Formula to calculate output power of class A amp? - Page 1
Forum section: electronics / projects

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 02:36:29 pm
content: 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.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 09:54:42 pm
content: 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 / RLwhere RLis the speaker impedance. Please let me know if this seems correct. Also I'm wondering how I could reverse it and determine VCCand the quiescent collector current from this?

author: NiHaoMike
author_rank: Super Contributor
timestamp: March 31, 2024, 10:10:55 pm
content: Quote from: 7AudioBB on March 31, 2024, 02:36:29 pmModern headphones get on the order of 90-100dBA/dBm, giving them 33dBm will get you 123-133dBA, which would destroy your hearing very quickly.

author: Benta
author_rank: Super Contributor
timestamp: March 31, 2024, 10:15:59 pm
content: Question is far too vague.Like "I want to build a house with two bedrooms. How should I proceed?"Think it over again.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 10:27:54 pm
content: Quote from: NiHaoMike on March 31, 2024, 10:10:55 pmI see, then what would be your recommendation in terms of maximum power? @NiHaoMike

author: pqass
author_rank: Super Contributor
timestamp: March 31, 2024, 10:36:50 pm
content: Quote from: 7AudioBB on March 31, 2024, 09:54:42 pmYes. Seeherefor a fuller explanation of how to calculate.You can also use a multimeter on ACV across the speaker output terminals.QuoteSince P=Vrms2/R we can rearrange it to Vrms=√(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, seehere(scroll down a bit) for warning on of headphone amp power.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 10:38:36 pm
content: Quote from: Benta on March 31, 2024, 10:15:59 pmLooking 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

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 10:41:02 pm
content: Quote from: pqass on March 31, 2024, 10:36:50 pmThank you @pqass, first proper answer to my questions.

author: tszaboo
author_rank: Super Contributor
timestamp: March 31, 2024, 10:51:06 pm
content: 2 watt for a headphone? Watt for? sorry the pun.https://nwavguy.blogspot.com/2011/09/more-power.html

author: DimitriP
author_rank: Super Contributor
timestamp: March 31, 2024, 10:55:27 pm
content: Quote from: tszaboo on March 31, 2024, 10:51:06 pmMight 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!

author: pqass
author_rank: Super Contributor
timestamp: March 31, 2024, 10:58:58 pm
content: 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 ofthisdesign or R13 inthisclass-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).

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 11:04:31 pm
content: Quote from: DimitriP on March 31, 2024, 10:55:27 pmYou 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

author: ejeffrey
author_rank: Super Contributor
timestamp: March 31, 2024, 11:19:37 pm
content: 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.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 11:22:21 pm
content: Quote from: ejeffrey on March 31, 2024, 11:19:37 pmI suggest you to move on to the next questions if you haven't got any relevant answers. @ejeffrey

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 11:39:39 pm
content: Quote from: pqass on March 31, 2024, 10:58:58 pmThank 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:

author: John B
author_rank: Frequent Contributor
timestamp: March 31, 2024, 11:48:45 pm
content: 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

author: pqass
author_rank: Super Contributor
timestamp: March 31, 2024, 11:49:47 pm
content: Quote from: 7AudioBB on March 31, 2024, 11:39:39 pmI've updated my blurb (#10) to include a link to a provenclass-A design.You really should look over the wholesiteas it is a goldmine of everything audio.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 11:54:24 pm
content: Quote from: John B on March 31, 2024, 11:48:45 pmThank 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.

author: 7AudioBB
author_rank: Contributor
timestamp: March 31, 2024, 11:57:06 pm
content: Quote from: pqass on March 31, 2024, 11:49:47 pmThanks for the useful resources @pqass, I'm gonna check them out.

author: John B
author_rank: Frequent Contributor
timestamp: March 31, 2024, 11:58:06 pm
content: Quote from: 7AudioBB on March 31, 2024, 11:39:39 pmA 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.

author: BrianHG
author_rank: Super Contributor
timestamp: March 31, 2024, 11:58:09 pm
content: 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.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 12:05:40 am
content: Quote from: John B on March 31, 2024, 11:58:06 pmYes @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.

author: BrianHG
author_rank: Super Contributor
timestamp: April 01, 2024, 12:08:12 am
content: 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.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 12:15:35 am
content: Quote from: BrianHG on March 31, 2024, 11:58:09 pmThanks for breaking it down for me @BrianHG. I was expecting some large values, but it still surprised meI'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.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 12:21:42 am
content: Quote from: BrianHG on April 01, 2024, 12:08:12 amYes, 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.

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 01, 2024, 12:34:12 am
content: 2W is a lot for headphones. But, but. As a very general consideration, headphones with a higher impedance (such as your 300 ohm figure) absolutely do not need that much power anyway. Possibly, *some* "high-grade" headphones, with low impedance and a relatively low efficiency (low dB SPL/W @ xxx m), can require 1W to 2W RMS to get loud, but those are really niche devices.Trying to design a "universal" headphones amp that can accomodate such a wide impedance range (say from 4 ohm to 300 ohm) is not necessarily a good idea. Just a thought.

author: NiHaoMike
author_rank: Super Contributor
timestamp: April 01, 2024, 01:56:36 am
content: The easiest solution is to just use a chip specifically designed to drive headphones, such as the TPA6120A2.

author: DimitriP
author_rank: Super Contributor
timestamp: April 01, 2024, 12:18:27 pm
content: Quote from: 7AudioBB on March 31, 2024, 11:04:31 pmCome to think of it, I wasn't "talking to you" in the first place. So excuse you.2W headphone amp without any reasoning invites speculation.Sometimes you get the answers you need, not the answers you want.

author: CaptDon
author_rank: Super Contributor
timestamp: April 01, 2024, 01:45:59 pm
content: Your single transistor solution to a 2 watt class A amplifier is bullocks and certainly not done with a 2n3904!!! A class A amplifier nearly mandates the output stage dissipates 1/2 of the total output power at rest. Calculate the peak to peak voltage required to get 2 watts across 300 ohms. Your output power supply voltage will have to be at least that amount plus about 10 to 15% to prevent clipping. With the coupling capacitor you can use a single sided power supply even with most any off the shelf output integrated circuit that is typically class B push pull topology. You don't need Class A AudioPhool design. Just buy off the shelf audio output I.C.'s for you project. Some don't even need external support devices, only an input level control and power.

author: Leuams
author_rank: Contributor
timestamp: April 01, 2024, 01:57:49 pm
content: Douglas Self has a few great books on audio amplifier design if you wish to go down that rabbit hole. I also found sound-au.com as a great source.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 04:32:43 pm
content: Quote from: SiliconWizard on April 01, 2024, 12:34:12 amMy initial goal was to cover all kinds of headphones but I agree, it might not be a good or even a safe option. Thanks for your reply.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 04:37:11 pm
content: Quote from: NiHaoMike on April 01, 2024, 01:56:36 amThanks, but then I guess I would miss out on the fun of learning, designing and building a class a circuit.

author: BrianHG
author_rank: Super Contributor
timestamp: April 01, 2024, 04:50:48 pm
content: You cannot create a Class A audio amp designed to drive a 300ohm headphone at 2 watts and expect it to be an appropriate design to drive at standard 32ohm headphone at the same time.Your pull-down resistors will be the wrong value by an order of magnitude and your power supply will be bonkers at 120v DC.If you need class A, you will have to either design for the 16ohm to 32ohm headphones, or a completely different design for 300ohm headphones.And a warning about having 2 watt capabilities for 32 ohm headphones, driving a 16 ohm headphones means you will be delivering up to 6 watts into each ear (depending on design). Someone will sue if you blow their expensive headphones let alone melting their ears.Remember, for 2 watts per channel, at 32 ohms, you will need ~+/-9v supply if you have a rail-to-rail output driving amp.For 2 watts into 16 ohms, you will only need ~+/-6v.For simple class A, you will need to double the negative voltage for the pull-down resistor unless you use an active constant current sink for your pull-down like the high end Pass Labs Class A amplifiers. However, some argue that this additional transistor element on the output counts as a 'gain' element, but you efficiency goes up quite a bit, almost 25%, as your power supply needs drop by 25%. This means only needing +/-9v for this class A amp driving 2 watts into a 32 ohm headphone load just like a Class-AB or Class-D amp. (You might need an extra 3-4v for the constant current source to begin operation at the bottom end, but not double your source voltage. And the current buffer gain/load curve is will also be linear...)

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 05:05:00 pm
content: Quote from: BrianHG on April 01, 2024, 04:50:48 pmThank you for your insights Brian. I'm going to consider this and give the design a narrower target range in terms of headphone impedance.

author: 7AudioBB
author_rank: Contributor
timestamp: April 01, 2024, 05:32:01 pm
content: Quote from: Leuams on April 01, 2024, 01:57:49 pmThank you @Leuams, amplifiers seem to be a good entry point for me to learning circuit design. I'm gonna check those sources out!