GATE 2018 EC Q.3: solution and answer

Official answer: DSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Definition. A transimpedance (transresistance) amplifier takes a current in and gives a voltage out. Its gain is $R_m=\frac{v_o}{i_i}$, measured in $\text{V/A}=\Omega$. The ideal version is a current-controlled voltage source.

2. Input side (current sensing). Model the signal source as a Norton current source $i_s$ with a shunt resistance $R_s$. By the current divider, $i_i=i_s\frac{R_s}{R_s+R_{in}}$. For $i_i\to i_s$ we need $R_{in}\ll R_s$. So the ideal input impedance is low, with $R_{in}\to 0$. The input behaves like an ammeter, which must present almost no resistance.

3. Output side (voltage delivery). Model the output as a Thevenin source $R_m{i}_i$ in series with $R_{out}$, driving a load $R_L$. By the voltage divider, $v_o=R_m{i}_i\frac{R_L}{R_L+R_{out}}$. For $v_o$ not to depend on the load we need $R_{out}\ll R_L$. So the ideal output impedance is low, with $R_{out}\to 0$, as for any ideal voltage source.

4. Feedback check. A practical transimpedance amplifier uses shunt-shunt feedback, which samples the output voltage and mixes at the input as a current. With loop gain $A\beta$, $R_{in,f}=\frac{R_{in}}{1+A\beta }$ and $R_{out,f}=\frac{R_{out}}{1+A\beta }$. Both impedances are reduced by the factor $1+A\beta$, which matches the ideal from steps 2 and 3.

5. A good transimpedance amplifier therefore has low input impedance and low output impedance, which is option (D).