Inverting and Non-Inverting Amplifiers Questions and Answers

Answer: A

The transfer function is \(-\left[\frac{R_{2}(R_{1}C_{s} + 1)}{R_{1}}\right]\), where \(C_{s}\) means C times s.

For an inverting amplifier the gain is \(-Z_{f}/Z_{in}\). Here \(Z_{f} = R_{2}\), and the input element is \(R_{1}\) in parallel with C.

\[ \begin{aligned} Z_{in} &= \frac{R_{1}}{1 + sR_{1}C} \\ \frac{V_{2}}{V_{1}} &= -\frac{R_{2}}{Z_{in}} \\ &= -\frac{R_{2}(sR_{1}C + 1)}{R_{1}} \end{aligned} \]

Option B swaps \(R_{1}\) and \(R_{2}\). Options C and D drop the minus sign of the inverting amplifier, and D also changes the sign inside the bracket.

Answer: A

The output is 10 V in magnitude.

An op-amp integrator gives \(v_{o} = -\frac{1}{RC}\int v_{i}\, dt\). With a constant input the output ramps linearly:

\[ \begin{aligned} RC &= 100 \times 10^{3} \times 1 \times 10^{-6} = 0.1\ \text{s} \\ |v_{o}| &= \frac{1 \times 1}{0.1} = 10\ \text{V} \end{aligned} \]

For the usual inverting integrator the output is -10 V, which is within the swing of a typical supply. 5 V and 15 V come from wrong time constants, and 0 V would mean no integration at all.

Answer: B

For millivolt-level signals a precision rectifier is used.

An ordinary diode rectifier passes nothing until the input exceeds the diode's cut-in voltage of about 0.7 V (silicon), so a signal of a few millivolts is lost entirely. A precision rectifier places the diode inside the feedback loop of an op-amp. The op-amp's large open-loop gain divides the diode drop by that gain, so the circuit rectifies even very small signals as if the diode were ideal.

An active filter and an amplifier do not rectify. A bridge rectifier uses two diodes in each conduction path and so has an even larger threshold, about 1.4 V.

Answer: B

A square wave is converted into a triangular wave by an integrator.

The output of an op-amp integrator is proportional to the time integral of its input, \(v_{o} = -\frac{1}{RC}\int v_{i}\,dt\). During each half-cycle a square wave is a constant, and the integral of a constant is a straight line, so the output ramps up during one half-cycle and down during the other. That makes a triangular wave.

A differentiator does the opposite; applied to a square wave it gives a train of spikes at the edges. A Schmitt trigger squares up a waveform rather than producing ramps, and a monostable gives a single pulse of fixed width when triggered.

Answer: B

An instrumentation amplifier is best suited for detection of output from a load cell.

A load cell uses strain gauges in a Wheatstone bridge. Its output is a small differential voltage, a few millivolts, riding on a large common-mode voltage of about half the bridge supply. An instrumentation amplifier has a high input impedance on both inputs, a precisely set differential gain and a very high common-mode rejection ratio, which is exactly what is needed to amplify the bridge output accurately.

A photodiode gives a current and is normally read with a transimpedance amplifier. Signals from distant stars and RF transponders need low-noise radio-frequency receivers, not a low-frequency differential amplifier.