# How do you do power transformer calculations?

Discussion in 'Shock Brother's DIY Amps' started by Dennis Perusse, Sep 10, 2015.

1. ### Dennis PerusseTele-Meister

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Hello everyone,

Question for anyone who can explain this to me as I haven't found any place that can explain this simply. My google-fu has not yielded me any places that can show me what's going on plus my attempt to send a letter to Merlin Blencowe has failed also, hence me asking here. It is a little unorthodox for this project, as this is a hi-fi project and not a guitar/bass amp, but any help here is greatly appreciated.

Enclosed is a single ended 6aq5 stereo amp that puts out an estimated 4 watts. I had acquired it from the "Wade's Audio and Tube" site and have been wanting to build it since I have most of the parts to build it, but the layout gives me a few questions that I'd love to have someone comment on.

Now obviously I have done a few build threads here so it isn't my first attempt with tube amps so that isn't in question but take a look at this design for a second as I am a bit lost here.

1. On the layout drawing provided you see the power settings in terms of the PT as being 290- 0 - 290 volts. Is that accurate and safe knowing that the 6aq5 tubes handle 250 volts as its maximum if I've read the datasheet correctly?

2. On the datasheet for a 5y3 GT tube it says that the maximum capacitance is 20 uf. Now this layout has 40 uf and I'm puzzled by this. Is it because their are two channels and it is getting halved to feed capacitance to both channels or do you think that it is being done just to increase the filtering since it is a common power supply? I mean I have seen this where it has been done commercially. The Grommes LJ-2 does this but uses 6v6 tubes in a push pull circuit.

3. I have figured out that if I add up the tube amperage 2x 6aq5 @ .45= .90 + 1x12ax7 @ .3 + 1x pilot light @ .15= 1.35amps. So a 2 amp rating or higher would be good.

4. I also know that the 5y3 needs 5 volts @ 2 amps so a 5v 2amp rating is spot on as another winding for the PT.

5. Lastly, I have no clue how you calculate the rating for this design in terms of amperage of the HT/B+. How does one calculate it and what does one look for to calculate it? I realize that sites do exist that calculate such things but I have no idea how a design like this is figured out where it is two channel and uses a common power supply. Plus I'd rather know how to do it in case I ever decide to design my own circuits using tubes that I want to use instead of using already pre-made designs. Plus my mathematical skills are about as good as the tooth fairy in a town full of meth-heads so I'll need help understanding what is going on math wise. ;-)

I'll be using the classictone 40-18031s in case anyone was wondering. They are a 5 k primary to 4/8/16 ohm taps to give me multiple types of speakers to try out. They are also 15 watt rated each so I won't have to worry about distortion from saturation either. As mentioned if anyone can can school me on how to figure this out I would gladly appreciate it at this time.

Dennis

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2. ### metalicasterTele-Afflicted

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Bump.

I asked a similar question and got squat in the form of replies.

3. ### JSC_ltdTDPRI Member

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1. The rectifier will provide a voltage drop, as will the power filtering. I don't really math, either, but I would guess that the voltage is in the neighborhood of 250V by the time it gets to the 6AQ5.

2. I've noticed a lot of apparent violations of a rectifier's max capacitance in commercial designs, too. I can't explain it, but my instinct is that if it's good enough for Fender, it's good enough for me.

3. Yes. I think current is the most important calculation because you get smoke if you draw too much of it.

4. Yes.

5. Didn't you do this back in step 3? Am I missing something?

4. ### robrobPoster ExtraordinaireAd Free Member

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First note that the 290-0-290v is AC RMS (root mean square average DC equivalent). This voltage is called HT AC (high tension). The rectifier doesn't use the RMS value to create DC, it uses the peak AC voltage.

A conventional, dual plate tube (or two diode) full wave rectifier uses only the 0 to 290v to convert to DC (not 290 x 2 that a bridge rectifier would use). This is because the transformer secondary winding is divided in half by the center tap at 0 volts.

So we convert that 290v AC RMS to AC peak voltage by multiplying by the square root of 2 which is 1.414, so 290V AC RMS x 1.414 = 410 volts peak AC (410Vp).

The 410Vp AC is what is flowing into our rectifier to be converted to DC. All rectifiers lose or drop some voltage during the conversion. A lowly diode only drops 0.7v DC and we use two of them so 1.4v total voltage drop. We would expect 410Vp -1.4V = 408.6v DC out. This is what we call B+ or B+1 voltage.

The 5Y3 tube rectifier drops about 60v so we would expect 410Vp - 60v = 350v DC out.

The first filter capacitor acts as a reservoir and will fill up under no electrical load to the DC voltage put out by the rectifier, 408v DC for a solid state rectifier or 350v DC for the 5Y3. This is what happens when we start up our new amps with only the rectifier tube installed. There's no load on the filter caps so they fill up to the voltage put out by the rectifier.

Once the rest of the tubes are installed their current draw will pull the voltage down and the filter caps won't stay full. How much voltage drop depends on how much current the power transformer pumps into the rectifier to "refill" the filter caps. You'll get less voltage drop with a higher current rated power transformer.

Paul-T likes this.
5. ### robrobPoster ExtraordinaireAd Free Member

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2. As JSC said amp designers exceed rectifier filter cap ratings all the time with seemingly no consequence.

Your #3 and #4 are correct.

5. I like to use this power transformer calculator which shows it's formulas: http://thesubjectmatter.com/calcptcurrent.html

It's output using 300-0-300v, 5Y3 rectifier, 1 EL84 power tube, 1 12AX7:

The formulas used on the web page:

So the calculator uses rough approximations of current from the preamp data sheets because the values are so low but uses calculations to generate the much more important power tube current. The calculations for a push-pull Class AB amp are much more complicated.

Last edited: Sep 11, 2015
6. ### Fred MertzTele-Afflicted

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Rectifier tubes drop voltage due to internal resistance of the tube. The voltage drop will vary depending upon the current drawn by the circuit. The greater the current demand, the greater the voltage drop.

Tube data sheets will often contain graphs that I find useful in estimating rectified DC voltage. The graph on the bottom of page 4 of the data sheet for the 5Y3GT reflects the operating characteristics for a full wave capacitor input power supply. The curved lines represent the AC volts supplied to the rectifier plates. The horizontal axis shows the current drawn under load. Find the current under load for the circuit on the horizontal line and extend a vertical line to intersect with the curved line representing the AC supplied to the rectifier plates. This point will fall on a horizontal line that reflects the DC voltage as indicated by the vertical axis on the left.

http://www.r-type.org/pdfs/5y3gt.pdf

The other variable in calculating/estimating rectified voltages is the DC resistance of the power transformer. There will be IR voltage loss that can be calculated if one knows the DC resistance of the transformer and the current drawn by the circuit. Hammond is the only manufacturer that I'm aware of that publishes the DC resistance of their transformers. Other transformer manufacturers/suppliers may provide a no load voltage specification and a voltage specification at a specified load. You can get a good estimate of the DC resistance, if you subtract the voltage under load from the no load voltage and divide the remainder by the specified current load.

7. ### Fred MertzTele-Afflicted

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Forgot to mention, if you want to use a resistor to simulate the sag effect of a rectifier, you can determine the appropriate resistor value using one of these graphs for the specific rectifier that you want to emulate.

Find the no load DC voltage on the vertical axis on the left of the graph that coincides with the AC volts supplied by the transformer. Follow the curved AC line to its intersection with the boundary line on the right hand side of the graph and find the horizontal DC voltage line that this coincides with on the vertical axis to the left. From the intersection of the curved AC line with the boundary line, find the vertical current line that coincides with on the horizontal axis at the bottom of the graph. Subtract the voltage under load from the no load voltage and divide the remainder by the current amount in amperes. This will give you the approximate resistor value in ohms to emulate the rectifier.

8. ### robrobPoster ExtraordinaireAd Free Member

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Great info Fred.

9. ### Dennis PerusseTele-Meister

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Hello everyone,

Thank you to everyone who had replied. A lot of info for me to digest as I look more into this layout. I also hope that Metallicaster got what he needed also. Whilst searching around I went to edcor and saw this type of tube PT. https://www.edcorusa.com/xpwr189

Whilst I will try to do the math which is what I want to do just to check stuff does anyone think that that Edcor model PT is too much, in amperage,for the project? It looks like it meets all of my requirements electrically.

As for the subject matter site it looks interesting. I'd like to run some numbers on that using what the layout says using their formula shown and see what numbers I get when I get a moment.

Dennis

10. ### robrobPoster ExtraordinaireAd Free Member

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That Edcore looks like it will work fine for you.

11. ### Bill HicklinTDPRI Member

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OK, does this make sense?

This is an older one, late 40s. But I don't understand how their 325-0-325 PT can only have 305vDC coming off the far side of the 5U4 rectifier tube, and a very modest 6L6 plate voltage of 285.

The calcptcurrent page says the B+ ought to be over 400V!

12. ### robrobPoster ExtraordinaireAd Free Member

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My only guess would be an under spec'd power transformer that drops excessive voltage due to the circuit's current demand.

13. ### Bill HicklinTDPRI Member

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Since it's a transformer-coupled circuit, might it be sucking up more current than conventional direct coupling? I can't imagine that the single 6J5 preamp tube is consuming much.

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