Topic: Analog Circuits
The zener diode shown in following figure is temperature compensated and current gain $\beta$ of transistor is very high. If current through 100 ohm resistor is 55 mA at $25^{\circ}C$, what is the approximate current through it at $65^{\circ}C$?
- A. 55 mA
- B. 54 mA
- C. 56 mA
- D. 100 mA
Answer: C. 56 mA
Explanation: $V_{base} = 6.2\text{ V}$, $I(25^{\circ}C) = 55\text{ mA}$.
$V_{BE}$ decreases by $2.5\text{ mV}$ with every $1^{\circ}C$ rise in temperature.
So, $V_{BE}(65^{\circ}C) = V_{BE}(25^{\circ}C) - 2.5 \times 10^{-3} \times 40 = 0.7 - 0.1 = 0.6\text{ V}$.
Voltage across $100\text{ }\Omega$ resistor $V_{100\text{ }\Omega} = 6.2 - 0.6 = 5.6\text{ V}$.
Current $I = \frac{5.6}{100} = 56\text{ mA}$.