How to determine What VA Line Isolation transformer to get

[nosaj]

I have a Silvertone 1448 customer wants to add a Line Isolation transformer to it. I would like to know how to determine what what VA rating to purchase.
Rather than someone just say buy this I'd like to know how to know what to buy. It looks like the amp draws 40watts by the schematic

Thanks,
Jason

silvertone_1448.pdf

[Enzo]

I see 40 watts also. SO find an 80-100 watt iso.

It is a class A circuit, so even a 60 watt would do.

[nosaj]

I see 40 watts also. SO find an 80-100 watt iso.

It is a class A circuit, so even a 60 watt would do.

So really all you do is just double the 40watts or whatever the circuit wattage draw is?

i think I know the answer on this, I do not think the customer will buy a $65 transformer, but we will see.

Thanks,
nosaj

[Enzo]

Arbitrary on my part, I am not aware of any rules. I just see no reason to buy a part and then run it right on its edge. I don't want it to get hot.

[nosaj]

Arbitrary on my part, I am not aware of any rules. I just see no reason to buy a part and then run it right on its edge. I don't want it to get hot.

So is Watts the same as VA?

nosaj

[Enzo]

For your purposes, yes. VA is volt-amps.

[Chuck H]

For your purposes, yes. VA is volt-amps.

Just for clarity... "For your purposes" was prefaced earlier when Enzo recognized that it was class A, single ended. Which means the determined current won't increase significantly with signal conduction as it would with an AB or B amp.

[Dave H]

So is Watts the same as VA?

VA is the “Apparent Power” if there is phase shift between the current and voltage the “Real Power” (Watts) will be less. Watts = VA x Power Factor. The Power factor is the cosine of the phase angle. If the phase angle is zero then its cosine is 1 and VA = Watts.

[Chuck H]

VA is the “Apparent Power” if there is phase shift between the current and voltage the “Real Power” (Watts) will be less. Watts = VA x Power Factor. The Power factor is the cosine of the phase angle. If the phase angle is zero then its cosine is 1 and VA = Watts.

Nerd

[nosaj]

Just for clarity... "For your purposes" was prefaced earlier when Enzo recognized that it was class A, single ended. Which means the determined current won't increase significantly with signal conduction as it would with an AB or B amp.

Ok in laymans terms if it were class AB how would I determine what VA to buy or were there any transformerless AB amps?

Thanks,
nosaj

[Chuck H]

Ok in laymans terms if it were class AB how would I determine what VA to buy or were there any transformerless AB amps?

Thanks,
nosaj

That's an excellent question. And one I thought might follow my post. Though I didn't hope for it

If I didn't already have a VA indicated on the amp or an available schematic I would do it the same way I choose a PT. I would total the current required for the filament circuit and the power amp circuit. The filament circuit/s are a known quantity listed in the tube data. A very close approximation of the power amps requirements can usually be determined by the operating conditions. Plate voltage, bias current, etc. For a PT you would also have to allocate the sums to their individual windings of course. Then, lacking the necessary education to determine "the cosine of the phase angle" (thanks Dave) I would just add some safety margin.

[R.G.]

Good advice Chuck! That's the way to deal with an unknown amp that you can't just run and measure. Say, one that hasn't been built yet. And I especially like the advice to add safety margin. Watts/VA in an isolation transformer are cheap.

In this somewhat unusual case where the amp already exists and runs (I think, from the description) the straightforward thing to do might be to just hook it up and measure the AC line current. There are some caveats to this. To get good results, you have to run the amp at maximum power draw from the AC line (as noted in the Class A versus Class AB items) and you also need to run it at the maximum AC line voltage that will be applied to the amp.

This last is important for really old amps where the AC line voltage used to be 110 to 112 when they were designed. Today's AC line runs 120 to 130. My house is way out on the end of the distribution network, but we commonly get 122-125, and I've measured 128. This means that transformers from the 1960s designed for 110 could be right on the edge of saturation (and hence, taking more AC line current) at 120-125. That would be a really good thing to know in choosing an isolation transformer.