Calculating required transformer voltage specs for 100w marshall

[Ray Barbee]

Given I want my rectified B+ at X volts, what is the calculation for xformer secondary voltage where the center tap isn't grounded, but goes in the middle of the first two series filter caps like on a 100w Marshall?

[Smitty]

Start here: https://www.hammfg.com/electronics/tran ... /rectifier

You are describing a capacitor input full wave rectifier. The center tap connected to the junction of the stacked resistors help keep the voltage evenly distributed across the series capacitors.

Factors which impact the actual B+ include transformer secondary winding resistance and total current draw at idle and full power.

[Roe]

originals had plate voltage at 460-500v normally, but some amps had even higher voltages. with a 1k75 Z, the plates should sag to 425v (or lower) at full power to avoid redplating

[cdemike]

That Hammond document is a great resource.

Assuming you keep the same arrangement as the originals with full-bridge rectification and the CT connecting to the junction between the series reservoir capacitors, you would treat it like a full-wave bridge rectification of a non-center tapped PT. So for a 175-0-175 PT, you'd functionally be rectifying 350VAC with a full-wave bridge rectifier going into a capacitive load. Your equation would therefore be 350VAC * 1.414 = 494.9VDC unloaded; taking into account loading effects, you'll generally see figures closer to what Roe described. What is your B+ goal?

[Ray Barbee]

That Hammond document is a great resource.

Assuming you keep the same arrangement as the originals with full-bridge rectification and the CT connecting to the junction between the series reservoir capacitors, you would treat it like a full-wave bridge rectification of a non-center tapped PT. So for a 175-0-175 PT, you'd functionally be rectifying 350VAC with a full-wave bridge rectifier going into a capacitive load. Your equation would therefore be 350VAC * 1.414 = 494.9VDC unloaded; taking into account loading effects, you'll generally see figures closer to what Roe described. What is your B+ goal?

Thanks! I'd like to be about 425v on a quad of 6v6s.

[Roe]

That Hammond document is a great resource.

Assuming you keep the same arrangement as the originals with full-bridge rectification and the CT connecting to the junction between the series reservoir capacitors, you would treat it like a full-wave bridge rectification of a non-center tapped PT. So for a 175-0-175 PT, you'd functionally be rectifying 350VAC with a full-wave bridge rectifier going into a capacitive load. Your equation would therefore be 350VAC * 1.414 = 494.9VDC unloaded; taking into account loading effects, you'll generally see figures closer to what Roe described. What is your B+ goal?

Thanks! I'd like to be about 425v on a quad of 6v6s.

try a set of 50w transformers then

[Ray Barbee]

That Hammond document is a great resource.

Assuming you keep the same arrangement as the originals with full-bridge rectification and the CT connecting to the junction between the series reservoir capacitors, you would treat it like a full-wave bridge rectification of a non-center tapped PT. So for a 175-0-175 PT, you'd functionally be rectifying 350VAC with a full-wave bridge rectifier going into a capacitive load. Your equation would therefore be 350VAC * 1.414 = 494.9VDC unloaded; taking into account loading effects, you'll generally see figures closer to what Roe described. What is your B+ goal?

Thanks! I'd like to be about 425v on a quad of 6v6s.

try a set of 50w transformers then

B+ probably too high, heater and HT current may be under rated.

[Ray Barbee]

Start here: https://www.hammfg.com/electronics/tran ... /rectifier

You are describing a capacitor input full wave rectifier. The center tap connected to the junction of the stacked resistors help keep the voltage evenly distributed across the series capacitors.

Factors which impact the actual B+ include transformer secondary winding resistance and total current draw at idle and full power.

Doesn't make sense. According to the hammond sheet, vdc= .71 secondary AC, which is approximately the same as the calculation for a grounded center tap where vdc=1.4x one rail with solid state rectifier.

Marshall 100W PT spec for 475v B+ is about 185v each side to center tap, or with the full wave capacitive load, secondary AC = 370v. How do we get from there to 475vdc B+? Does the 1.41x end up at 1.31x from loading?

[B Ingram]

... what is the calculation for xformer secondary voltage where the center tap isn't grounded, but goes in the middle of the first two series filter caps like on a 100w Marshall?

You are describing a capacitor input full wave rectifier. The center tap connected to the junction of the stacked resistors help keep the voltage evenly distributed across the series capacitors.

Doesn't make sense. According to the hammond sheet, vdc= .71 secondary AC, which is approximately the same as the calculation for a grounded center tap where vdc=1.4x one rail with solid state rectifier.

Marshall 100W PT spec for 475v B+ is about 185v each side to center tap, or with the full wave capacitive load, secondary AC = 370v. How do we get from there to 475vdc B+?

It doesn't make sense because there are multiple errors/wrong-assumptions present.

For Smitty: the Marshall amp uses a full-wave bridge rectifier; saying only "full-wave rectifier" means only half of the transformer secondary is conducting at any moment, rather than all of the transformer secondary.

For Ray Barbee: I don't know enough about vintage Marshall transformers to answer you question fully. However, either the winding AC voltage is lower than your information, or the idle B+ is very much higher than 475v DC and sags downward when the amp is driven hard.

- If you're planning a quad of 6V6s and want 425v DC, start by assuming no-sag.
- AC Volts rectifies to a DC Voltage equal to the Peak AC Voltage, minus some amount as the amp drains filter caps of their charge.
- Peak AC Volts = AC Volts x √2 = AC Volts x 1.414 = AC Volts / 0.7071

Working backwards, multiply the desired DC Voltage by 1/√2 to get the AC Volts before rectification:
- 425v DC x 1/√2 = 425v DC / 1.414 = 425v DC / 0.7071 = 300v AC

- I'm showing the conversion-term (√2 and 1/√2) in several forms so that maybe you recognize why you see certain numbers over & over in these discussions.


Since you appear to want to use the bridge rectifier & use the center-tap for voltage-sharing across stacked filter caps, that "300v AC" implies a 150-0-150v AC winding or "300v CT". That might have to be a custom-wound part, because common Power Transformers with a center-tapped 300v winding (like Hammond's 269BX) only have enough current for 2x 6V6s.
- Which reiterates why people build "the same stuff as everyone else": it's expensive to have custom-made transformers that deviate much from commonly made amps.

[cdemike]

I'm using a stand-up mounted version of Mojotone's JTM45 PT in one of my builds which runs 3x 12AX7 and 2xKT88 requiring a total of 4.5A of heater current whereas the transformer is rated for 5A. Assuming you're still keeping it at 3x 12AX7, total heater draw with 4x 6V6 would be 2.9A. Despite being close to maximum spec for heater current, my PT stays nice and cool, but the B+ was actually much lower than my 470-520V B+ goal at 434V. Sounds like something similar may actually be about right for your build, unless you want something that resists sagging (which, to me at least, is part of the Marshall "thing"). If I were to do the build over again, I'd want to pick a power transformer that's spec'd right for the heater current draw and is roughly in the neighborhood for B+ current draw while prioritizing predicted B+ voltage. From there, it's relatively simple to dial in the desired response with filtering values, especially compared to getting the B+ voltage right.

Start here: https://www.hammfg.com/electronics/tran ... /rectifier

You are describing a capacitor input full wave rectifier. The center tap connected to the junction of the stacked resistors help keep the voltage evenly distributed across the series capacitors.

Factors which impact the actual B+ include transformer secondary winding resistance and total current draw at idle and full power.

Doesn't make sense. According to the hammond sheet, vdc= .71 secondary AC, which is approximately the same as the calculation for a grounded center tap where vdc=1.4x one rail with solid state rectifier.

Marshall 100W PT spec for 475v B+ is about 185v each side to center tap, or with the full wave capacitive load, secondary AC = 370v. How do we get from there to 475vdc B+?

185-0-185VAC would be probably a little on the high side, but your unloaded B+ would be 1.414*370 = ~523VDC. Loaded, I'd predict that'd be closer to 510VDC with 4x EL34 in the output section, but that'd still be something closer to a JTM45/100 spec PT than the typical JMP PT. I like referencing Mojotone's transformers since a lot of them are included in their kits, and their kit schematics include voltages and that provides a more predictable sense of what to expect in terms of B+ voltage. Their 100W PT is rated at 178-0-178VAC, which works out to 503VDC estimated unloaded B+, whereas the kit schematic associated with that PT lists B+ at 476VDC at the first node due to loading effects dragging down B+ from predicted (see https://www.mojotone.com/Amp_Kits/Briti ... jIuOC4wLjA.). 4x 6V6 (or 6V6S) would draw less current than 4x EL34, raising B+ in comparison. Assuming both versions are otherwise identical and biased at 70% (17.5W for the EL34 amp and 8.4W for the 6V6 amp), current through the output section would be reduced to 48% of original. At 476V, we can estimate the current where P=IV, i.e., 4 * 17.5 = 476 * [plate current] giving an estimate output section plate current of 147.1mA. Preamp current draw wouldn't be significantly different, and using Ohm's law we can estimate it to be 7.2mA based off the voltage drop at the series 10K resistors feeding the phase inverter node. So the total amp's idle current demand would be ~154.3mA. Since the current draw for 6V6s would be about 48% of the EL34s, we can estimate that to be 70.6mA, and with preamp current draw estimated at 7.2mA, total draw would be ~77.8mA, i.e. 50.4%. Since predicted unloaded B+ with that PT would be 503VDC and loaded B+ is 476VDC, we know the load reduced B+ by 27V. With only 50.4% of relative current draw using 6V6s, we can take 50.4% of 27V to get an estimate voltage drop of 13.6V and an unloaded B+ of 489V using that PT. You can get more granular and also include screen current using voltage drop across those 1K resistors.

Of course, that's much higher than your goal, but the same procedure works for any transformer used in an amp with measured B+ voltages. Some datasheets make it even easier and give you actual measurements for voltage drop under load (Antek's datasheets come to mind). To Roe's point, the math actually works out really well if you check 50W kits' B+ measurements and provides B+ voltages very similar to your target.

[Roe]

Thanks! I'd like to be about 425v on a quad of 6v6s.

try a set of 50w transformers then

B+ probably too high, heater and HT current may be under rated.

No, many of the 50w amps have voltages around 420v. and suffient current. a bit of can be good really

[Roe]

Here's two PTs:
https://www.tubetown.net/ttstore/en/tor ... power.html
Pri: 0-220-230-240 V (schwarz-gelb-lila-rot / black-yellow-purple-red)
Sec 1: 0 - 300 V @ 0,15 A (grün-rot / green-red)
Sec 2: 0 - 300 V @ 0,15 A (weiß-blau / white-blue)
Sec 3: 50 V @ 0,05 A (orange)
Sec 4: 5 V @ 3 A (braun/brown)
Sec 5: 6,3 V @ 5 A (schwarz/black)

https://www.tubetown.net/ttstore/en/tor ... power.html
Pri: 0-220-230-240 V (blue-black-orange-yellow)
Sec 1: 150-0-150 V @ 0,4 A (Blue-white/green-red)
Sec 2: 6,3 V @ 7,0 A (purple-black/CT: brown)
Sec 3: 60 V @ 0,1 A (grey)
Sec 4: 15 V @ 1,0 (orange)

[Ray Barbee]

I'm using a stand-up mounted version of Mojotone's JTM45 PT in one of my builds which runs 3x 12AX7 and 2xKT88 requiring a total of 4.5A of heater current whereas the transformer is rated for 5A. Assuming you're still keeping it at 3x 12AX7, total heater draw with 4x 6V6 would be 2.9A. Despite being close to maximum spec for heater current, my PT stays nice and cool, but the B+ was actually much lower than my 470-520V B+ goal at 434V. Sounds like something similar may actually be about right for your build, unless you want something that resists sagging (which, to me at least, is part of the Marshall "thing"). If I were to do the build over again, I'd want to pick a power transformer that's spec'd right for the heater current draw and is roughly in the neighborhood for B+ current draw while prioritizing predicted B+ voltage. From there, it's relatively simple to dial in the desired response with filtering values, especially compared to getting the B+ voltage right.

Start here: https://www.hammfg.com/electronics/tran ... /rectifier

You are describing a capacitor input full wave rectifier. The center tap connected to the junction of the stacked resistors help keep the voltage evenly distributed across the series capacitors.

Factors which impact the actual B+ include transformer secondary winding resistance and total current draw at idle and full power.

Doesn't make sense. According to the hammond sheet, vdc= .71 secondary AC, which is approximately the same as the calculation for a grounded center tap where vdc=1.4x one rail with solid state rectifier.

Marshall 100W PT spec for 475v B+ is about 185v each side to center tap, or with the full wave capacitive load, secondary AC = 370v. How do we get from there to 475vdc B+?

185-0-185VAC would be probably a little on the high side, but your unloaded B+ would be 1.414*370 = ~523VDC. Loaded, I'd predict that'd be closer to 510VDC with 4x EL34 in the output section, but that'd still be something closer to a JTM45/100 spec PT than the typical JMP PT. I like referencing Mojotone's transformers since a lot of them are included in their kits, and their kit schematics include voltages and that provides a more predictable sense of what to expect in terms of B+ voltage. Their 100W PT is rated at 178-0-178VAC, which works out to 503VDC estimated unloaded B+, whereas the kit schematic associated with that PT lists B+ at 476VDC at the first node due to loading effects dragging down B+ from predicted (see https://www.mojotone.com/Amp_Kits/Briti ... jIuOC4wLjA.). 4x 6V6 (or 6V6S) would draw less current than 4x EL34, raising B+ in comparison. Assuming both versions are otherwise identical and biased at 70% (17.5W for the EL34 amp and 8.4W for the 6V6 amp), current through the output section would be reduced to 48% of original. At 476V, we can estimate the current where P=IV, i.e., 4 * 17.5 = 476 * [plate current] giving an estimate output section plate current of 147.1mA. Preamp current draw wouldn't be significantly different, and using Ohm's law we can estimate it to be 7.2mA based off the voltage drop at the series 10K resistors feeding the phase inverter node. So the total amp's idle current demand would be ~154.3mA. Since the current draw for 6V6s would be about 48% of the EL34s, we can estimate that to be 70.6mA, and with preamp current draw estimated at 7.2mA, total draw would be ~77.8mA, i.e. 50.4%. Since predicted unloaded B+ with that PT would be 503VDC and loaded B+ is 476VDC, we know the load reduced B+ by 27V. With only 50.4% of relative current draw using 6V6s, we can take 50.4% of 27V to get an estimate voltage drop of 13.6V and an unloaded B+ of 489V using that PT. You can get more granular and also include screen current using voltage drop across those 1K resistors.

Of course, that's much higher than your goal, but the same procedure works for any transformer used in an amp with measured B+ voltages. Some datasheets make it even easier and give you actual measurements for voltage drop under load (Antek's datasheets come to mind). To Roe's point, the math actually works out really well if you check 50W kits' B+ measurements and provides B+ voltages very similar to your target.

Ok so we're looking at 1.414xtotal secondary voltage, i.e. not what that hammond sheet said. Minus load drop.

You're looking at a 100w Marshall PT with 34s vs 6v6s. What I'm looking at is a 50W marshall pt with 4x6v6 vs 2x el34, set up the same way as the 100w would be.

So heater current is .9W higher with 4x6v6 than 2x34s, HT power tube power with el34 is 35W, with 4x6v6 JJ it's 4x(14x.7)=39.2W. Add a couple watts per screen gets to 47.2W. Given that, and that the pt in question runs the 34s at 475 ish, I'd expect a little more drop with the 6vs, but not enough. So yes, looking at a custom pt. Regular 50w marshall pt wont cut it due to the extra taps I need; this is not a 1 channel amp. Also it has to be robust enough for the extra amp of heater current.

[cdemike]

So heater current is .9W higher with 4x6v6 than 2x34s, HT power tube power with el34 is 35W, with 4x6v6 JJ it's 4x(14x.7)=39.2W. Add a couple watts per screen gets to 47.2W. Given that, and that the pt in question runs the 34s at 475 ish, I'd expect a little more drop with the 6vs, but not enough. So yes, looking at a custom pt. Regular 50w marshall pt wont cut it due to the extra taps I need; this is not a 1 channel amp. Also it has to be robust enough for the extra amp of heater current.

Unfortunately, I think you're making some mistakes with the math. You're correct that 4x 6V6S would make for an amp that idles at 39.7W in terms of combined plate and screens dissipation, though the heater current is a separate item entirely. Taking the datasheet figure of 0.5A of heater current per 6V6S, would be only 2A for a quad of 6V6S's (JJ datasheet: https://www.jj-electronic.com/images/st ... f/6v6s.pdf)

Compare to a duet of EL34's. which, per the Mullard datasheet, draw 1.5A each in heater current for a total of 3A in heater current: https://frank.pocnet.net/sheets/129/e/EL34.pdf (JJ's datasheet echoes the same heater current figure, so it does not vary when discussing new production EL34's: https://www.jj-electronic.com/images/st ... 4_e34l.pdf).

In other words, your total heater draw for an amp with 4x 6V6S and 3X 12AX7 (datasheet: https://frank.pocnet.net/sheets/049/1/12AX7A.pdf) would be 2.9A. Even if you doubled the number of 12AXX7's, your heater draw would only be 3.8A, which is well within either the 5A or 7A figures described in Roe's last post.

Ok so we're looking at 1.414xtotal secondary voltage, i.e. not what that hammond sheet said. Minus load drop.

Another math error unfortunately. The relevant reference in the Hammond document is the 3rd item in the right-hand column:


We're describing a center-tapped transformer where no current flows through the center tap since there is no circuit stemming from the junction of the series reservoir caps. Therefore, the B+ reaching the rectifier is 2x the voltage rating described using the center tapped convention (e.g., 180 - 0 - 180VAC becomes 360VAC). B Ingram's post describes why the peak voltage figure is the figure of interest and consequently why the top formula is the formula of interest in the Hammond document. A version that's closer to what guitar amplifiers usually use for rectification is here:

IMG_1446.PNG

So really no need for a custom PT based on the information available so far, which raises the question: what extra taps do you need? Are you operating relays with them? If so, relays draw very little current and would be well within range for any of the PT's described so far. If you wanted something stouter still than the 50W PTs mentioned previously, Antek makes at least two transformers which well exceed your current requirements, and it's very easy to know exactly what your predicted B+ would be since their datasheets provide voltage readings at different levels of current draw:
https://www.antekinc.com/as-1t300-100va ... ansformer/
Twin 300V secondaries which if wired in parallel provide 300VAC at 286mA (would be extremely resistant to sag: given previous estimate of ~78mA idle current, idle B+ would be approx 434.5VDC, sagging to only 418.5VDC at 400mA current draw)
Twin 6.3V secondaries which if wired in parallel provide 6A of heater current

https://www.antekinc.com/as-2t300-200va ... ansformer/
Twin 300V secondaries which if wired in parallel provide 300VAC at 107mA (would also be extremely resistant to sag: est unloaded B+ is approx 425.4VDC which would sag to 408.6V at 500mA current draw. Under these conditions, the B+ rail would be consuming 204.3W of power which I can't imagine would happen under realistic circumstances)
Twin 6.3V secondaries which if wired in parallel provide 8A of heater current

Since resistance to sag appears to be very important to this build, I'd add that a very large reservoir capacitor would probably be a more economical and tune-able way to dial that in. Also given the wide range of commercially available capacitors rated for this B+ range with very high capacitance ratings, I think you could probably make an amp that would be suitable even for bass use with nearly any of the PTs described in this thread with enough filtering (probably even more sag resistant than a lot of bass players would want); just by way of example, using 500uF at the reservoir and filtering nodes (https://www.amplifiedparts.com/products ... ectrolytic) would vastly exceed any Ampeg or Hiwatt power supply I'm aware of. Adjusting the current capacity of the PT is pretty difficult and involves solutions like sag resistors which produce a lot of heat, but spec'ing a transformer closer to your amp's projected current needs makes the amp more responsive to filtering changes without unacceptable noise/hum levels if you decide you wanted some sag to be available down the road.

[Ray Barbee]

It was mentioned above to look at Full Wave Capacitor input, which is different than Full Wave Bridge Capacitor input, so I was pointing out that was wrong. You'll notice under the former, the hammond sheet mentions .71x total secondary.

Yes on heater current; had that figure doubled in my head for some reason.

HARD NO on 0 sag, over-filtered amps. If I wanted that kind of feel, I'd play a modeler or a home stereo.

What I was getting at was, in that configuration, we're looking at 1.41xtotal secondary voltage? Which is yes. That's all the answer I needed.