Question

Consider an NPN transistor amplifies in common-
emitter configuration the current gain of the
transistor is 100. If the collector current changes by
1mA, what will be change in emitter current?​

Answer 1

Answer:

ΔIe  = 1.01 mA

Explanation:

The formula for current gain in case of npn transistor with common emitter configuration is;

Current Gain = Change in Collector current / change in base current

I gain = ΔIc / ΔIb

We are give;

100 = ΔIc / ΔIb

ΔIb = 1/100 mA = 0.01 mA

ΔIe = ΔIc + ΔIb = 1 + 0.01

ΔIe  = 1.01 mA

Change in emitter current is 1.01 mA

Answer 2

If the collector current changes by  1 mA, the change in emitter current is 1.01 mA.

Explanation:

The current again common- emitter configuration is given by the formula:

μ = ΔIc/ΔIb

From question, current gain of transistor = 100

ΔIb = ΔIc/μ

On substituting the values, we get,

ΔIb = 1mA/100

ΔIb = 10⁻² mA

In common- emitter configuration, we know that,

ΔIe = ΔIb + ΔIc

Now, on substituting known values, we get,

ΔIe = 10⁻² mA + 1 mA

ΔIe = 0.01 mA + 1 mA

ΔIe = 1.01 mA