Question:

Please convert the below content to rendered equations without any changes and errors.This is a **MOS differential pair with PMOS active loads** and a **resistor (R_1) between the two output nodes**. Let’s solve each part step-by-step. --- # (a) Differential input short-circuit transconductance (G_m) ### Step 1: Differential excitation For differential input [ v_{in1} = +\frac{v_{id}}{2}, \qquad v_{in2} = -\frac{v_{id}}{2} ] Transistors **M1 and M2** form the differential pair. Small-signal drain currents: [ i_{d1} = g_{m1}\frac{v_{id}}{2} ] [ i_{d2} = -g_{m2}\frac{v_{id}}{2} ] Assuming matched devices: [ g_{m1} = g_{m2} = g_m ] --- ### Step 2: Differential output current The differential output current is [ i_{out} = i_{d1} - i_{d2} ] Substitute: [ i_{out} = g_m\frac{v_{id}}{2} ------------------- \left(-g_m\frac{v_{id}}{2}\right) ] [ i_{out} = g_m v_{id} ] --- ### Step 3: Short-circuit transconductance [ G_m = \frac{i_{out}}{v_{id}} ] [ \boxed{G_m = g_m} ] --- # (b) Differential output resistance (neglect (R_1)) In differential mode, the circuit is symmetric. Each output node sees: * (r_{o2}) from **M2** * (r_{o4}) from **M4** Thus single-side resistance: [ r_o = r_{o2} \parallel r_{o4} ] For **differential output resistance** [ R_{out,diff} = 2(r_{o2} \parallel r_{o4}) ] --- ### Final answer [ \boxed{R_{out,diff} = 2(r_{o2} \parallel r_{o4})} ] --- # (c) Differential gain including (R_1) Now (R_1) connects the two output nodes. For differential signals, **the resistor splits into two halves**: [ R_1 \rightarrow \frac{R_1}{2} \text{ to ground (each half circuit)} ] --- ### Step 1: Effective output resistance Single-side resistance becomes [ R_{out,half} = (r_{o2} \parallel r_{o4} \parallel \frac{R_1}{2}) ] --- ### Step 2: Differential gain Differential gain: [ A_{vd} = G_m \cdot R_{out,diff} ] But differential resistance is [ R_{out,diff} = 2(R_{out,half}) ] Substitute: [ A_{vd} ====== g_m \cdot 2\left(r_{o2} \parallel r_{o4} \parallel \frac{R_1}{2}\right) ] --- ### Final result [ \boxed{ A_{vd} ====== 2g_m \left( r_{o2} \parallel r_{o4} \parallel \frac{R_1}{2} \right) } ] --- # Final Answers ### (a) [ \boxed{G_m = g_m} ] ### (b) [ \boxed{R_{out,diff} = 2(r_{o2} \parallel r_{o4})} ] ### (c) [ \boxed{ A_{vd} ====== 2g_m \left( r_{o2} \parallel r_{o4} \parallel \frac{R_1}{2} \right) } ] --- If you'd like, I can also show the **half-circuit method (the trick professors expect in exams)** which makes this problem **10× faster to solve**.