# mystery of emitter follower output impedance

#### acelectr

Joined Aug 28, 2010
73
Hi, I'm working on the emitter follower amplifier and I've got confused at some place. Till now had no problems in determining other type of amplifiers' output impedances but the emitter follower is different I gues. The circuit schematic is attached. I've understood where the input impedance came from but why the output impedance is only r(e) and why not r(e)+[Rsig/(B+1)].

Appriciate any concern #### mjhilger

Joined Feb 28, 2011
119
Follow the emitter current. The only device it can source current to (no load) is the emitter resistor. If a load is present the current is shared only between the emitter resistor and the load. If you follow the current loop of the base, it is supplying the base resistor(s), but the loop also flows through the emitter resistor for the additional voltage seen at the base (plus the Vbe). In contrast the voltage at the emitter is the voltage across the emitter resistor.

#### acelectr

Joined Aug 28, 2010
73
I'm sorry I still do not get it. You're saying it supplies the base resistors but also flows to the emitter. That is the problem, than why not also including the base resistors? What's the voltage at the emitter matter has to do with this situation exactly? Did not exactly understand??? Is there any equation that can prove this?

#### Wendy

Joined Mar 24, 2008
21,848
Think in terms of impedance. If you have 10ma out the emitter the base is approximately...

IB = IE / β

If β = 200 the Ib = 50µa

This equation is much more an approximation than yours, but it gets the concept across.

If the emitter has a 1KΩ resistor, the base to ground resistance is approximately...

Rb = β Re

I use the dickens out it, it is a very useful characteristic.

For example... This will barely load a CMOS circuit (a very good thing) while lighting up the LED.

#### acelectr

Joined Aug 28, 2010
73
So what are you trying to say that passing the resistor(at the base) to the emitter will result a very small amount of resistance so that it'll be negligible???? for instance Rb/B being so small thus neglecting it because of the very small current in the base?(i.e from the base- emitter currents relations)
don't know if I wrote my opinions clear but still some confusion there is !!!???

Joined Dec 26, 2010
2,147
Depending on the conditions, r(e) alone might or might not be a reasonable approximation. For a typical device running at about 1mA, you may expect this to be about 26 ohms.

If the current gain was say 100, and the signal source resistance was only 100 ohms, the second term Rsig/(B+1) would only add about an other ohm to the total. On the other hand, with a signal resistance of 2600 ohms, you would get about an extra 26 ohms, doubling the total to 52.

The value of the external emitter resistor RE could then be added in parallel, but normally this won't make much difference.

If Rsig was very big, the effect of base bias resistors in parallel might be significant, so you would need to find the parallel equivalent.

By the time you were taking all this into account, you might also want to think about ohmic resistances in the base and emitter... in fact you would probably be better off using a full transistor model, perhaps in a simulation program

#### mjhilger

Joined Feb 28, 2011
119
No, there is no approximation. Re is supplied current only by the emitter. Impedance is still Z = V/I. The only V at the emitter is Re * Ie (true for both DC analysis and in your case AC analysis as there is no emitter bypass cap), so Z as seen by a load is Ve / Ie which is Re. Calculating impedance from the base looking into the circuit V = Rb*Ib + Vbe + (Beta +1)Ib * Re. The Re is in the base impedance calculation because it is part of the loop voltage as seen by the input.

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#### mjhilger

Joined Feb 28, 2011
119
Part of the problem is the picture you have supplied. While it shows Ib entering the node between the controlled current source (diamond) and Re, the only current at Re is Ie. The node at the top of Re should be considered V=0 as the controlled current source is ideal and provides no voltage gain above ground.

Joined Dec 26, 2010
2,147
Well, I can't see that. If this is meant to be an emitter follower, the output impedance is most definitely a function of the emitter current and the signal source impedance.

If we take this to the extreme case of a current-fed base, the output impedance rises to approximate the common-emitter value.

#### mjhilger

Joined Feb 28, 2011
119
Can we agree that output impedance can be determined by changing the load impedance until the maximum power is transfered to the load? If so, go through the equations and determine where the second derivative dP/dr (delta power / delta load resistance) = 0. You will find that this occurs when Rl = Re.
Keep in mind that the schematic shown is an idealized model meant to show the breakdown for measurements. In the real world other things come into play that are not shown on the model. And there may be some small connection to the base impedance, but for this model, there is not.

#### PRS

Joined Aug 24, 2008
989
Think of the base to emitter junction as a diode, which it actually is. And what is the resistance of a diode? It is called re. It is a parameter of the material from which the transistor is made from. For a silicon BJT this parameter is experimentally determined as VT/Ic where VT is obtained experimentally, but Ic is a choice of the circuit designer. Experimentally, at room temperature, VT is 26mV per degree celsius. So, by re = VT/Ic, if Ic is 1 mA, then re = 26 ohms at room temperature.

When you make a transistor operate in the Emitter Folower mode, you are putting this low resistance of re in parallel with RE, the resistor biasing the voltage at the emitter. Hence at 1 mA the resistance is less than re.

Joined Dec 26, 2010
2,147
The signal impedance has a very definite effect, even with this simple model. We are given that ib = ie / (β + 1), also Rin = (β + 1)*(re + RL).

If then we change the load current by a small amount delta IL, the input current will change by delta IL / (β+ 1), and the base voltage will change by Rsig*delta IL / (β+1)

The total change in emitter voltage is therefore delta IL *re + delta IL * Rsig /(β+ 1). This gives us an incremental resistance of re + Rsig/(β + 1).

#### acelectr

Joined Aug 28, 2010
73
Guys , it has nothing to do with the maximum power transfer plus no approximation. There is something that is missed. I've tried to add the early effect of the transistor and so on but still can not catch what I've missed. I can say that there is deffinitely no approximation becaue my own circuit teacher at the university once made the eplanation of it but kinda forgat it .Now that I'm going back to the same concept I just can not remember how he proved it to me. There has to be a good explanation with the equations. The loop equation does not work I guess. Because than the base resistor should be also accounted in. I think somehow a high resistance stays in parallel with the internal emitter resistance, thus the equivalent result again still be quite close to that internal emitter resistance. Still waiting for a better explanation.
Though still thanx for the concerns. #### Wendy

Joined Mar 24, 2008
21,848
The emitter resistor (not the emitter resistance) is a form of negitive feedback. The gain of the transistor multiplies the resistance seen on the emitter, this is pretty simple overall. You do have to take into account the voltage drop of the emitter, but it's resistance does not enter too heavily into the Base to ground resistance.

The resistance is not negligable, just large. Try expanding it with a base bias network and see how negligable it is (it ain't).

BTW, you haven't even gotten to how the emitter current translates into a constant current on the collector.

#### acelectr

Joined Aug 28, 2010
73
There are no emitter resistaces only Rl which is the load the r(e) I am referring is the internal resistance of the component. Still waiting for clear proofs....

#### Wendy

Joined Mar 24, 2008
21,848
Try building one if you need proof. The equations I gave you are both valid and verified. Proof is in the reality, not the paperwork. You could build one and verify the equations, hmmm?

Of course, you might doubt the most basic of all equations, the mother of them all, that forms all the rest.

Ic = β Ib

Problem is, β is not a constant. It is an approximation.

A common misconception is just because you can not wrap your head around concepts doesn't mean no one can.

#### acelectr

Joined Aug 28, 2010
73
No, look, Ro being the emitter resistance can be proved, on paper, ideally. I am not talking about the reality. B may change of course maybe with temperature or with other stuffs, but ideally this thing is proved on paper, with not accounting any external complication such temperature and so on. What I am asking is that this proof was not made clearly on the book. This has nothing to do with B changing. Consider B is constant. The circuit is attached. Everything is on paper #### Jony130

Joined Feb 17, 2009
4,969
Well you book assume Vi = 0V (grounding the base).
So we find output resistance of a BJT (alone). We look into emitter terminal with
grounded base.
So if we assume Vi = 0V then Ro ≈ re

But if we analysis the hole amplifier and we find Ro for Vsig = 0V
then we get
Ro = re + Rsig/(β+1)

So I hope that this solve the mystery of Ro

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#### acelectr

Joined Aug 28, 2010
73
Thnx, this is also one of the things that I've thought of, hope the book assumed Vi=0V But we'll see whether the mystery is solved or not

#### Jony130

Joined Feb 17, 2009
4,969
hope the book assumed Vi=0V But we'll see whether the mystery is solved or not
There is no other way to assume Vi=0V and rce=∞ If we want Ro equal to re.