# Emitter Follower Theory Help

Discussion in 'General Electronics Chat' started by tobyw, Sep 13, 2013.

1. ### tobyw Thread Starter Member

Mar 21, 2013
37
4
I'm still plugging away trying to understand the theory and practice of transistors as amplifiers.

I can get a common emitter circuit to amplify the voltage from an LC resonant circuit, enough to drive headphones and pick up radio.

What I want to do now is create an emitter follower, which will allow me to drive a speaker from this source.

The voltage from my first transistor stage seems to be about 1Vpp from my scope readings.

The problem I have is understanding how to design my emitter follower to drive the speaker. I know the theory is to bias the base in order to supply a sufficient amperage to the emitter, but I don't know amps I should be aiming for - it's a 8ohm .5W small speaker.

Would anyone be able to show me a very basic set up where a emitter follower transistor arrangement can drive a small speaker like this from a 9v battery? Everything I can find on Google has loads of complicated bootstraps and feedback etc which make it hard to 'see' what is happening.

Does it matter what the current is coming out of my voltage amplifier (CE) or is it just the voltage that matters?

Thanks

2. ### MrChips Moderator

Oct 2, 2009
17,736
5,540
Current and voltage are interrelated via Ohm's Law I = V/R.
Change one and the other changes.

3. ### tobyw Thread Starter Member

Mar 21, 2013
37
4
Thanks - but I don't understand how that answers the question, probably because I know less than you are assuming...

For example, I can work out how to bias a transistor to amplify the radio signal. I choose a required collector current, and a quiescent dc voltage, calculate the required resistor values on the collector, emitter, and the two biassing resistors.

So I think I can still have the voltage swinging between, say, 1 and 5 volts, with a peak current through the collector of 1ma. But can't I also have the voltage swinging between 1 and 5 volts, with a peak current of 2ma? Or 3?

Is that not true? And in which case, isn't the current different while the voltage swing is the same? So are you just saying that it's the voltage on the next stage which determines the current always, and never the actual current coming out of the previous stage?

So what I'm not clear about is, when I take the output from the collector of my CE, and connect it to the next stage, is the current coming 'through' it irrelevant? Or is there no current in that sense?

4. ### #12 Expert

Nov 30, 2010
18,076
9,676
Run the math.
P = Esquared/R
1/2 = Esquared/8
E = 2
That's 2 volts RMS
You will need 5.66 volts peak to peak for a 1/2 watt sine wave.
With a single emitter follower transistor operating in Class A, you will need .7 amps peak collector current. (A 9 volt battery can't do that.)

Now that we have established that you were thinking in the wrong area when you said one or two milliamps, you can work on things like the current gain of your emitter follower transistor to find out how much current the previous stage will have to supply to its base.

5. ### tobyw Thread Starter Member

Mar 21, 2013
37
4
Thanks #12. So based on your maths, if I have a 8ohm speaker with a 0.2W rating, will this mean

P = ExE/R
0.2 = ExE/8
V = 1.26V RMS
V = 3.6Vpp
I = 1.26/8 = 158mA

So does this mean I need to bias the emitter follower with a quiescent current of 158mA, and a quiescent current of 1.26V? Or a swing of 3.6Vpp? I can find loads of stuff about how to determine the value for a common emitter transistor, but I can't find out anything on this..

BTW - when I was talking about 1 or 2ma, I mean on the initial stage - ie. the quiescent current on the first, common collector transistor. What I was trying to find out was whether choosing a higher quiescent current on this stage is necessary to drive a higher current on the next stage. It sounds like you are saying yes - but is this still true if the first stage is not directly coupled to the next? If I am biasing the emitter follower, then does this determine the current, or does the output current from the first transistor combine with the base current provided by the bias on the input of the second stage to drive the output current?

Last edited: Sep 13, 2013
6. ### ian field AAC Fanatic!

Oct 27, 2012
6,365
1,156
You won't get any voltage gain from an emitter follower - in fact you lose just a little, but the current gain is high.

You get the most voltage gain from common base, but the current gain is less than unity and they have very low input resistance.

The jack of all trades is the common emitter - you can make a fairly neat direct coupled amplifier with a cascode stage (common emitter driving common base) and buffer the output with a complementary pair of emitter followers.

7. ### adam555 Active Member

Aug 17, 2013
858
40
I was also studying amplifiers last week and the book I'm reading says the same as Ian above; the voltage gain is less than the unity with emitter follower amplifiers. There was a table at the end that might help you chose:

File size:
35.3 KB
Views:
271
8. ### tindel Well-Known Member

Sep 16, 2012
638
211
Emitter Follower amplifiers, assuming they are properly biased, will have a voltage gain of ~1... the output will be approximately 0.7V below the input using silicon technology. The current gain will be beta+1.