How to calculate for voltage dividers and grid leak resistors

[Steveo]

Hello, I made this basic schematic that I want to build. I've seen what other people have used for voltage dividers and grid leak resistor values in similar circuits but how does one come up with their own values that work? Is it a matter of keeping voltages in check? I plan to use about 280-300vdc for the b+ in the preamp.

[xtian]

Voltage dividers: Yes, to keep AC (signal) voltages in check. For example, I built a tweed-style, simple amp, and thought, naw, I don't need this voltage divider here. In fact, there was WAY too much gain. I put the divider back in, and ended up dumping 85% of the signal to ground!

In your circuit, you've got two pots serving as your voltage dividers, so you're set! You will probably have way too much gain with those pots dialed up, but you can tweak values later.

Grid ref resistors: typically, we see a max of 1M grid to ground on preamp stages. If there's no reference to ground, there's trouble. But I don't know more than that.

[MakerDP]

Well there's a voltage divider just before the third gain stage... 470k/470k... that will drop your voltage by 50% going through there, but curiously it comes AFTER a 1M pot... so the 1M pot is setting a value and then the voltage divider is cutting what the pot is already cutting by some amount in half. In practice, that sounds a little funny maybe but it sets the MINiMUM amount of voltage reduction to 50% when that pot is wide-open --- that is, the amount of voltage cut will never go BELOW 50%. (I hope that makes sense... I have a bad headache tonight )

Three ways to approach it really...
1) Experiment with different values, preferably where they add-up to as close to 1M as possible. You can experiment with an online voltage divider calculator to see the effects of different values. You can make things easier to wrap your brain around if you put in "100" for the input or source voltage. Then you know for example if the output voltage is 40V, you know a divider using those values drops your signal by 60%.

2) Put a 1M pot in there. Experiment with it's setting until you get what you are after. Measure the values... Pin 1 to Pin 2, Pin 2 to Pin 3... replace the pot with a voltage divider using values as close to those as you can get.

3) Put a "trim pot" inside the amp. Set it where you want it, put some red LocTite on it so it won't rotate under normal circumstances and leave it at that.

#1 is cleaner once everything is done but takes the most work. #2 is much easier but you likely won't get the EXACT values of the pot's setting so there is some give-and-take there. #3 is ideal... always set exactly to how you want it and a much cleaner install. This is how Mr. Dumble does it if you go look at the D* schematics you can see how he cuts the signal level going into the OD channel when it is switched on.

[Steveo]

Cool, thanks for your help. I've heard about the reference to ground but I'll have to read up on that to understand it.
I didn't know Dumble did that with his amps, that's a neat idea. I think I'll have to use the calculator for an estimate and really just tune it by ear after I assemble it. Probably a lot of other factors in how it will react anyways. Thanks

[xtian]

Cool, thanks for your help. I've heard about the reference to ground but I'll have to read up on that to understand it.

Grid reference to ground is a utility thing. If a triode does NOT have a reference to ground, its DC voltage can creep upward, changing the bias of the stage, and causing blocking distortion, or total signal cutoff.

[Ten Over]

Well there's a voltage divider just before the third gain stage... 470k/470k... that will drop your voltage by 50% going through there, but curiously it comes AFTER a 1M pot... so the 1M pot is setting a value and then the voltage divider is cutting what the pot is already cutting by some amount in half. In practice, that sounds a little funny maybe but it sets the MINiMUM amount of voltage reduction to 50% when that pot is wide-open --- that is, the amount of voltage cut will never go BELOW 50%.

That 470K-470pF/470K is commonly called a treble peaker. It is there for tone-shaping purposes. Can't get the "Marshall sound" without that little figure.

[Ten Over]

If you don't have a grid reference to ground (grid leak), the charge on the coupling capacitor from the previous stage will determine your bias. If the grid end of that capacitor has a large enough positive charge, grid current will flow and the positive charge will be reduced. If you apply a signal that is large enough to cause grid current, the capacitor will charge negative which makes the overall bias colder. With no discharge path, the capacitor will hold that negative charge and keep the bias cold even with a reduced signal.

Data sheets place the maximum grid leak for self-bias at 2M, presumably because larger values would introduce some grid-leak bias. But it turns out that grid-leak bias is only a problem if the cathode self-bias resistor is small. With a cathode resistor of 1.5K, a 10M grid leak resistor is no problem.

[Steveo]

I have 1meg ohm pots as my grid leak resistors now. Would it hurt to put in 1M grid leak resistors as well in case something bad happens to my 1M pot? Then the bias wont go crazy with a floating ground if the pot malfunctions.

[pompeiisneaks]

Generally the safety for a pot if it's being used as a variable resistor, is to tie one side to the wiper, so that in case the wiper loses connection, you get the full resistance of the pot as a protection.

Is that what you're referring to? Or more the pot literally failing open? I think that's a very rare failure state, but I could be wrong.

~Phil

[xtian]

I have 1meg ohm pots as my grid leak resistors now. Would it hurt to put in 1M grid leak resistors as well in case something bad happens to my 1M pot? Then the bias wont go crazy with a floating ground if the pot malfunctions.

No, don't worry about it. If pot fails, amp will just sound bad. No harm will come to your tubes or anything else. We're not talking about power tubes, here, just preamp tubes.

[Steveo]

Yeah i was thinking of failing open. Ok thanks to both of you.

[Steveo]

Three ways to approach it really...
1) Experiment with different values, preferably where they add-up to as close to 1M as possible. You can experiment with an online voltage divider calculator to see the effects of different values. You can make things easier to wrap your brain around if you put in "100" for the input or source voltage. Then you know for example if the output voltage is 40V, you know a divider using those values drops your signal by 60%.

2) Put a 1M pot in there. Experiment with it's setting until you get what you are after. Measure the values... Pin 1 to Pin 2, Pin 2 to Pin 3... replace the pot with a voltage divider using values as close to those as you can get.

3) Put a "trim pot" inside the amp. Set it where you want it, put some red LocTite on it so it won't rotate under normal circumstances and leave it at that.

#1 is cleaner once everything is done but takes the most work. #2 is much easier but you likely won't get the EXACT values of the pot's setting so there is some give-and-take there. #3 is ideal... always set exactly to how you want it and a much cleaner install. This is how Mr. Dumble does it if you go look at the D* schematics you can see how he cuts the signal level going into the OD channel when it is switched on.

Ive been reading about this some more and ive been getting a little confused.
If you were to make an amp would there be an ideal percentage of voltage right before the power tubes compared to your input voltage? Say i use 100mv as my input voltage and i end up with 500mv right before the power tubes (el34s in this case). I would have increased my voltage 500% I know its a lot more but just showing an example.

I know there is a lot of increasing gain voltage and attenuating gain voltage during amp gain stages. Would the answer be associated within spec sheet of the power tube or tubes your using along with the b+ voltage you supply it? Inside the load line calculations somewhere?

How do you know you are feeding your power tubes a big enough signal to be efficient or not too big of a signal where sound quality diminishes from being overloaded so to speak?

[martin manning]

To fully drive the power tubes, the peak signal at the grids has to be equal to the bias voltage (to reach Vg1 of zero). Given the gain of the PI, that might only require a couple of volts peak at the PI input.

[Steveo]

To fully drive the power tubes, the peak signal at the grids has to be equal to the bias voltage (to reach Vg1 of zero). Given the gain of the PI, that might only require a couple of volts peak at the PI input.

Thanks, your reply helped a lot. https://www.ampbooks.com/mobile/classic ... power-amp/

[MakerDP]

Yeah, what Martin said.

Try and find some schematics that show tested voltages at various points in the amp and you can get a good sense of how that all plays out.