The GATE Grind

GATE 2016 EC – Question 23

Analog Circuits · Op-amp Circuits · 1 mark · Multiple choice

Consider the constant current source shown in the figure below. Let $\beta$ represent the current gain of the transistor.

[Figure: A reference voltage $V_{ref}$ from a Zener diode is applied to the non-inverting input of an op-amp. The inverting input is connected to the emitter of a PNP transistor, whose emitter connects to $+V_{CC}$ through a resistor $R$. The op-amp output drives the base through $R_2$, and the collector feeds the load $R_L$ to ground. $R_1$ sets the bias of the Zener.]

The load current $I_0$ through $R_L$ is

Diagram for GATE 2016 EC question 23
  1. $I_0 = \left(\frac{\beta + 1}{\beta}\right)\frac{V_{ref}}{R}$
  2. $I_0 = \left(\frac{\beta}{\beta + 1}\right)\frac{V_{ref}}{R}$
  3. $I_0 = \left(\frac{\beta + 1}{\beta}\right)\frac{V_{ref}}{2R}$
  4. $I_0 = \left(\frac{\beta}{\beta + 1}\right)\frac{V_{ref}}{2R}$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (B) $I_0 = \left(\frac{\beta}{\beta + 1}\right)\frac{V_{ref}}{R}$

Explanation

The op-amp forces the emitter node to match the reference, so the voltage across $R$ is $V_{ref}$ and the emitter current is $I_E = \frac{V_{ref}}{R}$. The load current is the collector current, $I_0 = I_C = \frac{\beta}{\beta + 1} I_E = \left(\frac{\beta}{\beta + 1}\right)\frac{V_{ref}}{R}$.