Output resistance of the emitter-follower

Thread Starter

epsilonjon

Joined Feb 15, 2011
65
Hi.

I am just learning about the emitter-follower amplifier, but i'm struggling to understand the output resistance. My book says that it is roughly

\[R_{out}\approx (\frac{R_s}{\beta_{ac}})||R_E \approx \frac{R_s}{\beta_{ac}}\]
but I don't understand why it's not just \[R_{out} = \frac{V_e}{I_e} = R_E\]?

The voltage gain is

\[A_v = \frac{V_b}{V_e} = \frac{I_eR_E}{I_e(r'_e+R_E)}\]

which is roughly equal to 1 when the emitter resistor is much larger than the internal emitter resistance. If I now connect a load resistor \[R_L\] between the emitter and ground, we will have to replace \[R_E\] by \[R_E||R_L\] in the gain equation above, so it will be the load resistor's size relative to the emitter resistor which will determine how much the gain is decreased by.

But I thought the whole point in knowing the output resistance was so you know how small a load you can connect without affecting the circuit too much? If I follow my book's equation

\[R_{out}\approx \frac{R_s}{\beta_{ac}}\]

then I would think I could connect a much smaller load without affecting the gain?

Could someone help me to understand a little better?

Thanks :)

 

t_n_k

Joined Mar 6, 2009
5,448
If you are considering the actual gain of the emitter follower stage you would probably consider the overall gain going from the source [Vs] to the emitter. The gain equation you show considers the gain only from the base to the emitter. There's a difference between the two gain values.

We can link the small signal source & small signal base voltages with the relationship

\[v_b=\frac{R_{in}}{R_s+R_{in}}v_s\]

where

\[R_{in}=R_1||R_2||(1+\beta_{ac})(r_e+R_E||R_{load})\]
 
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Thread Starter

epsilonjon

Joined Feb 15, 2011
65
Ah yeah thanks, I forgot about that. But I still don't see how you get

\[R_{out}\approx (\frac{R_s}{\beta_{ac}})||R_E\].

Say I could somehow choose a voltage source with \[R_s=0\], then this equation above says that the output resistance would be roughly zero too (or certainly very small)? So I could connect a really small load and it would not change the circuit too much? But in this case there would be no attenuation so the voltage gain would be as in my original post, and again it would only be the load resistor's size relative to the emitter resistance which would determine how much the gain is decreased by. So clearly I cannot connect a really small load and expect the same gain.

Thanks.
 

t_n_k

Joined Mar 6, 2009
5,448
You can prove the output resistance relationship by finding the Thevenin equivalent at the emitter (relative to common / ground). That's a circuit analysis exercise.

It might also help if you try putting some actual numbers into the relationship and observe what happens as the ac load resistance varies. But keep in mind if you change the effective RE by adding a parallel AC coupled load resistance you also change the input resistance.
 

Ron H

Joined Apr 14, 2005
7,012
Why is r'e not in the output resistance equation? It is dependent in Ie, and is significant unless Rs is very high.
 

Ron H

Joined Apr 14, 2005
7,012
I realize this is outside the scope of the original problem, but the input impedance, and consequently the overall gain (including Rs) can be improved significantly by bootstrapping the base bias resistor. This requires an additional resistor and capacitor.
 

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KL7AJ

Joined Nov 4, 2008
2,208
Ah yeah thanks, I forgot about that. But I still don't see how you get

\[R_{out}\approx (\frac{R_s}{\beta_{ac}})||R_E\].

Say I could somehow choose a voltage source with \[R_s=0\], then this equation above says that the output resistance would be roughly zero too (or certainly very small)? So I could connect a really small load and it would not change the circuit too much? But in this case there would be no attenuation so the voltage gain would be as in my original post, and again it would only be the load resistor's size relative to the emitter resistance which would determine how much the gain is decreased by. So clearly I cannot connect a really small load and expect the same gain.

Thanks.
One factor that somewhat complicates the matter is that you get degenerative (current) feedback.

Eric
 

Jony130

Joined Feb 17, 2009
4,930
This diagram should help you find Rout

Rout = Vin/ Iin

Ie = Ib + Ic = (β+1) * Ib

And if you do all the math correct you end up with this solution

Rout = RE|| (re + (Rs||RB)/(β+1) )
 

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Thread Starter

epsilonjon

Joined Feb 15, 2011
65
Thanks for your help guys. The book i'm reading (Electronic Devices by Floyd) does give the equation in my original post (without derivation for some reason) so I guess they must be assuming Rs is very high, which I agree seems silly.

I took your advice Jony130 and derived the correct equation, and I see what is going on more clearly now.

Thanks again :)
 
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