I tried to calculate the RMS-Power of an amp at given DC-Driver voltage.
But I always get higher values than in the description of the (DIY) amp is given.
Well reasonable I must reduce the output Umax by the "Saturation" voltage of the amp but I don't think that an 400W amp working with 70V only give out 56 V.
I added my calculations.
This Probelm hounded me the last day's I hope someone can help.
Andi
But I always get higher values than in the description of the (DIY) amp is given.
Well reasonable I must reduce the output Umax by the "Saturation" voltage of the amp but I don't think that an 400W amp working with 70V only give out 56 V.
I added my calculations.
This Probelm hounded me the last day's I hope someone can help.
Andi
Attachments
Hi,
#1 is correct, #2 is not. RMS means 'root mean square' and you ommit the root.
I don't know what you mean with #3 nor the rest, but perhaps that adding that root solves your problem.
If I remember correctly than the Vrms of a sinus is 1/2*sqrt(2)*sin(x), and you can than square that voltage, divided by the number of Ohms should give you the output power.
Remco Poelstra
#1 is correct, #2 is not. RMS means 'root mean square' and you ommit the root.
I don't know what you mean with #3 nor the rest, but perhaps that adding that root solves your problem.
If I remember correctly than the Vrms of a sinus is 1/2*sqrt(2)*sin(x), and you can than square that voltage, divided by the number of Ohms should give you the output power.
Remco Poelstra
In #1
I mean not U=Umax I mean U=Urms
now you get #3 with Urms=Umax*1/sqr(2)
sin(x)*Umax is the voltage at the "angel" 0-360 deg respectively 0-2pi.
After the first equation I have to square the voltage -> (sin(x)*Umax)² and divide it with the impendance r.
now I add all pieces Power together and divide it with 2pi.
right ?
The Education should be right ? !
I mean not U=Umax I mean U=Urms
now you get #3 with Urms=Umax*1/sqr(2)
sin(x)*Umax is the voltage at the "angel" 0-360 deg respectively 0-2pi.
After the first equation I have to square the voltage -> (sin(x)*Umax)² and divide it with the impendance r.
now I add all pieces Power together and divide it with 2pi.
right ?
The Education should be right ? !
Attachments
Simple fact is you are disregarding losses in the amplifier,
power supply average voltage droop and the the extra
ripple below the average voltage droop.
You cannot estimate 4 ohm output power from nominal rail voltage.
(nor 8 ohms for that matter but its nearer)
What you can do is estimate losses for 8 ohms max power,
which again will be lower than predicted by rail voltages
and double them for 4 ohms.
🙂 sreten.
power supply average voltage droop and the the extra
ripple below the average voltage droop.
You cannot estimate 4 ohm output power from nominal rail voltage.
(nor 8 ohms for that matter but its nearer)
What you can do is estimate losses for 8 ohms max power,
which again will be lower than predicted by rail voltages
and double them for 4 ohms.
🙂 sreten.
there is no way to estimate the Power of an amp, before you have measured it ?
Isn't it possible to calculate/ name circa values for all the loss that occur ?
and take that in the estimation.
Well,
if I use a verry good/ powerfull supply, with big capacitances.
What loss will occur except the Saturation and some mV in the connection lines ?
Isn't it possible to calculate/ name circa values for all the loss that occur ?
and take that in the estimation.
Well,
if I use a verry good/ powerfull supply, with big capacitances.
What loss will occur except the Saturation and some mV in the connection lines ?
Class B amp's only have a efficiency of +-60% no matter what powersupply or wires you use, it's just the way they are build. Class A gets closer to 30%. Thus for every Watt you put in you'll only get 0.6 Watt back. 612*0.6=367 which is close to enough to 400 to rate the amp at 400 I think, taking into consideration that the calculations are very rough.
Remco Poelstra
Remco Poelstra
@ Thoru
but with this education you get much lower values as given in the DIY-Amps.
And I thik my solution should be right because you get the same as you get if you use #1 and Urms=Umax*1/sqr(2)
but with this education you get much lower values as given in the DIY-Amps.
And I thik my solution should be right because you get the same as you get if you use #1 and Urms=Umax*1/sqr(2)
tab30 said:
A perfect supply is hardly the point.
You want output with a sensible cost effective supply.
And yes intelligent estimations can be made.
But they are not based on nominal no load rail voltage.
🙂 sreten.
@ Thoru second message
But the efficiency of 60% means only you need an PowSup witch delivers 1,6 * (Pwanted out)
isn't it so ?
But the efficiency of 60% means only you need an PowSup witch delivers 1,6 * (Pwanted out)
isn't it so ?
Amplifiers start to distort long before the output devices saturate. You can't calculate maximum undistorted power using power supply voltage and transistor saturation values.
@ Sreten
the main problem is the stability of the PowerSup. Voltage ?
This means the Voltage of an normal PowerSup vary for example betwen 70V the example from above down to 56 in burden case.
result:
the power of an amp depends strong on the stability of his PowerSup ?
the main problem is the stability of the PowerSup. Voltage ?
This means the Voltage of an normal PowerSup vary for example betwen 70V the example from above down to 56 in burden case.
result:
the power of an amp depends strong on the stability of his PowerSup ?
maylar said:
Sorry but you've lost me on this one.
Transistor amplifiers distort usually by running out of rail voltage,
or more to the point rail voltage - working voltage of the amplifier.
🙂 sreten.
I suppose you mean that #3 follows from #2?
That #3 and #1 gives the same result is logical, they are exacly the same, just another notation. But #3 doesn't follow from #2 (at least not if #2 would be the correct RMS equation). I could write it down for you if I only had the goniometric tables here.......
Remco Poelstra
That #3 and #1 gives the same result is logical, they are exacly the same, just another notation. But #3 doesn't follow from #2 (at least not if #2 would be the correct RMS equation). I could write it down for you if I only had the goniometric tables here.......
Remco Poelstra
@ all
......... I see
I'm happy to see that the problem is difficult that means that I'm not too stupid because I don't get the solution in thinking about the problem for two day's
🙂
......... I see
I'm happy to see that the problem is difficult that means that I'm not too stupid because I don't get the solution in thinking about the problem for two day's
🙂
tab30 said:
Not necessarily.
Its usually the power supply but the amplifier topology can become
the limiting factor also, rather far too easily in the case of Mosfets.
If the amplifier is the limiting case you have an overspecified
power supply and/or or a Mosfet output stage.
🙂 sreten.
@ Thoru
#3 is simply the solution from #2 , at least mean this my math-Programm.
and with u=Umax it is the right "Prms education" I think.
#3 is simply the solution from #2 , at least mean this my math-Programm.
and with u=Umax it is the right "Prms education" I think.
Hi Tab!
You can roughly estimate the following.
400W at 4 Ohms means 10Arms, means about 14A peak.
Your output transistors (let's estimate BJTs in darlington configuration)might be able to work until 3V voltage drop.
Additional there is probably a resistance in series to the emitter.
May be 0.15 Ohms? ==> another 2.1V voltage drop.
Makes together already more than 5V drop.
Additional there is the saging of the DC rail itself. This depends very much on the transformer and the electrolytic caps.
At high frequencies this will be less worse and you could calculate
with the discharge magnitude of the cap by the arithmetic average half wave current of the audio signal.
At I=14.1Apeak this makes about 14.1A x 0,318 = 4.5A.
Time between charge current peaks is typically around
9ms at 50Hz mains (depends a little bit on the size of caps and the
transformer). What capacitance do your caps have?
30 000uF for each rail?
discharge magnitude = (i_average x t) / C
= (4.5A x 9ms) / 30 000µF = 1.35V
At low frequencies you will have to calculated with a value close
to the peak current ==> worst case: (14.1A x 9ms) / 30 000µF = 4.2V.
Makes already more than 9-10V voltage drop at worst case.
The remaining 4-5 V will easily drop at the internal impedance of the
transformer. Additional there might be also some small current depending voltage drop in the rectifier....
Well, the component values above are just some estimations, you may recalculate with the values of your amp.
But in fact it would not be unexpected if an amp which has
+/-70V DC rails at no load condition, might start clipping
at 56V peak output into 4 Ohms.
Cheers to my home town!!!!
Markus
You can roughly estimate the following.
400W at 4 Ohms means 10Arms, means about 14A peak.
Your output transistors (let's estimate BJTs in darlington configuration)might be able to work until 3V voltage drop.
Additional there is probably a resistance in series to the emitter.
May be 0.15 Ohms? ==> another 2.1V voltage drop.
Makes together already more than 5V drop.
Additional there is the saging of the DC rail itself. This depends very much on the transformer and the electrolytic caps.
At high frequencies this will be less worse and you could calculate
with the discharge magnitude of the cap by the arithmetic average half wave current of the audio signal.
At I=14.1Apeak this makes about 14.1A x 0,318 = 4.5A.
Time between charge current peaks is typically around
9ms at 50Hz mains (depends a little bit on the size of caps and the
transformer). What capacitance do your caps have?
30 000uF for each rail?
discharge magnitude = (i_average x t) / C
= (4.5A x 9ms) / 30 000µF = 1.35V
At low frequencies you will have to calculated with a value close
to the peak current ==> worst case: (14.1A x 9ms) / 30 000µF = 4.2V.
Makes already more than 9-10V voltage drop at worst case.
The remaining 4-5 V will easily drop at the internal impedance of the
transformer. Additional there might be also some small current depending voltage drop in the rectifier....
Well, the component values above are just some estimations, you may recalculate with the values of your amp.
But in fact it would not be unexpected if an amp which has
+/-70V DC rails at no load condition, might start clipping
at 56V peak output into 4 Ohms.
Cheers to my home town!!!!
Markus
It's it's it's........
Could have know it since #1 is totally true and #3 is the same, but #2 is strange. If you would calculate a Vrms with that it would be wrong (you have to ommit r then of course). But since your are calculating a power here the root doesn't have to be there. Strange... I'm certainly going to check this out if someone here can't enlighten me.
Remco Poelstra
Could have know it since #1 is totally true and #3 is the same, but #2 is strange. If you would calculate a Vrms with that it would be wrong (you have to ommit r then of course). But since your are calculating a power here the root doesn't have to be there. Strange... I'm certainly going to check this out if someone here can't enlighten me.
Remco Poelstra
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