 1
000 ===
( ) harmonic lfundamenta - sin 101 ϕω +ta
( ) ( ) harmonicshigher - ,...3sin ,2sin 303202 ϕωϕω ++ tata
a0 – dc part (dc offset) of the signal
Spectrum of the signal: the set of all harmonics.
The non-periodic signals can be represented also as a sum of sinusoids with very small amplitudes and with very close frequencies – Fourier transform. The set of these sinusoids forms the spectrum of the non-periodic signal.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 29
Example: A train of symmetrical rectangular pulses (with 50% duty cycle):
Figure 1.35 Periodic square wave and the sum of the first five terms of its Fourier series.
The Fourier series of the train has only odd harmonics and is
( ) ( ) ( )
+++= ...5sin
513sin
31sin4)( 000 tttAtv ωωω
π(1.17)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 30
Example 1.8 Determining Voltage Gain as a Complex Number
The input voltage of a certain amplifier isComplex Gain ( ) ( ) 302000cos1.0 0−= ttvi π
and the output voltage isFourier transform permits to investigate and describe the amplifier only in terms of its response to a sinusoidal input signal instead to investigate it about every particular input signal.
( ) ( ) 152000cos10 0+= ttvo π
Find the complex voltage gain of the amplifier and express the magnitude of the gain in decibels.
Phasor, corresponding to a sinusoid Solution: The phasors of the input and output signals are( ) ( )
( ) phasor its - expsignal the- cos
ϕϕω
jVtVtv
=+=
V ( ) ( )00 15exp10 ;30exp1.0 jj oi =−= VV
( )( ) ( )0
0
0
45exp10030exp1.0
15exp10 jj
j
i
ov =
−==
VVAComplex gain of the amplifier
i
ov V
VA = (1.18) dB40100log20log20 1010 === vdBv AA
1. Introduction 8307011 Basic Analog Circuits Spring 2005 31
Gain as Function of Frequency. AC Coupling Versus Direct Coupling
Figure 1.36 Gain versus frequency.
Figure 1.37 Capacitive coupling prevents a dc input component from affecting the first stage, dc voltages in the first stage from reaching the second stage, and dc voltages in the second stage from reaching the load.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 32
The High-Frequency Region. Half-Power Frequencies and Bandwidth
Figure 1.32 Gain versus frequency for a typical amplifier showing the upper and lower half-power (3-dB) frequencies (fH and fL ) and the half-power bandwidth B.
Half power frequency (-3dB frequency): the frequency at which the gain magnitude is times the midband gain magnitude.
21
Wideband Versus Narrowband Amplifiers
Wideband amplifier: dc coupled amplifier or amplifier with lower half power frequency that is a small fraction from the upper half power frequency.
Narrowband (bandpass) amplifier: the bandwidth B is small, compared with the central frequency.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 33
Pulse Response Rise Time
Figure 1.42 Rise time of the output pulse. (Note: No tilt is shown. When tilt is present, some judgement is necessary to estimate the amplitude Vf.
Figure 1.41 Input pulse and typical ac-coupled broadband amplifier output.
B.tr350
≅ (1.19)
1. Introduction 8307011 Basic Analog Circuits Spring 2005 34
1.11 Differential Amplifier
Differential amplifier: • Two inputs and one output;• Output voltage proportional to the difference
between both input voltages: Differential input signal:
(1.21)21 iiid vvv −=[ ])t(vA)t(vA
)t(v)t(vA)t(v
idid
iido
21
21−=−=
(1.20)Differential gain:
(1.22)iddo vAv =
)vv(v iiicm 2121
+=
Common-mode input signal:
(1.23)
Figure 1.43 Differential amplifier with input sources.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 35
Common - Mode Rejection Ratio
icmcmiddo vAvAv += (1.24)
cmd
AA
logCMRR 20= (1.25)
Figure 1.44 The input sources vi1 and vi2 can be replaced by the equivalent sources vicm and vid.
1. Introduction 8307011 Basic Analog Circuits Spring 2005 36
Example 1.9 Determination of CMRR Specification
Find the minimum CMRR for the electrocardiograph amplifier if the differential gain is 1000, the desired differential input signal is 1mV peak, the common - mode input signal is a 100V peak 60Hz sine wave, and it is desired that the output contain a peak common - mode contribution that is 1% or less of the peak output caused by the differential signal.
Solution: Peak output of the desired signal
The common - mode gain is
dB8010100
01.0 4
cm
cm −==== −
i
ocm V
VA
(The common - mode gain actually amounts to attenuation.)
dB140101000log20log20CMRR 410
cm10 === −A
Ad
V1001.01000 =×== idvdod VAV
To meet the required specification, the common -mode output signal must have a peak value of
V01.0101.001.0cm =×=×= odo VV
