Maybe it's as simple as this, but I need to know for sure...
Using the 40W/4 ohm design example at LM3886's datasheet, the power supply required is:
Since the amplifier is connected to the transformer, the load saw by transformer will be simply Ohm's Law?
I'm having a hard time trying to design my power supply...
Using the 40W/4 ohm design example at LM3886's datasheet, the power supply required is:
+/- 21V @ 4,5 A(peak)
Since the amplifier is connected to the transformer, the load saw by transformer will be simply Ohm's Law?
42 = R x 4,5
R=9,33 ohm
R=9,33 ohm
I'm having a hard time trying to design my power supply...
Impedance is not a useful concept here, as the voltage and current have quite different waveforms.
Use Duncan's PSUD2 to model your PSU. Compare your results with theory.
Use Duncan's PSUD2 to model your PSU. Compare your results with theory.
Modelers and online calculators are fine tools which save time but it's nice to know where the results come from.
Or maybe I'm just an old fashioned guy who studied Engineering in late 60's , early 70's and we used slide rules 😱 (yup, not even digital calculators were available, my first one was a Bowmar around '72), so I'm quite used to calculating stuff on the back of a napkin, estimating, interpolating, using graphs, etc. 🙂
In that case, you need to know what the equivalent load will be, meaning what would be the resistor which draws as much current as the amp will, at full load.
In the case of a split power supply, with a sinewave driving the amp to just clipping, meaning it's delivering its full RMS power, each rail will se a load of Pi*Rl
So in your case:
Pi*4 ohms=3.14*4=12.6 ohms
So average current pulled from each rail will be: 21V/12.6 ohms=1.7A .
Twice that for a stereo amp.
Calculating for 4.5A will provide a nice safety margin.
EDIT: so yes, you do use Ohm's Law, just not with straight 4 ohms but with a scaling factor, which in this case happens to be Pi .
Your 4 ohm speaker is not always connected to the + or - rails, but voltage and current are fed to it, reproducing an audio signal.
Or maybe I'm just an old fashioned guy who studied Engineering in late 60's , early 70's and we used slide rules 😱 (yup, not even digital calculators were available, my first one was a Bowmar around '72), so I'm quite used to calculating stuff on the back of a napkin, estimating, interpolating, using graphs, etc. 🙂
Maybe you meant the amplifier is connected to a power supply, which is more than just the transformer.
In that case, you need to know what the equivalent load will be, meaning what would be the resistor which draws as much current as the amp will, at full load.
In the case of a split power supply, with a sinewave driving the amp to just clipping, meaning it's delivering its full RMS power, each rail will se a load of Pi*Rl
So in your case:
Pi*4 ohms=3.14*4=12.6 ohms
So average current pulled from each rail will be: 21V/12.6 ohms=1.7A .
Twice that for a stereo amp.
Calculating for 4.5A will provide a nice safety margin.
EDIT: so yes, you do use Ohm's Law, just not with straight 4 ohms but with a scaling factor, which in this case happens to be Pi .
Your 4 ohm speaker is not always connected to the + or - rails, but voltage and current are fed to it, reproducing an audio signal.
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Yes, that is exactly what I want to know, so I can define how much current the amplifier will need from my PSU and, thus, my transformer's VA power, right?
So the total current draw by the LM3886 (mono) from the PSU will be 3.4A (average)?
Why Pi? Something to do with energy or power of the signal?
That 4.5A are average, right? This always confuses me...I hardly know when the datasheet means peak, average or RMS
If average:
Since
Vpk=Vav/0,637 and Vpk=Vrms/0,707
Vrms=Vav*1,109
Is that right?
I'm trying to understand all the steps too! I'm an undergraduate in electrical engineering but I'm not familiar with electronic circuits yet...
Thanks for all the anwsers!
Sorry, I misunderstood you. I though you were asking about the load imposed on the transformer by the amplifier PSU. I now understand you are asking about the load imposed on the PSU by the amplifier.luismoro said:
The pi which has appeared comes from calculating the average value of a half-wave rectified sine wave. (For full-wave rectification the result is 2/pi = 0.6366)
A good datasheet will normally say whether it is using mean, RMS or peak - unless it is obvious from the context. People differ on what they consider obvious!
Note that you can't simply convert from 'amplifer draws X watts' to 'transformer can be X VA'. The ratio depends on the type of PSU, and the amplifier duty cycle.
Agree and add.
For day to day use (not while writing a Thesis, of course 😉 ) we often apply a so called "rule of thumb" , meaning some estimation which has been used by many and in practical use works reasonably well.
One such rule of thumb is that in most cases, specially Home Hi Fi, a transformer VA rating 150% (1.5X) amplifoer RMS output is fine.
That should be the absolute minimum, of course.
VA rating 2X the total RMS power is safe even for DJ use, Musical Instruments and such.
For day to day use (not while writing a Thesis, of course 😉 ) we often apply a so called "rule of thumb" , meaning some estimation which has been used by many and in practical use works reasonably well.
One such rule of thumb is that in most cases, specially Home Hi Fi, a transformer VA rating 150% (1.5X) amplifoer RMS output is fine.
That should be the absolute minimum, of course.
VA rating 2X the total RMS power is safe even for DJ use, Musical Instruments and such.
JMFahey,
So you're saying 80VA for a 40W/4 ohm amplifier should suffice?
Still about your previous explanation, 1,7A should be the current supplied by my PSU for each rail?. Ergo, considering a center tapped transformer and current from both rails (3.4A), my transformer should be rated at something about 32Vrms in the secondary (aprox. 44Vdc after rectfied) and 32 x 3.4 VA, which is aprox. 110VA.
Is it right or not?
So you're saying 80VA for a 40W/4 ohm amplifier should suffice?
Still about your previous explanation, 1,7A should be the current supplied by my PSU for each rail?. Ergo, considering a center tapped transformer and current from both rails (3.4A), my transformer should be rated at something about 32Vrms in the secondary (aprox. 44Vdc after rectfied) and 32 x 3.4 VA, which is aprox. 110VA.
Is it right or not?
The load on the transformer is the discharged psu capacitance. So it's not a constant load like a resistor, more like a resistor that's switched at120hz. Makes things more complicated. So the current is pulsed.
Your PSU will supply 44*1.7=74.8W so 80VA is fine.
In fact both halves of the PSU are in series so currents do not add.
Effective current is 1.7A
To be more precise, *if* you had a single 44V supply (your 2 22V rails in series) equivalent load would be 2*Pi*4 ohms=25 ohms and total current would be: 44/25=1.76A , same as before.
As I mentioned earlier as a rule of thumb , VA>=2*RMS out is safe always (of course you can improve that for better response) ; 1.5*W RS usually works well for home use, even if you listen to Rock, because otherwise distortion becomes unbearable.
As in: distortion on the *guitar* is nice; on bass *sometimes* acceptable, just don't abuse; on voice, keyboards or drums is ugly and on everything together is vomit inducing, because different instruments intermodulation is just a mess impossible to understand.
So Music is listened at no more than 25 to 50% of average power ... and that is a very dirty level.
That is a wrong assumption and I forgot to correct you earlier.
In fact both halves of the PSU are in series so currents do not add.
Effective current is 1.7A
To be more precise, *if* you had a single 44V supply (your 2 22V rails in series) equivalent load would be 2*Pi*4 ohms=25 ohms and total current would be: 44/25=1.76A , same as before.
As I mentioned earlier as a rule of thumb , VA>=2*RMS out is safe always (of course you can improve that for better response) ; 1.5*W RS usually works well for home use, even if you listen to Rock, because otherwise distortion becomes unbearable.
As in: distortion on the *guitar* is nice; on bass *sometimes* acceptable, just don't abuse; on voice, keyboards or drums is ugly and on everything together is vomit inducing, because different instruments intermodulation is just a mess impossible to understand.
So Music is listened at no more than 25 to 50% of average power ... and that is a very dirty level.
> R=9,33 ohm
Many-many decades ago, Dan "SWTP" Meyer set me straight on this. The math is chewable, but you have to think it all the way through.
JMFahey's answer is correct.
A totem-pole audio power amp, driven to full power sine wave, will put a load on its DC supply which is very nearly _6_ (*) times the audio load impedance.
8 Ohm adio load, 50 Ohms on the DC supply.
4 Ohm audio load, 25 Ohms on the DC supply.
quote JMFahey
> Effective current is 1.7A
> single 44V supply
> 4 ohms
> equivalent load ... 25 ohms
4 times 6 is 25 Ohms.
If you are testing a 2*22V DC supply for a 4 Ohm audio load, you want two 12.56 Ohm 38.5 Watt resistors. (It is handy to have piles of large 50 Ohm resistors to parallel down to your intended loading; this is how I proved to myself that Mr Meyer knew his stuff.)
(*) This takes the simplification pi=3. This is close-enough for any loudspeaker-amp purpose. (pi=3 is 5% off, loudspeakers never get within 20% of their marked impedance, and are usually/always far further out at some frequencies.)
This assumes "perfect" devices. Some, especially MOSFETs won't come anywhere near the supply rail. If the output only swings to 75% or 0.75 of supply, the output power is less and the equivalent DC load on the supply is 133% of the "6 times" rule.
_____________________
I would use the assumption that transformer VA should be *double* the intended sine RMS output audio power. This comes from too much big-system work, where if it aint clipping steady then you didn't figure the job right. In most home systems VA = 1.5*Watts will be fine. JMFahey's figure of 20% is quite high for "nice" reproduction, 10% is usually as loud as you go without rude clipping.
I was writing to Meyer when debugging his Tiger, which used a power transformer VA about 1.5 times the audio output. Further amp-abuse proved that his PT was plenty big for ANY home music application, most Pro work, and just a bit shy for long-term bench-test (which was not a consideration in 1969).
_____________________
> load on the transformer is the discharged psu capacitance
The caps are part of the load at switch-on. Actually for any audio-reasonable caps the "load" is mostly the transformer's internal resistance, which must be small compared to the load of the amplifier. A 200VA tranny may be delivering 2,000VA to charge the caps. However the switch-on surge is a part-second, and it takes many-many seconds to burn up the transformer. Big power supplies may dim the lamps, and strain the rectifier, but it is not a problem for the transformer.
Many-many decades ago, Dan "SWTP" Meyer set me straight on this. The math is chewable, but you have to think it all the way through.
JMFahey's answer is correct.
A totem-pole audio power amp, driven to full power sine wave, will put a load on its DC supply which is very nearly _6_ (*) times the audio load impedance.
8 Ohm adio load, 50 Ohms on the DC supply.
4 Ohm audio load, 25 Ohms on the DC supply.
quote JMFahey
> Effective current is 1.7A
> single 44V supply
> 4 ohms
> equivalent load ... 25 ohms
4 times 6 is 25 Ohms.
If you are testing a 2*22V DC supply for a 4 Ohm audio load, you want two 12.56 Ohm 38.5 Watt resistors. (It is handy to have piles of large 50 Ohm resistors to parallel down to your intended loading; this is how I proved to myself that Mr Meyer knew his stuff.)
(*) This takes the simplification pi=3. This is close-enough for any loudspeaker-amp purpose. (pi=3 is 5% off, loudspeakers never get within 20% of their marked impedance, and are usually/always far further out at some frequencies.)
This assumes "perfect" devices. Some, especially MOSFETs won't come anywhere near the supply rail. If the output only swings to 75% or 0.75 of supply, the output power is less and the equivalent DC load on the supply is 133% of the "6 times" rule.
_____________________
I would use the assumption that transformer VA should be *double* the intended sine RMS output audio power. This comes from too much big-system work, where if it aint clipping steady then you didn't figure the job right. In most home systems VA = 1.5*Watts will be fine. JMFahey's figure of 20% is quite high for "nice" reproduction, 10% is usually as loud as you go without rude clipping.
I was writing to Meyer when debugging his Tiger, which used a power transformer VA about 1.5 times the audio output. Further amp-abuse proved that his PT was plenty big for ANY home music application, most Pro work, and just a bit shy for long-term bench-test (which was not a consideration in 1969).
_____________________
> load on the transformer is the discharged psu capacitance
The caps are part of the load at switch-on. Actually for any audio-reasonable caps the "load" is mostly the transformer's internal resistance, which must be small compared to the load of the amplifier. A 200VA tranny may be delivering 2,000VA to charge the caps. However the switch-on surge is a part-second, and it takes many-many seconds to burn up the transformer. Big power supplies may dim the lamps, and strain the rectifier, but it is not a problem for the transformer.
Thanks for all the anwsers, guys!
It was really helpful and clarified a lot of things for me.
This forum is awesome!
It was really helpful and clarified a lot of things for me.
This forum is awesome!
Calculate ripple, or calculate how much ripple is acceptable?
To calculate ripple you need to work out how much charge will leave the reservoir capacitor between charging pulses. Then use V=Q/C. Or find an online calculator, or use PSUD2.
To calculate how much ripple is acceptable you need to know the PSRR of your circuit, and how sensitive you and your speakers are to hum.
To calculate ripple you need to work out how much charge will leave the reservoir capacitor between charging pulses. Then use V=Q/C. Or find an online calculator, or use PSUD2.
To calculate how much ripple is acceptable you need to know the PSRR of your circuit, and how sensitive you and your speakers are to hum.
Yes, I want to calculate how much ripple is acceptable. How do I use PSRR at LM3886's datasheet to calculate it?
Absolutely subjective Rule of thumb, around 10% ripple is acceptable, and practically can't be heard (being that it's buried behind loud music) .
Remember you'll have that high ripple at full power and under a continuous audio signal (such as a test sinewave) ; under normal listening conditions it will be, say, 10% of that or less.
Now, when the amp clips , al PSRR goes out the window, as the ripple will straight make up the upper and lower edges of your output clipped squarewave ... but I guess you don't listen to music that loud 😉
Remember you'll have that high ripple at full power and under a continuous audio signal (such as a test sinewave) ; under normal listening conditions it will be, say, 10% of that or less.
Now, when the amp clips , al PSRR goes out the window, as the ripple will straight make up the upper and lower edges of your output clipped squarewave ... but I guess you don't listen to music that loud 😉
> how much ripple would be acceptable for my PSU?
Depends on the amplifier?
Some shoot PS ripple right to the output. Others reject as much as they can. No amplifier can reject ripple when hard-clipping.
> PSRR at LM3886's datasheet to calculate it?
If an amp is pretty good, you can use some rules-o-thumb.
The '3886 is very good at rejecting ripple. (One thing that chip-designers excel at.)
With a pretty good or better amp, one 8 Ohm load:
* 1,000ufd total (two 2,000uFd split) is kinda-maybe OK.
* 10,000ufd total (two 20,000uFd split) is pretty over-kill.
Double for stereo.
Double for 4 Ohm loads.
I have no dispute with folks who use 2X470uFd. Fashions change. But 1000++uFd has been a good guide over the decades. (Back to the 1956 Lin amp, which used two 1,000uFd for one 16 Ohm load, in days when 1000uFd was custom-order very expensive.)
{echoing Fahey again....}
Here's what's happening. Most audio power amps (all chip-amps) run near class B. Their idle current is much smaller than their full-load current. Ripple is proportional to current. So when the music is quiet, ripple is low. Unless the amp is junk, it will reject this low ripple down below its hiss-floor. This plus ear-curve means you should not hear buzz at low output.
At high output the ripple rises, but so does the music. You won't hear small ripple behind LOUD music. With a good amp, you can hardly measure the ripple in the output on a test-bench.
When in steady clipping, all amps will pass the ripple. However you already have gross distortion, so what is a little buzz under the fuzz? (This does become important in guitar amps.)
> to calculate it?
First you need goals. How much 100/120Hz buzz can you tolerate at the speaker? Soft? Loud?
Then you do the math. Ugh.
Say you calculate that 678uFd will meet a -88dB goal. There's no 678uFd caps in the catalog. There is 680uFd, but the fine print maybe says -50% tolerance, though today more likely -20%. Look for a cap 25% bigger, 848uFd. Maybe there is a 850uFd. However look down. The 1,000uFd is nearly the same price and size (cap costs do not go up as fast as uFd), and more is surely better? And what if you move to a quieter room and need -90dB ripple goal?
No point in a sharp-pencil "exact" analysis.
Two 4,700uFd do not cost much today and cover nearly all cases. If cash/space is tight, two 2,200uFd will do fine. Less is no big disaster, but don't push it.
Depends on the amplifier?
Some shoot PS ripple right to the output. Others reject as much as they can. No amplifier can reject ripple when hard-clipping.
> PSRR at LM3886's datasheet to calculate it?
If an amp is pretty good, you can use some rules-o-thumb.
The '3886 is very good at rejecting ripple. (One thing that chip-designers excel at.)
With a pretty good or better amp, one 8 Ohm load:
* 1,000ufd total (two 2,000uFd split) is kinda-maybe OK.
* 10,000ufd total (two 20,000uFd split) is pretty over-kill.
Double for stereo.
Double for 4 Ohm loads.
I have no dispute with folks who use 2X470uFd. Fashions change. But 1000++uFd has been a good guide over the decades. (Back to the 1956 Lin amp, which used two 1,000uFd for one 16 Ohm load, in days when 1000uFd was custom-order very expensive.)
{echoing Fahey again....}
Here's what's happening. Most audio power amps (all chip-amps) run near class B. Their idle current is much smaller than their full-load current. Ripple is proportional to current. So when the music is quiet, ripple is low. Unless the amp is junk, it will reject this low ripple down below its hiss-floor. This plus ear-curve means you should not hear buzz at low output.
At high output the ripple rises, but so does the music. You won't hear small ripple behind LOUD music. With a good amp, you can hardly measure the ripple in the output on a test-bench.
When in steady clipping, all amps will pass the ripple. However you already have gross distortion, so what is a little buzz under the fuzz? (This does become important in guitar amps.)
> to calculate it?
First you need goals. How much 100/120Hz buzz can you tolerate at the speaker? Soft? Loud?
Then you do the math. Ugh.
Say you calculate that 678uFd will meet a -88dB goal. There's no 678uFd caps in the catalog. There is 680uFd, but the fine print maybe says -50% tolerance, though today more likely -20%. Look for a cap 25% bigger, 848uFd. Maybe there is a 850uFd. However look down. The 1,000uFd is nearly the same price and size (cap costs do not go up as fast as uFd), and more is surely better? And what if you move to a quieter room and need -90dB ripple goal?
No point in a sharp-pencil "exact" analysis.
Two 4,700uFd do not cost much today and cover nearly all cases. If cash/space is tight, two 2,200uFd will do fine. Less is no big disaster, but don't push it.
This is a very useful thread, thank you! How does the effective load resistance change with amplifier class? If the rule of thumb says effective resistance is ~6 x amplifier load for a Class AB amp, is there a lower effective resistance for Class A and a higher effective resistance for Class D?
If so, how does this effect the recommendation of transformer VA being 1.5-2 times the amplifier output power?
If so, how does this effect the recommendation of transformer VA being 1.5-2 times the amplifier output power?
Class A draws current equal to the peak signal current all the time. You lose the factor of pi. You also lose the advantage of duty cycle. The transformer VA rating (for a normal cap input supply) may have to be 6 times the amplifier power output.
> Class A draws current equal to the peak signal current
There are class-A designs which do that. ""Usually"" at higher power we design so the idle current is half of the peak current, with peaks (both 2X and zero) flowing from the filter caps.
A perfect Class A design is 50% efficient (measured sine-wave) so the equivalent battery must support twice the "RMS" output.
In real life, 40% efficiency may be tough so DC Watts = 2.5X "RMS" output.
Cap-input rectification is very hard on the transformer, and we usually need to up-rate the VA significantly from the DC Watts. Factor may be 1.5, 1.8, even 2.2. So VA can be around 5 times the "RMS".
Heat is already a big problem in class-A work so we want to be generous with transformer capacity. "May have to be 6 times" is probably a good guide.
There are class-A designs which do that. ""Usually"" at higher power we design so the idle current is half of the peak current, with peaks (both 2X and zero) flowing from the filter caps.
A perfect Class A design is 50% efficient (measured sine-wave) so the equivalent battery must support twice the "RMS" output.
In real life, 40% efficiency may be tough so DC Watts = 2.5X "RMS" output.
Cap-input rectification is very hard on the transformer, and we usually need to up-rate the VA significantly from the DC Watts. Factor may be 1.5, 1.8, even 2.2. So VA can be around 5 times the "RMS".
Heat is already a big problem in class-A work so we want to be generous with transformer capacity. "May have to be 6 times" is probably a good guide.
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