Let's solve the question step by step: --- ### **Part (a): Find the mirror ratio error** #### **1. Given Data** - \(\mu_n C_{ox} = 400 \mu \text{A}/\text{V}^2\) - \(V_{OV} = V_{GS} - V_{th} = 0.2 \text{ V}\) - \(V_{th} = 0.5 \text{ V}\) - Channel length modulation coefficient, \(\lambda = 0.5 \text{ V}^{-1}\) - Neglect channel length modulation effects (first part) - Reference current \(I_{REF} = 5\mu\text{A}\) (from the current source) - Target output current \(I_o = 10\mu\text{A}\) Assume all MOSFETs are identical. #### **2. Current Mirror Equation** For a MOSFET operating in saturation: \[ I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 \] For a simple current mirror (without channel length modulation), the current ratio (ignoring λ) is determined by the aspect ratios (W/L) of the MOSFETs. \[ \frac{I_o}{I_{REF}} = \frac{(W/L)_o}{(W/L)_{REF}} \] #### **3. With Channel Length Modulation** When considering channel length modulation: \[ I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2 (1 + \lambda V_{DS}) \] The current mirror ratio will be affected by the output voltage (\(V_{DS}\)) differences between the reference and output transistors: \[ \frac{I_o}{I_{REF}} = \frac{1+\lambda V_{DS,o}}{1+\lambda V_{DS,ref}} \] #### **4. Error in Mirror Ratio** The error in the mirror ratio due to channel length modulation is: \[ \text{Error} = \frac{I_o - I_{REF}}{I_{REF}} \] #### **5. Calculate \( V_{DS} \) values** - For \(M_1\) (reference): \(V_{DS,ref}\) = \(V_{GS}\) - For \(M_3\) (output): \(V_{DS,o}\) can be different (not given, but typically assumed same unless stated). But here, **the main error comes from channel length modulation**, so let's use the general formula for the error: \[ \frac{I_o}{I_{REF}} = \frac{1 + \lambda V_{DS,o}}{1 + \lambda V_{DS,ref}} \] Assume \(V_{DS,o} = V_{DS,ref}\) (if not otherwise specified), the ratio is 1, so error is zero. But if, say, \(V_{DS,o} = 2V_{DS,ref}\), the error is: \[ \text{Error} = \frac{I_o}{I_{REF}} - 1 = \frac{1+\lambda V_{DS,o}}{1+\lambda V_{DS,ref}} - 1 \] But **since values of \(V_{DS,ref}\) and \(V_{DS,o}\) are not given**, let's compute using typical values. - \(V_{GS} = V_{th} + V_{OV} = 0.5 + 0.2 = 0.7 \text{ V}\) - Assume \(V_{DS,ref} \approx V_{GS} = 0.7 \text{ V}\) - Suppose \(V_{DS,o} = 2 \text{ V}\) (since output could be higher). \[ \frac{I_o}{I_{REF}} = \frac{1 + 0.5 \times 2}{1 + 0.5 \times 0.7} = \frac{1 + 1}{1 + 0.35} = \frac{2}{1.35} \approx 1.48 \] \[ \text{Error} = 1.48 - 1 = 0.48 = 48\% \] But **if \(V_{DS,o} = V_{DS,ref}\), error is zero**. So, the error is **dependent on the difference between \(V_{DS,o}\) and \(V_{DS,ref}\)**. ### **Summary of Steps:** 1. Write down saturation current equation with channel length modulation. 2. Write the current ratio including λ. 3. Estimate typical \(V_{DS,o}\) and \(V_{DS,ref}\) values. 4. Calculate the error using the ratio formula above. --- ### **Part (b): Output Resistance of the Current Mirror** #### **1. Output Resistance Formula** The output resistance of a MOSFET in saturation: \[ r_o = \frac{1}{\lambda I_D} \] For the current mirror, the output resistance seen at the output is mainly due to the output transistor \(M_3\): \[ r_{out} \approx r_{o3} = \frac{1}{\lambda I_o} \] Given: - \(\lambda = 0.5 \text{ V}^{-1}\) - \(I_o = 10\mu \text{A}\) \[ r_{out} = \frac{1}{0.5 \times 10 \times 10^{-6}} = \frac{1}{5 \times 10^{-6}} = 0.2 \times 10^6 = 200\, k\Omega \] --- ## **Final Answers** ### **(a) Error in Mirror Ratio** \[ \boxed{ \frac{I_o}{I_{REF}} = \frac{1 + \lambda V_{DS,o}}{1 + \lambda V_{DS,ref}} } \] - For example, if \(V_{DS,o}=2\,V\) and \(V_{DS,ref}=0.7\,V\), then the error is approximately 48%. --- ### **(b) Output Resistance** \[ \boxed{ r_{out} \approx \frac{1}{\lambda I_o} = 200\, k\Omega } \] --- **Let me know if you want a more detailed calculation or explanation for any part!**