Let's solve the problem step by step: --- ## Given: - \((W/L)_1 = 50/0.5 = 100\) - \(R_D = 2\,k\Omega\) - \(R_S = 200\,\Omega\) - \(I_D = 0.5\,mA = 0.0005\,A\) --- ## (a) **Small-signal voltage gain calculation** ### Step 1: **Find \(g_m\) (transconductance)** For a MOSFET in saturation: \[ g_m = \sqrt{2 K_n I_D} \] where \[ K_n = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} \] Alternatively (commonly used): \[ g_m = \frac{2 I_D}{V_{GS} - V_{th}} \] But if \(V_{GS} - V_{th}\) is not given, the following is often used: \[ g_m = \frac{2 I_D}{V_{ov}}, \quad V_{ov} = V_{GS} - V_{th} \] But you can also use: \[ g_m = \frac{I_D}{V_{ov}} \] But if \(K_n\) is not given, use: \[ g_m = \sqrt{2 \mu_n C_{ox} \frac{W}{L} I_D} \] Assuming standard process \(K_n\) is not given, we'll leave as \(g_m\). But for small-signal gain, you can use: \[ g_m = \frac{2 I_D}{V_{ov}} \] but let's proceed with the small-signal model. --- ### Step 2: **Small-signal model** The gain is: \[ A_v = \frac{v_{out}}{v_{in}} = - \frac{g_m R_D}{1 + g_m R_S} \] #### **Derivation:** - The source resistor \(R_S\) provides negative feedback. - **Output resistance** \(r_{o}\) is not considered since \(\lambda = 0\) unless otherwise specified. #### **Plug in values:** We need \(g_m\). If not given, use the approximation: \[ g_m \approx \frac{2 I_D}{V_{ov}} \] But if \(V_{ov}\) is not given, let's use typical values. #### **For calculation, let's use:** \[ g_m = \frac{2 I_D}{V_{ov}} \] Assume \(V_{ov} \approx 0.2\,V\) (a standard value for strong inversion). \[ g_m = \frac{2 \times 0.5\,mA}{0.2\,V} = \frac{1\,mA}{0.2\,V} = 5\,mA/V \] #### **Calculate gain:** \[ A_v = -\frac{g_m R_D}{1 + g_m R_S} \] \[ g_m = 5\,mA/V = 0.005\,A/V \] \[ R_D = 2\,k\Omega = 2000\,\Omega \] \[ R_S = 200\,\Omega \] \[ A_v = -\frac{0.005 \times 2000}{1 + 0.005 \times 200} \] \[ = -\frac{10}{1 + 1} = -\frac{10}{2} = -5 \] --- ## **(a) Final Answer:** \[ \boxed{A_v = -5} \] --- ## (b) **Input voltage at the edge of saturation (\(\lambda = 0\))** ### **At the edge of saturation:** \[ V_{DS} = V_{GS} - V_{th} \] - **Source voltage:** \(V_S = I_D R_S\) - **Drain voltage:** \(V_D = V_{DD} - I_D R_D\) - **Gate voltage:** \(V_G = V_{GS} + V_S\) - **Input voltage:** \(V_{in} = V_G\) #### **At the edge:** \[ V_{DS} = V_{GS} - V_{th} \] But, \[ V_{DS} = V_D - V_S \] \[ V_{GS} = V_G - V_S \] \[ V_{DS} = V_D - V_S = (V_{GS} + V_S) - V_S = V_{GS} \] But for the edge, \[ V_{DS} = V_{GS} - V_{th} \] So, \[ V_{GS} = V_{GS} - V_{th} \implies V_{th} = 0 \] which is not reasonable unless \(V_{th}\) is specified, usually about \(0.4 - 0.7\,V\). But let's write in terms of \(V_{th}\). --- ### **Find \(V_{in}\) at the edge:** 1. \(V_{GS}\) at the edge = \(V_{DS,sat} + V_{th}\) 2. \(V_{DS} = V_{GS} - V_{th}\) 3. \(V_S = I_D R_S\) 4. \(V_G = V_{GS} + V_S\) 5. \(\boxed{V_{in,edge} = V_{GS,edge} + I_D R_S}\) But we need \(V_{GS}\) at the edge; for a given \(I_D\): \[ I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 \] Let \(\beta = \mu_n C_{ox} \frac{W}{L}\): \[ I_D = \frac{1}{2} \beta (V_{GS} - V_{th})^2 \implies V_{GS} - V_{th} = \sqrt{\frac{2 I_D}{\beta}} \] So, \[ V_{GS,edge} = V_{th} + \sqrt{\frac{2 I_D}{\beta}} \] If \(\beta\) is not given, express in terms of \(\beta\): \[ V_{in,edge} = V_{GS,edge} + I_D R_S = V_{th} + \sqrt{\frac{2 I_D}{\beta}} + I_D R_S \] --- ### **Gain at the edge:** At the edge, \(V_{DS} = V_{GS} - V_{th}\), but the small-signal gain is still: \[ A_v = -\frac{g_m R_D}{1 + g_m R_S} \] where \[ g_m = \sqrt{2 \beta I_D} \] If you substitute for \(I_D\) at the edge, you can find \(g_m\). --- ## **Summary Table** | Parameter | Value / Expression | |----------------|-------------------| | \(A_v\) (a) | \(-5\) | | \(V_{in,edge}\) (b) | \(V_{th} + \sqrt{\frac{2 I_D}{\beta}} + I_D R_S\) | | \(A_v\) at edge | \(-\dfrac{g_m R_D}{1 + g_m R_S}\) with \(g_m = \sqrt{2 \beta I_D}\) | --- ### **If you provide \(V_{th}\) and process parameters (\(\mu_n C_{ox}\)), I can calculate specific numbers!** --- **Let me know if you have those values, or if you want to proceed with these symbolic answers!**