# Common-Collector Amp Current

Discussion in 'General Electronics Chat' started by seanlikeskites, Jan 16, 2010.

1. ### seanlikeskites Thread Starter New Member

Jan 13, 2010
10
0
Im having trouble grasping what happens with the current in a common collector amp.

If i have read correctly the current at the emitter is (β+1) times the current at the base. Where does this current go. In my mind it should flow through the load resistor and round to the collector. But if it does this surely the voltage across the load resistor will be much greater than that at the input when it should be slightly less.

2. ### Audioguru Expert

Dec 20, 2007
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The base current goes to the emitter and the transistor amplifies the current.
The emitter voltage is 0.6V to 0.7V less than the base voltage.
Ohm's Law determines the current in the load resistor. Then the base current is determined by the emitter current, you had it backwards.

A common-collector transistor is also called an emitter-follower because the voltage at the emitter follows the voltage at the base (minus 0.6V to 0.7V).

3. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
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You did not state whether your common-collector amplifier was an NPN or PNP. It is worth noting that audioguru's description applies specifically to an NPN based common-collector amplifier.

hgmjr

4. ### ifixit Distinguished Member

Nov 20, 2008
649
117
hgmjr wrote...
"You did not state whether your common-collector amplifier was an NPN or PNP. It is worth noting that audioguru's description applies specifically to an NPN based common-collector amplifier. "

Why does it only apply to NPN?

Regards,
Ifixit

P.S. How do I turn on "HTML code"?

5. ### Audioguru Expert

Dec 20, 2007
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I said that the emitter voltage is less than the base voltage which describes an NPN common-collector transistor. A PNP has the emitter voltage higher (more positive) than the base voltage.

6. ### seanlikeskites Thread Starter New Member

Jan 13, 2010
10
0
So say the load resistor was 1kΩ and had 5V across it. The current at the emitter would be 5mA. So the base current would be 5mA/(β+1)?

Feb 17, 2009
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Dec 20, 2007
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Yes it is.

Jan 28, 2005
9,030
218
Precisely.

hgmjr