Can anyone tell me how many Class A watts I have if my AB amp is rated 80 watts but draws 125 watts from the wall at idle? Is there an easy equation for this? Thanks in advance.
Its not that simple.... Do you know what the bias setting is on your amp?
Even better can you pull a rail fuse and put your meter inline to measure DC current at idle in one channel?
Even better can you pull a rail fuse and put your meter inline to measure DC current at idle in one channel?
Hi Chamberman, I'm not sure. My Tech told me it was drawing 125 and it made me wonder if there was a way to get a ball-park figure. I found a Pass article on it once but I couldn't for the life of me work out the maths.
What amp is it? Model & manufacturer? If a service manual or schematic is available then the amount of Class A can be determined. Most AB amps do not have a lot of Class A power available even with high wattage draws from the wall. I would be surprised if its more than a couple of watts.
I'll give an example. Say your 80wpc rated AB amp has +/-50VDC on its DC rails. If you were to set the quiescent current to 1A p-p total per channel (thats 500ma on the negative rail and 500ma on the positive rail) you'd have 8watts peak (or about 4watts RMS) of class A power into an 8 ohm load. Now that one channel will be dissipating 50W at idle into the heatsink and that's a lot of heat so the heatsinks on your amp better be large. Multiply that by 2 and you're pulling 100W from the wall and this doesn't take into account losses elsewhere in the transformer and power supply.
I'll give an example. Say your 80wpc rated AB amp has +/-50VDC on its DC rails. If you were to set the quiescent current to 1A p-p total per channel (thats 500ma on the negative rail and 500ma on the positive rail) you'd have 8watts peak (or about 4watts RMS) of class A power into an 8 ohm load. Now that one channel will be dissipating 50W at idle into the heatsink and that's a lot of heat so the heatsinks on your amp better be large. Multiply that by 2 and you're pulling 100W from the wall and this doesn't take into account losses elsewhere in the transformer and power supply.
Here’s the math you’re looking for, is towards the bottom of this page;
Adjusting the Bias Of Your Amp [English]
Adjusting the Bias Of Your Amp [English]
Sweet, thanks man. It's a vintage Exposure IV from the eighties so no schematic unfortunately. I only use about one watt TBH with the odd ten watt peak. It sounds great so I don't actually want it rebiased. Was just curious if there was an easy equation to determine the AB crossover point.
This ballpark is beyond ridiculous.
Nothing much can be said from the idle draw if the rail voltage is unknown. If it is high, then even a modest bias can lead to such a power draw.
Example:
100v rails
315mA bias
63W idle dissipation per channel, or 126W for both (only for the outputs with no additional losses)
Class A 0.8W RMS /4ohms
The author has no idea what is a single ended amp but that doesn't stop him from using the term. Greatness is so subjective 🙄
I just pulled up the Exposure IV amplifier images online. Those heatsinks are very small and do not look promising for a lot of Class A power. It seems that Exposure made several different models of this amp. Is your model one of the regulated versions? If so some of the wattage you're using from the wall will be due to losses in the regulators especially if these are used for the DC rails in the output stage.
125W total consumption means something like 110W dissipation from the outputs. At 40v supply rails, deduced from the 80W/8ohms, but possibly higher in reality, this means that 80v is responsible for 55W of dissipation, hence the bias current is about 0.69A. Which in turn means 7.6W class A into 8ohms or
3.8W into 4ohms. If the actual supply rails are higher the class A power would be lower.
The math in that TNT article is wrong.
200mA p-p (100mA pos & 100mA neg) of total bias per channel works out to 0.32W into 8 ohms and that's peak power, RMS power will be basically half of that.
400mA of total bias per channel is 1.28W into 8 ohms peak or about 0.65W RMS.
Certainly. I must have missed Exposure being mentioned.
Those amps had no sinks to talk of. With or without regulators there is no way these could dissipate 125W at idle.
It is quite simple really........ 😉
Output stage Bias current squared times your load impedance gives you your operating in Class A output voltage range of the amplifier.
Cheers !!
jer 🙂
Output stage Bias current squared times your load impedance gives you your operating in Class A output voltage range of the amplifier.
Cheers !!
jer 🙂
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Actually, it's bias current squared times half the load impedance if you want RMS power.
Recall, I_RMS = I_peak/sqrt(2). P = I_RMS^2*R -> P = (I_peak/sqrt(2))^2*R -> P = I_peak^2/2 * R -> P = I_peak^2*R/2.
Similarly, if you want to find the max RMS output power for a given rail voltage:
V_RMS = V_peak/sqrt(2). P = V_RMS^2/R -> P = (V_peak/sqrt(2))^2/R -> P = V_peak^2/2/R -> P = V_peak^2/(2R).
Naturally this assumes that the output stage can actually drive the output to the full rail voltage. For a bipolar output stage that's almost true. You can probably get within a volt or two of the rail voltage, assuming the driver operates from the same rails as the output stage. For a MOS output, you'll be hard pressed to get closer than 5-6 V to the rail. Vgs > Vbe.
Tom
Recall, I_RMS = I_peak/sqrt(2). P = I_RMS^2*R -> P = (I_peak/sqrt(2))^2*R -> P = I_peak^2/2 * R -> P = I_peak^2*R/2.
Similarly, if you want to find the max RMS output power for a given rail voltage:
V_RMS = V_peak/sqrt(2). P = V_RMS^2/R -> P = (V_peak/sqrt(2))^2/R -> P = V_peak^2/2/R -> P = V_peak^2/(2R).
Naturally this assumes that the output stage can actually drive the output to the full rail voltage. For a bipolar output stage that's almost true. You can probably get within a volt or two of the rail voltage, assuming the driver operates from the same rails as the output stage. For a MOS output, you'll be hard pressed to get closer than 5-6 V to the rail. Vgs > Vbe.
Tom
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i would guess the railvoltage is somwhere around +/-45Vdc. There will be about 20% loss in transformer/PSU and amp curcuit. That leves around 100W (50W pr ch)
If 45V rail, you have a bias current of about 0,55A. That lewves you with close te 5W rms class A at 8ohm.
If 45V rail, you have a bias current of about 0,55A. That lewves you with close te 5W rms class A at 8ohm.