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Yet another way of looking at it...

67.164.165.165

In Reply to: RE: Optimum bias point? posted by rick_m on May 04, 2008 at 20:13:38

I just remembered another, simpler way to look at this problem.

At idle, since both transistors are conducting, the output impedance is (re + RE) || (re + RE)

= 1/2 (re + RE)

where re = 26 mV / IE and RE is the emitter resistor.

At higher currents, one transistor turns off and re approaches zero, so the output impedance is approximately RE.

Setting these two expressions equal (equal output impedance for high and low currents) gives re = RE. So for 0.22 Ohm emitter resistors,

re = 0.22 = 26mV/IE
or IE = 118 mA, which is close to Self's optimum.

Follow Ups:

RE: Yet another way of looking at it... - rick_m 17:41:03 05/10/08 (1)
- RE: Yet another way of looking at it... - andy_c 18:28:09 05/10/08 (0)

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