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

 

Homework?

 

Homework?

No I am an old retired "fart" trying to learn something new.
"Remember the human - Be courteous when replying to others."
Or don't reply at all.

 

hi David,
You may find this link helpful, the calculations and presentation are excellent.

https://www.electronics-tutorials.ws/amplifier/common-collector-amplifier.html

E

 

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

 

hi David,
You may find this link helpful, the calculations and presentation are excellent.

https://www.electronics-tutorials.ws/amplifier/common-collector-amplifier.html

E

Thanks E, I will check this out.
David

 

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

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

 

No I am an old retired "fart" trying to learn something new.
"Remember the human - Be courteous when replying to others."
Or don't reply at all.

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.

 

If you want to design an emitter-follower power amplifier that supplies 562mW (real power)
to an 8Ω load,

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

 

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.

It is ignorant. End of discussion.

 

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

thanks Jony

 

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

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?

 

but how did you determine that Vcc should be > 12V?

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

https://forum.allaboutcircuits.com/...single-ended-follower-amplifier-pg-91.120993/

 

In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at emitter.

View attachment 173888

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

https://forum.allaboutcircuits.com/...single-ended-follower-amplifier-pg-91.120993/

12 volts would be enough for Vcc or would you recommend bumping it up a little?

 

It is ignorant. End of discussion.

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.

 

In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at emitter.

View attachment 173888

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

https://forum.allaboutcircuits.com/...single-ended-follower-amplifier-pg-91.120993/

Since the maximum output swing is ≈ 3 volts peak would any VE higher than 3 volts dc not work?

 

What the heck are you talking about? I never even heard of this site until last May.
Get your facts straight!

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

 

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

 

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

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

 

It is ignorant. End of discussion.

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.

Not exactly your first thread with unjustified attitude.

AND - taking a swing at one of the mods ! ! !

Really?

ak