Block C: Amplifiers - City University of Hong Kong
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Transcript Block C: Amplifiers - City University of Hong Kong
Block D Unit 2 Outline
Op-amp circuits with resistors only
> Inverting amplifier
> Non-inverting amplifier
> Summing amplifier
> Differential amplifier
> Instrumentation amplifier
Op-amp circuits with reactive components
> Active filters (Low pass, High pass, Band pass)
> Differentiator & Integrator
Physical limits of practical op-amps
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Source follower
Vin
+
Rs
Vout
-
RL
Find the gain of the above circuit
The key features of the source follower are:
1) Large input resistance
2) Small output resistance
3) Unity gain (i.e. gain of close to one)
It is therefore commonly used as a buffer between a load and source where the
impedances are not well matched
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Inverting op amp
R2
R1
Vin
Note that the inverting input (node X) is at ground
R1 and R2 are in series
-
Applying KCL at X
Vout
+
Vin Vout
0
R1 R2
Closed-loop gain
Vout
R2
Vin
R1
Negative sign indicates a 1800
shift in the phase
Gain is set by the resistor values
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Non-inverting amplifier
iin
Note that since iin = 0:
R
+
1) v+ = vin (no voltage drop across R)
2) We can apply voltage divider rule
to RS & RF
+
vin _
RS
vRS
vout
RS RF
+
_
vout
iin
+
RS
vRS
-
RF
-
But we also note that A is infinite, so:
1) vin = vRS
2) Hence we then obtain: G
EE2301: Block D Unit 2
vout
RF
1
vin
RS
4
Summing Amplifier
RS1
Apply NVA at the inverting input
terminal:
vS1 vS 2 vout
0
RS1 RS 2 RF
v
vout
v
S 1 S 2
RF
RS 1 RS 2
+
vS1 _
-
RS2
+
vS2 _
Hence the form of the gain relation can
be described by:
Vout = -(A1vS1 + A2vS2)
A1 and A2 are set by the resistor values
chosen.
EE2301: Block D Unit 2
RF
+
+
vout
-
We can extend this result to write a
general expression for the gain:
N
R
vout F vSn
n 1 RSn
5
Differential amplifier
R2
V1
V2
R1
One way of analyzing this circuit is to apply
superposition (find Vout for V1 or V2 only)
-
R1
Vout
+
If we short V2 first, we obtain:
V1
R2
+
R1
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R2
Vout1
vout1
[R2/(R1+R2)]V2
Vout2
R2
+
If we short V1 now, we obtain:
-
R1
vout 2
R2
v2
R1
R2
v1
R1
Hence, finally: Vout = (R2/R1)(v2 – v1)
Output is the difference between the
inputs amplified by a factor set by the
resistor values.
6
Difference amplifier
In the case whereby all the resistors are different, as shown in the circuit below,
while vout1 remains unchanged, vout2 now becomes:
vout 2
R2 R4
v2
1
R1 R3 R4
R2
Hence the overall gain expression is given by:
R2 R4
R
v2 2 v1
vout 1
R1
R1 R3 R4
This is similar to the form of the summing
amplifier except that we take the difference
between the two inputs:
Vout = A2V2 – A1V1
A2 and A1 are set by the resistor values
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V1
V2
R1
R3
+
Vout
R4
7
Diff Amp: Example 1
Find vout if R2 = 10kΩ and R1 = 250Ω
R2
V1
V2
R1
R1
+
Vout
R2
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Diff Amp: Example 2
For the same circuit in the previous example, given v2 = v1:
Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively
R2 R4
R2
v2 v1
vout 1
R1
R1 R3 R4
R2
V1
V2
500Ω R1
R1
250Ω
+
R2
Vout
Now: R2 = 10kΩ, R1 = 750Ω,
R3 = 500Ω, R4 = 10kΩ
vout = 13.65 v2 – 13.33v1
We can see that as a result of the source resistances:
1) Differential gain has changed
2) vout is not zero for v2 = v1
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Instrumentation Amplifier
V1
+
-
R3
R1
R2
vo1
RX
-
V2
R1
vo2
+
-
Vout
R2
+
R3
The instrumentation amplifier takes care of this problem by including a
non-inverting amplifier (which possesses an infinite input resistance)
between the differential amplifier an the inputs.
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Instrumentation Amplifier
V1
+
-
Vo1
R3
Vo1
R1
+
RX/2
-
1st stage comprises a pair of non-inverting
amplifiers:
Vout
Vo2
R3
Vo1 = (2RX/R1 +1)V1
Closed-loop gain:
2nd stage is a differential amplifier:
Vout = (R3/R2)(2RX/R1 + 1)(V2 - V1)
Vout = (R3/R2) (Vo2 - Vo1)
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Op amp example 1
Problem 8.5
Find v1 in the following 2 circuits
What is the function of the source follower?
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Op amp example 1 solution
Fig (a):
6Ω || 3Ω = 2Ω
v1 = {2 / (2+6)}*Vg = 0.25Vg
Fig (b):
Voltage at the non-inverting input = 0.5Vg
Voltage at output = Voltage at non-inverting input = 0.5Vg
This slide is meant to be blank
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Op amp example 2
Problem 8.7
Find the voltage v0 in the following circuit
Transform to Thevenin
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Op amp example 2 solution
Thevenin equivalent circuit:
Rth = 6 + 2||4 = 22/3 kΩ
Vth = {4/(2+4)}*11 = 22/3 V
Closed loop gain expression:
A = - 12kΩ / Rth (With Vth as input source)
Vout = 12/(22/3) * (22/3) = 12V
This slide is meant to be blank
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Op amp example 3
R2
Find the closed-loop gain
R1
vin
-
Consider:
Current through RB = 0A (Infinite input
resistance of op amp)
Voltage across RB = 0V
+
RB
+
vout
-
Voltage at inverting input = 0V
Same as inverting op amp:
A = -R2 / R1
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Op amp example 4
Find the closed-loop gain
R2
Consider:
-
Current through R2 = 0A (Infinite
input resistance of op amp)
Voltage across R2 = 0V
Vout = VVout = Vin (Infinite open loop
gain of op amp)
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+
vin
-
+
+
vout
-
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Common and differential mode
A differential amplifier should ideally amplify ONLY DIFFERENCES between
the inputs. That is to say identical inputs should give an output of zero.
Ideal differential amplifier: Vout = A(V2 – V1)
In reality this is not the case as we have seen in a previous example. The output of
the amplifier is more accurately described by:
Vout = A2V2 + A1V1
A2: Gain when V1 = 0 (V2 is the only input)
A1: Gain when V2 = 0 (V1 is the only input)
It is then useful to describe the performance of a differential amplifier in terms
of the gain when both input are identical (common mode) and when the inputs
are equal and out-of-phase (differential)
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Common mode rejection ratio
Common mode gain (Acm): Gain when both inputs are exactly the same
(V1 = V2 = Vin)
Vocm = (A2 + A1)Vin Voltage output for common input
Acm Vocm Vin
A2 A1
Common mode gain (Adm): Gain when both inputs are equal but out-of-phase
(V2 = -V1 = Vin) V = (A - A )V Voltage output for differential input
odm
2
1
in
Acm Vodm 2Vin
A2 A1 2
Divide by 2 since the difference of a pair of differential inputs is twice that of each input
The common mode rejection ratio (CMRR) is simply the ratio of the differential
mode gain (Adm) over the common mode gain (Acm):
CMRR = Adm/Acm
A large CMRR is therefore desirable for a differential amplifier
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Active filters
Range of applications is greatly expanded if reactive components are used
Addition of reactive components allows us to shape the frequency response
Active filters: Op-amp provides amplification (gain) in addition to filtering
effects
Substitute R with Z now for our analysis:
Vout
ZF
j
VS
ZS
ZF and ZS can be arbitrary
(i.e. any) complex impedance
Vout
ZF
j 1
VS
ZS
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Active low pass filter
Vout
ZF
j
VS
ZS
ZS = RS
Z F RF || C F
R
1
F RF
jC F
jC F
RF
1 jC F RF
Amplification
Vout
j RF RS
VS
1 jCF RF
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Shaping of frequency
response
21
Active high pass filter
Vout
ZF
j
VS
ZS
ZF = RF
Z S RS 1 jCS
1 jCS RS jCS
Amplification
Vout
j jCs RF
VS
1 jCS RS
ω→0, Vout/VS → -RF/RS
EE2301: Block D Unit 2
Shaping of frequency response
22
Active band pass filter
Vout
ZF
j
VS
ZS
Z S RS 1 jCS
1 jCS RS jCS
Z F RF || C F
RF
1 jC F RF
Vout
jCs RF
j
1 jCS RS 1 jCF RF
VS
ZS: CS blocks low frequency inputs but lets high frequency inputs through
High pass filter
ZF: CF shorts RF at high frequency (reducing the gain), but otherwise looks just like
a high pass filter at lower frequencies
Low pass filter
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Active band pass filter
Vout
jCs RF
j
1 jCS RS 1 jCF RF
VS
ωHP = 1/(CSRS) – Lower cut-off frequency
ωLP = 1/(CFRF) – Upper cut-off frequency
For ωHP < ωLP: Frequency response curve is shown below
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Second-order Low Pass Filter
Vout
ZF
j
VS
ZS
Z F R2 || C
Z S R1 jL
R2
1 jCR2
Vout
R2
j
1 jCR2 R1 jL
VS
R1, R2, C and L are specially chosen so that: ω0 = 1/(CR2) = R1/L.
The frequency response function then simplifies to:
Vout
R2
j
2
VS
R1 1 j 0
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Second-order Low Pass Filter
R1, R2, C and L are specially chosen so that:
ω0 = 1/(CR2) = R1/L.
H v j
Vout
R2
j
2
VS
R1 1 j 0
Above ω0, Hv is reduced by a factor of 100 for a ten fold
increase in ω (40dB drop per decade)
First-order filter: Hv is reduced by a factor of 10 for a
ten fold increase in ω (20dB drop per decade)
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Ideal integrator
KCL at inverting input: iS = -iF
Virtual ground at inverting input: iS = VS/RS
For a capacitor: iF = CF[dVout/dt]
VS
dVout
C
RS
dt
Output is the integral
of the input
dVout
VS
dt
CF RS
1
Vout (t )
VS (t )dt
CF RS
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Ideal differentiator
KCL at inverting input: iS = -iF
Virtual ground at inverting input: iF = Vout/RF
For a capacitor: iS = CS[dVS/dt]
Vout
dVS
CS
RF
dt
Vout RF CS
Output is the timedifferential of the input
EE2301: Block D Unit 2
dVS
dt
28
Physical limit: Voltage supply limit
The effect of limiting supply voltages is that
amplifiers are capable of amplifying signals
only within the range of their supply voltages.
RS = 1kΩ, RF = 10kΩ, RL = 1kΩ;
VS+ = 15V, VS- = -15V; VS(t) = 2sin(1000t)
Gain = -RF/RS = -10
Vout(t) = 10*2sin(1000t) = 20sin(1000t)
But since the supply is limited to +15V and 15V, the op-amp output voltage will saturate
before reaching the theoretical peak of 20V.
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Physical limit: Frequency response limit
So far we have assumed in our ideal op-amp model that the open loop gain AV(OL)
is infinite or at most a large constant value. In reality, AV(OL) varies with a
frequency response like a low pass filter:
ω0: frequency when the
response starts to drop off
The consequence of a finite bandwidth is a fixed gain-bandwidth product
If closed loop gain is increased, -3dB bandwidth is reduced
Increasing the closed loop gain further results in a bandwidth reduction
till the gain-bandwidth produce equals the open-loop gain
Gain bandwidth product = A0ω0
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