Question

When the switch $S_2$ is closed, the gain of the programmable gain amplifier shown in the following figure is:

• A. 0.5
• B. 2
• C. 4
• D. 8

Hint:

For programmable gain amplifiers, remember that the gain is determined by the ratio of the feedback resistor to the input resistor.

Answer 1

The Correct Option is B

For a programmable gain amplifier, the gain is determined by the resistors in the feedback and input paths. The general formula for the gain of an inverting amplifier is: \[ \text{Gain} = -\dfrac{R_f}{R_{in}} \] where $R_f$ is the feedback resistor and $R_{in}$ is the input resistor.
In this case, when switch $S_2$ is closed, the resistors connected are:
- $R_f = 2k\Omega$ (feedback resistor),
- $R_{in} = 1k\Omega$ (input resistor).
Thus, the gain of the amplifier is: \[ \text{Gain} = -\dfrac{2k\Omega}{1k\Omega} = -2 \] Since the gain is negative for an inverting amplifier, the magnitude of the gain is 2. Therefore, the correct answer is 2.