# How to calculate the output impedance of the BJT in a Common Collector Amplifier?

The book I am reading from (Electronic Devices and Circuits by David A Bell) analyzes the Common Collector circuit by using a H-parameter model as shown.

My confusion is in understanding the output impedance of the transistor. By looking at the circuit, I find the output impedance of the BJT to be RE(since [RE||{1/hoc}]~RE). But the book seems to disagree. To find the output impedance of the transistor it looks like the AC input source has to be shorted, the current due to hrcvo (which is ~vo since hrc~1) should be taken as Ib now and then the Ie corresponding to this Ib is used to arrive at the output impedance as shown below.

Why should we do all of this and not just take RE as the output impedance? I understand that there is a feedback from the output to the input of the amplifier but isn't the value of RE (and hence Ze) independent of that since we choose it to properly fix the operating point of the BJT?

The trick with the CC circuit is that there's local feedback going on.

Look at this schematic:

The output impedance is an indication of how much the voltage at $V_{out}$ would change if I draw a little bit of extra current out of that point.

What happens when I do that?

I'm drawing some extra current from the emitter. What would the emitter voltage do? It is not a hard voltage, it cannot supply unlimited current so the voltage at the emitter will decrease a little.

Now what happens?

That drop of the emitter voltage increases $V_{BE}$ a little. Increasing $V_{BE}$ means that the transistor is "opened further", it will supply more current from the collector.

This extra current from the collector is crucial as it flows to the emitter and tries to compensate for the extra current I was drawing there. This causes the CC circuit to have a very low output impedance. When designed properly this output impedance will be much lower than the value of $R_L$ (in your drawings it is $R_E$).

So you must take this effect of the collector current into account because it is significant.

As an experienced circuit designer I already know that the output impedance is roughly equal to $\frac{1}{gm} = \frac{1}{40I_c}$. At $I_c$=1mA that would give 25 ohms, which is generally much lower than the value of $R_L$

You might also want to look at this derivation which does not use the $h$ parameters.

• In the schematic, say I took the R<sub>L</sub> and put it at the collector. This would make a CE amplifier (if my understanding is right). Now if I draw extra current at this R<sub>L</sub> what would happen? – VenkiPhy6 Sep 29 '17 at 1:51
• In a CE circuit there is be no such feedback. In a CE (with the transistor not being in saturation) the voltage at the collector does not influence $V_{be}$ (in first order, I'm ignoring miller effect etc.). Remember that the collector (in a CE circuit) behaves sort of like a current source without feedback. And the emitter (in a CC circuit) behaves more like a voltage source due to the feedback. – Bimpelrekkie Sep 29 '17 at 6:04
• @Bimpelrekkie...I cannot follow oyur explanation. You are "drawing some extra current from the emitter"? How are you doing this? Which kind of feedback are you referring to? Current-controlled voltage feedback provided by Re? Do you remember that the input resistance of the CB stage is identical to the output resistance of the CC stage (1/gm) - ignoring the large parallel RL? As far as the output resistance of the CC stage is concerned, the only feedback influence is RL which has very small influence on r,out only (neglected by you). – LvW Jan 31 '20 at 9:49
• @VenkiPhy6..Do you really think that your question has been answered? – LvW Jan 31 '20 at 10:35

A Common Collector amplifier, also can be refereed to as an emitter follower.

An emitter follower provides no voltage gain. Typical voltage gain of CC amplifier is just slightly below unity.

However, it can provide current gain. This is why you will see the output stage of most amplifiers contain CC buffer amplifier. In the case of an opamp you will see 2, one to source current and one to sink current into the load.

$$R_o = \dfrac{\delta V}{\delta I}$$

So if an amplifier can provide high output current with little voltage sag, it must have a low output resistance.

As the author of the book illustrates, Ro of a CC-amp is mostly a function of the transistor not the emitter resistor.

Re sets the DC bias current, but the output signal current is sourced mostly through the transistor.

• Quote: "So if an amplifier can provide high output current with little voltage sag, it must have a low output resistance." Does this (questionable) assertion also to operational transconductance amplifiers (OTAs) ? – LvW Jan 31 '20 at 10:49

The formula or Ze above is ignoring a couple of things.

hfc = - (1+hfe) which gives him a negative Ze at first glance.

he might be assuming an absolute value for hfc but this gives me pause.

At the risk of stating the flaming obvious, the RE across which the OP signal is derived is in parallel with the parameter hoc which is normally in admittance units. If we invert RE to convert an admittance and add to hoc (BUT Measured at the quiescent collector current of the circuit, not some arbitrary datasheet value) then we have the output admittance. INvert again and we have the output Z in resistance units. Note that the current controlled current source on OP has infinite small signal impedance because it drives the programmed current into the load regardless of the small signal load, within limits.

There is no discrepancy here. Ic will control what hoc turns out to be so the bias resistors in the input circuit are involved in the output calculation as long as hoc is calculated . Hoc= hoe which is the slope of the Ic versus VCE curve at the bias current in question. You may need to use LTspice to plot the family of C for you because most dsheets are light on for this data/curve set.

I have just been doing this for a 2n 2369 cct where the Ic is 6.16 mA dc. RE is 390R. hoc is 117E-6 Siemens from LTspice simulation. The parallel combination is 373 ohms, only a little less than RE. That's a sanity check on the answer, it still has to be calculated. This problem is why I came here for inspiration.

Bimpelrekkie - fundamental flaw in argument. Apart from fact that hybrid pi analysis really doesn't work well especially in terms of the accuracy of the r-pi resistor. Re' which you calculated is in series with a parallel combination of the bias resistors (if we assume the source Z to be high) so the number 25 ohms is a wonderful calculation but does not solve the physical schematic problem. It is WRONG.

Derivation of the dynamic (small signal) output resistance r,o (based on the schematic diagram as shown in Bimpelrekkie´s answer):

1) Assuming a test voltage vo to be applied at the emitter node. The corresponding current into this node is io=ir-ie (conventional direction for ie)

2) As a first step, the resistance into the emitter r,e is calculated (later: RL is in parallel):

r,e=Vo/(-ie)=(veb-ib* Rin)/(-ie)=(-vbe-ib*Rin)/(-ie)=(vbe/ie)+Rin(beta+1)

r,e=(1/gm) + Rin/(beta+1) (1/gm=transconductance; ie/ib=beta+1)

r,e=(1/gm)[1+Rin/hie]

(This last expression shows the influence of current-controlled feedback, which - in accordance with feedback theory - enlarges the output resistance.)

3) Total output resistance ro=r,e||RL