To solve this circuit problem, let's first break down the information and approach required.

Given Information:

1. Transistors (W/L)$_{1,2}$ = (25/0.5): The width-to-length ratio for both transistors M1 and M2.
2. $\mu_n C_{ox} = 50 \, \mu A/V^2$: Process transconductance parameter.
3. $V_{TH} = 0.6 \, V$: Threshold voltage of both transistors.
4. Voltage across $R_{SS} = 0.5 \, V$: The voltage across the source resistance $R_{SS}$.
5. $A_v = 5$: Voltage gain.
6. $V_{DD} = 3 \, V$: Supply voltage.
7. $I_D = 1 \, mA$: Drain current through each transistor.

Approach:

We need to calculate the input and output common-mode voltages for this differential pair circuit.

Step 1: Calculate $R_{SS}$

The total current through the tail current source is $I_{SS} = 1 \, mA$, and the voltage across $R_{SS}$ is given as 0.5 V. Using Ohm's law, we can calculate $R_{SS}$ as: $R_{SS} = \frac{V_{R_{SS}}}{I_{SS}} = \frac{0.5 \, V}{1 \, mA} = 500 \, \Omega$

Step 2: Calculate $V_{GS}$

The gate-source voltage $V_{GS}$ can be calculated using the equation for the MOSFET in saturation: $I_D = \frac{1}{2} \mu_n C_{ox} \left( \frac{W}{L} \right) (V_{GS} - V_{TH})^2$ Substituting the known values: $1 \, mA = \frac{1}{2} \times 50 \, \mu A/V^2 \times \frac{25}{0.5} \times (V_{GS} - 0.6)^2$ Simplifying: $1 \, mA = 1250 \, \mu A/V^2 \times (V_{GS} - 0.6)^2$ $(V_{GS} - 0.6)^2 = \frac{1 \, mA}{1250 \, \mu A/V^2} = 0.0008 \, V^2$ $V_{GS} - 0.6 = \sqrt{0.0008} \approx 0.0283 \, V$ $V_{GS} \approx 0.6283 \, V$

Step 3: Input Common-Mode Voltage $V_{in,cm}$

The input common-mode voltage $V_{in,cm}$ is simply the gate-source voltage $V_{GS}$: $V_{in,cm} = V_{GS} = 0.6283 \, V$

Step 4: Output Common-Mode Voltage $V_{out,cm}$

The output common-mode voltage is taken as the voltage at the drains of the transistors, which can be calculated as: $V_{out,cm} = V_{DD} - I_D R_D$ We are not given $R_D$ directly, but we can use the voltage gain formula. Since the small-signal voltage gain $A_v = 5$, we know: $A_v = \frac{g_m R_D}{2}$ Where $g_m$ is the transconductance of the MOSFET. The transconductance $g_m$ is given by: $g_m = \sqrt{2 \mu_n C_{ox} \frac{W}{L} I_D}$ Substituting values: $g_m = \sqrt{2 \times 50 \, \mu A/V^2 \times \frac{25}{0.5} \times 1 \, mA} = \sqrt{125 \, mA/V} = 11.18 \, mA/V$ Now, using the gain formula: $5 = \frac{g_m R_D}{2} \implies R_D = \frac{5 \times 2}{g_m} = \frac{10}{11.18} \approx 0.894 \, k\Omega$ Now calculate $V_{out,cm}$: $V_{out,cm} = V_{DD} - I_D R_D = 3 \, V - 1 \, mA \times 0.894 \, k\Omega \approx 3 \, V - 0.894 \, V = 2.106 \, V$

Final Answers:

1. Input common-mode voltage $V_{in,cm}$ = 0.628 V
2. Output common-mode voltage $V_{out,cm}$ = 2.106 V

Would you like further details or explanations?

Related Questions:

1. What is the role of $R_{SS}$ in this circuit, and how does it affect common-mode rejection?
2. How would the common-mode voltages change if the supply voltage $V_{DD}$ were increased?
3. What is the significance of the small-signal gain $A_v = 5$ in determining the output?
4. How does the width-to-length ratio $W/L$ influence the transistor’s behavior in this circuit?
5. How does varying the tail current $I_{SS}$ affect the common-mode voltages?

Tip: In differential pairs, the input common-mode voltage is typically close to the gate-source voltage, which depends on the biasing of the transistors.