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# Current Amplification Help (page 2)

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By McGraw-Hill Professional
Updated on Sep 11, 2011

## Current Amplification Practice Problems

#### Problem 1

A bipolar transistor has a current gain, under ideal conditions, of 23.5 at an operating frequency of 1,000 Hz. The alpha cutoff is specified as 900 kHz. What is the maximum possible current gain that the device can have at 900 kHz?

#### Solution 1

Multiply 23.5 by 0.707 to obtain 16.6. This is the maximum possible current gain that the transistor can produce at 900 kHz.

#### Problem 2

Suppose that the peak-to-peak (pk-pk) signal input current in the aforementioned transistor is 2.00 mA at a frequency of 1,000 Hz. Further suppose that the operating conditions are ideal and that the transistor is not driven into the nonlinear part of the characteristic curve during any part of the input signal cycle. If the frequency is changed to 900 kHz, what will be the pk-pk signal output current?

#### Solution 2

First, note that the current gain of the transistor is 23.5 at a frequency of 1,000 Hz. This means that the pk-pk output signal current at 1,000 Hz is 2.00 μA × 23.5 = 47.0 μA. At 900 kHz, the pk-pk output signal current is thus 0.707 × 47.0 μA = 33.2 μA.

Practice problems of these concepts can be found at: Semiconductors Practice Test

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