B+ Parallel dropping resistors

[Jon434]

I have a 12k 3W resistor dropping B+ for the PI and then onto the preamp, and I want to lower the value. But with 3W metal oxide resistors that I'm using they usually come in set values like 8.2K or 10K.

I'd like to use 9K by paralleling in a 39K resistor. Would there be any downside to this, and is there a minimum wattage I should aim for?

Ordinarily I'd use any wattage but since it involves high voltages and B+, I thought that it's better to consider the options.

Would a 1W be ok, or would 3W be better? It seems wrong somehow if you could parallel in a 0.5W resistor that has 420V going through it.

[guitician]

Current is what is used to figure the power in a resistor. The larger resistor will be drawing a proportionate amount of current and can be less wattage proportionately.

[Mick Bailey]

The current in a parallel circuit is divided between the resistors, and the voltage across the resistors is the same for both. If 3W is adequate for your circuit as it stands, then consider that with the 39k you'll end up with very roughly a third of the current through that, and two-thirds through the 12K. 1W is OK.

Voltage rating of resistors varies according to wattage and material technology.

The resistor won't have 420v across it, otherwise your existing circuit would be dissipating 14.7W through that 12K resistor. It may have 420v at one end, but the voltage drop across it depends on current draw and may only be a few tens of volts.

Measure the voltage across the existing resistor and use Ohm's law to calculate the current. A good thermal safety margin for a dropper resistor is to use one three times larger than needed. It still dissipates the same amount of heat overall, but it's distributed over a larger area with a bigger resistor.

[Chuck H]

I just guestimate the wattage. I think I know the formulas but I'd be embarrassed to get it wrong. But the first part is easy...

Say your dropping 100V across that 12k resistor (measured from one end to the other). Divide the voltage by the resistance for .00833 that's your current in amps. Then multiply that by the voltage and you get .833 that's your watts. So that resistor is dissipating just under a watt of power without considering fluctuations. So, say 1 watt. Over rate and you'd probably go to 2 watts. But two watts isn't a common spec so just use the 3W resistor. Now... You want to parallel a 39k resistor across that 12k for 9k. The voltage differential won't be large enough to bother refiguring since you're generously over rated with a 3W unit. 12k is about a third of 39k and will therefor be dissipating about two thirds of the power. So the 39k will be dissipating something close to 1/3 of 1 watt. Considering fluctuations again, while a 1/2 watt unit would probably be fine, a 1 watt unit would be better.

[guitician]

For Reference.. If you measure the voltage across the 39k and square it, then divide it by 39k, you get the power.

[Enzo]

9k is an off value, but 9.1k would be a standard value. Close enough, and you don't have a parallel cobble. They have both wirewound and metal film in stock in that value at 3w. I didn't look at 5w, but there may be even more options there.

AC05000009101JAC00 Vishay / Draloric | Mouser


283-9.1K-RC Xicon | Mouser

[Chuck H]

9k is an off value, but 9.1k would be a standard value. Close enough, and you don't have a parallel cobble. They have both wirewound and metal film in stock in that value at 3w. I didn't look at 5w, but there may be even more options there.

+1

I've used wire wound for power supplies many times without problems. Their max voltage spec is often too close for comfort or even a little low but I've never had a problem.

The current trend seems to be metal oxide since they usually have a suitable voltage spec. But how many old Fenders were made with big, drifty CC's in the power supply and worked for thirty years before anyone started measuring and changing them!?! It's easy to get too boutiquey about all this. I'm guilty.

[Jon434]

Thanks guys

I just guestimate the wattage. I think I know the formulas but I'd be embarrassed to get it wrong. But the first part is easy...

Say your dropping 100V across that 12k resistor (measured from one end to the other). Divide the voltage by the resistance for .00833 that's your current in amps. Then multiply that by the voltage and you get .833 that's your watts. So that resistor is dissipating just under a watt of power without considering fluctuations. So, say 1 watt. Over rate and you'd probably go to 2 watts. But two watts isn't a common spec so just use the 3W resistor. Now... You want to parallel a 39k resistor across that 12k for 9k. The voltage differential won't be large enough to bother refiguring since you're generously over rated with a 3W unit. 12k is about a third of 39k and will therefor be dissipating about two thirds of the power. So the 39k will be dissipating something close to 1/3 of 1 watt. Considering fluctuations again, while a 1/2 watt unit would probably be fine, a 1 watt unit would be better.

The numbers you used match up exactly with what I've been doing (strange coincidence?). I'd been using ampbooks RC Ripple calculator and came up with 8.3ma using that, because I AM dropping 100V through that resistor but it's nice to see it match up with what you've described and to see the manual calculation in a simple way. It all makes a lot of sense. I wasn't sure about the voltage and current sharing in terms of the 12K vs the 39K.

Whether I used 1W or just get a 9K resistor, it's nice to know that either would work depending on the situation and how to calculate what would be required. Sometimes the value might not be available etc. Thanks again.

[Chuck H]

I wasn't sure about the voltage and current sharing in terms of the 12K vs the 39K.

Think of like you would a speaker load. That is, if you plug in an 8 ohm cabinet and a 16 ohm cabinet parallel. The 16 ohm cab will only be handling one third of the wattage while the 8 ohm cabinet carries the bulk of the load. Same thing, and easy to guestimate with only that experience since 12k and 39k are in similar proportion.

BTW, I utterly guessed with the 100V figure. It seemed like a likely enough and round number

[Enzo]

Voltage rating usually applies to end to end voltage. If one end is 400v and the other end is 300v, there is only 100v across the resistor, not 400.

Drifty parts? Remember, all those old Fender schematics said right on them, everything should be considered +/-20%.

[Chuck H]

Voltage rating usually applies to end to end voltage. If one end is 400v and the other end is 300v, there is only 100v across the resistor, not 400.

Drifty parts? Remember, all those old Fender schematics said right on them, everything should be considered +/-20%.

Right! Which is insane by modern standards. Imagine a modern high gainer with an 80k or 120k plate load where it should be 100K.?. There's a very good reason some of those amps were magic while most were just amps with a common "workman like" tone. I don't think modern amps capture enough of the magic of the best but I'll bet it's a damn site better than the average tone back in the day. And I'm talking about the standard stuff like the Peavey Classics and Fender HR series amps. They hint at that magic clean tone way better than the average BF did. But then there's that one or two amps in a thousand (+/- 20%) that sounded exactly right. So the legend lives.

[Jon434]

Voltage rating usually applies to end to end voltage. If one end is 400v and the other end is 300v, there is only 100v across the resistor, not 400.

This confused me in the beginning when trying to spec out resistors but it makes more sense now. I'm not sure if this is the right topic to go into it in depth but I'll try to keep it as short/simple as possible, in terms of the way I see it. Hopefully there's no oversights.

It seems like a wording thing for me more than anything else, but maybe there's more to it.

If I take a jumper wire and substitute that for a power resistor, there'd be 400V on one side and 400V on the other side. To me, there's 400V going through it. But by the way you are talking and the way resistors are spec'd there might be 0V going through it. One way is measuring the voltage across the resistor, and the other is measuring it's voltage to ground.

When buying 18AWG wire for example, you can get 300V rated wire or 600V rated wire. Which is important for us because if we are using HT it could have 350-500V going through it and it might only be 0.2A or 0.4A but still. You wouldn't judge the wire spec by the voltage that's being dropped through the wire.

But with resistors, there whole purpose is to restrict some of the current/voltage and the higher the value, the more heat will be generated. Energy being restricted and stored as heat energy etc. On the site I order the resistors from, the specs usually call this "limiting element voltage". Which is usually somewhere between 250V and 500V depending on the wattage/type/brand of resistor. And then on top of that there'd be a maximum overload/surge rating that it can handle during brief periods but not constantly like at startup (but my amps have thermistors so I'm not sure even have that surge).

To go one step further, I believe that Russell Aiken's site has an article on it where he talks about how he likes to overspec everything to the point where the "limiting element voltage" is equal to or more than the actual B+ voltage. Starting out, a lot of this stuff can be very confusing. A lot of amp makers or brands don't do this and just use the voltage being dropped by the resistor as the guideline. So like in the OP, I'm dropping 100V through a 3W 12K resistor, which is well below 350V.

I don't know if any of this makes sense, but if using a jumper wire, or a low value resistor that drops very little voltage/current, I'm thinking of the voltage in both ways. The amount of voltage that's coming out the other side, like you would when picking out wire guages, but also the voltage being dropped. Where as a lot of people only like to talk about the latter when talking resistors. If I were jumpering at that voltage, I'd rather use a spare lead from a 600V OD cap or a 3W resistor for example, rather than a 1/2W resistor.

I don't know if it's wrong to do this or not, it's just a habit. And I always like the idea of having bigger/more whenever it's possible. For example, if there's 400V coming out the other side, I always want the leads to be thick, and not the thin bendable ones you'd see in a pedal or cathode resistor of a preamp. Even if that resistor would "technically" be suitable and be able to handle it in terms of the wattage dissipated and limiting element voltage. They also give something easier and more stable to clip onto with alligator clips if measuring, because for me they are usually on a PCB and bending down into the board. I guess this is my long winded way of saying that I'd want to use a 1W in this case above, even if 1/2W would be applicable and had the same limiting element voltage rating if it were for the B+ voltage. It could be worrying about nothing but just feels better somehow.

[Enzo]

Don't over-think it. A wire has zero ohms, essentially. Ohm's Law tells us Voltage equals current times resistance. So no matter how much current flows through the wire, there will still be zero volts dropped across it. There isn't 400v flowing through it. It has 400v ON it. Current flows though it, not voltage.

Your wire has a voltage rating for the insulation. In other words if there is 400v on the wire, AND the wire is lying on the grounded chassis, then there is 400v across the insulation. If you have 300v wire, there is the chance the voltage will jump through the insulation, but with 600v wire, it won't. resistor voltage ratings are not insulation related. Well not like wire anyway. A film resistor for example has an insulated core with a spiral resistive stripe around it, like a barber pole. Each turn around the core that spiral makes puts one part of the stripe next to another. The more voltage across the resistor, the more likely it is to jump between "windings".

Ever see a 10,000 volt power line overhead, and birds are sitting on the wire? The voltage doesn't bother them because current is not flowing through their bodies. If they sat ther and could touch something grounded, then 10,000v would be across the bird, and he'd fry. Otherwise, he sits there happy as a clam with 10,000v on him. Voltage to ground doesn't matter until you have a path to ground for it.

A bare wire? OK, lets say you have a pair of tube sockets next to each other and you want to connect the pin 3 on each one together. Maybe that has 400v on it. Now you could use insulated wire. How about the 300v wire? If I lay the wire on the chassis between the sockets, I might be concerned. But what if I ran it direct across the space, in mid-air?Now my 300v wire is not a concern. Go a step further. Like many old point to point chassis, why not just use a piece of bare wire? That has zero insulation, and yet it never touches chassis and is just fine.

[Jon434]

Makes sense, thanks.

[guitician]

I've used those tapped wire wound resistors, the ones that are bare one one side with a slider, for PS nodes in amps. You can nail the voltages you after without a lot of fuss.