Amplifiers and Operational Amplifiers module quiz: voltage gain, the transistor amplifier, op-amp configurations and the comparator (O-Level Electronics) quiz

The output voltage minus the input voltage

The output current times the input voltage

The input voltage divided by the output voltage

The ratio of output voltage to input voltage

$20$

$0.05$

$3.8$

$0.8$

$2\ \text{dB}$

$40\ \text{dB}$

$20\ \text{dB}$

$100\ \text{dB}$

To stop it drawing any current

To convert it into a diode

To set a steady operating point so the whole signal stays in the active region without clipping

To make it store charge

$A_v = R_\text{in}/R_f$

$A_v = -R_f/R_\text{in}$

$A_v = 1 + R_f/R_\text{in}$

$A_v = R_f \times R_\text{in}$

$A_v = -R_f/R_1$

$A_v = 1 + R_f/R_1$

$A_v = R_1/R_f$

$A_v = 1 - R_f/R_1$

$+4$

$+5$

$-0.25$

$-4$

It sets a precise, stable gain determined by the resistor values

It makes the gain unstable

It removes the output entirely

It turns the amplifier into a rectifier

It has a gain of exactly 1

Its very high open-loop gain drives the output fully high or fully low

It produces no output

It acts as a fixed resistor

The size of the feedback resistor

The supply frequency

Which of the two input voltages is larger

The colour code of the resistors

It gives a continuously varying analogue output

It amplifies the sensor signal by a precise factor

It stores the sensor reading

It gives a clean high or low output that switches at a chosen threshold

$20\ \text{dB}$

$10\ \text{dB}$

$1\ \text{dB}$

$40\ \text{dB}$

By storing charge on the base

By controlling a much larger collector current that flows through the collector load resistor

By rectifying the input

By reducing the supply voltage