Technical/math question re: Fender 6G16 schematic

[RobBozic]

Hi guys,

Regarding the Fender 6G16 Brownface Vibroverb, after the second gain stage this circuit has the 22K/82K split load plate resistors, a .02 cap and then a 470K/470K voltage divider before it heads into the 3rd gain stage.

http://www.schematicheaven.com/fenderamps/vibroverb_6g16_schem.pdf

Now, if I replaced the split load with a single 100K plate resistor obviously the gain would increase, however if I increased the value of the first 470K resistor one could compensate for this increased gain.

How does one calculate this increased resistor value?

The reason I want to do this is because I want to parallel a pot over this resistor to have a variable gain pot controlling how hard I hit the third gain stage.

Thanks
Rob

[HotBluePlates]

... then a 470K/470K voltage divider before it heads into the 3rd gain stage.

The "2nd voltage divider" is not intentionally a divider; said another way, it aims to divide down both channels by the same amount because the true intention is mixing and isolation of the 2 channels.

But assume there is division going on, and that you want to simulate the amount of division present at the split plate load. stock, you have 82k and 22k, and the 22k is the portion from which the signal is taken. If you imagine the B+ cap as a ground, you'll see this as a divider much like a volume control, with the output signal taken across the 22k to "ground".

22k / (82k + 22k) = 0.2215

So we need to reduce output by about 80% (really 1 - 0.2215...). Cross-multiply to find the right resistor value.

470k / 0.78846 = X / .21154 -> Multiply the numerator of one side by the denominator of the other side, then divide by the denominator of the original side to get the unknown numerator. Refer to the equation above, then look at the numbers below to see what this really means.

470k * 0.21154 = 99423.8k, and 99423.8k / 0.78846 = ~126.1k

So your divider would be made of a 470k resistor and a 126k resistor, but the closest standard value is 120k. My use of the math approach is a little loose, so you have to do some thinking to figure what side needs the smaller valued resistor; it will be the side that should be stronger.

The split plate load is still probably a more efficient way to do this, as the reverb pot will have some effect on the division of the channels with the lowered value.

[RobBozic]

Thanks HBP,

From what I just read you lowered the value of the reverb mixer resistor from 470K to 126K. Please correct me if I'm wrong here.

This would reduce the strength of the dry signal coming from the 2nd triode into the 3rd triode and also increase the reverb gain. I don't want to increase the strength of the reverb in this instance.

Therefore the other way to do it would be to leave the 470K reverb mixer resistor alone but increase the value of the 470K attenuating resistor.

So using your math below it should be as follows

82K / (82K + 22K) = .788
1 - .788 = .2115
470K x .788 = 370360 / .2115 = 1750K

Is this correct?

Thanks
Rob

[HotBluePlates]

Yep, that looks right.

It still might be just as easy to leave in the original split plate load, unless you're working on a board for another amp without that space for an extra part.

[RobBozic]

HBP,

My amp is a Heritage Patriot which is like a 6G16 except the 3rd gaiin stage is set as per the Komet 60 & Trainwreck amps with the 10K cold biased cathode resistor. I put in the split load switch (fast / gradual) but I wanted to make it variable with a pot so I had time yesterday & came up with this, I'm hoping it will work. There's plenty of room to do this mod & even install a new pot on the front panel.

Thanks for your help with the math.
Rob

[RobBozic]

I'm not sure if I drew it correctly but I want to have the 'drive' pot go from 620K to 120K, instead of the original value of 470K.

Rob