# ac analysis emitter follower

Discussion in 'General Electronics Chat' started by soedar, Apr 7, 2011.

1. ### soedar Thread Starter New Member

Apr 7, 2011
3
0
Hello,
I'm confused . How did they get to the following equation for the output impedance of an emitter follower? It's on page 270 of "Electronic devices and circuit theory", Robert L.Boylestad and Louis Nashelsky, 9th edition.

Equation 5.64: Zo=ro//RE//(βre/(β+1))

I would only get to Zo=ro//RE//βre

I have attached the ac equivalent circuit.

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2. ### saqib altaf New Member

Apr 7, 2011
2
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Dear u have written the wrong value of Z0 its like

Zo=ro//RE//Bre/(B+1)
1 can be ignored as B is a very high value so

Zo=ro//RE//re

as ro is very greater than re than we hav

Zo=RE//re

for example ro//re if ro=2ok ohm and re=12 ohmthan re//ro is nearly =re thats y we ignore ro.so our resultant ro is

Z=RE//re

3. ### soedar Thread Starter New Member

Apr 7, 2011
3
0
Hello Saqib,

I have attached a scan of the page. Check it out...

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4. ### soedar Thread Starter New Member

Apr 7, 2011
3
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The question is: How did they get to equation 5.64?

In theory to establish the output impedance Zo, the Vi should be set to zero, the base current Ib will then be zero. Further more the current source should be replaced by an open circuit.

So if I now look into the circuit from the output side, after doing all the things described above, I would get Zo=ro//RE//βre.

But they get Zo=ro//RE//(βre/(β+1)). How did they get the βre/(β+1)???

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,175
1,188
Ib will not gonna to be 0A
Because in general we can calculate output impedance by applying a voltage source Vin to the output node.
Zo = Vin/Iin

So I hope that now is is clear to see that
Ib = Vin/(βre+RB)

And also remember that Ie = Ib + Ic = (β+1)*Ib

Zo = Vin/Iin

Iin = Iro+IRe + Ie

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Last edited: Apr 7, 2011