# Common Collector Transistor Question - Voltage?

#### cube01

Joined Mar 10, 2011
16
I've been working through the tutorials here on this site - what a great resource, and well written. My sincere thanks to the writers...

There is one thing that is bothering me though, concerning Common Collector transistor circuits and their output voltage:

Given the below circuit (taken from the tutorials here) I understand why there is current gain and why the voltage across Rload would be lagging behind by .6volts if there is no voltage gain from V1. What I don't understand is, why is there no voltage gain from V1? Why don't I see voltage from V1, at all, across Rload?

My multimeter says it is all dropped across Q1, but why? Shouldn't I still see some of it across Rload? It being in series and all?

-Cube

#### Jony130

Joined Feb 17, 2009
5,230
Voltage gain for Common Collector is smaller than one.
Vout = Vin - Vbe = Vin - 0.6V (for DC voltage)
So if we say that Common Collector has no voltage gain.
This will simply mean that voltage gain is less then one.

#### jimkeith

Joined Oct 26, 2011
540
While the DC 'gain' is less than unity, the incremental or AC gain is very, very close to unity.

#### cube01

Joined Mar 10, 2011
16
I understand all that... What I don't understand is why we don't see voltage across Rload from V1.

Since Rload is in series with V1, shouldn't we see some of V1's voltage across it?

#### jimkeith

Joined Oct 26, 2011
540
99% (depending upon beta or hFE) of the current through RL is the transistor collector current--the remaining 1% is the base current coming from Vin. If things are not so, you must have an error somewhere such as wrong pin-out or wrong type of transistor.

Joined Dec 26, 2010
2,148
Most of the load current comes from V1, but it is not possible for the Rload voltage to rise even as high as Vin let alone rise above it, assuming that the transistor is working normally: not broken down with excessive voltage, nor leaking severely.

The reason is that the base-emitter junction of the transistor requires forward bias for the transistor to conduct. Any tendency for the emitter voltage to rise above about Vin-0.6V (for a typical silicon transistor) would immediately* choke off the emitter current.

Edit: *Not the most fortunate choice of words , given that transistor action involves delays - what I meant here is that reduction of the current would follow as a direct consequence. This is a form of negative feedback, and results in the emitter voltage "following" the base, but with the negative voltage drop required to turn on the base-emitter junction.

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#### cube01

Joined Mar 10, 2011
16
The reason is that the base-emitter junction of the transistor requires forward bias for the transistor to conduct. Any tendency for the emitter voltage to rise above about Vin-0.6V (for a typical silicon transistor) would immediately* choke off the emitter current.

Edit: *Not the most fortunate choice of words , given that transistor action involves delays - what I meant here is that reduction of the current would follow as a direct consequence. This is a form of negative feedback, and results in the emitter voltage "following" the base, but with the negative voltage drop required to turn on the base-emitter junction.

Ahhh... I see.
Very good explanation, and answers my question perfectly... I had assumed it was some form of negative feedback.