Question related to voltage drop

[MichaelJHuman]

Hello, I have been trying to understand the exact technical details of voltage drop under power supply load.

My curiosity stemmed from modern audio video receivers. When they bench test these, their power with one channel driven is often much higher than their power output with all channels driven. I have seen power drop by 1/3 with between one channel driven and all channels driven in published benchmarks.

My understanding is that transformer size is important. But why?

I understand that there are physical limits to current draw from a transformer. But say, for example, we are not drawing enough to burn up the transformer.

Then what? Copper losses are understandable as I understand resistance increases as current increases. But is there more?

Iron losses seem more or less constant.

How about flux? My understanding is that maximum flux is at NO load. But as current draw increases does IR increase due to some factor? Back EMF or something like that?

Thank you for any help.

[ra7]

I would refer to the excellent text written by Paul Horowitz: The Art of Electronics.

The first chapter will answer many of your questions.

[MichaelJHuman]

Bit pricey, unless there's an online sample from the book explaining the answer.

I was hoping for a slightly more helpful answer, no offense.

[woody]

You have a combination of things happening. The power transformer voltage will
drop a little under load and as it heats up it will loose a little voltage, the diode bridge will also have a little more voltage drop under load and the cap bank will not be as effective at higher curent loads. All of these losses add up. And rember a 10% reduction in the dc voltage could easly show up as a 30% power loss depending on the amp.

[MichaelJHuman]

How about voltage across the secondary? As the resistance from all of the output devices would decrease under heavy load, couldn't the load resistance become close enough to the resistance of the secondary to cause a substantial drop in voltage?

Say -

V1 + V2 = 70 volts (rail voltage0
R1 = secondary resistance
R2 = load of the output devices in an amplifier

R1 = 100
I = .697
V2 = 69.7 (very little proportional drop across the secondary)

But if the amplifier load increases -

R1 = 5 ( This does not seem unreasonable in a heavily loaded 7 channel amp to me)
I = 70 / (5 + .5) = 12.73 amps
V2 = 64 volts (A fair drop)

And this assume secondary resistance is not increasing under load.

Does any of that make sense? (I don't know a lot about electronics, so I could be very wrong.)