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Question: In a silicon transistor, the base current is changed by \( 20 \) change of \( 0.02V \) in base to em...

In a silicon transistor, the base current is changed by 2020 change of 0.02V0.02V in base to emitter voltage and a change of 2mA2mA in the collector current.
Find the input resistance, βac{\beta _{ac}} .
If the transistor is used as an amplifier. Find the voltage gain of the amplifier with the load resistance 5k5k amplifier.
A. a)100 \\\ b) - 500 \\\
B. a)1000 \\\ b) - 500 \\\
C. a)2000 \\\ b) - 5000 \\\
D.a)10 \\\ b) - 5 \\\

Explanation

Solution

Hint : Here first we first convert the units into proper form, then use the formula to find the input resistance first then find out the current gain for ac alters by collector and base currents. Then voltage gain can be calculated by the additional load that is given as 5k5k .
For input resistance, Ri=ViIb{R_i} = \dfrac{{\vartriangle {V_i}}}{{\vartriangle {I_b}}} .
For alter beta current gain; βac=IcIb{\beta _{ac}} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {I_b}}} , and voltage gain,
Av=βacR0Ri{A_v} = - {\beta _{ac}}\dfrac{{{R_0}}}{{{R_i}}} .

Complete Step By Step Answer:
In order to solve this question first see what we have,
We have, change in collector current that is 2mA2mA and base current which is 20μA20\mu A .
We have also changed the input voltage as 0.02V0.02V .
Now,
Ri=ViIb=0.02V20μA=1000Ω{R_i} = \dfrac{{\vartriangle {V_i}}}{{\vartriangle {I_b}}} = \dfrac{{0.02V}}{{20\mu A}} = 1000\Omega
To find the alter ac current gain we have;
βac=IcIb=2mA20μA=100{\beta _{ac}} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {I_b}}} = \dfrac{{2mA}}{{20\mu A}} = 100
Also,
{g_m} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {V_i}}} = \dfrac{{2mA}}{{0.02V}} = \dfrac{{2 \times {{10}^{ - 3}}}}{{0.02}} = 0.1{\Omega ^{ - 1}} \\\ \\\
In the third part of the question, when it is used as an amplifier;
Load gain is given as 5k5k .
So, voltage gain;
Av=βacR0Ri=100×5kΩ1kΩ=500{A_v} = - {\beta _{ac}}\dfrac{{{R_0}}}{{{R_i}}} = - 100 \times \dfrac{{5k\Omega }}{{1k\Omega }} = - 500
Since gm{g_m} is equal to βacn\dfrac{{{\beta _{ac}}}}{n} .
Option A is the correct answer.

Note :
Here note that before applying the formula it is required to convert the units. The solution above in the current gain reveals that for an ac input at the base, the collector current will be above hundred times the magnitude of the base current. Generally AC and DC betas tend to have similar magnitudes at lower levels.