# Wiring dpdt gain switch help

#### matt117

##### Tele-Holic
Hey,
Looking for some help with wiring up a dpdt switch to toggle a resistor that is part of a voltage divider to grid input.
On this dumble/two rock build there is a 68k tied to a 1m making a voltage divider between v1b and v4a.
The 68k dumps a lot of gain to ground.
I’d like to put the stock 68k in the center off position, and have a 220k and 100k or 120k as the other option.

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#### andrewRneumann

##### Tele-Afflicted
The DPDT only has two positions. Your layout only gives you the choice of 68K+150K or 68K+220K. Is that what you were thinking? If you want to have a stock 68K only option, I think you will need a SP3T switch.

#### matt117

##### Tele-Holic
The DPDT only has two positions. Your layout only gives you the choice of 68K+150K or 68K+220K. Is that what you were thinking? If you want to have a stock 68K only option, I think you will need a SP3T switch.
I’d like to have 3. Stock 68k in middle then the up and down higher values. I think it works with my on/off/on if I do this.. I just always get confused with these switches. It would put the other values in series with the 68k to achieve approximately a 150k and 220k. Let me know if I’m thinking clearly.

#### andrewRneumann

##### Tele-Afflicted
I’d like to have 3. Stock 68k in middle then the up and down higher values. I think it works with my on/off/on if I do this.. I just always get confused with these switches. It would put the other values in series with the 68k to achieve approximately a 150k and 220k. Let me know if I’m thinking clearly. View attachment 998389

Sorry that jumper across the middle lug means the 100K and 150K will always be bypassed. You’ll end up with 68K in all 3 positions.

#### matt117

##### Tele-Holic
Sorry that jumper across the middle lug means the 100K and 150K will always be bypassed. You’ll end up with 68K in all 3 positions.
Is this Better ?

#### dogmeat

##### Friend of Leo's
if you can be happy with a different order.....

make the center off position the 220k. simply tie the center pin to the high side of the 220k and let the other end of the resistor go to chassis.

on one throw, put another 220 so when activated, the parallel resistors will give you RT of 110k

on the other throw put a 100 so the Rt will be about 70ish

this can be done with a spdt switch. I did something like this on a modified blackface build & used a Telecaster switch. if you want the original order, I think you'll need another set of contacts so that the 220k can be eliminated

#### andrewRneumann

##### Tele-Afflicted
Here's how I would do it with an ON-OFF-ON switch. @dogmeat was on to the same idea.

With this, get rid of the 68K on the board altogether.

#### matt117

##### Tele-Holic
Here's how I would do it with an ON-OFF-ON switch. @dogmeat was on to the same idea.

With this, get rid of the 68K on the board altogether.
View attachment 998424

Thanks so much. That makes a lot of sense. I was wondering if it would happen to be parallel resistance to make this work. I did the math on that and those options are great.

#### andrewRneumann

##### Tele-Afflicted
Have you heard of the "product over sum" for parallel resistors? It says that if you want to find the equivalent value of 2 resistors in parallel, you can divide the product of the two resistors by their sum:

$R_{equiv}=&space;\frac{{R_1&space;*&space;R_2}}{{R_1&space;+&space;R_2}}&space;$

But in your case, you want to find the R2 that gives you a desired Requiv. In this case, use the "product over difference" rule. Same idea, but looks a little different. We know Requiv and R1 but we want to find R2.

The "Product Over Difference" Rule

$R_2=&space;\frac{{R_1&space;*&space;R_{equiv}}}{{R_1&space;-&space;R_{equiv}}}&space;$

For example, you have 220K and you want to add a parallel to be equivalent to 68K.

$R_2=&space;\frac{{220&space;*&space;68}}{{220&space;-&space;68}}=&space;\frac{14,960}{152}=98.4&space;\approx&space;100K&space;$

#### D'tar

##### Friend of Leo's
Another option, DP3T, on-on-on, 3 resistors of your choice. Green wire input

#### matt117

##### Tele-Holic
Have you heard of the "product over sum" for parallel resistors? It says that if you want to find the equivalent value of 2 resistors in parallel, you can divide the product of the two resistors by their sum:

$R_{equiv}=&space;\frac{{R_1&space;*&space;R_2}}{{R_1&space;+&space;R_2}}&space;$

But in your case, you want to find the R2 that gives you a desired Requiv. In this case, use the "product over difference" rule. Same idea, but looks a little different. We know Requiv and R1 but we want to find R2.

The "Product Over Difference" Rule

$R_2=&space;\frac{{R_1&space;*&space;R_{equiv}}}{{R_1&space;-&space;R_{equiv}}}&space;$

For example, you have 220K and you want to add a parallel to be equivalent to 68K.

$R_2=&space;\frac{{220&space;*&space;68}}{{220&space;-&space;68}}=&space;\frac{14,960}{152}=98.4&space;\approx&space;100K&space;$
Thank you very much for your help and time teaching me. Really appreciate it!

#### andrewRneumann

##### Tele-Afflicted
Another option, DP3T, on-on-on, 3 resistors of your choice. Green wire input View attachment 998568

Yes--that's a DPDT ON-ON-ON wired like a SP3T. Makes resistor selection much easier. OP seemed to only have an DPDT ON-OFF-ON switch though. Many ways to skin this cat.

#### 2L man

##### Tele-Afflicted
Other power tube pair does not have 1R bias measure resistors.

That bias circuit is not good. Trimmer should be between ground and bias feed and then if trimmer fails the bias voltage drive tubes to cutoff.

#### D'tar

##### Friend of Leo's
Other power tube pair does not have 1R bias measure resistors.

That bias circuit is not good. Trimmer should be between ground and bias feed and then if trimmer fails the bias voltage drive tubes to cutoff.
Wiper tab connected to bias input tab should do the trick.

#### matt117

##### Tele-Holic
Other power tube pair does not have 1R bias measure resistors.

That bias circuit is not good. Trimmer should be between ground and bias feed and then if trimmer fails the bias voltage drive tubes to cutoff.
Do they usually each get their own 1R on the bias board? Where the wire comes off from the diode, how would you lay it out to improve that using this arrangement? You’re not talking about the 1r to grounds for measuring tube current, correct?
Wiper tab connected to bias input tab should do the trick.
Hey, where would the current center lug wire coming from the 220k after the caps from plates of the pi go?

#### D'tar

##### Friend of Leo's
Do they usually each get their own 1R on the bias board? Where the wire comes off from the diode, how would you lay it out to improve that using this arrangement? You’re not talking about the 1r to grounds for measuring tube current, correct?

Hey, where would the current center lug wire coming from the 220k after the caps from plates of the pi go?
To the wiper. The pot and limit resistor would go to ground. You can vary the resistance but if the wiper/pot/resistor fails you have full negative bias voltage that would bias your tubes cold or cutoff instead of zero voltage allowing full saturation.

#### matt117

##### Tele-Holic
Sorry to not understand, is it just jumpered or physically move the wire over to the center?

#### D'tar

##### Friend of Leo's
Just add red jumper to connect them. Good to go.

#### matt117

##### Tele-Holic
Just add red jumper to connect them. Good to go.
Thanks again. If I’ve understood correctly, what you’re saying is having the wiper and input tied together is a fail safe to prevent a tube from melting down from saturation (too much current). If the pot fails now (with wiper and input tied) all of the negative bias would go into the tubes and make them cold/cut off but prevent a burn down. Where as the old way sends all the bias to ground allowing a tube to melt down if the bias pot fails.
Thanks so much for your help

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