Differential Amplifiers Questions and Answers

Q: A 5 mV, 1 kHz sinusoidal signal is applied to the input of an op-amp circuit shown below. Find the output voltage :

The output is \(\frac{1}{40\pi}(\text{cos}\, 2000\pi t - 1)\) mV.The circuit is an inverting integrator, \(V_{o} = -\frac{1}{RC}\int_{0}^{t} V_{in}\,dt\), with \(RC = 10^{5} \times 10^{-6} = 0.1\) s. For \(V_{in} = 5\,\text{sin}(2000\pi t)\) mV starting at t = 0:\[ \begin{aligned} \int_{0}^{t} 5\,\text{sin}(2000\pi t)\,dt \\ &= \frac{5}{2000\pi}(1 - \text{cos}\, 2000\pi t) \\ V_{o} &= -\frac{10 \times 5}{2000\pi}(1 - \text{cos}\, 2000\pi t) \\ &= \frac{1}{40\pi}(\text{cos}\, 2000\pi t - 1)\ \text{mV} \end{aligned} \]The constant \(-1\) inside the bracket is the integration constant fixed by starting from zero output. The other coefficients correspond to other values of RC (0.05 s gives \(1/20\pi\), 0.15 s gives \(1/60\pi\) and 0.2 s gives \(1/80\pi\)), not to the 100 k\(\Omega\) and 1 \(\mu\)F shown.

Q: For the following differential amplifier circuit, the output voltage \(V_{out}\) is :

The output is +10.0 V.The non-inverting input sits at the voltage divider value of \(V_{2}\):\[ V_{+} = 0.5 \times \frac{10}{1 + 10} = 0.455\ \text{V} \]For an ideal op-amp the output is the sum of the non-inverting and inverting gains, with \(R_{F}/R_{1} = 10\):\[ \begin{aligned} V_{out} &= \left(1 + \frac{R_{F}}{R_{1}}\right)V_{+} - \frac{R_{F}}{R_{1}}V_{1} \\ &= 11(0.455) - 10(-0.5) \\ &= 5.0 + 5.0 = +10.0\ \text{V} \end{aligned} \]Because \(R_{3}/R_{2} = R_{F}/R_{1}\), this is the standard difference amplifier, \(V_{out} = 10(V_{2} - V_{1}) = 10(1.0)\). The 100 kΩ load does not change an ideal op-amp's output. -10 V reverses the inputs, and ±1.0 V forgets the gain of 10.

Q: For the following OP-amp circuit, R1 = R2 = R3 = RF =1.0 kΩ, the output voltage Vout is :

The output is +1.0V.The circuit is an inverting summing amplifier. With equal resistors, the output is the negative sum of the inputs:\[ \begin{aligned} V_{out} &= -R_{F}\left(\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}\right) \\ &= -(1.0 - 5.0 + 3.0) = +1.0\ \text{V} \end{aligned} \]-1.0 V forgets the inversion. The 0.5 V values would follow if \(R_{F}\) were half the input resistors, which it is not.

111.

A 5 mV, 1 kHz sinusoidal signal is applied to the input of an op-amp circuit shown below. Find the output voltage :

Inverting op-amp integrator: input resistor R = 100 kilohm to the inverting input, feedback capacitor C = 1 microfarad, non-inverting input grounded, Vin = 5 mV at 1 kHz, output Vo

A.\(\frac{1}{20\pi}(\text{cos}\, 2000\pi t - 1)\) mV
B.\(\frac{1}{40\pi}(\text{cos}\, 2000\pi t - 1)\) mV
C.\(\frac{1}{60\pi}(\text{cos}\, 2000\pi t - 1)\) mV
D.\(\frac{1}{80\pi}(\text{cos}\, 2000\pi t - 1)\) mV

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The output is \(\frac{1}{40\pi}(\text{cos}\, 2000\pi t - 1)\) mV.

The circuit is an inverting integrator, \(V_{o} = -\frac{1}{RC}\int_{0}^{t} V_{in}\,dt\), with \(RC = 10^{5} \times 10^{-6} = 0.1\) s. For \(V_{in} = 5\,\text{sin}(2000\pi t)\) mV starting at t = 0:

\[ \begin{aligned} \int_{0}^{t} 5\,\text{sin}(2000\pi t)\,dt \\ &= \frac{5}{2000\pi}(1 - \text{cos}\, 2000\pi t) \\ V_{o} &= -\frac{10 \times 5}{2000\pi}(1 - \text{cos}\, 2000\pi t) \\ &= \frac{1}{40\pi}(\text{cos}\, 2000\pi t - 1)\ \text{mV} \end{aligned} \]

The constant \(-1\) inside the bracket is the integration constant fixed by starting from zero output. The other coefficients correspond to other values of RC (0.05 s gives \(1/20\pi\), 0.15 s gives \(1/60\pi\) and 0.2 s gives \(1/80\pi\)), not to the 100 k\(\Omega\) and 1 \(\mu\)F shown.

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112.

For the following differential amplifier circuit, the output voltage \(V_{out}\) is :

Op-amp: V1 = -0.5 V to the inverting input through R1 = 1.0 kOhm with feedback RF = 10 kOhm; V2 = +0.5 V to the non-inverting input through R2 = 1.0 kOhm, with R3 = 10 kOhm to ground; load R1 = 100 kOhm

A.+10.0 V
B.-10.0 V
C.-1.0 V
D.+1.0 V

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The output is +10.0 V.

The non-inverting input sits at the voltage divider value of \(V_{2}\):

\[ V_{+} = 0.5 \times \frac{10}{1 + 10} = 0.455\ \text{V} \]

For an ideal op-amp the output is the sum of the non-inverting and inverting gains, with \(R_{F}/R_{1} = 10\):

\[ \begin{aligned} V_{out} &= \left(1 + \frac{R_{F}}{R_{1}}\right)V_{+} - \frac{R_{F}}{R_{1}}V_{1} \\ &= 11(0.455) - 10(-0.5) \\ &= 5.0 + 5.0 = +10.0\ \text{V} \end{aligned} \]

Because \(R_{3}/R_{2} = R_{F}/R_{1}\), this is the standard difference amplifier, \(V_{out} = 10(V_{2} - V_{1}) = 10(1.0)\). The 100 kΩ load does not change an ideal op-amp's output. -10 V reverses the inputs, and ±1.0 V forgets the gain of 10.

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113.

For the following OP-amp circuit, R₁ = R₂ = R₃ = R_F =1.0 kΩ, the output voltage V_out is :

V1 = +1.0 V, V2 = -5.0 V and V3 = +3.0 V are applied through R1, R2 and R3 to the inverting (-) input; RF connects Vout back to the inverting input; the non-inverting (+) input goes to ground through a resistor

A.+1.0V
B.-1.0V
C.+0.5V
D.-0.5V

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The output is +1.0V.

The circuit is an inverting summing amplifier. With equal resistors, the output is the negative sum of the inputs:

\[ \begin{aligned} V_{out} &= -R_{F}\left(\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}}\right) \\ &= -(1.0 - 5.0 + 3.0) = +1.0\ \text{V} \end{aligned} \]

-1.0 V forgets the inversion. The 0.5 V values would follow if \(R_{F}\) were half the input resistors, which it is not.

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114.

If differential amplifier has a differential gain of 20,000. CMRR = 80 dB, then common mode gain is

A.2
B.1
C.\(\frac{1}{2}\)
D.0

Exam Year(s)

UGC NET - 2012,

Answer: A

The common mode gain is 2, option (A). The CMRR in decibels is \(20\log_{10}(A_{d}/A_{c})\), so 80 dB means a ratio of \(10^{4}\).

\[ \text{CMRR} = 10^{80/20} = 10^{4} \]

\[ A_{c} = \frac{A_{d}}{\text{CMRR}} \]

\[ = \frac{20{,}000}{10{,}000} = 2 \]

Option (B), 1, would need a CMRR of 20,000. Option (C), one half, would need a CMRR of 40,000. Option (D), 0, is the common mode gain of an ideal amplifier with an infinite CMRR, which a real one with 80 dB does not have.

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115.

CMRR (Common Mode Rejection Ratio) for a differential amplifier should be

A.Zero
B.Unity
C.Small
D.Large

Exam Year(s)

UGC NET - 2013,

Answer: D

The common mode rejection ratio is the differential gain divided by the common mode gain, \(CMRR = A_d/A_c\). A good differential amplifier amplifies the difference between the inputs but rejects signals common to both, such as noise and drift, so \(A_c\) should be as small as possible and CMRR as large as possible, typically 80 to 100 dB for a good op-amp. A CMRR of zero or unity would mean that the amplifier cannot tell the wanted signal from the interference, and a small value means poor rejection. The answer is large.

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