Hello, I would like to ask the following:
It is about a 100-watt amplifier, with an 8-ohm load and a +-40v power supply. To get the current of the power source, we divide 40v by 8 ohms, and get 5 amps. It is not clear to me if this 5 amps is common to both /+ and –/ halves, because if it is, then multiplying the current 5 amps by the voltage 40 volts, we get 200 watts. If we divide these 5 amps by 2 /for the + and – halves/, then 2.5 amps at 40 volts will give us 100 watts of power for each arm. In this regard, am I to understand that the power supply will be 2x40V at 2x2.5 amps? I'm sorry for the bad English, but I use Google translator.
It is about a 100-watt amplifier, with an 8-ohm load and a +-40v power supply. To get the current of the power source, we divide 40v by 8 ohms, and get 5 amps. It is not clear to me if this 5 amps is common to both /+ and –/ halves, because if it is, then multiplying the current 5 amps by the voltage 40 volts, we get 200 watts. If we divide these 5 amps by 2 /for the + and – halves/, then 2.5 amps at 40 volts will give us 100 watts of power for each arm. In this regard, am I to understand that the power supply will be 2x40V at 2x2.5 amps? I'm sorry for the bad English, but I use Google translator.
80 volts into an 8 ohm load at the output gives us 10 amps at both ends, does that mean that from each arm to neutral we'll have 5 amps, or 2 x 40 volts at 2 x 5 amps? Are the calculations I listed above correct?And then 80 volts times 10 amps gives us 800 watts, right?
Your assumption is that the 100W is output all the time, constantly. It isn't.
Music is perhaps 5% that (unless you listen to AC/DC).
So although your supply has to be able to provide that max current, it only needs to do it once in a while.
Jan
Music is perhaps 5% that (unless you listen to AC/DC).
So although your supply has to be able to provide that max current, it only needs to do it once in a while.
Jan
The current needed to drive your speakers is much less. 100watts output will be voltage output x current output. And it will be AC voltage while the Power supply will be DC voltage. So if in theory your amp can output close to VCC + say 36 volts AC, 100 watts divided by 36 volts equals 2.778 amps
As already said, a 100W amplifier will only deliver 100W peaks at the output and the average power is given by the crest factor which can be anywhere between 2 and 10. It will be 1 if you only listen to continuous sine, which is not valid for music.
Finally, you have to take into account that the theoretical maximum efficiency of a class B stage is 78%, but because it has to work in class AB and because it also consumes the other components in the amplifier, a 50% efficiency is taken into account .
So if the amplifier outputs 100W of music, a source capable of delivering 100W should be sufficient. ((100W/crest factor 2)*50%efficiency=100W).
Between the power peaks, the filter capacitors are charged and deliver power to the output at the peaks.
For various reasons, a transformer capable of delivering between 1.5 and twice the output power is usually chosen. Higher transformer powers are chosen only for purely audiophile reasons and usually do not help too much.
Finally, you have to take into account that the theoretical maximum efficiency of a class B stage is 78%, but because it has to work in class AB and because it also consumes the other components in the amplifier, a 50% efficiency is taken into account .
So if the amplifier outputs 100W of music, a source capable of delivering 100W should be sufficient. ((100W/crest factor 2)*50%efficiency=100W).
Between the power peaks, the filter capacitors are charged and deliver power to the output at the peaks.
For various reasons, a transformer capable of delivering between 1.5 and twice the output power is usually chosen. Higher transformer powers are chosen only for purely audiophile reasons and usually do not help too much.
If your amp is class (A)B ist isn't the case. Each polarity will furnisch one half wave polarity, the other supply polarity will provide the other half wave.
So each polarity only delivers power half the time.
It gets complicated to exactly calculate the power needed for 100W, but Duncans PSUD will help.
And don't overthink it.
For a 100W misuc amp, a power supply that can deliver a couple of amps will be fine.
(Why you are looking so shocked now? Yes, engineering is often the art of getting away with it ;-).
Jan
So each polarity only delivers power half the time.
It gets complicated to exactly calculate the power needed for 100W, but Duncans PSUD will help.
And don't overthink it.
For a 100W misuc amp, a power supply that can deliver a couple of amps will be fine.
(Why you are looking so shocked now? Yes, engineering is often the art of getting away with it ;-).
Jan
I'm not shocked at all, if I understand correctly in this case we have +36 volts and 2.778 amps for one half wave and -36 volts and 2.778 amps for the other half wave. Then 36x2.778 gives us 100 watts for one half-wave, and we need as much for the other half-wave, i.e. after all, for the two half-waves we need 2x36 volts and 2x2.778 amps, right?
No!
Only one half wave will deliver power to the load at one moment in time.
So you will need a transformer with 2 secundaries of 36Vac/2.8A, but will be barely enough. Better go to a 4A transformer, 2x36V/4A.
Only one half wave will deliver power to the load at one moment in time.
So you will need a transformer with 2 secundaries of 36Vac/2.8A, but will be barely enough. Better go to a 4A transformer, 2x36V/4A.
Your transformer will deliver 40x2x5=400va
Regardless of how you use the secondaries, parallel or series.
Parallel will give you one 40v/10amps
Series will give you 2x40vols/5amps
Regardless of how you use the secondaries, parallel or series.
Parallel will give you one 40v/10amps
Series will give you 2x40vols/5amps
Usually a transformer with a power double of the output power is enought.
For 2x100W is enought a 400W transformer.
For 2x100W is enought a 400W transformer.
I have a transformer but the question is whether I have correctly calculated the current at +–40 volts and an 8 ohm load, namely 5 amps, and since I have two 40 volt halves, is it 5 amps each or 2.5 amps for each?
Rod elliott has published an extensive article regarding power supplies for amps:
https://sound-au.com/power-supplies.htm
from the summary:
https://sound-au.com/power-supplies.htm
from the summary:
Last edited:
Peak current is 40V / 8ohm = 5A. Effective current for sine is peak current / 1.41. In this case 5A/1.41= 3.55A.
Hence max power from 40V supply is 3.55A * 3.55A * 8ohm = 100W.
Due to limitations of active devices, peak output voltage is always somewhat lower than supply voltage. So 40V supply could be good for perhaps 38V peak output voltage that yields 90W.
Transformer delivers current in short intense pulses 100 or 120 times per second. Since transormers are rated for delivery of continuous sine current, there is no simple relation between transformer's rated power (and current) and the capability to deliver current in impulse mode.
Rule of tumb for AB class amps is that rated transformer's power should be about twice the output power. For 2 x 100W, trans of 400W is recommended.