Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Determine the minimum value of current gain $\beta$ (up to nearest integer) required to put the transistor in saturation when ${V_{BB}} = + 5\,V$ . Assume that in saturation state, ${V_{BE}} = 0.8\,V$ and ${V_{CE}} = 0.12\,V$.

Last updated date: 21st Jun 2024
Total views: 394.8k
Views today: 9.94k
Verified
394.8k+ views
Hint: In this problem we need to find the value of current gain. Current gain is the ratio of the collector current to the base current. In order to find the collector current and the base current, apply Kirchhoff’s voltage law in the base-emitter loop and in the collector loop.

Complete step by step solution:
Let us understand current gain and voltage gain first. The current gain in the common-base configuration is defined as the change in collector current divided by the change in emitter current, when base-to-collector voltage is constant. Similarly, the voltage gain for the common base amplifier is the ratio of output voltage to the input voltage.
Let ${I_B}$ and ${I_C}$ be base current and collector current respectively, taking $+ 5V$ as input voltage ${V_{in}}$ , the summation of voltages in the base-emitter loop gives
$8 \times {10^3} \times {I_B} + {V_{BE}} - 5V = 0$
We assume that the transistor is in saturation, this implies
${V_{BE}} = {V_{BE(sat)}} = 0.8V$
$\Rightarrow 8 \times {10^3} \times {I_B} + 0.8V - 5V = 0$
$$\Rightarrow {I_B} = \dfrac{{4.2V}}{{80 \times {{10}^3}}} = 0.0525mA$$ --equation $$1$$
The base current is $$0.0525mA$$
Applying Kirchhoff’s voltage law in the collector loop, we have
$$\Rightarrow 5 \times {10^3} \times {I_C} + {V_{CE}} - 12\,V = 0$$
We assume that the transistor is in saturation, this implies
${V_{CE}} = {V_{CE(sat)}} = 0.12V$
Substituting this value in the above equation, we get
$$\Rightarrow 5 \times {10^3} \times {I_C} + 0.12V - 12\,V = 0$$
$$\Rightarrow {I_C} = \dfrac{{12 - 0.12}}{{5 \times {{10}^3}}} = 2.376mA$$
The emitter current is $$2.376mA$$
The current gain is given as
$\beta = \dfrac{{{I_c}}}{{{I_B}}}$
Substituting the values, we get
$$\beta = \dfrac{{2.376}}{{0.0525}} \approx 45.25$$

The minimum value of current gain $\beta$ (up to nearest integer) is $$45$$.

Note: In saturation mode, the transistor acts like a short circuit between the collector and the emitter. In saturation mode both the diodes in the transistor are forward biased. Be careful of the sign conventions when applying Kirchhoff’s voltage law. Remember that the current gain is the ratio of the collector current to the base current.