Option 2 : 10 dB

__Concept__:

CMRR (Common mode rejection ratio) is defined as the ratio of differential-mode voltage gain (Ad) and the common-mode voltage gain (AC).

Mathematically, in dB this is expressed as:

\(CMRR(dB)= 20\log \left| {\frac{{{A_d}}}{{{A_c}}}} \right| \)Ad = Differential gain

Ac = Common mode gain

Using the property of log, we can write:

CMRR (dB) = 20log (A_{d}) - 20log (A_{c})

CMRR (dB) = A_{d}(dB) - A_{c}(dB)

__Calculation__:

Given:

Ad(dB) = 110 dB

CMRR (dB) = 100 dB

We can write:

CMRR (dB) = Ad(dB) - Ac(dB)

100 dB = 110 dB - Ac(dB)

Ac(dB) = 110 dB - 100 dB

**Ac(dB) = 10 dB**

Calculate the expression of output voltage for the given circuit:

Option 1 : 41 (V1 - V2)

**Concept:**

Let us consider a circuit given below:

__\({V_0} = \frac{{{R_4}}}{{{R_3}}}\left( {1 + \frac{{2{R_2}}}{{{R_1}}}} \right)\left( {{V_1} - {V_2}} \right)\)__

__Calculation:__

R_{4} = 7 kΩ, R_{3} = 7 kΩ, R_{2} = 7 kΩ, R_{1} = 350 Ω

\({V_0} = \frac{{{R_4}}}{{{R_3}}}\left( {1 + \frac{{2{R_2}}}{{{R_1}}}} \right)\left( {{V_1} - {V_2}} \right)\)

\({V_0} = 1\left( {1 + \frac{{2 \times 7 \times 1000}}{{350}}} \right)\left( {{V_1} - {V_2}} \right)\)

V_{0} = 41 (V_{1 }– V_{2})

** Note**:

Take care of V1 and V2.

A_{3} op-amp work like a difference Amplifier from its inverting and non-inverting terminal we can get V_{1} and V_{2}.

Option 3 : Dual input unbalanced output

**Explanation:**

The functions of various stages of op-amp are:

- Input Stage (Dual Input Balanced Output Differential Amplifier)

- Provides most of the voltage gain
- Provides High input resistance to op-amp
- Provides high CMRR
- +VCC & - VEE Supply are due to this stage

**Intermediate Stage**__(Dual Input Unbalanced output)__

**Provides gain to achieve overall high gain for op-amp**

- Level Shifter stage

- Because of direct coupling, output of Intermediate stage has DC voltage above ground
- Function of level shifting stage is to shift the dc level at the output of the intermediate stage down to zero
- Emitter follower in common collector configuration is used for BJT
- Source follower or common drain is case of FET/MOSEFT

- Output stage (Complementary Symmetry Class B Push – Pull amplifier)

- High voltage swing
- High current supply capacity
- Short circuit protection

Option 2 : A_{v}/A_{c}

__ CMRR:__.

Common Mode Rejection Ratio (CMRR) is defined as the ratio of differential-mode gain to common-mode gain taken in magnitude

CMRR stands for Common Mode Rejection Ratio It is the ability of an operational amplifier to reject the common-mode signals at the input terminals

Mathematically, this is expressed as:

\(CMRR = \frac{A_v}{A_c}\)

A_{v} = Differential gain

A_{c} = Common mode gain

For practical operational amplimers, The Common Mode Rejection Ratio (CMMR) should be 80 to 100 dB.

For Ideal operational amplimers, The Common Mode Rejection Ratio (CMMR) should be infinity.

Option 2 : V/μs

__Slew Rate__:

- Slew rate is the maximum rate of change of output voltage with respect to time.
- Slew rate limits the maximum frequency of operation of op-amp
- The slew rate is usually measured in volts per microsecond.

Mathematically:

Vin = Vm sin 2πfmt and

V0 = AVin

∴ The Slew rate will be:

\({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}} = 2{\rm{\pi }}{{\rm{f}}_{{\rm{max}}}}{{\rm{V}}_0}_{{\rm{max}}}{\rm{\;}}\)

Since the slew rate is the ratio of voltage and time, they are expressed as V/sec.

**Also, since the time period is usually very small (in μsec), the slew rate is commonly expressed as V/μsec.**

Option 2 : increases

**Explanation:**

__CMRR:__.

Common Mode Rejection Ratio (CMRR) is defined as the ratio of differential-mode gain to common-mode gain taken in magnitude

CMRR stands for Common Mode Rejection Ratio It is the ability of an operational amplifier to reject the common-mode signals at the input terminals

Mathematically, this is expressed as:

\(CMRR = \frac{A_v}{A_c}\)

Av = Differential gain

Ac = Common mode gain

**Hence if Common mode gain(A _{c}) decreases, CMRR increases.**

__Important Points__

For practical operational amplimers, The Common Mode Rejection Ratio (CMMR) should be 80 to 100 dB.

For Ideal operational amplimers, The Common Mode Rejection Ratio (CMMR) should be infinity.

You are looking for an op-amp to be used for large signal amplification of high frequency signals. Name just one specification you would look for first of all

Option 3 : Slew rate

- The output of an Op-Amp can only change by a certain amount in a given time. This limit is called the slew rate of the Op-Amp.
- Op-Amp slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
- This usually happens at high frequencies, i.e. when the rate of change of signals is large.
- When the rate of change is large, it becomes difficult for the Op-Amp to track such fast variations.

This is best explained with the help of the given waveform:

The Slew-Rate of an Op-Amp for a give sinusoidal input is given by:

Slew Rate = 2π.f.V

Where,

f = The Highest signal frequency component (in Hz).

V = the maximum peak voltage of the signal.

Option 4 : how fast its output voltage can change

**Slew rate:**

- The slew rate of an operational amplifier indicates
**how fast its output voltage can change**. - The output of an Op-Amp can only change by a certain amount in a given time. This limit is called the slew rate of the Op-Amp.
- Op-Amp slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
- This is best explained with the help of the given waveform:

__Mathematical Analysis__:

The maximum rate of change of the output voltage in response to a step input voltage is defined as the slew rate of the op-Amp.

Mathematically, the slew rate is defined as:

\(S.R = {\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}\)

When the input is a sinusoid given as:

Vi (t) = Vm sin2πf t

The rate of change of input:

\(\frac{{d{V_i}}}{{dt}} = {{\rm{V}}_m}2{\rm{\pi }}{{\rm{f}}}{\rm{cos}}2{\rm{\pi }}{{\rm{f}}}{\rm{t}}\)

The maximum rate of change will be:

|Vm 2πf cos2πf t|max

\({\left. {\frac{{{d_{Vi}}}}{{dt}}} \right|_{max}} = {V_m}2\pi {f} =Slew~Rate\)

Option 4 : 1.50 V/μs

**Concept:**

Slew rate is defined as the maximum rate of change of output voltage per unit of time under large-signal conditions.

i.e. \(SR = {\left. {\frac{{d{V_o}}}{{dt}}} \right|_{max}}V/\mu s\)

**Calculation:**

Given that, i_{max} = 150 μA

C = 100 pF

Considering a charging current of a capacitor i_{max}

\({i_{max}} = C\frac{{dV}}{{dt}}\)

Hence, \(\frac{{dV}}{{dt}} = \frac{{{i_{max}}}}{C} = \frac{{150\mu }}{{100p}} = 1.5\;V/\mu s\)

The slew rate of a capacitor is 1.5 V/μs.Option 3 : Slew rate

Slew rate is the maximum rate of change of output voltage with respect to time.

Slew rate limits the maximum frequency of operation of op-amp

The slew rate is usually measured in volts per microsecond.

Mathematically,

Vin = Vm sin 2πfmt and

V0 = AVin

then, Slew rate = \({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}} = 2{\rm{\pi }}{{\rm{f}}_{{\rm{max}}}}{{\rm{V}}_0}_{{\rm{max}}}{\rm{\;}}\)

∴ We can say that, the signal bandwidth fm is limited by the Slew rate.

Option 2 : 250

__Concept:__

**Slew rate**

It is the maximum rate of change of output voltage for all possible input signals.

It is defined as:

\(SR = {\left| {\frac{{d{V_0}}}{{dt}}} \right|_{max}}\frac{{volts}}{{\mu sec}}\)

__Calculation:__

Given slew rate is 5 V/ms and the rate of change of the input signal is 0.2 V in 10 ms

\(\frac{{d{V_0}}}{{dt}} = \frac{{5V}}{{ms}}\)

\(\frac{{d{V_{in}}}}{{dt}} = \frac{{0.2}}{{10\;ms}}\)

Required voltage gain is

\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{\frac{{d{V_0}}}{{dt}}}}{{\frac{{d{V_{in}}}}{{dt}}}}\)

\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{d{V_0}}}{{dt}} \times \frac{{dt}}{{d{V_{in}}}}\)

\(\frac{{{V_o}}}{{{V_{in}}}} = \frac{5}{{1\;ms}} \times \frac{{10\;ms}}{{0.2}}\)

\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{50}}{{0.2}}\)

Option 2 : common-mode rejection

CMRR (Common mode rejection ratio) is defined as the ratio of differential-mode voltage gain (Ad) and the common-mode voltage gain (Ac).

Mathematically, this is expressed as:

\(CMRR = \frac{A_d}{A_c}\)

Ad = Differential gain

Ac = Common mode gain

** Note: **Ideally, the common-mode gain of the Op--Amp should be zero., i.e. it must give a zero output for common input at both the inverting and non-inverting terminal. ∴ The CMRR of an ideal Op-Amp is infinity.

Option 4 : 10^{7}

**Concept:**

**Common mode rejection ratio(CMRR): **It is the “ratio of the differential voltage gain (A_{d}) to the common-mode voltage gain (A_{cm})”.

The higher the value of CMRR the better is the matching between two input terminals and smaller is the output common-mode voltage.

CMRR = \( \frac{{{A_d}}}{{{A_{cm}}}}\)

It is infinity ∞ for the ideal op-amp.

**Calculation:**

Given that, common-mode gain = 0.01

Differential mode gain = 105

CMRR=\(\frac{{{{10}^5}}}{{0.01}}\) = 10^{7}.

Option 4 : 102 dB

**Concept:**

Two or more amplifiers circuits can be cascaded to increase the gain of an ac signal.

The overall gain Av can be calculated by simply multiplying each gain together. This is as shown:

Mathematically, this is defined as:

\({{\rm{A}}_{\rm{v}}} = {{\rm{A}}_{{{\rm{v}}_1}}}.{{\rm{A}}_{{{\rm{v}}_2}}}.{{\rm{A}}_{{{\rm{v}}_3}}} \ldots .\)

In dB, the overall gain is defined as:

\({A_v}\left( {dB} \right) = 20\;log{A_v}\)

This can be written as:

\({A_v}\left( {dB} \right) = 20\;log{A_v} = 20\;{\rm{log}}({A_{{v_1}}}.{A_{{v_2}}}.{A_{{v_3}}} \ldots .\;\)

\({A_v}\left( {dB} \right) = 20log{A_{v1}} + 20\;log{A_{v2}} + \;20\;log{A_{v3}}\; + \ldots \)

Hence,

\({A_v}\left( {dB} \right) = {A_{v1\left( {dB} \right)}} + {A_{v2\left( {dB} \right)}} + \;{A_{v3\left( {dB} \right)}}\; + \; \ldots \)

∴ If the gain of each amplifier stage is expressed in decibels (dB), the total gain will be the sum of the gains of individual stages.

**Calculation:**

The voltage gain of each amplifier = 50

Voltage gain of the combined amplifier = 50 × 50 × 50 = 125000

= 101.9 dB

The output offset voltage of the given circuit is (given V_{IO} = 1.2 mV):

Option 4 : 91.2 mV

__Concept:-__

**Output offset voltage (V _{oo})-** It is the output voltage of op-amp when both inputs are zero. V

**Input offset voltage (V _{IO})-** It is a non-zero voltage that should be applied as difference input to make the output of op-amp zero.

**Feedback factor (β)-** The feedback factor-beta is defined as the portion of the output voltage that is fed back to the differential op-amp input (input source grounded).

**\({\rm{β }} = \frac{{{{\rm{R}}_{{\rm{in}}}}}}{{{{\rm{R}}_{\rm{f}}} + {{\rm{R}}_{{\rm{in}}}}}}\) **-----(1)

\({{\rm{V}}_{{\rm{oo}}}} = {{\rm{V}}_{{\rm{IO}}}}\left( {\frac{1}{{\rm{β }}}} \right)\) -----(2)

__Calculation:-__

**Given**

Input offset voltage (V_{IO})= 1.2 mV

Input Resistance (R_{in})= 2 kΩ

Feedback Resistance (R_{f})= 150 kΩ

From equation 1, the value of β is

\({\rm{β }} = \frac{{2{\rm{k\Omega }}}}{{\left( {150 + 2} \right){\rm{k\Omega }}}}\)

β = 0.01316

∴ \(\frac{1}{β } = 76\)

Now, From equation 2, the output offset voltage is-

V_{oo}= 1.2× 76 mV

V_{oo}= 91.2 mV

The slew rate for a 741 is 0.5 V / μ second. The combination of maximum frequencies for an undistorted sine-wave output of 10 V peak and 1 V peak is approximately:

(1) 8 kHz and 80 kHz

(2) 48 kHz and 4.8 kHz

(3) 20 kHz and 2 kHz

(4) 2 kHz and 20 kHz

Option 1 : 1

__Slew Rate__:

- The output of an operational amplifier can only change by a certain amount in a given time this limit is called the slew rate of the op-amp.
- Operational amplifier slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
- It can distort the waveform and prevent the input signal from being faithfully represented at the output if the slew rate is exceeded.

Mathematically, Slew Rate (SR) is calculated as:

\(SR = \frac{{dV}}{{dt}}\)

For sinusoidal signal, the slew rate is defined as:

**Slew rate = 2π × f × V**

f = frequency of the signal

V = The maximum peak voltage of the signal

__Calculation:__

Given input signals are 10 V and 1 V peak.

The maximum frequency allowed at the output is

\({{\rm{f}}_{\max }} = \frac{{slew\;rate}}{{2\pi V}}\)

**For V = 10 V**

\({{\rm{f}}_{\max }} = \frac{{0.5 \times {{10}^6}}}{{2\pi 10}}\)

\({{\rm{f}}_{\max }} = \frac{{0.5 \times {{10}^6}}}{{20\pi}}\)

**f _{max }= 7957.74 Hz**

**f _{max} ≈ 8 kHz**

**For V = 1 V:**

\({{\rm{f}}_{\max }} = \frac{{0.5 \times {{10}^6}}}{{2\pi }}\)

**f _{max }= 79557.47 Hz**

**f _{max} ≈ 80 kHz**

__Note__:

Slew rate-distortion can be represented as shown:

Option 1 : 1.0 MHz

__Concept:__

Slew rate is the maximum rate of change of output voltage with respect to time.

Slew rate limits the maximum frequency of operation of op-amp

The slew rate is usually measured in volts per microsecond.

Mathematically,

Vin = Vm sin 2πfmt and

V0 = AVin

then, Slew rate = \({\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}} = 2{\rm{\pi }}{{\rm{f}}_{{\rm{max}}}}{{\rm{V}}_0}_{{\rm{max}}}{\rm{\;}}\)

∴ We can say that the signal bandwidth fm is limited by the Slew rate.

__Calculation:__

Slew rate \(= \frac{{{\rm{\Delta }}{V_0}}}{{{\rm{\Delta }}{V_i}}} × \frac{{{\rm{\Delta }}{V_i}}}{{{\rm{\Delta }}t}} = {A_{CL}}\left( {2\pi {f_m}{V_m}} \right)\)

Given that, slew rate = 62.8 V/μsec

62.8 × 10^{6} = 6.28 × 10 × f_{m}

f_{m} = 1 MHz

Option 2 : 164.8 mV

__Concept__:

For an op-amp, the output is given by:

Vo = Ad Vd + Ac Vc

Vd is the differential voltage at the input of the Op-Amp, i.e.

V_{d} = V1 – V2

V_{c} is the common-mode input of the Op-Amp, i.e.

\({V_c} = \frac{{{V_1} + {V_2}}}{2}\)

Also, the CMRR (Common mode rejection ratio) for a differential amplifier is defined as:

\(CMRR=\frac{A_d}{A_c}\)

__Calculation__:

Given: A_{d} = 4000, CMRR = 150

\(150=\frac{4000}{A_c}\)

A_{c} = 26.66

With V_{1} = 200 μV and V_{2} = 160 μV:

Vd = V1 – V2 = 40 μV

\({V_c} = \frac{{{200μ} + 160μ}}{2}=180μ V\)

The output voltage for the given Op-Amp will be:

Vo = Ad Vd + Ac Vc

V_{0} = (4000)(40μ) + (26.66)(180μ)

V_{0} = (160 + 4.8) μV

**V0 = 164.8 μV**

Option 1 : \(\frac{1}{{12\pi }}\;MHz\)

**Concept:**

The bandwidth of the amplifier is given by f_{o} which is

\({f_o} = \frac{{slew\;rate \times {{10}^6}}}{{2 \times \pi \times {V_m} \times {A_{CL}}}}\)

Where V_{m} = peak output voltage

A_{CL} = Close loop gain

**Calculation:**

Slew rate = 2 V/μs

V_{m} = 12 V

Now, slew rate SR ≥ \(\frac{{{A_{CL}}{V_m}{\omega _o}}}{{{{10}^6}}}\)

Now, assuming A_{CL} = 1

\({f_o} = \frac{{SR \times {{10}^6}}}{{2\pi {V_m} \times {A_{CL}}}} = \frac{{2 \times {{10}^6}}}{{2\pi \times 12}} \)

\(= \frac{1}{{12\pi }}\;MHz\)

Option 4 : Step input

**Slew rate:**

- The output of an Op-Amp can only change by a certain amount in a given time. This limit is called the slew rate of the Op-Amp.
- Op-Amp slew rate can limit the performance of a circuit if the slew rate requirement is exceeded.
- This is best explained with the help of the given waveform:

__Mathematical Analysis__:

The maximum rate of change of the output voltage in response to a **step input voltage** is defined as the slew rate of the op-Amp.

Mathematically, the slew rate is defined as:

\(S.R = {\left( {\frac{{d{V_0}}}{{dt}}} \right)_{max}}\)