# Emitter-Follower Design

Discussion in 'Analog & Mixed-Signal Design' started by elec_eng_55, Apr 1, 2019.

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1. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
Hi:

If you want to design an emitter-follower power amplifier that supplies 562mW (real power)
to an 8Ω load, what steps would you follow to determine the required supply voltage,
supply current and Vce? Assume load and RE are both 8Ω, I did come across a site that had such a guide but I can't find it any more.

David

Jun 4, 2014
7,901
1,949
Homework?

3. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
No I am an old retired "fart" trying to learn something new.
"Remember the human - Be courteous when replying to others."

Last edited: Apr 1, 2019

Jan 29, 2010
8,099
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5. ### BobTPH Senior Member

Jun 5, 2013
1,801
470
Are you talking about a single emitter follower, or a push-pull output stage, which is two emitter followers back to back?

A single emitter follower would be an odd configuration for a power amp.

Bob

May 13, 2018
214
3
7. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
Hi Bob.
I thought that I would just study the single e/f first then take another look at the push-pull.
David

8. ### MrChips Moderator

Oct 2, 2009
18,831
6,024
Hello David,

Don't be such a curmudgeon.

Asking if it is Homework is a perfectly valid question on AAC forums.

If you were a student seeking assistance with school homework, AAC forum responses can be custom tailored to assist students in better understanding course assignments.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,367
You want around 0.6W of power at 8Ω speaker. So we need I = √0.6W/8Ω = 274mA of an RMS current or 386mA peak current.
This correlates with a voltage drop across the speaker around 3V peak.
So, for RE = RL you need Vcc larger than 12V (single supply emitter follower) with the quiescent current around Ieq ≥ 3V/RE||RL = 3V/4Ω = 750mA

10. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
It is ignorant. End of discussion.

May 13, 2018
214
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thanks Jony

12. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
Ok. That makes good sense. The sim seems to work with the 12 volt supply but how did you determine that Vcc should be > 12V?

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,877
1,367
In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at emitter.

The RE and RL form a voltage divider so that the maximum negative voltage across is equal to:

VL = Veq * RL/(RL + RE)

Hence for RE = RL, we have VL = Veq *0.5

That means to get desired VL voltage across RL we need emitter voltage (Veq) to be equal to:

Veq = VL*2 = 3V*2 = 6V therefore Vcc = 2*Veq = 12V

14. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
12 volts would be enough for Vcc or would you recommend bumping it up a little?

15. ### wayneh Expert

Sep 9, 2010
16,099
6,212
No, it isn't. A significant portion of the inquiries we get here are homework, and a lot of those inquiries fail to mention that fact or omit that detail on purpose, hoping someone will produce an easy answer. The general policy around here is to not provide only the final answers to homework questions, but instead lead the student through the thinking process so they can solve such problems on their own.

Your first question is fine but is basic enough and oddly specific enough to sound like a homework problem. It's perfectly reasonable for folks here to ask.

16. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
Since the maximum output swing is ≈ 3 volts peak would any VE higher than 3 volts dc not work?

17. ### sghioto Well-Known Member

Dec 31, 2017
1,039
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From your attitude and defensive replies you certainly sound like this person who blew a fuse when asked the same question "is this homework". I apologize if you are not the same person but you need to check your attitude.
SG

18. ### BobTPH Senior Member

Jun 5, 2013
1,801
470
Because it has to go + and - 3V or 6V peak to peak and half the voltage is dropped across the emitter resistor.

Edit: Notice that the emitter resistor will carry a quiescent current of 750 mA. At 12V, that is 8 Watts of power wasted to get 1/2 a Watt into the speaker. This is why we would not use that configuration for a power amplifier.

Bob

19. ### elec_eng_55 Thread Starter Member

May 13, 2018
214
3
That is very inefficient 9 watts in and 0.5 watts out.

This came about because I was looking at the small signal emitter-follower and thought that it would be interesting to look at a "large signal" version. I suppose you would drive this circuit with another emitter-follower or a common-emitter or perhaps a CE using a darlington? Or would it never be a good idea?

David

Last edited: Apr 1, 2019
20. ### AnalogKid AAC Fanatic!

Aug 1, 2013
7,920
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Really? Let's review:

Member for less than one year.

Over 200 postings, only 2 likes.

Asking a question that is entirely consistent with a classroom exercise, in that it has no practical value other than demonstrating a design technique no matter how impractical, and then getting cranky when the *obvious* question is asked.

Admitting that you are new to / inexperienced in electronics, and then getting upset at the responses you get. Note - the nature and wording of your questions is what determines the quality of the responses you get, not snapping at your (technical) elders.