Post on 03-Dec-2014
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Differential Amplifier A differential (or difference) amplifier is a circuit used for amplifying a voltage difference
between two input signals while rejecting signals that are common to both inputs.
Basic operation:
Modes of Signal Operation:
1. Single-ended input – input signal is applied to either input with the other input connected
to ground
2. Differential or double-ended input – two opposite polarity input signals are applied.
3. Common-mode input – same signal is applied to both inputs
1
2
‒
‒
Loop 1:
‒VBE – VE = 0
VE = ‒VBE = ‒ 0.7V
and IE1 = IE2
since both currents combine in RE,
IE1 = IE2 = IRE/2
Loop 2:
‒ IRERE + VEE + VE = 0
IRE =
based on approximation IC ≈ IE
then IC1 = IC2 = IRE/2
therefore, VC1 = VC2 = VCC – IC1R1
VEE + VE
RE
2
Single-ended Input:
Differential Input: Common-mode Input:
Common-mode signal ‒ signal that drives both inputs of a differential amplifier equally.
‒ these are interference, static and other kinds of undesirable signals
picked-up by the circuit.
Common Mode Rejection Ratio(CMRR):
‒ measure of an amplifier’s ability to reject common-mode signals.
CMRR =
CMRR = 20 log
Example: A certain differential amplifier has a differential voltage gain of 2000 and a common-
mode gain of 0.2. Determine the CMRR and express in dB.
CMRR = = 10,000 CMRR = 20 log 10000 = 80dB
Av(d)
Acm Av(d)
Acm
‒
2000
0.2
3
Differential gain:
substitute in (2)
vin2 = ie2(re’ + RE) + RE
simplify in terms of ie2 =
substitute ie2 in equation 1: vin1 = ie1(re’ + RE) + RE
which makes ie1 =
do the same to compute for ie2 =
at the output side: vout(d) = vc1 ‒ vc2
= RC (ic1 ‒ ic2) = RC (ie1 ‒ ie2)
= RC ‒
Simplify to obtain Av(d) = = *true for balanced output (vout(d) = vc1 ‒ vc2)
for unbalanced output (vc1 or vc2 only): Av(d) = =
vin(d) = vin1 ‒ vin2
From Loop 1:
vin1 ‒ ie1re’ ‒ (ie1 + ie2)RE = 0
vin1 = ie1(re’ + RE) + ie2RE (1)
vin2 ‒ ie2re’ ‒ (ie1 + ie2)RE = 0
vin2 = ie2(re’ + RE) + ie1RE (2)
express in terms of the currents:
from (1) ie1=
1 vin1 ‒ ie2RE
re’ + RE
vin1 ‒ ie2RE
re’ + RE
vin2 ‒ ie1RE
re’ + RE vin2 ‒ ie1RE
re’ + RE (re’ + RE) vin1 ‒ REvin2
(re’ + RE)2 ‒ RE
2
(re’ + RE) vin1 ‒ REvin2
(re’ + RE)2 ‒ RE
2
(re’ + RE) vin2 ‒ REvin1
(re’ + RE)2 ‒ RE
2
(re’ + RE) vin2 ‒ REvin1
(re’ + RE)2 ‒ RE
2
vout(d) RC
vin(d) re’
vout(d) RC
vin(d) 2re’
4
Common-mode gain:
vicm = Acm =
for common-mode, emitter currents ie1 = ie2
since the two transistors are matched, only one-half of the circuit may be considered:
Acm = =
Example: For the circuit shown, calculate:
(a) ICQ and VCEQ
(b) Av(d) and Acm
(c) CMRR
Solution:
(a) VE = ‒0.7V
IRE = = = 1.378mA
IE = IRE/2 = 0.689mA = ICQ
VCQ = VCC ‒ ICQRC = 9.726V Av(d) = = 90.95
VCEQ = VCQ ‒ VEQ = 9.726 ‒ (‒0.7) = 10.426V Acm = = 0.2
re’ = 25mV/IE = 36.28ΩΩΩΩ CMRR = 90.95/0.2 ≈ 453
vin1 + vin2
2
voutcm
vincm
voutcm RC
vincm re’ + 2RE
VEE + VE 12 ‒ 0.7
RE 8.2k
RC
re’
RC
re’ + 2RE
5
Introduction to Operational Amplifiers (Op-amp)
The operational amplifier is a direct coupled high gain amplifier and is used to perform a wide
variety of linear as well as non-linear functions. This circuit was originally used for carrying out
mathematical operations such as summation, differentiation, and integration on input signals.
Now, operational amplifiers are used for functions other than mathematical operations such as dc
as well as ac amplification, rectification, waveform generation, filtration, non-linear
waveshaping, etc.
Block Diagram of an op-amp
Input stage ‒ this stage provides most of the voltage gain and also establishes the input
resistance of the OPAMP.
Intermediate stage ‒ is another differential amplifier which is driven by the output of the
first stage.
Level shifting circuit ‒ used to shift the dc level at the output downward to zero with
respect to ground.
Output stage ‒ increases the output voltage swing and raise the current supplying
capability of the OPAMP, also provides low output resistance.
A simple op-amp arrangement:
6
Symbols and Terminals:
The standard operational amplifier symbol is shown. It has two input terminals, the inverting
input (-) and the non-inverting input (+), and one output terminal. The typical op-amp operates
with two dc supply voltages, one positive and the other negative.
Ideal Op-amp Characteristics:
Practical Op-amp Characteristics:
Op-amp Parameters:
Input Offset Voltage
It is desired that the dc voltage at the output is zero with no input voltage. But because of
the unequal amount of current drawn by the input transistors of the first differential
amplifier due to unbalance in the circuit, the output voltage will not become zero. Input
The ideal op-amp has infinite
voltage gain, infinite input
resistance, zero output impedance
and infinite bandwidth.
1. Very high voltage gain (~105)
2. Very high input impedance (~2MW)
3. Very low output impedance (~75W)
4. Wide bandwidth (0 – 1MHz)
5. Very high differential gain (~80dB)
6. Large CMRR (~80dB)
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offset voltage is the voltage required between the inputs to force the differential output to
zero volts. Typical values are in the range of 2mV or less.
Input Offset Voltage Drift with Temperature
The input offset voltage drift is a parameter related to Vos that specifies how much
change occurs in the input offset voltage for each degree change in temperature. Typical
values range anywhere from about 5mV per degree Celsius to about 50mV per degree
Celsius. Usually, an op-amp with a higher nominal value of input offset voltage exhibits a
higher drift.
Input Bias Current
The input bias current is the dc current required by the inputs of the amplifier to properly
operate the first stage. By definition, the input bias current is the average of input currents
and is calculated as follows:
Input Impedance
- Differential input impedance is the total resistance between the inverting and the non-
inverting inputs. It is measured by determining the change in bias current for a given
change in differential input voltage.
- Common-mode input impedance is the resistance between each input and ground and is
measured by determining the change in bias current for a given change in common-mode
input voltage.
Input Offset Current
Ideally, the two input bias currents are equal, and thus their difference is zero. In a
practical op-amp, however, the bias currents are not exactly equal.
The input offset current, Ios, is the difference of the input bias currents expressed as an
absolute value.
Output Impedance
The output impedance is the resistance viewed from the output terminal of the op-amp.
Common-mode Input Voltage Range
All op-amps have limitations on the range of voltages over which they will operate. The
common-mode input voltage range is the range of input voltages which, when applied to
both inputs will not cause clipping or other output distortion.
1 2BIAS
2
I II
+=
OS 1 2I I I= −
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Open-loop Voltage Gain
The open-loop voltage gain of an op-amp is the internal voltage gain of the device and
represents the ratio of output voltage to input voltage when there are no external
components. The open-loop voltage gain is set entirely by the internal design. Open-loop
voltage gain can range up to 200,000 and is not a well-controlled parameter. Data sheets
often refer to the open-loop voltage gain as the large-signal voltage gain.
Maximum Output Voltage Swing (Vo(pp))
The output voltage of an op-amp cannot be higher than the positive dc power supply
voltage (+VDC), and cannot be lower than the negative dc power supply voltage (-VDC).
Vo(pp) also varies with the load connected and increases directly with load resistance.
Common-mode Rejection Ratio (CMRR)
The CMRR is a measure of an op-amp’s ability to reject common-mode signals. A good
op-amp should have a very high value of CMRR, this enables the op-amp to virtually
eliminate interference signals from the output.
Slew Rate
The slew rate of an op-amp is the maximum rate of change of the output voltage in
response to a step input voltage. It is dependent upon the high-frequency response of the
amplifier stages within the op-amp.
The slew rate can be measured using the circuit given below:
( )CMRR
v d ol
cm cm
A A
A A= =
-the output voltage
cannot change
instantaneously when a
high frequency, large
amplitude signal is
applied at the input side.
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Example: What is the slew rate for the output signal shown in response to a step input?
The output goes from -10 V to +10 V in 25 µs.
Frequency Response
Ideally, an op-amp should have infinite bandwidth. This means the gain of an op-amp
must remain the same for all frequencies from 0 to infinite. Practical op-amps however
decrease its gain at higher frequencies. The dependence of gain on frequency is due
primarily to the presence of capacitive component in the equivalent circuit of the op-amp.
Maximum Operating Temperature.
The maximum temperature is the highest ambient temperature at which the device will
operate according to specifications with a specified level of reliability.
Minimum Operating Temperature.
This is the lowest temperature at which the device operates within specification.
Output Short-Circuit Duration.
This is the length of time the op-amp will safely sustain a short circuit at the output
terminal. Many modern op-amps can carry short circuit current indefinitely.
Maximum Output Voltage.
The maximum output potential of the op-amp is related to the DC power supply voltages.
Typical for a bipolar op-amp with ± 15 V power supply is about 13 V maximum and - 13
V minimum.
Negative Feedback
Negative feedback is the process whereby a portion of the output voltage of an amplifier
is returned to the input with a phase angle that opposes (or subtracts from) the input
signal. This method helps stabilize the gain and reduce distortion. It can also increase the
input resistance.
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Advantages of Negative feedback:
1. Stable voltage gain
2. Decreased output impedance
3. Increased/decreased input impedance depending on circuit
4. Decreased distortion
5. Increased bandwidth
Closed-loop Voltage Gain (Acl):
The closed-loop voltage gain (Acl) is the voltage gain of an op-amp with external
feedback. The gain can be controlled by external component values.
Concept of Virtual Ground:
When finding the gain, assume there is infinite impedance at the input (i.e. between the
inverting and non-inverting inputs. Infinite input impedance implies zero current at the
input. If there is no current at the input impedance, there is no voltage drop between the
inverting and non-inverting inputs. Thus, the voltage at the inverting input is zero. The
zero at the inverting input is referred to as virtual ground.
(a) Virtual ground (b) Iin=If and current I1 = 0
Op-amp Configurations with Negative Feedback
The Inverting Amplifier:
The inverting amplifier has the output fed back to the inverting input for gain control.
The gain for the inverting op-amp can be determined by the formula: V1 = V2 = 0
0V
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Example: Given the op-amp configuration in the figure, determine the value of Rf
required to produce a closed-loop voltage gain of -100.
The Non-inverting Amplifier:
The closed loop gain for a non-inverting amplifier can be determined by the formula
V1 = V2 = Vin
Example: Determine the gain of the amplifier in the figure. The open-loop voltage gain
of the op-amp is 100,000.
in
i
f
out
ioutfin
VR
RV
RVRV
−=
=−− 0
011 =−
+f
out
i R
VV
R
V
( ) 0=−+⋅ outinifin VVRRV
in
i
f
in
fi
out
outifiin
VR
RV
R
RRV
VRRRV
⋅+=⋅+
=
⋅=+
)1(
)(
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The Voltage Follower:
The voltage-follower amplifier configuration has the
entire output signal fed back to the inverting input. The
voltage gain is 1. This makes it useful as a buffer amp
since it has high input impedance and low output
impedance.
Effect of Negative Feedback on Open-loop Gain
Input impedance of the non-inverting amplifier
The input impedance of an op-amp without feedback is Zin. The input impedance with
negative feedback increases the value of Zin to a very large value that for all practical circuits it
can be considered infinite.
Vin = Vd + Vf
Substituting βVout for Vf
Vin = Vd + βVout
Since Vout≈AolVd
Vin = Vd + AolβVd
= Vd (1 + Aolβ)
Substituting IinZin for Vd
Vin = IinZin (1 + Aolβ)
Where Zin is the open-loop impedance
Vin/Iin = Zin (1 + Aolβ)
Vin/Iin is the overall input impedance of the closed-loop non-inverting configuration:
Zin(NI) = (1 + Aolββββ)Zin
Vf = Vout β = Vf / Vout
(attenuation)
Ri
Ri + Rf
Vout = Aol (Vin – Vf)
substituting βVout for Vf
Vout = Aol (Vin – βVout )
= = Acl(NI)
since 1<< Aol β then Acl(NI) ≈ =
VouT Aol
Vin 1+ Aol β
1 Ri + Rf
β Ri
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Output impedance for the non-inverting amplifier
The output impedance of an op-amp without feedback is Zout. Negative feedback
decreases this by a factor of (1 + AolB). This is so small that for all practical circuits it can be
considered to be zero.
Vout = AolVd ‒ ZoutIout
but Vd = Vin ‒ Vf
by assuming AolVd >>ZoutIout
then Vout ≈Aol(Vin‒Vf)
substituting βVout for Vf
Vout ≈ Aol (Vin ‒ βVout)
Vout ≈ AolVin ‒ AolbVout
AolVin ≈ Vout + AolβVout
≈ (1 + Aolβ) Vout
since output impedance Zout(NI) = Vout/Iout
AolVin ≈ (1+Aolβ) IoutZout(NI)
dividing both sides by Iout:
AolVin/Iout ≈ (1+Aolβ) Zout(NI)
AolVin/Iout ≈ (1+Aolβ) Zout(NI)
since AolVin = Vout and Vout/Iout =Zout
then
Zout = (1+Aolβ) Zout(NI)
thus,
Example: What are the input and output impedances and the gain of the non-inverting amplifier?
Assume the op-amp has Aol = 100,000, Zin = 2 MΩ, and Zout = 75 Ω.
The gain is
25
The feedback fraction is
The input impedance is 8GΩ
The output impedance is 0.019Ω
(I)
36 k1 1
1.5 k
f
cl
i
RA
R
Ω= + = + =
Ω
10.040
25B = =
14
Input impedance for the inverting amplifier
Output impedance for the inverting amplifier
The equation for the output impedance of the inverting amplifier is the essentially the
same as the non-inverting amplifier:
Example: What is the input resistance and the gain of the inverting amplifier? Assume the op-
amp has Aol = 100,000, Zin = 2 MΩ, and Zout = 75 Ω.
The gain is
‒24
The input impedance is Zin(I) = 1.5kΩ
The output impedance is
0.019Ω
Voltage Follower
The voltage-follower is a special case of the non-inverting amplifier in which Acl = 1. The input
resistance is increased by negative feedback and the output resistance is decreased by negative
feedback. This makes it an ideal circuit for interfacing a high-resistance source with a low
resistance load.
Zin(NI) = (1 + Aolβ)Zin
Recall that negative feedback forces the inverting
input to be near ac ground for the inverting
amplifier. For this reason, the input impedance of
the inverting amplifier is equal to just the input
resistor, Ri. That is, Zin(I) = Ri.
(I)
36 k
1.5 k
f
cl
i
RA
R
Ω= − = − =
Ω
15
Bias Current and Offset Voltage Compensation
Effect of an Input Bias Current
Ideally, if the input voltage is zero, there should be zero current coming into the inverting input
of the op-amp. However, there is a small bias current I1 that goes through Rf. This current
creates a voltage at the output equal to I1Rf known as the error voltage.
Bias current compensation:
Effect of Input Offset Voltage
In the voltage follower circuit shown, it is shown that
the output error voltage is –I1Rs.
(a) Voltage follower (b) non-inverting (c) inverting
The output voltage of an op-amp should be zero when the differential
input is zero. However, there is always a small output error voltage
present whose value typically ranges from microvolt to millivolts.
This is due to unavoidable imbalances within the internal op-amp
transistors aside from the bias currents previously discussed.
VOUT(error) = AclVIO
since Acl for the voltage follower is 1,
VOUT(error) = VIO
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Input Offset Voltage Compensation
Open-loop vs. Closed-loop Voltage Gain
Open-loop voltage gain is set entirely by the internal design, whereas the closed-loop
voltage gain is determined by the external component values.
3-dB Open-Loop Bandwidth
Unity-Gain Bandwidth
In the bode plot of the Open-loop amplifier, the gain steadily decreases to a point where it
is equal to 1 (0 dB). The value of the frequency at which this unity gain occurs is the unity gain
bandwidth.
•The bandwidth of an AC Amplifier is
the frequency range between the
points where the gain is 3dB less than
the midrange gain.
•In general, the bandwidth equals the
upper critical frequency (fCU) minus
the lower critical frequency (fCL).
•Since fCL for an op-amp is zero, the
bandwidth is simply equal to the
upper critical frequency.
BW = fC(OL)
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Phase Shift
An RC lag circuit found in an op-amp stage causes a propagation delay from input to output,
thus creating a phase shift between the input signal and the output signal.
• Phase Shift (θ) is expressed as: θ = -tan-1
(f/fC)
• The negative sign indicates that the output lags the input.
• The math expression shows that the phase shift increases with frequency and approaches
-90° as f becomes much greater than fC.
Example. Calculate the phase shift for an RC lag circuit for each of the following frequencies,
and then the curve of phase shift versus frequency: (a) f = 1 Hz (b) f= 10 Hz (c) f = 100 Hz (d)
f = 1000 Hz (e) f = 10,000 Hz. Assume fc = 100 Hz.
(a) θ = -tan-1
(f/fC) = ‒tan-1
(1/100)
= ‒ 0.5730
(b) θ = -tan-1
(10/100) = ‒ 5.70
(c) θ = -tan-1
(100/100) = ‒ 450
(d) θ = -tan-1
(1000/100) = ‒ 84.30
(e) θ = -tan-1
(10,000/100) = ‒ 89.40
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Example. Determine Aol for the following values of f: (a) f = 0 Hz (b) f = 10 Hz (c) f = 100 Hz.
Assume fc(ol) = 100 Hz and Aol(mid) = 100,000.
Open-Loop Frequency and Phase Response
Example: A certain op-amp has three internal amplifier stages with the following gains and
critical frequencies:
Stage 1: Av1 = 40dB , fc1 = 2000Hz
Stage 2: Av2 = 32dB , fc2 = 40kHz
Stage 3: Av3 = 20dB , fc3 = 150kHz
Determine the open-loop midrange gain in decibels and the total phase lag when f = fcl.
Solution: Aol(mid) = Av1 + Av2 + Av3 = 92dB
θ(tot) = ‒ tan-1
‒ tan-1
‒ tan-1
= ‒ 48.60
2k 2k 2k
2k 40k 150k
19
Closed-Loop vs. Open-Loop Gain
Effect of Negative Feedback on Bandwidth:
• The closed-loop critical frequency of an op-amp is: fC(CL) = fC(OL) ( 1 + βA (mid) )
• The bandwidth of a closed loop amplifier is: BW(CL) = BW(OL) ( 1 + βA (mid) )
Gain-Bandwidth Product
• An increase in closed loop gain causes a decrease in the bandwidth and vice versa, such
that product of gain and bandwidth is constant.
• Condition is true as long as the roll-off rate is fixed at
• -20dB/decade.
• The gain bandwidth product is always equal to the frequency at which the op-amp’s open
loop gain is unity (unity gain bandwidth).
• AC(OL) fOL = AC(CL) fCL = unity gain bandwidth
Example. Determine the BW of each of the amplifiers below. Both op-amps have an open-loop
gain of 100dB and a unity-gain bandwidth of 3MHz.
(a) Acl ≈ = 1 + = 1 + = 67.7
fc(cl) = BWcl =
BWcl = = 44.3kHz
(b) Acl = ‒ = ‒ 47
BWcl = = 63.8kHz
1 Rf 220k
β Ri 3.3k
Unity-gain BW
Acl
3MHz
67.7
Rf
Ri
3MHz
47
20
Positive Feedback and Stability
Positive Feedback
With negative feedback, the signal fed back to the input of an amplifier is out of phase
with the input signal, thus subtracting from it and effectively reducing the voltage gain. As long
as the feedback is negative, the amplifier is stable.
When the signal fed back from output to input is in phase with the input signal, a positive
feedback condition exists and the amplifier can oscillate.
Oscillation is an unwanted voltage swings on the output when there is no signal present
on the input.