AC * 1.4 = DC ???

[Luthierwnc]

I've always figured 1.21X rule of thumb. Depending on how stiff the supply is, a lot of it depends on the current draw of the power tubes. If you leave the volt meter on the B+ and diddle the bias adjustment, you'll see the voltage swing. In the spark-testing phase, if the volts seem way low I immediately check if the tubes are red-plating.

I also use the Duncan power supply II software with a couple different load resistors. Those KT66's like to run hot so be sure to lower the resistance to reflect the current needs. Say you have 42 ma per tube (at idle) plus another 10 ma for the preamp tubes you are drawing 94 ma total. At 390 volts that's 4150 ohms. Makes a difference from the 5k that automatically populates the field. Actually play the thing and it pulls more. sh

[gingertube]

The exact value of the DC will depend upon the load. Full wave vs 1/2 wave rectifier etc. etc.

It can be calculated mathematically but it is a pain.

May I suggest you grab page 11 from the Hammond Catalog here:
http://www.hammondmfg.com/pdf/5C08.pdf
THis is a good "cheat sheet".

It shows all of those rules of thumb for currents and voltages for the various rectifier/filter combinations.

For a full wave rectifier (Solid State) AVERAGE output voltage is typically 1.27 x the AC RMS Voltage.

Cheers,
Ian

[Gaz]

For a full wave rectifier (Solid State) AVERAGE output voltage is typically 1.27 x the AC RMS Voltage.

Cheers,
Ian

Ian, where/how did you come up with my experience. IME it has never been that low.

Anyway, it was not too long ago someone tried to make a thread with a list of PTs and their specs, including loaded DC voltage out. Didn't really catch on, but was a good idea, I thought, because every PT seems different.

[rdjones]

As Cliff points out, the 1.414 ratio is an established constant (based on the square root of 2) and does not take any other losses into account.
Any other number used as a factor is simply a derived estimation based on possibly undefined circuit conditions.

If you measure the AC voltage and the resultant filtered DC output under the same loaded condition, and subtract measurable voltage drops (diode drop, wiring and ground return loss if any, ripple, capacitor, etc) you should be closer to the theoretical.
The voltage has to go somewhere and you should be able to measure or account for it.

Something else that may effect readings would be an imperfect (distorted) sine wave. The peak/rms ratio is based on a true sinusoidal wave.

Differences between loaded and unloaded or lightly loaded voltage outputs of a transformer are rated as degree of regulation expressed as a percentage.
IIRC a transformer with 5% regulation would be excellent, 10% typical and 15% mediocre. Anything worse than that is probably being used outside it's range of capability.

rd

[Luthierwnc]

rdjones got it right

You just don't know the losses from looking at the thing. I got a 50-watt aftermarket Marshall power transformer for a Dumble build -- supposed to be around 700VCT. I got maybe 415 VDC at idle. Slapped a Bassman trannie in there (flipped on its side with an ad hoc bracket and some new holes -- hate that) and got my 450 without breaking a sweat. Same manufacturer for both and wound to spec. There is an old thread where I asked the forum what was going on and it was just that the transformer couldn't stand up to the power supply.

My luck has gone the other way too. The power transformers in the 6V6 M-series Hammond organ amps will easily drive a pair of 6L6s at 450 volts and not even get warm. Those are the newer ('60's) silver ones. I can't say about the 1950's black ones.

FWIW, Skip

[10thTx]

I use this cheat sheet. It is at least a starting place to make guesstimates when designing an amp.

With respect, 10thtx

[Gaz]

That's cool 10thtx, you seem to have a sheet for everything

[sliberty]

Actually that sounds about right to me. The 1.414 figure is for unloaded. Under load 1.2 is usually ballpark. So 293*1.2 = 352.

I haven't worked on this amp in a few days, but I put it back on teh bench last night, pulled all tubes, and measured the DC just after the diodes - 370V. Should have been closer to 420V.

Tried new diodes with no improvement.

If I disconnect the diodes from the rest of the amp (filters and beyond), and measure again at the diodes, what should I expect? Is it valid to measure with nothing connected like that?

[LeftyStrat]

Actually that sounds about right to me. The 1.414 figure is for unloaded. Under load 1.2 is usually ballpark. So 293*1.2 = 352.

I haven't worked on this amp in a few days, but I put it back on teh bench last night, pulled all tubes, and measured the DC just after the diodes - 370V. Should have been closer to 420V.

Tried new diodes with no improvement.

If I disconnect the diodes from the rest of the amp (filters and beyond), and measure again at the diodes, what should I expect? Is it valid to measure with nothing connected like that?

Yes, and that will be the true unloaded value. If you have any bleeder resistors still attached it will load the ps. Disconnection at the diodes and measure and it should be 420v.

[Jana]

For a diode bridge rectifier, I use 0.66 * VAC on the HT to get the DC voltage. I came upon this number through experience and it is usually quite close to the actual result. I used to try all the "official" formulas but they never seemed to work out. So, after years of measuring what I actually got for voltages, this number is the result. Multiply it by 2 for a FWB.

[cbass]

I use this cheat sheet. It is at least a starting place to make guesstimates when designing an amp.

With respect, 10thtx

10thtx,Your sheet has the heater current for an el34 listed at 1 amp.Shouldn't it be more around 1.5

[VacuumVoodoo]

CLiff, it's all because the new breed of ees (they don't deserve capital letters) don't know how to apply the basics.
Do you know more than one EE who can calculate peak-peak ripple voltage on a napkin with a pencil only, not using a calculator? (The "one" being yours truly )

[katopan]

Do you know more than one EE who can calculate peak-peak ripple voltage on a napkin with a pencil only, not using a calculator?

That's where it's at!

[David Root]

Hammond publishes a cheat sheet on rectification that seems to cover all the bases. Good reference sheet for me.

[Phil_S]

For a diode bridge rectifier, I use 0.66 * VAC on the HT to get the DC voltage. I came upon this number through experience and it is usually quite close to the actual result. I used to try all the "official" formulas but they never seemed to work out. So, after years of measuring what I actually got for voltages, this number is the result. Multiply it by 2 for a FWB.

For solid state, I usually figure there is a loss of 4%-6% after using the root 2 number (.707 or 1.414, depending). You are reporting 6.65% loss, which is not at all unreasonable and not that far from what I've seen. The actual loss should correlate to how "soft" the PT is. Primary DCR and core size can sometimes be goods hint for that.