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AoE 2 edition emitter follower biasing question

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todorp:
Hi all, I was reading AoE 2e chapter on BJT transistors. When discussing the emitter follower configuration the authors state the following (p.71 - Follower with split supplies):
"Because signals often are near ground it is convenient to use symetrical positive and negative supplies. This simplifies biasing and eliminates coupling capacitors.
Warning: You must always provide a dc path for base bias current, even if it goes only to ground."

Could somene please clarify the warning for me: why is this necessary?

Thanks,
    Todor
 

Paul Rose:
In the attached schematic, C1 is a DC blocker, and R3 to ground establishes the DC voltage (zero in this case) that the AC voltage is centered around.  This allows you to figure out the emitter voltage.

If you don't have R3, what is the "center" voltage of your input waveform?

Paul Rose:
It might help to think of what would happen if you turned the amplitude of your AC source down to zero.   It would be like removing the AC source and blocking capacitor.  Then (unless you have R3) the base would be connected to nothing.  No base current, so no emitter current.  The output voltage would be the negative rail (due to the emitter resistor). 

You want to establish a stable DC system.   Base current can flow from ground (0v) through R3, through the B-E junction to the negative rail (-15v).

BTW: The R3 value I picked was too big given the emitter resistor.   Use the design rules from the book to pick a better value.

todorp:
Ok, I get it. Thanks a lot for the clear explanation.
    Tod

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