1. Thank you to all members! what a great board this is. i have been able to answer many questions i have had for years just by reading this board! Thank You!
2. I have been wanting to test some amplifier Power supplies under load as would be in a power amplifier.
If my thinking is correct. then each half of the power supply Positive and negative would "see" the speaker/cable resistance and any resistance in the output transistors etc.
so,
wouldnt it be safe to assume, that each half of the power supply would "see" 8 ohms+ or 4ohms+ etc? (granted i have simplfied this greatly and i am leaving out the time/under the curve factor etc.)
I have a massive load bank of 8x 100 watt A-B resistors mounted on a 3' long finned heatsink extrusion wired to provide dual 8/4 ohm connections @ 400watts++++ of dissapation.
Couldnt i simply connect my load bank across the power supply to "load" the supply down as would be seen under max power use? or is my thinking seriously flawed?
I would like to determin load drop and temp rise under max power conditions.
Thanks
Zero Cool
2. I have been wanting to test some amplifier Power supplies under load as would be in a power amplifier.
If my thinking is correct. then each half of the power supply Positive and negative would "see" the speaker/cable resistance and any resistance in the output transistors etc.
so,
wouldnt it be safe to assume, that each half of the power supply would "see" 8 ohms+ or 4ohms+ etc? (granted i have simplfied this greatly and i am leaving out the time/under the curve factor etc.)
I have a massive load bank of 8x 100 watt A-B resistors mounted on a 3' long finned heatsink extrusion wired to provide dual 8/4 ohm connections @ 400watts++++ of dissapation.
Couldnt i simply connect my load bank across the power supply to "load" the supply down as would be seen under max power use? or is my thinking seriously flawed?
I would like to determin load drop and temp rise under max power conditions.
Thanks
Zero Cool
What????
Connect the power supply to a load resistance equal to the speaker???
Thevenin equivalent circuit for any amp is a current source in series with an output impeadence, in series with the load impeadence or speaker. In other words, with the correct speaker impeadence half of the power to the amp circuit is disapated as heat in the outputs, the other half goes to the load. If 400W RMS is what the amplifier outputs, then you have to use 800W for the power supply. The peak current of the signal is going to be at a higher current level than the continuous output of the power supply. But if the power supply will hold the RMS current (whatever you have calculated based on the voltage) and the voltage doesn't drop, then it should drive the amp.
IOW, use 2X(speaker impeadence) to test the rail voltage for one side, either +V or -V.
Connect the power supply to a load resistance equal to the speaker???
Thevenin equivalent circuit for any amp is a current source in series with an output impeadence, in series with the load impeadence or speaker. In other words, with the correct speaker impeadence half of the power to the amp circuit is disapated as heat in the outputs, the other half goes to the load. If 400W RMS is what the amplifier outputs, then you have to use 800W for the power supply. The peak current of the signal is going to be at a higher current level than the continuous output of the power supply. But if the power supply will hold the RMS current (whatever you have calculated based on the voltage) and the voltage doesn't drop, then it should drive the amp.
IOW, use 2X(speaker impeadence) to test the rail voltage for one side, either +V or -V.
Cunningham,
I don't understand your answer. In the first place I think you meant a voltage source in series with et cetera, instead of current source. In the second place it is not always the case that with a correct speaker impedance half the power is dissipated in the load and half in the amplifier. For some cases of class A, but not for class AB or B.
An ideal class B amplifier has an efficiency for sinewaves at clipping level of 78% (pi/4), so 78% of the power is in the load, only 22% is in the amplifier. For square waves at clipping level 100% of the power is in the load, 0% is in the amplifier (ideal case, looks like a class D amplifier).
One could say that for the load resistance as seen by the power supply take 1/0.78 = 1.27 x speaker resistance. This is based on sine wave load at clipping level, which is definitely a heavier load than with normal music or speech at clipping level. The worst case situation is when the amplifier would have been driven by a squarewave (clipping level), then take speaker impedance as the power supply load, though this is not a realistic situation.
In practice the load could even be higher, because speakers have a back-EMF that can be out of phase and even higher currents are demanded from the amplifier. But these are only short current peaks and the average value is much lower.
Steven
I don't understand your answer. In the first place I think you meant a voltage source in series with et cetera, instead of current source. In the second place it is not always the case that with a correct speaker impedance half the power is dissipated in the load and half in the amplifier. For some cases of class A, but not for class AB or B.
An ideal class B amplifier has an efficiency for sinewaves at clipping level of 78% (pi/4), so 78% of the power is in the load, only 22% is in the amplifier. For square waves at clipping level 100% of the power is in the load, 0% is in the amplifier (ideal case, looks like a class D amplifier).
One could say that for the load resistance as seen by the power supply take 1/0.78 = 1.27 x speaker resistance. This is based on sine wave load at clipping level, which is definitely a heavier load than with normal music or speech at clipping level. The worst case situation is when the amplifier would have been driven by a squarewave (clipping level), then take speaker impedance as the power supply load, though this is not a realistic situation.
In practice the load could even be higher, because speakers have a back-EMF that can be out of phase and even higher currents are demanded from the amplifier. But these are only short current peaks and the average value is much lower.
Steven
Music isn't like that though. You will be building very big power supplies to be able to continuously supply the peak power. I don't know the "crest factor" for music, but I think it has been discussed here before.
There are of course many opinions on this, but apparently music is only like a continuous sine wave when sizing power supplies.
There are of course many opinions on this, but apparently music is only like a continuous sine wave when sizing power supplies.
I was being very general for the power supply requirements. It does certainly matter what type of amp circuit is being powered. Judging from the breif description, this is probably a class AB or B. Anyway, if more than one channel is powered, and the power supply can drive 2xoutput power or close to it, the power supply shouldn't have much trouble running without showing significant ripple voltage.
Alright let me see if i can clear this up a bit.
I have way oversimplified things here to get to the basics. so let;s forget crest factor, back emf, etc etc.
What i want to do, is load a power supply down to test for % of load drop as would be seen in normal every day use and under worst case scenarios. I want to measure ripple and regulation etc.
So assuming/pretending our speaker load is a purely resistive 8 ohm load. and assuming that the amplifier is at clipping. again ingoring crest factor etc. then what resistance is seen by the power supply?
Effectivly each bank Positive or negative is only on one at a time so each bank would then see the 8 ohm resistance of the speaker plus any internal resistance of the output devices etc correct??
So for oversimplified testing a very large 8 ohm resitor across the + to ground and - to ground should simulate what the power supply will see under worst case conditions correct???
I understand actual use will be different.
Or let me put it this way. IF i wanted to load a power supply down, what resistance should i use to simulate max power worst case scenarios as would be seen with a nominal 8 ohm speaker??
The load would only be connected for short peiods of time just to take measurments etc. and i understand real world performance will be different. but i just wanted to get something close to the real thing without the modulation effects the amp would have on the PSU Rails when driven to full power.
Does that make any sense? or have i convaluted it up more???
I have way oversimplified things here to get to the basics. so let;s forget crest factor, back emf, etc etc.
What i want to do, is load a power supply down to test for % of load drop as would be seen in normal every day use and under worst case scenarios. I want to measure ripple and regulation etc.
So assuming/pretending our speaker load is a purely resistive 8 ohm load. and assuming that the amplifier is at clipping. again ingoring crest factor etc. then what resistance is seen by the power supply?
Effectivly each bank Positive or negative is only on one at a time so each bank would then see the 8 ohm resistance of the speaker plus any internal resistance of the output devices etc correct??
So for oversimplified testing a very large 8 ohm resitor across the + to ground and - to ground should simulate what the power supply will see under worst case conditions correct???
I understand actual use will be different.
Or let me put it this way. IF i wanted to load a power supply down, what resistance should i use to simulate max power worst case scenarios as would be seen with a nominal 8 ohm speaker??
The load would only be connected for short peiods of time just to take measurments etc. and i understand real world performance will be different. but i just wanted to get something close to the real thing without the modulation effects the amp would have on the PSU Rails when driven to full power.
Does that make any sense? or have i convaluted it up more???
I think that if you are going to the trouble of measuring something, then the results should be meaningful, and the limitations of the measurement should be known.
The best way to measure the power supply is to use it to power an amp, driving a load. A resistor could be used for the amp load if the sound level would be very loud with a speaker load. Then the ripple could be observed while music is playing. Use a worst case music signal. This would tell you exactly what you want to know, and the results would be meaningful. After doing this, compare the results with a resistor load across the power supply, to see how different the results are.
The best way to measure the power supply is to use it to power an amp, driving a load. A resistor could be used for the amp load if the sound level would be very loud with a speaker load. Then the ripple could be observed while music is playing. Use a worst case music signal. This would tell you exactly what you want to know, and the results would be meaningful. After doing this, compare the results with a resistor load across the power supply, to see how different the results are.
I now see what you are trying to do.
You should be able to calculate it.
First of all, even if the amp is at clipping and the full rail voltage is droped across the 8Ohm resistor, there is going to be half positive and half negative. So the RMS current would be only half of the peak current for a square wave, or completely clipped amplifier. current spikes is what filter caps are for.
For 100W RMS across 8Ohms, you would have 3.5A continuous AC current out. About 5A peak, but that doesn't really matter if the filter caps are large enough. 3.5A continuous current out is being driven by two transistors. So each transistor has to output 1.75A. If your power supply can hold the rail voltages (relativly, there will always be some ripple) at 1.8A then you should be fine.
This is for one channel thogh so for 4, it would be 7.2A.
Just determain what size resistor needed for the rail voltages, from rail to rail, that would conduct 7.2A. \
Personally I would use seperate power supplies for each channel and commonly connect all of the grounds together as circuit ground. This will give you less chance for crosstalk and ground loops.
You should be able to calculate it.
First of all, even if the amp is at clipping and the full rail voltage is droped across the 8Ohm resistor, there is going to be half positive and half negative. So the RMS current would be only half of the peak current for a square wave, or completely clipped amplifier. current spikes is what filter caps are for.
For 100W RMS across 8Ohms, you would have 3.5A continuous AC current out. About 5A peak, but that doesn't really matter if the filter caps are large enough. 3.5A continuous current out is being driven by two transistors. So each transistor has to output 1.75A. If your power supply can hold the rail voltages (relativly, there will always be some ripple) at 1.8A then you should be fine.
This is for one channel thogh so for 4, it would be 7.2A.
Just determain what size resistor needed for the rail voltages, from rail to rail, that would conduct 7.2A. \
Personally I would use seperate power supplies for each channel and commonly connect all of the grounds together as circuit ground. This will give you less chance for crosstalk and ground loops.
This is almost correct, (referring to Cunninghams calculations) except that the current drawn from the power supply is the average current, not the RMS current.
For 100W in 8 ohm, peak current is 5A, (2 x SQRT(100/8)),
RMS current is peak current divided by SQRT2 or 5/1.4142 = 3.54 A, average current of a Sine wave is RMS current/1.1 so 3.54/1.1 or 3.2A. There are 2 transistors so each draw current half the time giving 1.6A. not so much difference compared to RMS current but still important. Note! This is only valid for sinewaves.
There is an explanation on this page about difference between RMS and average current http://sound.westhost.com/efficiency.htm
Regards Hans
For 100W in 8 ohm, peak current is 5A, (2 x SQRT(100/8)),
RMS current is peak current divided by SQRT2 or 5/1.4142 = 3.54 A, average current of a Sine wave is RMS current/1.1 so 3.54/1.1 or 3.2A. There are 2 transistors so each draw current half the time giving 1.6A. not so much difference compared to RMS current but still important. Note! This is only valid for sinewaves.
There is an explanation on this page about difference between RMS and average current http://sound.westhost.com/efficiency.htm
Regards Hans
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