Princeton Reverb adjustable bias mod question?

[kgstewar]

Hello,

I'm building a Mojotone Blackface Princeton Reverb kit and plan on installing an adjustable bias control. As most of you know, this circuit comes with a non-adjustable bias, controlled by a single 22K resistor, which I assume would yield a plate current near 20mA.

Most of the mods I have seen for the adjustable bias include a 10K pot with a 22K or 25K fixed resistor, which yields a minimum resistance of 22K or 25K and a maximum resistance of 32K or 35K.

It's my understanding that the lower the resistance, the lower the resulting plate current. Using the values in the previous sentence, then, plate current could be a minimum of ~20mA (total resistance = 22K) but turning the pot would create higher resistance and higher plate current.

Given the fixed resistor values in most of the mods I've seen (22k or 25K), the adjustable bias circuit is capable of creating higher (>20mA) plate currents, but not lower. Does it make sense to use a lower value for the fixed resistor (18K or 20K?) in this mod so that the adjustment can create plate currents that are potentially lower or higher than the plate currents that result from the non-adjustable 22K resistor in the original Princeton circuit?

Many thanks in advance, and there is a great probability that I am completely misunderstanding this part of the circuit.

[stratomaster]

You're misunderstanding the circuit.  That part of the circuit is essentially a voltage divider.  There are a number of ways to skin this cat, and the one that you mentioned with the 10k pot will get you there.

The point of this part of the circuit is to provide a negative voltage reference to the grid that will set the proper operating point of the tube given the other circuit parameters--including the specific properties of the tube.  This is why we adjust bias so suit a new set of the same tube type.

The point of the fixed resistor along with the pot is to set the range of available negative voltages to accommodate the expected tubes but prevent the user from going too positive and damaging the tubes. 

In the case of the 10kpot and 22k resistor you should be able to adjust from maximum negative to about a third of the way to 0.  So if maximum negative is -70v then you should be able to adjust to about -47v before running out of range.  While this may not be enough range, it's not because the sum of the resistance exceeded the original value but because the ratio of adjustable resistance was too low. At that point you can either increase the pot to 20k or decrease the resistor to say 15k.  Since resistors are cheaper than pots, that is usually what people do. 

[sluckey]

It's my understanding that the lower the resistance, the lower the resulting plate current.

That's wrong. Lower resistance will give HIGHER plate current.

My preference would be to change the 22K resistor to 15K and put a 25K pot in series. Then the resistance swing would be 15K to 40K and will allow you to set the plate current above and below what you would have with just a 22K resistor.

[stratomaster]

To avoid confusion.  Sluckey's answer refers to replacing the existing resistor with the pot and resistor to ground with the bias voltage taken off the input to the pot.  There is no voltage divider in this arrangement. 

My answer takes the voltage off the wiper.

His way is more reliable in that it will fail safe, but relies on the values of the resistor/pot.

My way is how it was traditionally done in Fenders and relies more on the ratios.   

[PRR]

...I assume would yield a plate current near 20mA....

Why assume? Did you measure?

> no voltage divider in this

There "is a divider" hidden in the 100k and 22k parts. Not simple to calculate because one leg has DC and the other is pulsating half-AC. Of course making the 22k smaller (or tapped) runs the 6V6es hotter.

We should have posted the schematic in the beginning (there's more than one way to do it). I will include a snip of the Mojotone docs.

But be sure you want and need to run the 6V6es hotter!! They are already far past the original limits. Many-many tube amps (not just guitar, not just Fender) have lived long happy lives with un-adjustable bias. More than a few tubes, and some transformers, have been melted when bias goes wrong.

[kgstewar]

...I assume would yield a plate current near 20mA....

Why assume? Did you measure?

> no voltage divider in this

There "is a divider" hidden in the 100k and 22k parts. Not simple to calculate because one leg has DC and the other is pulsating half-AC. Of course making the 22k smaller (or tapped) runs the 6V6es hotter.

We should have posted the schematic in the beginning (there's more than one way to do it). I will include a snip of the Mojotone docs.

But be sure you want and need to run the 6V6es hotter!! They are already far past the original limits. Many-many tube amps (not just guitar, not just Fender) have lived long happy lives with un-adjustable bias. More than a few tubes, and some transformers, have been melted when bias goes wrong.

Many thanks for this and other answers.

No, have not measured the plate current (still building the amp). I assumed around 20mA plate current because the 6V6 tubes from mojotone have 20mA written on them (matched?) and so I figured that was the target plate current with their circuit. Maybe not.

My desire is to not run the tubes hotter, but rather to be able to adjust the bias easily if I need new tubes at some point or if I wish to run the existing tubes cooler than the plate current that results from the fixed 22K resistor.

Greatly appreciate all the answers and realize I need to take a step back and ask a more-basic question: Would a -60V grid voltage result in a higher or lower plate current in this amp than say a -50V grid voltage? I had thought -60V would yield a lower plate current (more negative results in more blocking of electrons resulting in lower current) but based on previous answers, I'm not thinking about this correctly.

[sluckey]

Greatly appreciate all the answers and realize I need to take a step back and ask a more-basic question: Would a -60V grid voltage result in a higher or lower plate current in this amp than say a -50V grid voltage? I had thought -60V would yield a lower plate current (more negative results in more blocking of electrons resulting in lower current) but based on previous answers, I'm not thinking about this correctly.

That is correct. The part you had wrong was "It's my understanding that the lower the resistance, the lower the resulting plate current." LOWER resistance means LOWER negative bias voltage which means MORE plate current.

Now if we were talking about the 100K (many of us adjust this resistor instead of the 22K) bias range resistor, just the opposite is true. Lower resistance means more negative bias voltage which means less plate current. Hoffman's schematic may be easier to read. Take a look...

     https://el34world.com/Hoffman/files/Hoffman_PrincetonReverb.pdf

[kgstewar]

Many thanks for your reply and the beautiful schematic. Now I think I'm getting it, and apologies in advance for my casual, non-electronic terminology in what follows.

The 100K resistor, if increased, diminishes the grid voltage since it's inline with the bias tap. It's a porous dam and the less porosity (higher resistance) the less grid voltage reaches the tube, which results in higher plate current.

The 22K resistor is also a porous dam, but it siphons off voltage to ground. The more porous (lower resistance), the more voltage gets siphoned off before reaching the grid. Lower grid voltage yields higher plate current.

I realize that my conclusions are just what you said (thanks!) but I wanted to see if I could come up with a model (which may be wrong!) that explains the behavior of the two resistors.