# Power Supply B+ voltage?

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#### losacco

Are you able to check power supply B+ voltage without connection to the rest of the circuit? I need B+ to be 250v but getting 385v measured at he last PS capacitor. I think a resistor needs to be added but not sure where and what value.

#### Vinylsavor

Hi!

The higher the output impedance of your PSU, the more the voltage will rise if you test it without load. If your PSU has a choke input filter, there is an even higher voltage rise if the current draw is below critical value.

In general it is good practice to have a bleed resistor across the PSU to discharge it when you turn it off. If you want to test it under normal load condittions but without the circuit attached. Use a dummy resistor to simulate your load. Calculate the resistor using Ohm's law so that it draws the same current as the circuit which will be attached.

Best regards

Thomas

#### SemperFi

+1 to what Thomas said. Ideally u should load the PSU as if intended ciruit is hooked up and drawing power, though drawing some 50% of intended is probably close enough. That is still a lot of power for a resistor, so use several in series/parallel. You could perhaps use a lightbulb or two as well.

#### Brouilly

Calculate the resistor using Ohm's law so that it draws the same current as the circuit which will be attached.

So if my amp should draw 200mA when running, and I have an unloaded B+ of 350, do I take the unloaded value or the B+ value I want to end up with?

Thanks

Charlie

#### DF96

The loaded value, because that is what it should be when loaded. However, except for a choke input PSU or one with large smoothing resistors, the difference won't be that great. Fortunately most valve circuits aren't too fussy about exact supply voltage.

#### tomchr

Paid Member
Here's my suggestion. Measure the output impedance of the power supply.

Measure the output voltage unloaded. Then measure with a dummy load of, say 33 kOhm. Note that the dummy load will dissipate about 5 W (385^2/33000) so you may use something like 3 x 100 kOhm in parallel.

The output impedance forms a voltage divider with the dummy load. Hence, you can calculate the output impedance from:

<-->

So lets say the supply voltage measures 385 V without load and 350 V when loaded by 33 kOhm. Rout would then be:

Rout = 33000 * ((385/350) -1) = 3300 ohm

Above numbers were just picked out of thin air and not representative of any particular supply or amplifier.

Once you know the output impedance of your power supply, you can estimate its output voltage when loaded by the amplifier. From there you can calculate any series resistance needed to bring the output voltage down to the desired value.

For a bleeder, I'd use 330 kOhm, 2 W across the main reservoir cap. The purpose of this is to discharge the reservoir cap in a relatively short amount of time so you don't zap yourself when working on the amp.

~Tom

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#### aardvarkash10

or just use PSUDII

#### Brouilly

or just use PSUDII

isnt supported on macs unfortunately...

... maybe a cheap pc should be on the cards

#### Miles Prower

Are you able to check power supply B+ voltage without connection to the rest of the circuit? I need B+ to be 250v but getting 385v measured at he last PS capacitor. I think a resistor needs to be added but not sure where and what value.

Doanworrydouddit. Any PS (including batteries) will show a higher voltage unloaded than it will under load. The only reason to worry about the unloaded voltage is to avoid poofing the filter capacitors while waiting for the rest of the heaters to warm up.

VT circuits are a good deal more "forgiving" than most solid state, and so deviations from design nominal values have relatively less effect. If your loaded output voltage is within +/- 10% of design nominal, that's good enough.

If you need that 250Vdc to be accurate, then the answer is active regulation, either series pass or parallel, depending on the current range and PS capabilities.

#### DF96

Power supplies often have non-linear output impedance, so measure at three points just to check whether they lie on a straight line. Typically you will find that the output impedance reduces slightly with extra load i.e. the voltage drops more slowly than you expect. This is because the charging pulse occupies a greater proportion of the cycle.

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