Question

The potential difference across the collector is a transistor, used in common emitter mode is $ 1.5V $ , with the collector resistance of $ 3k\Omega $ . Find (i) the emitter and (ii) the base current, if the d.c gain of the transistor is 50.

Answer 1

Hint: In a common emitter mode, the emitter is connected in external circuits with both the base and the collector of the transistor. The emitter current will be the sum of the collector and the base current in this case.

Formula used: In this solution, we will use the following formula
Current flowing across a resistance: $ I = \dfrac{V}{R} $ where $ V $ is the potential difference and $ R $ is the resistance.
- Current gain in a common-emitter configuration: $ \beta = \dfrac{{{I_C}}}{{{I_B}}} $ where $ {I_C} $ is the collector current and $ {I_B} $ is the base current

Complete step by step answer:
We’ve been given that the potential difference across the collector is a transistor is $ 1.5V $ in a transistor that is connected in common emitter mode. We also know the collector resistance as $ 3k\Omega = 3 \times {10^3}\Omega $ . Then the collector current can be determined as
 $ {I_C} = \dfrac{{{V_{CE}}}}{{{R_C}}} $
 $ \Rightarrow {I_C} = \dfrac{{1.5}}{{3 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\,A $
Now, as the load resistance is connected in series with the collector in the common-emitter configuration, the current gain of the common emitter transistor configuration is the ratio of the current in the collector to the current in the base. So, we can write
 $ \dfrac{{{I_C}}}{{{I_B}}} = \beta $
Which can be rearranged to write
 $ \dfrac{{{I_C}}}{\beta } = {I_B} $
Substituting the value of $ {I_C} = 0.5 \times {10^{ - 3}}A $ and $ \beta = 50 $ , we get
 $ {I_B} = 0.01 \times {10^{ - 3}}A $
In this type of configuration, the current flowing out of the transistor must be equal to the sum of currents flowing into the transistor and the emitter. So, we can write
 $ {I_E} = {I_B} + {I_C} $
Which gives us
 $ {I_E} = 0.5 \times {10^{ - 3}} + 0.01 \times {10^{ - 3}} $
 $ \Rightarrow {I_E} = 0.51 \times {10^{ - 3}}A $
Or $ {I_E} = 51mA $
Hence the emitter current will be $ 51mA $ .

Note:
To answer such questions, we should be familiar with the construction and the working of a transistor connected in a common emitter mode. The potential difference of the collector will be applied in the external circuit connecting the collector and the emitter. Any small change in base current for this configuration will result in a large change in the collector current. This can help remember the relation of the current gain as the ratio of the current in the collector to the current in the base.