1

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)

7

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= −

8

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.

9

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.

10

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

11

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(

)(

12

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

13

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

16

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)

17

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

18

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.