# common collector amplifier

In common collector amplifier, we know that the collector is commons since it is connected to AC ground. Now, the characteristics of a common collector amplifier is that it has unity voltage gain, to be exact, it is less than 1 due to internal resistance of the transistor, and has high input impedance and low output impedance. Now, the problem is, whenever i try to add a 8 ohm load (my speaker), the voltage gain drops drastically. I thought common collector amplifier is immune to loading effects since it is use as a buffer. What is happening? (It appears to me that it is suffering the same fate as common emitter (which drops voltage when a load is connected to it directly))

I thought common collector amplifier is immune to loading effects

A CC is not immune to loading since it has a low, but non-zero, output impedance.

The small signal output impedance of a CC amplifier is roughly:

$\dfrac{V_T}{I_E} + \dfrac{R_{tb}}{\beta + 1}$

where

$V_T \approx 25mV$ (assuming room temperature)

and $R_{tb}$ is the equivalent AC resistance connected to the base.

For example, if $I_E = 1mA$, the output impedance is at least 25 ohms.

So, while the output impedance of a CC amplifier is much lower than a CE amplifier, it is not immune to loading effects.

• Since this post is the same as my answer that I was going to submit, I'll addendum this. Small Signal "guessing"/estimation should be ruled out if your V/Vt is roughly 1, since the truncated Taylor Series approximation will be invalid. As well, properly biasing the transistor is important in any configuration and pushing it into Saturation can have unintended/non-linear consequences that need to be either avoided or accounted for. Since in saturation the Beta will be different, a much different Rtb may be seen affecting the loading characteristics. – Jeff Langemeier Dec 19 '12 at 17:30
• @JeffLangemeier, just to be clear, I want to point out that it's $v_{be}$, not $v_b$, that must be much less than $V_T$. In the case of a CC, the (typically) large impedance connected to the emitter provides local negative feedback such that $v_{be}$ can be "small" even while $v_b$ is "large", i.e., the CC is a large signal amplifier. – Alfred Centauri Dec 19 '12 at 21:10

You are apparently thinking of a emitter follower amplifier. To be clear, here is what we are talking about:

Yes, that is basically a impedance buffer. However, that does not make it "immune" to loading effects, only that it reduces the output loading effects onto the input by roughly the gain of the transistor. Let's say the transistor has a gain of 100. If the output load is 8 Ω, then it will appear as around 800 Ω at the input. That can still be a substantial load for some circuits.

• can we put the 8 ohm speaker as R1? I dont think it will work since from my reasoning, speaker should be place with a bypass capacitor so that only ac signals will pass and dc signals will be block. what do you think? where should speaker be place according to your drawing. thanks – WantIt Dec 2 '12 at 14:28
• isn't it that we should only fed our speakers with ac signal? not dc signal (thats why we use bypass to block dc level) – WantIt Dec 2 '12 at 14:29
• @vvav: Yes, most speakers don't like net DC. However, if you just want to get a result and the total AC+DC is still within the speaker's power limits, then you can get away with this. It would be better to put a capacitor in series to block DC. However, you asked about "common collector" amplifiers, so that was not a relevant detail to your question. – Olin Lathrop Dec 2 '12 at 17:06

Voltage should not drop is current capability of the transistor is such that can provides sufficient current (let's say 1-2A)

You can online simulate common collector amplifier on

http://www.cirvirlab.com/simulation/npn_bjt_common_collector_amplifier_online.php

And read more on

http://www.cirvirlab.com/index.php/tutorials/91-npn-bjt-common-collector-amplifier.html

• Hello, capslock. – Connor Wolf Dec 19 '12 at 11:21