Formula to calculate output power of class A amp?

[7AudioBB]

My project is an integrated single ended Class A headphone amplifier.

It needs to have an output power of 2 watts RMS into max 300 ohms (capacitor coupled). I'm looking for a fairly simple formula for that, it doesn't have to be very accurate. I guess once I have this formula I can reverse it and calculate the quiescent collector current and start designing the circuit. So the output power would be the starting point for the project as that's the main requirement.

[7AudioBB]

What I found so far is to crank up the amplitude of a 1kHz input signal just below the clipping point of the output (into a dummy load), then read the RMS Voltage of the signal (with the oscilloscope), then to calculate the power we use this formula: PRMS = (VRMS)2 / RL where RL is the speaker impedance. Please let me know if this seems correct. Also I'm wondering how I could reverse it and determine VCC and the quiescent collector current from this?

[NiHaoMike]

My project is an integrated single ended Class A headphone amplifier.

It needs to have an output power of 2 watts RMS into max 300 ohms (capacitor coupled). I'm looking for a fairly simple formula for that, it doesn't have to be very accurate. I guess once I have this formula I can reverse it and calculate the quiescent collector current and start designing the circuit. So the output power would be the starting point for the project as that's the main requirement.

Modern headphones get on the order of 90-100dBA/dBm, giving them 33dBm will get you 123-133dBA, which would destroy your hearing very quickly.

[Benta]

Question is far too vague.
Like "I want to build a house with two bedrooms. How should I proceed?"

Think it over again.

[7AudioBB]

My project is an integrated single ended Class A headphone amplifier.

It needs to have an output power of 2 watts RMS into max 300 ohms (capacitor coupled). I'm looking for a fairly simple formula for that, it doesn't have to be very accurate. I guess once I have this formula I can reverse it and calculate the quiescent collector current and start designing the circuit. So the output power would be the starting point for the project as that's the main requirement.

Modern headphones get on the order of 90-100dBA/dBm, giving them 33dBm will get you 123-133dBA, which would destroy your hearing very quickly.

I see, then what would be your recommendation in terms of maximum power? @NiHaoMike

[pqass]

What I found so far is to crank up the amplitude of a 1kHz input signal just below the clipping point of the output (into a dummy load), then read the RMS Voltage of the signal (with the oscilloscope), then to calculate the power we use this formula: PRMS = (VRMS)2 / RL where RL is the speaker impedance. Please let me know if this seems correct.

Yes. See here for a fuller explanation of how to calculate.
You can also use a multimeter on ACV across the speaker output terminals.

Also I'm wondering how I could reverse it and determine VCC and the quiescent collector current from this?

Since P=V_rms²/R   we can rearrange it to   V_rms=√(P*R)
As for current, insert your MM (on DCmA) with amplifier input leads shorted.
For DC offset, measure across the speaker leads (DCmV) also with amplifier input leads shorted.

Also, see here (scroll down a bit) for warning on of headphone amp power.

[7AudioBB]

Question is far too vague.
Like "I want to build a house with two bedrooms. How should I proceed?"

Think it over again.

Looking at the question in your reply, I think you misunderstood mine. I asked for a specific formula, and calculating a very specific value. Didn't ask how to design an amplifier. @Benta

[7AudioBB]

What I found so far is to crank up the amplitude of a 1kHz input signal just below the clipping point of the output (into a dummy load), then read the RMS Voltage of the signal (with the oscilloscope), then to calculate the power we use this formula: PRMS = (VRMS)2 / RL where RL is the speaker impedance. Please let me know if this seems correct.

Yes. See here for a fuller explanation of how to calculate.
You can also use a multimeter on ACV across the speaker output terminals.

Also I'm wondering how I could reverse it and determine VCC and the quiescent collector current from this?

Since P=V_rms²/R   we can rearrange it to   V_rms=√(P*R)
As for current, insert your MM (on DCmA) with input leads shorted.
For DC offset, measure across the speaker leads (DCmV) also with input leads shorted.

Thank you @pqass, first proper answer to my questions.

[tszaboo]

2 watt for a headphone? Watt for? sorry the pun.
https://nwavguy.blogspot.com/2011/09/more-power.html

[DimitriP]

2 watt for a headphone? Watt for? sorry the pun.
https://nwavguy.blogspot.com/2011/09/more-power.html

Might be a class project, one of those "show your work" type projects and by using a semi-unrealistic value it may supposedly make it harder to find something to copy.
But I just made all this up. I could be waaaay off. Or not!

 

[pqass]

Regarding my "As for current, insert your MM (on DCmA) with amplifier input leads shorted" comment...

I just checked my multimeter. It has a 2R shunt in the 400mA post so it should be inconsequential with higher impedance headphones.

Crap! I'm really having a brain-fart today...

If you're asking about the current between +Vcc and -Vcc rails (with shorted amp inputs) then you'll have to measure (DCmV) across one of the two emitter resistors just prior to the +Vout output terminal; like the 10R (R7L) resistor in Figure 1 of this design or R13 in this class-A design.  The voltage reading is then divided by the emitter resistor value to get current.   You won't be able to insert your MM (DCmA) since it would likely distort the true value too much (even with a 2R shunt).

[7AudioBB]

2 watt for a headphone? Watt for? sorry the pun.
https://nwavguy.blogspot.com/2011/09/more-power.html

Might be a class project, one of those "show your work" type projects and by using a semi-unrealistic value it may supposedly make it harder to find something to copy.
But I just made all this up. I could be waaaay off. Or not!

 

You are way off indeed. Why don't you go somewhere else to troll and spam the chat if you don't have any answers to the question Dimitri? @DimitriP

[ejeffrey]

When you ask for answers here you don't get to dictate how people answer.  It can definitely be annoying when people go off on a tangent or tell you your question is wrong, but you don't get to insult people for it.

  Also your question is extremely... odd to say the least.  It's likely you are either don't understand your own question (likely if you are asking for confirmation about a very simple power calcuation) or you are deliberately obfuscating for whatever reason.  Which is fine to do if you really want to but it is 100% to be expected that people will then speculate about why you have asked this question that doesn't make sense.  If you don't like that I suggest you hire someone to answer just the questions you ask rather than asking for free advice on the Internet.

[7AudioBB]

When you ask for answers here you don't get to dictate how people answer.  It can definitely be annoying when people go off on a tangent or tell you your question is wrong, but you don't get to insult people for it.

  Also your question is extremely... odd to say the least.  It's likely you are either don't understand your own question (likely if you are asking for confirmation about a very simple power calcuation) or you are deliberately obfuscating for whatever reason.  Which is fine to do if you really want to but it is 100% to be expected that people will then speculate about why you have asked this question that doesn't make sense.  If you don't like that I suggest you hire someone to answer just the questions you ask rather than asking for free advice on the Internet.

I suggest you to move on to the next questions if you haven't got any relevant answers. @ejeffrey

[7AudioBB]

Regarding my "As for current, insert your MM (on DCmA) with amplifier input leads shorted" comment...

I just checked my multimeter. It has a 2R shunt in the 400mA post so it should be inconsequential with higher impedance headphones.

Crap! I'm really having a brain-fart today...

If you're asking about the current between +Vcc and -Vcc rails (with shorted amp inputs) then you'll have to measure (DCmV) across one of the two emitter resistors just prior to the +Vout output terminal; like the 10R (R7L) resistor in Figure 1 of this design.  The voltage reading is then divided by the emitter resistor value to get current.   You won't be able to insert your MM (DCmA) since it would likely distort the true value too much (even with a 2R shunt).

Thank you pqass. I'm only going to have one output transistor per channel biased to class A, but I'm going to apply your notes to my circuit, which is going to look similar to the one I designed below:

[John B]

Here are some real world figures tested with some Beyerdynamic DT 770 250ohm headphones:

~400mV RMS is what I would consider louder than comfortable listening. That is less than 1mW of power per channel

~3V RMS is unbearably loud, yet still only comes to 36mW per channel

[pqass]

Thank you pqass. I'm only going to have one output transistor per channel biased to class A, but I'm going to apply your notes to my circuit, which is going to look similar to the one I designed below:

I've updated my blurb (#10) to include a link to a proven class-A design.
You really should look over the whole site as it is a goldmine of everything audio.

[7AudioBB]

Here are some real world figures tested with some Beyerdynamic DT 770 250ohm headphones:

~400mV RMS is what I would consider louder than comfortable listening. That is less than 1mW of power per channel

~3V RMS is unbearably loud, yet still only comes to 36mW per channel

Thank you for those figures @John B, it really helps me to have a better idea of the required power. It looks like I went wrong with 2 watts.

[7AudioBB]

Thank you pqass. I'm only going to have one output transistor per channel biased to class A, but I'm going to apply your notes to my circuit, which is going to look similar to the one I designed below:

I've updated my blurb (#10) to include a link to a proven class-A design.
You really should look over the whole site as it is a goldmine of everything audio.

Thanks for the useful resources @pqass, I'm gonna check them out.

[John B]

Regarding my "As for current, insert your MM (on DCmA) with amplifier input leads shorted" comment...

I just checked my multimeter. It has a 2R shunt in the 400mA post so it should be inconsequential with higher impedance headphones.

Crap! I'm really having a brain-fart today...

If you're asking about the current between +Vcc and -Vcc rails (with shorted amp inputs) then you'll have to measure (DCmV) across one of the two emitter resistors just prior to the +Vout output terminal; like the 10R (R7L) resistor in Figure 1 of this design.  The voltage reading is then divided by the emitter resistor value to get current.   You won't be able to insert your MM (DCmA) since it would likely distort the true value too much (even with a 2R shunt).

Thank you pqass. I'm only going to have one output transistor per channel biased to class A, but I'm going to apply your notes to my circuit, which is going to look similar to the one I designed below:

A 10uF coupling capacitor on the output will give you terrible bass response. From my experience, a DC coupled amplifier is the better way to go (even directly powered by op amps), but if you want to stick with a class A amp, you should size the output caps a few orders of magnitude higher than what a basic RC filter calculation would suggest.

[BrianHG]

Doing a little math, to hit 2 watts peak, not RMS, into s 300 ohm load, you will need an output voltage swing of +25v to -25v.  That's a 50v supply if you made a class D amp assuming no losses.

Ok, now for class AB, to play it safe, you would need a +/-30v supply at least, that's 60v supply for a headphone amp.  Remember, for 2W RMS, you would need even more voltage.

Ok, here is the kicker, for a true class A amp, remember, approximate only ~20% of your supply will reach the speaker/headphone.  Perfectly biased, you would want a -60v supply and a +30v supply assuming you are using NPN or N-Channel mosfet to drive the output.  This means a 90v supply for your headphone amp.

Say you want 2w true RMS, your amp will need an ~120v supply, 0.4 amps per channel, so for a margin of safety, lets call it 1 amp for 2 channels.  This means a 120watt power supply.

This also means that on the pass transistor, your typical dissipation will be on the order of ~20 watts per channel totaling 40 watts.  Better have a really good large heat-sink.  (I'm sure you do not want a fan, and you need to make sure your product will operate in a good ambient 35 degree celsius environment without dying or causing a fire.)

Now, you will also need good 75 watt pulldown resistors.  These will radiate ~50 watts each for your 2 watt design.  They will need to be mounted on an even bigger heat-sink than the transistors.

Nothing wrong with the laid out specs.  This is doable.  Just making sure you understand what you are in for.

[7AudioBB]

Regarding my "As for current, insert your MM (on DCmA) with amplifier input leads shorted" comment...

I just checked my multimeter. It has a 2R shunt in the 400mA post so it should be inconsequential with higher impedance headphones.

Crap! I'm really having a brain-fart today...

If you're asking about the current between +Vcc and -Vcc rails (with shorted amp inputs) then you'll have to measure (DCmV) across one of the two emitter resistors just prior to the +Vout output terminal; like the 10R (R7L) resistor in Figure 1 of this design.  The voltage reading is then divided by the emitter resistor value to get current.   You won't be able to insert your MM (DCmA) since it would likely distort the true value too much (even with a 2R shunt).

Thank you pqass. I'm only going to have one output transistor per channel biased to class A, but I'm going to apply your notes to my circuit, which is going to look similar to the one I designed below:

A 10uF coupling capacitor on the output will give you terrible bass response. From my experience, a DC coupled amplifier is the better way to go (even directly powered by op amps), but if you want to stick with a class A amp, you should size the output caps a few orders of magnitude higher than what a basic RC filter calculation would suggest.

Yes @John B you are absolutely right, I noticed it when I tried it on a breadboard.  Then I added way bigger caps. Also I connected a capacitor in parallel with the emitter resistor and that improved gain a lot and it also seem to have an effect on trimming higher frequencies depending on the capacitance.

[BrianHG]

     Instead of a DC filter cap with resistor load driving the headphones, you should look at an audio transformer driven output with a center-tap.  These type of class A amps can achieve 50% efficiency instead of the lousy ~20% of the resistor loaded type.

     Though finding a good flat audio transformer today runs in the 50$ to 250$ per channel as such transformers are used in vacuum tube amps.  But, your source voltage will go from 120v DC down to ~30v DC and your amp will radiate something like 10 watts if waste heat per channel, not 50 watts per channel.

     What you loose its the possibility for very low frequencies below 35hz - 25hz depending on the transformer and if you use negative feedback in your design.

[7AudioBB]

Doing a little math, to hit 2 watts peak, not RMS, into s 300 ohm load, you will need an output voltage swing of +25v to -25v.  That's a 50v supply if you made a class D amp assuming no losses.

Ok, now for class AB, to play it safe, you would need a +/-30v supply at least, that's 60v supply for a headphone amp.  Remember, for 2W RMS, you would need even more voltage.

Ok, here is the kicker, for a true class A amp, remember, approximate only ~20% of your supply will reach the speaker/headphone.  Perfectly biased, you would want a -60v supply and a +30v supply assuming you are using NPN or N-Channel mosfet to drive the output.  This means a 90v supply for your headphone amp.

Say you want 2w true RMS, your amp will need an ~120v supply, 0.4 amps per channel, so for a margin of safety, lets call it 1 amp for 2 channels.  This means a 120watt power supply.

This also means that on the pass transistor, your typical dissipation will be on the order of ~20 watts per channel totaling 40 watts.  Better have a really good large heat-sink.  (I'm sure you do not want a fan, and you need to make sure your product will operate in a good ambient 35 degree celsius environment without dying or causing a fire.)

Now, you will also need good 75 watt pulldown resistors.  These will radiate ~50 watts each for your 2 watt design.  They will need to be mounted on an even bigger heat-sink than the transistors.

Nothing wrong with the laid out specs.  This is doable.  Just making sure you understand what you are in for.

Thanks for breaking it down for me @BrianHG. I was expecting some large values, but it still surprised me I'm learning about circuit design for the first time and it's great to see what to expect when thinking of a project with a class A amp. Now I understand why some integrated high-end class A amps weigh so much. 

[7AudioBB]

     Instead of a DC filter cap with resistor load driving the headphones, you should look at an audio transformer driven output with a center-tap.  These type of class A amps can achieve 50% efficiency instead of the lousy ~20% of the resistor loaded type.

     Though finding a good flat audio transformer today runs in the 50$ to 250$ per channel as such transformers are used in vacuum tube amps.  But, your source voltage will go from 120v DC down to ~30v DC and your amp will radiate something like 10 watts if waste heat per channel, not 50 watts per channel.

     What you loose its the possibility for very low frequencies below 35hz - 25hz depending on the transformer and if you use negative feedback in your design.

Yes, I read about it and saw them in tube amps. For now I might just stick with a coupling capacitor as it's easy to swap and experiment with, then when I know more and progress it would be great to try an audio transformer and improve efficiency like you said. Thanks Brian.