Quick resistor use question

WizardOfBoz · Sep 4, 2021

I'm doing some factory-suggested mods to a guitar amp (Fender Prosonic - a tube/valve amp). I need a 27K 1W resistor. There aren't any in stock at Mouser. So I bought 4 1/2W 27K resistors. Solder two assemblies of two resistors in series, then solder the two assemblies together in parallel.

My inclination is to not connect the center conductors together. Each assembly will have a likely lower variation in resistance and soldering the center taps together will slightly increase the likelihood of one resistor having a higher current. I see no upside to connecting the center taps.

As an aside, it probably doesn't matter. The assembly has a rating of 2W now.

But is my logic correct: better to not connect the center conductors?

AA+ · Sep 4, 2021

It really doesn't matter if you connect or not.
Best wishes --- Allen Anway

rons · Sep 4, 2021

You add 3 more points of failure which does not include the connections.
Just order the same resistor in a higher watt range. Over build and over ....

JST · Sep 4, 2021

Concur.

You can also parallel two of about double the resistance and half the wattage, if there is a space problem (never seems to be on older Fender amps).

If Mouser does not have something, try Digikey.

WizardOfBoz · Sep 4, 2021

Thanks all. I looked up the part on Digikey. Whereas Mouser was pretty easy to order onesies and twosies, Digikey wants to sell me 6000, minimum. Actually for the 1W design, minimum order was 2000. And they have nothing in stock. Even if they did, I'd feel goofy paying 8 bucks shipping for a single 16 cent part. But I don't have a choice sooo, I think my expedient approach is going to have to work. I don't think I'll have a problem doing high reliability solder joints. It's not in a signal path (its a filter resistor to smooth DC power for a relay) and so I don't feel the need to minimize lead length.

And yes, the chassis of Fender Tube Amps have no problem of space constraint. I could part my SUV in there.

What I plan on doing is to measure the resistance of each resistor, and pair the high and low together, and the two mid values together, and to not solder the center connector. This minimizes variabilty in the current flow and evens out power dissapation. I'm a little concerned with this design change as posted by Fender. They went from a 150K to a 27K resistance value and kept the power rating the same. I would have expected the power rating to go up. Which is a benefit to my ad hoc approach: I get a 2W rating out of this setup. I'll let you know how my amp sounds.

Heres some guy on youtube demoing the amp:
https://youtu.be/SHXMY6OKTNs

Scottl · Sep 4, 2021

If there is enough variation in the resistance values you are correct that connecting the centers MIGHT cause more current to flow in the one with the lowest value. A lot would depend on what the other exact values are.

If you want to balance the current in each leg as much as possible the best course would be to measure and record the actual resistances and then select series pairs such that each series pair would have similar total resistance to the other.

In the real world it probably won't matter given the 2 x wattage but doing as I've described might make you feel a little better about the finished job.

In simplest terms, if you have two plus tolerance resistors and two minus tolerance ones then put a low in series with a high for each leg.

Gordon Heaton · Sep 4, 2021

bosleyjr said:
. . .Digikey wants to sell me 6000, minimum. Actually for the 1W design, minimum order was 2000. . .

Interesting. When I look at Digikey I see many thousands in stock, and a large variety with minimum order of 1, about $.30. That's for a 27K 1W 5% standard axial resistor.
I usually start with Digikey, then Mouser and when I have to, Jameco.
The 'blocked' links below are not blocked.

Blocked

Blocked

rons · Sep 4, 2021

Here

https://www.newark.com/c/passive-co...e-resistors?resistance=27kohm&power-rating=1w

MetalCarnage · Sep 4, 2021

bosleyjr said:
I'm doing some factory-suggested mods to a guitar amp (Fender Prosonic - a tube/valve amp). I need a 27K 1W resistor. There aren't any in stock at Mouser. So I bought 4 1/2W 27K resistors. Solder two assemblies of two resistors in series, then solder the two assemblies together in parallel.

My inclination is to not connect the center conductors together. Each assembly will have a likely lower variation in resistance and soldering the center taps together will slightly increase the likelihood of one resistor having a higher current. I see no upside to connecting the center taps.

As an aside, it probably doesn't matter. The assembly has a rating of 2W now.

But is my logic correct: better to not connect the center conductors?

Good news, and bad news. First, the bad news - as I understand it, the assemblage you've put together or are about to won't affect the wattage they way you think it will. Putting two 4 1/2 watt resistors in series will still be capable of handling 4 1/2 watts, putting them in parallel will just basically double the amount of current (and thus wattage) they can carry (each leg would be able to handle 4 1/2 watts).

Now the good news - you shouldn't have to worry about it at all, just use the 4 1/2 watt resistor in place of the 1 watt resistor. You need only be concerned with the Ohms of the resistor and whether it can carry at least 1 watt.

FWIW

-Ron

Gordon Heaton · Sep 4, 2021

^^^Read more carefully what the OP posted. He's not talking about 4 and a half watt resistors. He's going with four, 1/2 watt resistors.

1yesca · Sep 4, 2021

there's going to be hell to pay when the thermster see this one

machinistrrt · Sep 4, 2021

bosleyjr said:
Thanks all. I looked up the part on Digikey. Whereas Mouser was pretty easy to order onesies and twosies, Digikey wants to sell me 6000, minimum. Actually for the 1W design, minimum order was 2000. And they have nothing in stock. Even if they did, I'd feel goofy paying 8 bucks shipping for a single 16 cent part. But I don't have a choice sooo, I think my expedient approach is going to have to work. I don't think I'll have a problem doing high reliability solder joints. It's not in a signal path (its a filter resistor to smooth DC power for a relay) and so I don't feel the need to minimize lead length.

And yes, the chassis of Fender Tube Amps have no problem of space constraint. I could part my SUV in there.

What I plan on doing is to measure the resistance of each resistor, and pair the high and low together, and the two mid values together, and to not solder the center connector. This minimizes variabilty in the current flow and evens out power dissapation. I'm a little concerned with this design change as posted by Fender. They went from a 150K to a 27K resistance value and kept the power rating the same. I would have expected the power rating to go up. Which is a benefit to my ad hoc approach: I get a 2W rating out of this setup. I'll let you know how my amp sounds.

Heres some guy on youtube demoing the amp:
https://youtu.be/SHXMY6OKTNs

As you have room and the application is not critical, how about 27K 5W wirewound? in stock at Mouser, about $3/ea.

Gordon Heaton · Sep 4, 2021

machinistrrt said:
As you have room and the application is not critical, how about 27K 5W wirewound? in stock at Mouser, about $3/ea.

He can buy a single 1W at Digikey for thirty cents + shipping.

WizardOfBoz · Sep 4, 2021

I couldn't find the 1W at digikey - they all specified a min order of like 2000. But others have said that Digikey is harder to search. I found this to be true. That said, a link to a 1W (or a 5W) would be great - thanks!

Mark Rand · Sep 5, 2021

try this link. it seems to work in IE and Firefox, but might not work for you.

Gordon Heaton · Sep 5, 2021

The first link i posted in #7 is a whole page of what you want. This one is right to a good resistor: Blocked
Both links work for me, do they not work for you? For me, Digikey is on the order of McMaster for easy search!

awander · Sep 5, 2021

MetalCarnage said:
... Putting two 4 1/2 watt resistors in series will still be capable of handling 4 1/2 watts...

How do you figure? Each resitor should be capable of the 4-1/2 watts, for a total of 9 watts, no?

DDoug · Sep 5, 2021

awander said:
How do you figure? Each resitor should be capable of the 4-1/2 watts, for a total of 9 watts, no?

I read the OP (bad technical writing et all...) as "qty (4) 1/2 watt resistors"

awander · Sep 5, 2021

digger doug said:
I read the OP (bad technical writing et all...) as "qty (4) 1/2 watt resistors"

Yeah, I get that, Digger.

What I'm questioning is the statement that using 2 resistors in series will not double the power-handling rating.

So to use the actual example, why would QTY (2) 1/2-watt resistors in series not be able to dissipate 1 watt?

markz528 · Sep 5, 2021

You guys are over-complicating it. Remember Ohm's law.

I don't care which configuration you put it in, the result is the same. If you have one resistor and its dissipating 4 watts, when you use 4 identical resistors in a series parallel equivalent combination, each resistor will dissipate 1 watt.

You are not changing the wattage capability of the resistor, just how much wattage the resistor sees.

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