power supply question

[david counter]

when you add a cap to a choke input supply before the chokes it increases the B+ can any one tell me how this works,

Thanks

[A Stuart]

I would think the rectifier snatches a large current at the peak of the sine-wave to top-up the capacitor to the peak voltage. The capacitor will discharge a proportion only of this voltage before the next top-up. Thus feeding the choke at near peak voltage all the time.

The choke on its own, in contrast, without the capacitor will only be fed by the rectified sine wave.

Correct me if I, myself, misunderstand.

Alastair

[NickG]

I will have a go.

In a choke input supply, the choke and following cap forms a high pass filter (the cap isn't becessary, but it makes it a 2nd order filter, so improves the result, it could be a choke into a resistor, and the same would still be true). By removing the 100hz and above harmonics from the rectoified sine wave, the result is the remaining DC component of the rectified sine wave. This gives the 0.9 figure for the voltage output of a choke input supply. Note that there is nothing in the above that makes the voltage dependednt on the load, hence the regulation provided by a choke input supply.

If you place a cap (ignore the choke for the moment), after the rectrifier, the cap will charge from the rectifier, and discharge into the load, if the cap is large and the load is low, the cap will remain mostly charged, and will be at the peak voltage of the incomming sine wave, or 1.414 of the average voltage (thats where this number comes from). As the load increases, the cap discharges more and more between cycles, so the average voltage will drop (hence the output from a cap input supply, is directly dependent on load, so poorer regulation). Also if the cap is smaller, it will hold less charge, so again will discharge faster, leading to a lower average voltage.

So, for a given load, the size of the cap after the rectifier will control the average DC voltage across the cap. Then if you add the two together, the low pass filter of the choke, will now not be filtering the rectified sine wave, with its DC component of 0.9, but the output of the cap, with its additional DC component from 0 (very very small cap), to 1.414 (very very large cap). So you can see (hopefully) that the size of the cap, will adjust the output from the choke from 0.9 times, (zero cap) to 1.414 times (very very big cap).

But then of course, the cap voltage is load dependent, so the regulaton of the combination will be somewhere betwen a pure choke input supply, and a cap input supply.

Bit of a long ramble, hope it makes sense.

[pre65]

Hi-so in a nutshell,is it beneficial to have the cap before choke ?

Rankin uses it in the 45 DC CC.

Phulip

[david counter]

thanks Alastair, Nick,
so If I understand this, in practical terms I gain B+ but at the expence of regulation,

As Philip says this is a favourite of Gordon Rankin,

is it's only benifit to allow you to use a tx with a lesser output and in no way improves things

[NickG]

Hi-so in a nutshell,is it beneficial to have the cap before choke ?

Depends, in the case of a pure choke input, you need to have a snubber cap somewhere in there, so something up to .22u can be a good thing.

Other than that, personally I don't think so, but then I like choke input supplies, as much for their regulation as the reduction in switching noise and large current pulses. For the latter reason I think they are worth having even if followed by a regulator. The downside is it places larger stess on the choke, so that needs to be bigger and heavier. The other big downside is making sure it stays in choke input, so you need to ensure the minimum current is maintained. Hence the large heat sinks on the 211 power supply, to allow 40ma through the dropper resistors. Its less of a issue with lower voltage. Just make sure the caps and other bits can deal with 1.414/0.9 (1.6) times your intended voltage, and then it doesn't matter if the valve stops conducting if the heater supply fails.

But thats just my take on it.

[Ianm2]

the cap charges to the peak of the ac waveform.

like nick says, 1.41...( square root of 2)

choke input is something like 0.9 of whatever is going in, I was never any good with understanding that, and am very rusty here.

differences are that choke input apparently gives better regulation, ( less voltage droop/sag/ resistive losses) under high current draw.

BUT, in psu designer, I have found that to get good smoothing with 1 choke and cap, it makes the power supply ring/overshoot, possibly even unstable.

No-one I have ever seen talks about that on the web, not to say it isn't there.
And that's only on idle, when the amp is drawing bias, goodness knows what happens under ac conditions. Perhaps Class A aren't affected as much by this.

In most transistor amps(and probably valve), half wave rectified signal voltages appear on the power rails. Not entirely sure what that means totally.

Rememer if there is no music going in, your amp is only drawing dc from the psu, music of course is ac and that's a bigger can. DC is simple errr.



There is an awful lot more to psu design than meets the eye, it will work if you copy one or chuck things in, but it really needs fully studying in depth.

That's why I am not really into building my own anymore, it gets far too complex to do right, and I dislike the idea of unstable amps/power supplies intensely.

Imagine what could happen when you start adding lower loads, reactive loads, sucking more current, phase changes, the mind boggles how these things actually work.

[NickG]

In most transistor amps(and probably valve), half wave rectified signal voltages appear on the power rails. Not entirely sure what that means totally.

In push pull class B thats the case. In push pull A in theory the power supply only see's a constant load. In single ended it see's the current required to replace lost flux in the output transformer, for inductive loads, and the signal current for resistive loads.

[A Stuart]

Originally Posted by NickG

In single ended it see's the current required to replace lost flux in the output transformer

Short of asking you to write a textbook on electrical theory, could you expand on this? Why is this different from saying more simply, "...sees the signal current allowed to flow by the output valves"?

Not disputing what you are saying. Simply trying to learn.

Alastair

[NickG]

Short of asking you to write a textbook on electrical theory

Good, coz I can't.

I don't know if what I have said is true, but I believe its what happens, maybe someone else could help.

Ok, this is what I believe happens. At idle (no signal) there is a constant flow of current through the transformer, so a constant magnetic flux is created in the core. As the flus is constant, there is no current induced in the secondary so (idnoring assorted losses) all is static, the current flowing into the transformer is the same as the current flowing through the valve. Now a +ve going signal arrives at the valve, this make the valve conduct more current. But the attempt to change the current in the transformer, will cause a change in the flux in the core, this change in flux will induce a current in the secondary and so produce output. But it will also produce the same effect in the primary, the result of this will be to attempt to oppose the change in current. So the extra current through the valve, will not be entirly taken from the power supply, some will come from the changing flux level in the core. Of course eventually the flux is replenished by the power supply, but if the signal is AC, then it may not need to be, it can be provided by the folowing cycle.

Of course I could be very wrong, but I think the use of an inductive load must alter the way the supply interacts compaired with a resistive load.