Hi all, this question has probably been asked before, but I am confused and would like some assistance please.
What is the RMS power that I can obtain from a push pull class A amplifier using the following perammiters:
1) plus and minus 24V rails ( loaded ) with 1.8 Amp per rail bias
2) plus and minus 35V rails ( loaded ) with 1.6 Amp per rail bias
The load impedance will be 8 Ohms
Also what heat dissapation for each case.
Thanks for your help.
Alan.
What is the RMS power that I can obtain from a push pull class A amplifier using the following perammiters:
1) plus and minus 24V rails ( loaded ) with 1.8 Amp per rail bias
2) plus and minus 35V rails ( loaded ) with 1.6 Amp per rail bias
The load impedance will be 8 Ohms
Also what heat dissapation for each case.
Thanks for your help.
Alan.
Peak rail voltage will loose a bit going through the output devices. This can be as low as 1 volt or less. The RMS voltage will be 1/2 times the inverse square root of 2 times the peak to peak voltage. So for 24 volt rails that might be as high as 46 volts peak to peak or around 16.25 volts RMS. An 8 ohm load would try to draw slightly more than 2 amps RMS from this. Now a real loudspeaker rated at 8 ohms could drop as low as 4 ohms and draw way more current. So a 1.8 amp current limit is too low.
The power dissipation with no output is the bias current times the rail voltage. Unless you are seriously clipping your system that will be slightly less than the in use dissipation. Although you could calculate the maximum power dissipation based on a full value sine wave into an 8 ohm load, you will never see that in real use.
The power dissipation with no output is the bias current times the rail voltage. Unless you are seriously clipping your system that will be slightly less than the in use dissipation. Although you could calculate the maximum power dissipation based on a full value sine wave into an 8 ohm load, you will never see that in real use.
Class A can be voltage limited or current limited. In your case it will be current limited, as the load impedance is too low. Average power output will be 8x1.8x1.8/2=12.96W. Heat (no signal) will be 24x1.8=43.2W. True Class A means the power draw from the supply remains unchanged with signal, but power gets moved from heat to the output.
Thanks for the replies Simon and DF96 but I am still confused. Perhaps if I ask the question in a different way I may be able to understand the answer.
Am I correct with :
Class A RMS power = Iq squared x 2 x load divide by 2. ( for push - pull )
So if Iq was 1.8 A and load was 8 Ohm then:
1.8 x 1.8 x2 x 8 /2 = 25.92 W
Is this correct so far ? If so then what I do not understand is the question of rail voltage. Where does that come into the calculation? How do I work out how many volts on the rails I require to get the above wattage . If the rails are at say plus and minus 24 volts and I Increase the bias to say 2 or 2.5A then surely I will not be able to get more power unless i increase the rail voltage. If this is so how do I calculate the voltage to say give me 40 watts into 8 ohms ( I can increase the bias current ).
Sorry to be so dense.
thanks for your help.
Alan
Am I correct with :
Class A RMS power = Iq squared x 2 x load divide by 2. ( for push - pull )
So if Iq was 1.8 A and load was 8 Ohm then:
1.8 x 1.8 x2 x 8 /2 = 25.92 W
Is this correct so far ? If so then what I do not understand is the question of rail voltage. Where does that come into the calculation? How do I work out how many volts on the rails I require to get the above wattage . If the rails are at say plus and minus 24 volts and I Increase the bias to say 2 or 2.5A then surely I will not be able to get more power unless i increase the rail voltage. If this is so how do I calculate the voltage to say give me 40 watts into 8 ohms ( I can increase the bias current ).
Sorry to be so dense.
thanks for your help.
Alan
I think I made a mistake in my calculation. Sorry. For push-pull Class A the peak signal current is twice the quiescent current, so 3.6A. That would give a peak signal voltage of 28.8V, which is greater than your DC supply. Hence you are voltage limited. Highest average output power is 24 x 24 / (2 x 8) = 36W. Peak signal current will be 3A.
The equations you need are Ohm's Law: V=IR, and P=I^2 R = V^2 /R - remembering that average power is half peak power for a sine wave signal.
No such thing as RMS power (except as a weird hi-fi standard in the 1970s?). When you use RMS voltage or current to calculate power, the result is mean/average power.
The equations you need are Ohm's Law: V=IR, and P=I^2 R = V^2 /R - remembering that average power is half peak power for a sine wave signal.
No such thing as RMS power (except as a weird hi-fi standard in the 1970s?). When you use RMS voltage or current to calculate power, the result is mean/average power.
Are you sure? I'd think that for both transistors to remain conducting, the current diverted to the load must be no higher than Iq, so you'd get 2Iq peak-peak at most, and that's an effective Iq/sqrt(2) for a sine.
One transistor can drop 1.8A, the other can add 1.8A, so the net peak signal current is 3.6A. On the opposite peak, the reverse is true so you get 3.6A again but in the opposite direction.
No point in confusing people with details like BJT collector saturation or non-linearity when they are still trying to understand the basics.
Thanks everyone for the help, much appreciated.
DF96, the penny finally dropped and I now understand (not the technicalities)the basics and can go forward from here.
I am still in the throes of building a Krell KSA50, but I don't want to dissapate too much heat, so I will use plus and minus 24v and bias at 1.5A per rail and dissapate 72watts of heat per channel and get about 30W at 8 Ohms, I hope. Sort of a KSA30.
Thanks for the help.
Jacco, you must read every thread, cheers
Alan
DF96, the penny finally dropped and I now understand (not the technicalities)the basics and can go forward from here.
I am still in the throes of building a Krell KSA50, but I don't want to dissapate too much heat, so I will use plus and minus 24v and bias at 1.5A per rail and dissapate 72watts of heat per channel and get about 30W at 8 Ohms, I hope. Sort of a KSA30.
Thanks for the help.
Jacco, you must read every thread, cheers
Alan
Conversing with my pencil head eraser is somewhat monotonous.
The crave for attention is so strong, at non-English fora, I beg folks to talk to me.
Figures are OK, a little extra never hurts.
The crave for attention is so strong, at non-English fora, I beg folks to talk to me.
Figures are OK, a little extra never hurts.
Yeah, but I think you could easily say try not to let the Vce drop below five (4, 3?) and the current below 70ma in each device,.... if you're serious about maintaining class A performance. That whittles quite a bit off the utrasimplified version of pwr-in/pwr-out for small class A amplifiers, but it's still pretty simple. If you're willing to allow increased distortion at the peaks, you don't have to allow so much to completely avoid cutoff and saturation.
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No. The OP clearly needed the simplest possible explanation. Once he understands that, and the concepts of voltage limiting vs current limiting, he can then start looking at non-ideal conditions in a real circuit.
Ok. I admit I didn't start reading from the beginning of the thread, and I'll still assume you got it covered then.
I really only wanted to add this slightly fatter rule of thumb because I think it's useful, right away, to know how power hungry actual class A is... Not meaning to confuse.
I really only wanted to add this slightly fatter rule of thumb because I think it's useful, right away, to know how power hungry actual class A is... Not meaning to confuse.
For illustration:
Iq is about 4.8 mA.
By the time one output transistor is down to zero, the other one has ramped up to 2*Iq, so that marks maximum output current delivery.
Iq is about 4.8 mA.
By the time one output transistor is down to zero, the other one has ramped up to 2*Iq, so that marks maximum output current delivery.
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