Question

For the closed-loop amplifier circuit shown below, the magnitude of open-loop low-frequency small signal voltage gain is 40. All the transistors are biased in saturation. The current source \(I_{SS}\) is ideal. Neglect body effect, channel length modulation, and intrinsic device capacitances. The closed-loop low-frequency small signal voltage gain \(\frac{v_{out}}{v_{in}}\) (rounded off to three decimal places) is: \begin{center} \includegraphics[width=6cm]{18.png} \end{center}

• A. 0.976
• B. 1.000
• C. 1.025
• D. 0.488

Hint:

For negative feedback amplifiers, use the closed-loop gain formula to relate the open-loop gain and feedback factor. The value of \(\beta\) often depends on the circuit's feedback configuration, so analyze it carefully.

Answer 1

The Correct Option is A

Step 1: Recognize the circuit configuration.
The circuit is a differential amplifier using a source-coupled pair and an ideal current source \( I_{SS} \). It operates with negative feedback to improve stability and control the gain. Step 2: Apply the closed-loop gain formula for negative feedback.
The closed-loop gain for a negative feedback amplifier is given by: \[ A_{CL} = \frac{A}{1 + A\beta}, \] where \(A\) is the open-loop gain, and \(\beta\) is the feedback factor. Step 3: Determine the feedback factor.
In this circuit, the feedback network directly connects the output to the input. Thus, the feedback factor \(\beta\) is approximately 1. Step 4: Calculate the closed-loop gain.
Substituting the given values, \(A = 40\) and \(\beta = 1\), into the formula: \[ A_{CL} = \frac{40}{1 + 40(1)} = \frac{40}{41} \approx 0.976. \] The closed-loop voltage gain is therefore approximately \(0.976\), which corresponds to the given option (1). Final Answer: \[ \boxed{0.976} \]