Art of Electronics, Emitter Follower output impedance

Discussion in 'General Electronics Chat' started by raymondxloutlookcom, Nov 20, 2016.

1. raymondxloutlookcom Thread Starter New Member

Nov 20, 2016
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Hi, I've been reading the Art of Electronics third version. On page 80 the author says that the output impedance of an emitter follower is Zsource/(1+beta). How can I derive that?

Thanks very much in advance!

Raymond

2. Jony130 AAC Fanatic!

Feb 17, 2009
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Notice that Ie = Ic + Ib additional we know that Ic = Ib*β therefore Ie = Ib*β + Ib = Ib*(β + 1) and Ib = Ie/(β + 1)
And because of this any base resistor will be seen at emitter as a RB/(β + 1) smaller resistor and any emitter resistor is seen at the base as a (β+1) larger. V1 = Ib*RB + Vbe + Ie*Re (1) and if we substitute Ib = Ie/(β + 1) we have

V1 = Ie/(β + 1) * RB + Vbe + Ie*Re

Ie/(β + 1) * RB + Ie*Re = V1 - Vbe

Ie = (V1 - Vbe)/(RB/(β + 1) + Re)

As you can see RB at emitter is seen as a RB/(β + 1) resistor.

Or try this Iin = Ie = Ib * (β + 1)

Ib = Vin/RB

Rout = Vin / Iin = ??

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3. raymondxloutlookcom Thread Starter New Member

Nov 20, 2016
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The part that confuses me is ' be seen as '. What exactly does it mean?
If we express the final equation using Ib, we have Ib= (V1 - Vbe)/(RB+ Re(β + 1) ), does it mean Re at base is seen as a Re(β + 1) resistor?

Thanks!

4. Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, exactly.
RB resistor from emitter point of view (when we are looking into emitter) has a resistance RB/(β + 1)

5. raymondxloutlookcom Thread Starter New Member

Nov 20, 2016
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Thanks very much! Now I understand it!