Electronic Devices - BJT Amplifiers - Discussion
Discussion Forum : BJT Amplifiers - General Questions (Q.No. 44)
44.
The dc emitter current of a transistor is 8 mA. What is the value of re?
320 
13.3 k
3.125
5.75
Discussion:
3 comments Page 1 of 1.
Upendar kadem said:
8 years ago
Transistor data sheets tell us that for a small signal bipolar transistor this internal resistance is the product of 25mV ÷ Ie (25mV being the internal volt drop across the Emitter junction layer), then for our common Emitter amplifier circuit above this resistance value will be equal to:
re = 25mV divided by ie.
re = 25mV divided by ie.
Chandu said:
1 decade ago
Can I have explanation.
Jiy@ said:
1 decade ago
25mV/8mA
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