Bias pot value theory

[Gaz]

I have no idea why the value of a bias pot is chosen, and I want to so badly want understand.

[Cliff Schecht]

It's determined based on the current draw and voltage that the bias circuit will see. Looking at the 1968 Fender Deluxe Reverb that's on my bench now as an example (this schematic: http://ampwares.com/schematics/deluxe_reverb_ab763.pdf), you want to choose the value of this pot so that it gives -35V somewhere in between the extremes of the bias pot (ideally near the middle). Maybe this is obvious.. How about some math .

So let's say the bias tap/rectifier/resistor/cap gives you -55V (as a guess). You would choose the bias pot so that with a small current draw you get that usable range of sweep. Just from experience I know the bias circuits usually draw anywhere from 1-5mA depending on who designed it and so let's pick 3mA as a nice median to work with. Simple Ohms law will tell you that with -55V (or 55V for our math) and 3mA of current draw gives an equivalent resistance of about 18.3k (55V/3mA=18.3kOhms).

But this isn't your final pot value, typically you also put some series resistance in to limit how much current can be drawn from this circuit as well as how much range the bias pot has. In our case the value chosen is 10k which when multiplied by our chosen current of 3mA gives you a "maximum" (least negative) voltage of -30V at the extreme travel of the pot. If you would perhaps want a bit more range then an 8.2k or 6.8k resistor can be subbed for that 10k resistor which would give a minimum bias voltage about 22V (with 6.8k). This number comes from 55V divided by the new equivalent resistance (10k + 6.8k) which gives draws about 3.25mA of current. Now multiply this (3.25mA) by the new resistor (6.8k in this example) and you get 22V. In practice the value might be a bit above this (you would have the 10k pot in series that drops some voltage) but this is how you choose the value for the earlier bias circuits.

All of this is working with Ohms law. You have to know how to work with this to do any kind of circuit design really. A basic circuits book and a circuit simulator will get you there pretty quickly .

[Gaz]

Thanks, Cliff, that was really useful. I know Ohm's law, but just have trouble applying it!

My other question is if you add another bias pot in parallel for a simple dual bias adjust, do you double the pot value to keep things equal?

[Reeltarded]

No! You double the bias circuit! hah

[Cliff Schecht]

Thanks, Cliff, that was really useful. I know Ohm's law, but just have trouble applying it!

My other question is if you add another bias pot in parallel for a simple dual bias adjust, do you double the pot value to keep things equal?

Not if you make the bias circuit a whole new branch. Instead each branch would draw equal current (assuming they are the same circuit) which to the bias tap would look like double the current. But since we are drawing a few mA off of a tap that can handle 10x this, it will drop negligible voltage at the diode/cap junction that feeds each bias circuit.

[roberto]

Are you asking why wth the same range of bias voltage control we use different values, or simply why I use a 15k pot with a 10k to ground after a diode on a 50V tap? You need a certain value of current available and a certain voltage on the output. This way you can choose bias circuit values.

EDIT
too many reply while I was look at too many forums. =)
Sorry!

[Gaz]

Are you asking why wth the same range of bias voltage control we use different values, or simply why I use a 15k pot with a 10k to ground after a diode on a 50V tap? You need a certain value of current available and a certain voltage on the output. This way you can choose bias circuit values.

EDIT
too many reply while I was look at too many forums. =)
Sorry!

I mean the former. Just different ways to skin a cat? I mean, why not use a 1M pot? or a 1K pot?

[Gaz]

Thanks, Cliff, that was really useful. I know Ohm's law, but just have trouble applying it!

My other question is if you add another bias pot in parallel for a simple dual bias adjust, do you double the pot value to keep things equal?

Not if you make the bias circuit a whole new branch. Instead each branch would draw equal current (assuming they are the same circuit) which to the bias tap would look like double the current. But since we are drawing a few mA off of a tap that can handle 10x this, it will drop negligible voltage at the diode/cap junction that feeds each bias circuit.

I just mean adding another pot, but use the same rectifier and filtering.

[Reeltarded]

You would be splitting the supply just by adding a parallel second trimmer.

Voltage divider.

That is why dual or even quad bias circuits are separate identical sections taking equal portions of the total load.

I think.

[Gaz]

Reeltarded, I may be realtarded, but you are hard to follow sometimes! So you think that's correct that a bias circuit that normally has a 25K pot should be split into two 50K pots?

[renshen1957]

Reeltarded, I may be realtarded, but you are hard to follow sometimes! So you think that's correct that a bias circuit that normally has a 25K pot should be split into two 50K pots?

Hi Gaz,

How many tubes are in the power amp circuit? Two.

If you are going from one pot biasing two tubes, to two pots (one for each tube) then add the same value pot in parallel with the first and half the resistor value (in the example above a value of 10k to 5k) This is equivalent to paralleling another resistor (10k in this case).

Adding the extra pot eliminates the need for matched tube sets (not that Fender or Marshall in their glory days ever matched tubes or purchased matched tubes). As DC balance will be better, the standing flux in the OT is reduced enhancing bass response. Less Electro magnetic radiation will be emitted to be pickup by your guitar.

Best Regards,

Steve

[Gaz]

Hi Steve,

Thanks for explanation. I've actually used the circuit a bunch of times, but never really understood how the pot value was chosen.

I recently acquired some nice 100K locking pots that I'd like to use this application, but feared that the value would be too large, but only because I couldn't find a schematic that used such a high value. I then realized that I didn't know why the value of the bias pot was selected!

I'm not certain, but I'm think I'm seeing that the bias pot is in series with with the power tubes grid leak resistors. If so, making the bias pot(s) too large would increase the grid leaks too much???

[roberto]

I mean the former. Just different ways to skin a cat? I mean, why not use a 1M pot? or a 1K pot?

You have usually two caps to be supplied, 8 to 47µF each, from a 0-50V 100mA tap. So you have to charge them fast enough to ensure bias voltage even if output tubes are overdriven, but to limit charging current well below the maximum current available. This is the reason you usually find total resistors' value around 50-100k for 6L6/EL34 amps with 50-60V bias tap.

[Gaz]

This is the reason you usually find total resistors' value around 50-100k for 6L6/EL34 amps with 50-60V bias tap.

Thanks, Roberto. I'm slowly piecing this together, but what do mean exactly by 'total; resistors'? The bias pots? Is there a particular bias schem you're thinking of? I'm having trouble visualizing how the what's what in the RC time constant. I think it'd be helpful to plug the values into a calculator to see how the charging time is affected.

I'm also baffled by this:

So you have to charge them fast enough to ensure bias voltage even if output tubes are overdriven, but to limit charging current well below the maximum current available.

Could you break that down a little bit more for me, please?

[Gaz]

For example, here's a bias circuit I've used before. I arrived the values by simply tweaking them. Honestly, I didn't know how to calculate the decoupling caps on the wipers, so I just guessed

I've observed that the caps take about 4 seconds to charge from power on, but I'd really like to know how to calculate the RC time constant, so I can see how high of pot values I can get away with.