Let’s solve the question **step by step**. --- ### **(a) Error in Mirror Ratio** #### **Given Data** - \( I_{REF} = 5\,\mu A \), \( I_O = 10\,\mu A \) (desired) - \( \mu_n C_{ox} = 400\,\mu A/V^2 \) - Overdrive voltage (\( V_{OV} \)) = 0.2 V - Threshold voltage (\( V_{th} \)) = 0.5 V - Channel length modulation coefficient (\( \lambda \)) = \( 0.5\,V^{-1} \) - **All MOSFETs are identical**. #### **Step 1: Current Mirror Operation** The current through a MOSFET in saturation: \[ I_D = \frac{1}{2} \mu_n C_{ox} \left( \frac{W}{L} \right) (V_{GS} - V_{th})^2 (1+\lambda V_{DS}) \] But first, **neglect channel length modulation** (\( \lambda = 0 \)): \[ I_D = \frac{1}{2} \mu_n C_{ox} \left(\frac{W}{L}\right) (V_{OV})^2 \] #### **Step 2: Find \(\frac{W}{L}\)** For \( M_1 \), \( I_{REF} = 5\,\mu A \): \[ 5 \times 10^{-6} = \frac{1}{2} \cdot 400 \times 10^{-6} \cdot \left(\frac{W}{L}\right) \cdot (0.2)^2 \] \[ 5 \times 10^{-6} = 200 \times 10^{-6} \cdot \left(\frac{W}{L}\right) \cdot 0.04 \] \[ 5 \times 10^{-6} = 8 \times 10^{-6} \cdot \left(\frac{W}{L}\right) \] \[ \frac{W}{L} = \frac{5}{8} = 0.625 \] --- #### **Step 3: Mirror Ratio with Channel Length Modulation** Now include channel length modulation (\( \lambda \neq 0 \)). Assume \( V_{DS1} \neq V_{DS2} \), but typically in a simple mirror, \( V_{DS1} \approx V_{DS2} \). But in this topology, \( M_1 \) and \( M_2 \) might have different \( V_{DS} \) values due to the stack. Let: - \( M_1 \): \( V_{DS1} \) (usually grounded at source, gate at 5uA) - \( M_2 \): \( V_{DS2} \) (source at ground, drain at \( V_{DS2} \), which is higher due to the stack) \[ \frac{I_O}{I_{REF}} = \frac{1 + \lambda V_{DS2}}{1 + \lambda V_{DS1}} \] ##### - **Find \( V_{DS1} \) and \( V_{DS2} \):** Assume \( M_1 \) and \( M_2 \) have the same \( V_{GS} \). - For \( M_1 \), \( V_{DS1} = V_{GS1} \) (since source is grounded). - For \( M_2 \), \( V_{DS2} = V_{GS2} + V_{DS3} \). \( M_3 \) is in diode connection, so \( V_{DS3} = V_{GS3} \). Let’s assume **ideal case**: \( V_{GS1} = V_{GS2} = V_{GS3} \). So: - \( V_{DS1} = V_{GS} \) - \( V_{DS2} = V_{GS} + V_{GS} = 2V_{GS} \) Since overdrive voltage is 0.2V, and \( V_{th} = 0.5V \): \[ V_{GS} = V_{th} + V_{OV} = 0.5 + 0.2 = 0.7\,V \] \[ V_{DS1} = 0.7\,V \\ V_{DS2} = 1.4\,V \] --- #### **Step 4: Calculate Mirror Ratio** \[ \frac{I_O}{I_{REF}} = \frac{1 + \lambda V_{DS2}}{1 + \lambda V_{DS1}} \] Plug in values: - \( \lambda = 0.5\,V^{-1} \) - \( V_{DS1} = 0.7\,V \) - \( V_{DS2} = 1.4\,V \) \[ \frac{I_O}{I_{REF}} = \frac{1 + 0.5 \cdot 1.4}{1 + 0.5 \cdot 0.7} = \frac{1 + 0.7}{1 + 0.35} = \frac{1.7}{1.35} \approx 1.26 \] --- ##### **Error in Mirror Ratio:** - **Ideal ratio:** 2 (since \( I_O = 10\,\mu A, I_{REF} = 5\,\mu A \)) - **Actual ratio:** 1.26 So, **percentage error:** \[ \text{Error} = \frac{\text{Actual ratio} - \text{Ideal ratio}}{\text{Ideal ratio}} \times 100\% \] \[ \text{Error} = \frac{1.26 - 2}{2} \times 100\% = -37\% \] **But, wait:** The ratio should be 1 if \( I_{REF} \) and \( I_O \) transistors are the same size. If you want \( I_O = 10\,\mu A \), then \( M_2 \) should be twice the size of \( M_1 \), i.e., \( (W/L)_2 = 2 \times (W/L)_1 \). Assuming this, the ratio becomes: \[ \frac{I_O}{I_{REF}} = \frac{\left(\frac{W}{L}\right)_2}{\left(\frac{W}{L}\right)_1} \cdot \frac{1 + \lambda V_{DS2}}{1 + \lambda V_{DS1}} \] \[ = 2 \cdot \frac{1.7}{1.35} \approx 2 \cdot 1.26 = 2.52 \] \[ \text{Error} = \frac{2.52 - 2}{2} \times 100\% = 26\% \] --- ### **Final Answers for (a):** #### **(a) Stepwise Summary** - **Without channel length modulation (\( \lambda = 0 \)), mirror ratio is exactly 2.** - **With channel length modulation (\( \lambda = 0.5\,V^{-1} \)), mirror ratio is 2.52.** - **Error in mirror ratio:** \( \boxed{26\%} \) (the output current is 26% higher than intended). --- ## **(b) Output Resistance of the Current Mirror** The output resistance is primarily due to the output transistor (\( M_2 \)): \[ r_{out} = \frac{1}{\lambda I_O} \] Plug in values: - \( \lambda = 0.5\,V^{-1} \) - \( I_O = 10\,\mu A \) \[ r_{out} = \frac{1}{0.5 \times 10 \times 10^{-6}} = \frac{1}{5 \times 10^{-6}} = 200,000\,\Omega = 200\,k\Omega \] --- ### **Final Answers** ### **(a)** - **Error in mirror ratio:** \( \boxed{26\%} \) ### **(b)** - **Output resistance of the current mirror:** \( \boxed{200\,k\Omega} \) --- If you need the derivation or clarification of any step, let me know!