Value of first power supply dropping resistor with different power transformer?

[Spectre]

Hi,

After building a few amps from schematics and layouts, this is my first foray into some technical questions. I'd really appreciate any advice!

I have a schematic that I'm working to that has a PT with 310-0-310VAC. The valve config is 2 x 6v6 and 2 x 12ax7. Rectifier is full wave silicon diode.

Total B+ current use is 66.3mA, power use is 26.2W.

B+1 Voltage under load is 395VDC. After the choke (5H, 120mA) it is 394VDC.

The first dropping resistor is 6.8K (2W) which gives me a B+2 of 340V (-54V) with 0.429W of power dissipation.

However, the PT I'm looking to use is 380-0-380VAC. This gives me a B+1 of 484VDC, B+ current use of 58mA and power use of 28.1W.

So, after the choke, I'd anticipate that the B+ would be 483VDC.

My questions are:

1) To get to a B+2 of 340VDC, what value do I need for the first dropping resistor?
2) How would I normally calculate this? I've been messing around with my limited knowledge of Ohms law and can't seem to figure it out.
3) Would I be better off putting a dropping resistor between the rectifier and the B+1 so that the B+1 is at 395VDC and if yes what value resistor would I need to go from the 483VDC coming out of the rectifier?

Thanks all!

[Phil_S]

I'm thinking, use a full wave bridge with choke input. That should get you about 340VDC at B1. Because the choke carries the full load of the circuit, I want it to be over rated. I think I'd choose a Hammond 159Q or 159R. Both are rated at 500V and 150mA, 200m/A, respectively. (If your 5H choke has an adequate voltage rating, you can move it to come after the bridge rectifier.) The Hammonds are hefty at about 2.2 lbs, so you'd need room on the chassis to mount it, but not too hard on the wallet at $20-$30. You'd want to eliminate the other choke and replace it with a resistor.

You could use a large power resistor to drop voltage. You want to drop 55V. Using Ohm's law, V=I*R, we know V=55V and I=66.3mA. R=V/I or 830Ω. 820Ω is the nearest standard value. I think you are looking for something like this: https://www.mouser.com/ProductDetail/Xi ... OtcgIVM%3D or https://www.mouser.com/ProductDetail/TE ... 31etBLk%3D I'd give it plenty of free air space because it is going to get hot.

Show us your schematic, or tell us where to find it. It might get better informed suggestions.

[Stevem]

A transformer of 380/380 and full wave rectified will produce 530 volts of V+, and you will need to drop that down a bunch to have 6V6 tubes live!

I have run JJ brand 6V6 outputs on 500 volts, but even that's pushing it a bit, not to mention that a different output transformer with a higher impedance will be needed.

[R.G.]

I'd be careful about going for a choke input filter. Choke input filters at low load have the same output voltage as a capacitor input filter does. As loading rises, the output voltage drops until it reaches a stable point for all higher load currents. Using a choke input filter means you have to design the filter caps and all the tube circuitry to withstand the full wave bridge voltage until the power tubes heat up and start conducting the normal operating current. Every power-on event becomes a high-voltage-and-wait event. It's do-able, but you have to plan for it.

[sluckey]

I'm thinking, use a full wave bridge with choke input. That should get you about 340VDC at B1.

No! Using a FWB will give you twice that amount of B+. You want to use a conventional (2 diode) FW rectifier. The formula would be 760 x .45 = 342. Here's a good pdf to add to your collection...

[Phil_S]

Sorry, again I muffed a transformer question by only looking at one leg. I'd push the brain reset button if I had one. I'll try for a new year's resolution to avoid answering this type of question for a while.

[chaccmgr]

Hi,
I would not change the topology. Calculating the drop resistor is no rocket science. But the core problem is, that B+ for 6V6es will be too high.

Adding a resistor after the rectifier before the first cap/B+ tap acts like a sag resistor, mimicing the behaviour of a rectifier tube (dropping voltage under load), the amount depends on the load. In case of a PP amp the load varies and thus the voltage drop is not predictable. The problem is, that with little or no load the voltage will still be too high.

Try find a suitabale transformer instead.

[Colossal]

A transformer of 380/380 and full wave rectified will produce 530 volts of V+, and you will need to drop that down a bunch to have 6V6 tubes live!

Steve, your use of the revised standard of V+ (versus B+) didn't go unnoticed!

I have run JJ brand 6V6 outputs on 500 volts, but even that's pushing it a bit, not to mention that a different output transformer with a higher impedance will be needed.

In the past, I've run The JJ6V6S at 440-460VDC and they held up fine. After a lot of abuse, they kind of lost their sparkle (you know, they were never the same) but they didn't fail. I did manage the screen voltage though.