Calculating the value for a safety resistor

[ted01]

Hey Gang,

In TUT, KO'C suggests the inclusion of a safety resistor strapped across the bias pot. How would I go about calculating the proper value for that resistor? Here's the bias circuit I'm using. I added the second pot to give me discrete control over the bias for each tube. I know many of you can just look at this and give me a safe value, but if you don't mind taking a minute, explain to me how you came up with that value. Pretend I'm dumber than dirt (not so far from true!) and take me step by step.

Thanks,
Ted

[Structo]

I believe the resistor is a safety feature in case the bias pot fails and puts a load on it to the tube to prevent runaway power tubes.

As far as how to calculate it I don't have a clue.

[Phil_S]

This is a basic application of Ohm's Law. Resistors wired in series add and in paralle they divide (look it up; it's more complex than that, the sum of the reciprocals = the reciprocal of total).

The pot will give you 5K on each side of the wiper and 10K in the middle. So you get a range from 0 to 10K to 0.

If you just have the pot + the fixed R, it is 10K + 15 K, giving you a range of 0K+15K=15K to 10K+15K=25K.

If you add a 5K resistor between one leg and the wiper, you are limiting the range of the pot. When you sweep between that leg and the mid point, you will have between 0 and 5K provided by the pot and 5K provided by the fixed R in parallel. For this purpose, the two divide (let's not split hairs, it is more complex), so your effective resistance on that side of the pot is a max of 2.5K. At the mid point, you also have the 5K from the other leg to the wiper, for a total max on the pot of 7.5K. As you move the wiper towards the fixed R, the resistance will decrease to as low as 2.5K total (what is provided by the other side of the pot and the fixed R to the wiper).

You will need to fool with the calculation to see what gives you the desired range of bias voltage. You may want to increase the fixed R to 22K to compensate. Or you may want to use 10K fixed on the pot which will give you 3.3K max on the side with the fixed R and total of 3.3K + 5K or 8.3K.

You can fool around with this out of circuit with a pot and some R's and your meter to help understand how it works. This wasn't so easy to write and I'm not sure how clear it is.

As for doing it? I wouldn't bother. The fixed R will always give you a minimum of 15K if the pot goes open and if it goes infinite, then your tube bias will be so cold it won't operate. The only possible problem I see is if the wire comes off the first leg of the pot so it loses it's bias resistance all together. If you make it that poorly, well,...I won't say something unkind.

[martin manning]

The failure that the safety resistor is protecting against is the pot's wiper losing electrical contact with its element. If that happens, the grid voltage is on average zero, and there will be excessive current flow through the power tubes and the OT.

In normal operation, the bias voltage is set by the voltage divider made up of the resistance from the diode to the wiper followed by the resistance from the wiper to ground (which includes the fixed resistor).

When the wiper is at the top of the pot, the safety resistor has no effect since it is shorted, and the bias voltage is equal to the voltage at the diode.

When the wiper is at the bottom of the pot, the resistance from the diode to the wiper is the parallel combination of the pot's element and the safety resistor, so the bias voltage (using the voltage divider equation) is the voltage at the diode times Rfixed/(Rpot||Rsafe + Rfixed), where Rpot||Rsafe is the parallel combination Rpot*Rsafe/(Rpot + Rsafe).

Now, if the pot fails by the wiper losing contact, the grid is still connected to the diode through the safety resistor, and the bias voltage will effectively be the voltage at the diode (max negative bias) regardless of the value of the safety resistor, since the current through it will be negligibly small.

Selecting a large value for the safety resistor will protect against failure of the bias pot and minimize the effect on the range of bias. If you choose a 100K, say, the parallel combination with a 10K pot's element will be 9K1.

Suppose there is -60V at the diode, a 10K pot with a 100K safety resistor, and a 15K fixed resistor. The range of bias voltage without the safety resistor is -36 to -60V, and with a 100K safety the range will be -37.3 to -60V.

MPM

[Structo]

Thanks Martin.
That was my take on it as well, I just can't explain the why very well.

Is the way that diagram drawn the correct way to implement the safety resistor?

And what do you think the odds are of a pot failing in that manner?

Have you seen a bias pot fail like that and take out tubes or transformers?

[martin manning]

I haven't looked it up in TUT, but it looks right. The failure is the same as when a pot becomes "noisy," so it certainly could happen. On the other hand, there are milions of amps out there that are not wired "fail-safe." Maybe one of the pros can comment on how often it happens.

There are other ways besides the one shown, but what I like about this one is that the bias circuit current is nearly constant throughout the bias adjustment range and current flows through the entire resistance element (not the case if you wire the pot as a variable resistor) . It's easy to implement and a ten-cent resistor seems like pretty cheap insurance.

MPM

Note I was a little sloppy above. The safety resistor will increase the current flowing to ground a little so the voltage at the diode will come up slightly, but not enough to worry about.

A much bigger effect results from adding a second bias pot and fixed resistor. That will halve the resistance from the diode to ground if the values used are the same as for the single control. The resistances of those parts should be doubled to keep the voltages and ranges more or less the same.

[ted01]

Hey Phil and Martin,

Thanks a zillion for your replies. It totally makes sense to me now. I'll be using just what you suggested Martin, a 100K resistor across the pot. After I sat down, digested what you said and then broke out a pencil and paper and ran through a whole mess of combinations using the appropriate formulae, the light really came on. Phil, I did what you suggested as well, got out several different pots and a nice random variety of values of fixed resistors and just played with 'em along with my DMM. I think this is one concept I can say with some degree of confidence "I got it!!". 'Course that doesn't mean much, odds are I'll forget if I don't use it more. I'm definitely one of those "use it or loose it" people.

Ted