Differential Amplifiers Questions and Answers

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.

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.

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.