B+ Dropping/Decoupling Resistor

[FourT6and2]

I'm playing around with filtering and B+ voltages on a gain stage in a modded Marshall. The B+ node is seeing 355v. The plate resistors that go from this node to the tube are the standard 1/2W (maybe 1W?) 100K Iskra carbon films Marshall used in older amps. My question is about what kind of dropping resistor I can get away with using for this node. Typical 2W metal oxides rated at 500v won't physically fit on the board. I do have a 2W metal oxide that will fit, but data sheet says the working voltage is 350v. Is this the max rating for the voltage the resistor will see in-circuit, or the max rating for the voltage drop ACROSS the resistor? In this circuit, with 355v at the node, the max voltage drop is less than 100v.

I do have some high-quality, modern metal films from PRP that will fit. 1/2W and 1W. Both rated for 500v. Is there any issue with using these on the B+ rail, considering the plate resistors themselves are only 1/2W carbon films?

[martin manning]

It’s the voltage across the resistor that the rating refers to. MF or MO will work, MF will produce less noise. 1W might be ok, but do the math.

[R.G.]

I'm playing around with filtering and B+ voltages on a gain stage in a modded Marshall. The B+ node is seeing 355v. The plate resistors that go from this node to the tube are the standard 1/2W (maybe 1W?) 100K Iskra carbon films Marshall used in older amps. My question is about what kind of dropping resistor I can get away with using for this node. Typical 2W metal oxides rated at 500v won't physically fit on the board. I do have a 2W metal oxide that will fit, but data sheet says the working voltage is 350v. Is this the max rating for the voltage the resistor will see in-circuit, or the max rating for the voltage drop ACROSS the resistor? In this circuit, with 355v at the node, the max voltage drop is less than 100v.

I do have some high-quality, modern metal films from PRP that will fit. 1/2W and 1W. Both rated for 500v. Is there any issue with using these on the B+ rail, considering the plate resistors themselves are only 1/2W carbon films?

1. How much power is dissipated by the dropping resistor? That is, what is the resistor value, and what it the value on the high side of that resistor before the 355V? The power is them
P = V*V/R. This tells you the actual power in the resistor; then you add to that a safety factor, generally at least 2 times the power, and round up to the next standard power rating.

Until you have a good estimate of the actual power the resistor generates, picking what's enough power rating is a pure guess.

2. Once you know the resistor and the power, you can play with the voltage rating. It is common practice in higher voltage circuits to use two resistors in series for both higher voltage rating and higher power. If the actual power wasted/dissipated in the dropping resistor is something under 1/2W, then a one watt total resistor rating works fine. If you can't find the right resistance and voltage rating in a single, divide the resistor into two resistors of half the value, in series. Two identical 1/2W resistors, each rated at 200V, should be fine for 355Vdc, and will dissipate 1W total, 1/2W in each.

If you have resistors of the right resistance value in 1/2W and 1W, and the voltage rating is 500V, then you only need to know if the actual power is 1/4W or less to use the 1/2W; or if the actual power is 1/2W or less you can use one of your 500V 1W devices.

But you have to have at least some idea what the actual power is; knowing the voltage just across the dropping resistor lets you calculate the power, then add a safety factor.

It’s the voltage across the resistor that the rating refers to. MF or MO will work, MF will produce less noise. 1W might be ok, but do the math.

Strictly speaking this is correct - it's the working voltage across the resistor that the rating refers to, and in this setup it's probably under 100V or so. I like to be more conservative if it's reasonable. If the node is 355V, and a tube further down the line shorts, more voltage will be put across the dropping resistor. This will always be smaller than the 355 node voltage, but may be higher than the actual normal working voltage. So using a 400+ voltage rating, even from two resistors in series, will always be enough.

Yes, it's way more than necessary, but then belts and suspenders were always a good combination.

[FourT6and2]

1. How much power is dissipated by the dropping resistor? That is, what is the resistor value...

I haven't decided yet. I have a 5W pot I'll use to tune the stage to the exact voltage I like and then I'll find the closest value resistor. But most likely 10K. Let's just go with that for this discussion.

and what it the value on the high side of that resistor before the 355V?

What do you mean by this? Are you asking what the voltage is before the resistor? It's currently 331v but my upper target is 350v, so I will adjust the droppers earlier on the B+ rail to get 350v.

So it's: 331v (or potentially 350v) —> 10K dropper with filter cap —> 100K plate resistors —> tube plates.

I don't know the current though. But I can say that the previous dropping resistors and voltage drops are as follows. Using Ohm's Law, I can at least estimate current.

B+ supply: 488v
After first 10K dropper: 400v (88v drop). Current = 8.8mA
After second dropper (8K2): 331v (69v drop). Current = 8.4mA
After third dropper (10K): 289v (42v drop). Current = 4.2mA
After fourth dropper (10K): 275v (14v drop). Current = 1.4mA

The node in question will be tapped after the second dropper (331v node). With my 5W pot set to 10K, I'm reading 317v resulting voltage = 14v drop. Current = 1.4mA.

Resistor power rating is suggested at 1/8th of a watt? Even with a 100v drop across a 10K resistor, a 1W resistor should suffice?

The power is them
P = V*V/R. This tells you the actual power in the resistor; then you add to that a safety factor, generally at least 2 times the power, and round up to the next standard power rating.

Until you have a good estimate of the actual power the resistor generates, picking what's enough power rating is a pure guess.

2. Once you know the resistor and the power, you can play with the voltage rating. It is common practice in higher voltage circuits to use two resistors in series for both higher voltage rating and higher power. If the actual power wasted/dissipated in the dropping resistor is something under 1/2W, then a one watt total resistor rating works fine. If you can't find the right resistance and voltage rating in a single, divide the resistor into two resistors of half the value, in series. Two identical 1/2W resistors, each rated at 200V, should be fine for 355Vdc, and will dissipate 1W total, 1/2W in each.

If you have resistors of the right resistance value in 1/2W and 1W, and the voltage rating is 500V, then you only need to know if the actual power is 1/4W or less to use the 1/2W; or if the actual power is 1/2W or less you can use one of your 500V 1W devices.

But you have to have at least some idea what the actual power is; knowing the voltage just across the dropping resistor lets you calculate the power, then add a safety factor.

It’s the voltage across the resistor that the rating refers to. MF or MO will work, MF will produce less noise. 1W might be ok, but do the math.

Strictly speaking this is correct - it's the working voltage across the resistor that the rating refers to, and in this setup it's probably under 100V or so. I like to be more conservative if it's reasonable. If the node is 355V, and a tube further down the line shorts, more voltage will be put across the dropping resistor. This will always be smaller than the 355 node voltage, but may be higher than the actual normal working voltage. So using a 400+ voltage rating, even from two resistors in series, will always be enough.

Yes, it's way more than necessary, but then belts and suspenders were always a good combination.

So 500v rating on a resistor should be fine. Now it's just the wattage that's in question. I have 1/2W 500v and 1W 500v metal films. But I do see some metal oxide that are 2W/500v that are small enough to fit.

[FourT6and2]

It’s the voltage across the resistor that the rating refers to. MF or MO will work, MF will produce less noise. 1W might be ok, but do the math.

Great, thank you.

If my math is correct (331v supply dropped to 317v via a 10K resistor), I'm only getting a 17v drop. So 500v rating is way overkill and even a 1/2w metal film resistor will work.

I mean the 50-year-old, 1/2W, 100K carbon film plate resistors are doing more work... At 100K, they're dropping closer to 100v and they're still going strong after half a century.

[jabguit]

been using these for power resistors with great results - they're very small for the power rating:

https://www.cedist.com/products/resisto ... metal-film


cheers,

[R.G.]

and what it the value on the high side of that resistor before the 355V?

What do you mean by this? Are you asking what the voltage is before the resistor? It's currently 331v but my upper target is 350v, so I will adjust the droppers earlier on the B+ rail to get 350v.

So it's: 331v (or potentially 350v) —> 10K dropper with filter cap —> 100K plate resistors —> tube plates.

I don't know the current though. But I can say that the previous dropping resistors and voltage drops are as follows. Using Ohm's Law, I can at least estimate current.

B+ supply: 488v
After first 10K dropper: 400v (88v drop). Current = 8.8mA
After second dropper (8K2): 331v (69v drop). Current = 8.4mA
After third dropper (10K): 289v (42v drop). Current = 4.2mA
After fourth dropper (10K): 275v (14v drop). Current = 1.4mA

The node in question will be tapped after the second dropper (331v node). With my 5W pot set to 10K, I'm reading 317v resulting voltage = 14v drop. Current = 1.4mA.

Resistor power rating is suggested at 1/8th of a watt? Even with a 100v drop across a 10K resistor, a 1W resistor should suffice?

Yes, that's exactly the path I was suggesting. Find the voltage on the high-voltage side of the resistor, find the voltage on the lower-voltage side, then subtract to get the voltage just across the resistor. Then square that voltage across the resistor, divide by the resistance, and you have the power. For your example, yes, 100V across a 10K resistor is 1W.

Then you add a safety factor to the resistor rating. A 1W resistor producing 1W will have a temperature rise of 75C to 100C; that's blister-your-finger hot, and it will have a short life. So conservative practice is to calculate the actual power, double it, and use the next standard power rating. For a 10K with 90V across it, you'd get 90*90/10K = 0.81W double that to 1.62W, pick the next standard rating, or 2W.

If your calculations say 1/8 watt, then 1/4W would be a good conservative choice and live long. For a power chain, I'd probably never use less than 1/2W just because those are easy and cheap as well as being a decent size. A whole lot of engineering involves calculating furiously to find out where the edge of the cliff is, then backing away a safe distance from the cliff. Further from the edge is fine, too, if it's easy and cheap.

So 500v rating on a resistor should be fine. Now it's just the wattage that's in question. I have 1/2W 500v and 1W 500v metal films. But I do see some metal oxide that are 2W/500v that are small enough to fit.

Yes, 500V would be fine. And note, like I mentioned - Martin is right, you only need to consider the normal working voltage across the resistor itself. I was being super extra-cautious in talking about the voltage the resistor floats on. So if the resistor only has a voltage directly across it of, say, 80V, then a 100V rating is OK. I was thinking what if a tube connected to the dropping resistor shorted, or a wire broke or something. In that -- admittedly disaster-case -- scenario, having a voltage rating even higher would keep the resistor from arcing over. And I wanted to point out that you can stack lower-voltage-rated resistors to get a higher total voltage rating if you don't have a higher-voltage-rated part at hand.

[FourT6and2]

and what it the value on the high side of that resistor before the 355V?

What do you mean by this? Are you asking what the voltage is before the resistor? It's currently 331v but my upper target is 350v, so I will adjust the droppers earlier on the B+ rail to get 350v.

So it's: 331v (or potentially 350v) —> 10K dropper with filter cap —> 100K plate resistors —> tube plates.

I don't know the current though. But I can say that the previous dropping resistors and voltage drops are as follows. Using Ohm's Law, I can at least estimate current.

B+ supply: 488v
After first 10K dropper: 400v (88v drop). Current = 8.8mA
After second dropper (8K2): 331v (69v drop). Current = 8.4mA
After third dropper (10K): 289v (42v drop). Current = 4.2mA
After fourth dropper (10K): 275v (14v drop). Current = 1.4mA

The node in question will be tapped after the second dropper (331v node). With my 5W pot set to 10K, I'm reading 317v resulting voltage = 14v drop. Current = 1.4mA.

Resistor power rating is suggested at 1/8th of a watt? Even with a 100v drop across a 10K resistor, a 1W resistor should suffice?

Yes, that's exactly the path I was suggesting. Find the voltage on the high-voltage side of the resistor, find the voltage on the lower-voltage side, then subtract to get the voltage just across the resistor. Then square that voltage across the resistor, divide by the resistance, and you have the power. For your example, yes, 100V across a 10K resistor is 1W.

Then you add a safety factor to the resistor rating. A 1W resistor producing 1W will have a temperature rise of 75C to 100C; that's blister-your-finger hot, and it will have a short life. So conservative practice is to calculate the actual power, double it, and use the next standard power rating. For a 10K with 90V across it, you'd get 90*90/10K = 0.81W double that to 1.62W, pick the next standard rating, or 2W.

If your calculations say 1/8 watt, then 1/4W would be a good conservative choice and live long. For a power chain, I'd probably never use less than 1/2W just because those are easy and cheap as well as being a decent size. A whole lot of engineering involves calculating furiously to find out where the edge of the cliff is, then backing away a safe distance from the cliff. Further from the edge is fine, too, if it's easy and cheap.

So 500v rating on a resistor should be fine. Now it's just the wattage that's in question. I have 1/2W 500v and 1W 500v metal films. But I do see some metal oxide that are 2W/500v that are small enough to fit.

Yes, 500V would be fine. And note, like I mentioned - Martin is right, you only need to consider the normal working voltage across the resistor itself. I was being super extra-cautious in talking about the voltage the resistor floats on. So if the resistor only has a voltage directly across it of, say, 80V, then a 100V rating is OK. I was thinking what if a tube connected to the dropping resistor shorted, or a wire broke or something. In that -- admittedly disaster-case -- scenario, having a voltage rating even higher would keep the resistor from arcing over. And I wanted to point out that you can stack lower-voltage-rated resistors to get a higher total voltage rating if you don't have a higher-voltage-rated part at hand.

Great, thank you.

So even if I raise that node to 350v, the drop across it won't ever be more than 50v. In that case, the power would be 0.25w and a 1/2W rating would be fine. So I can actually just use the 1/2W, 500v metal films I already have. This exercise was needed because space on my board is limited. I can't run multiple resistors in series and I can't use larger over-rated components.

[FourT6and2]

After tuning the stage with the pot, I settled on 10K with a 33uF filter cap. Supply node is 368v, dropping down to 350v. I used a PRP 10K metal film rated at 0.5W/500v. Seems to work great.