Interstage Voltage Dividers

[defaced]

What's the proper way to do an interstage voltage divider (starting at the top of the two resistor stack):
Signal in - R - Grid - R - Ground

or

Grid - R - Signal In - R - Ground

I see both all the time, but don't understand the practical implications of doing either way. Sometimes for a turret layout, one way is easier to layout than the other, so knowing if there's a big difference between the two would be nice.

[tubeswell]

Usually you knock down the voltage, and so it wired just like a vol pot. The input from the previous stage goes to the top/input; the output for the grid of the next stage comes from the knee; and the ground goes to the bottom.

input/top (from the coupling cap in AC coupled stages)
r
e
s
i
s
t
o
r
1
knee/output (sometimes you see a grid stopper between here and the grid pin)
r
e
s
i
s
t
o
r
2
ground

Depending on your taste, you can add a small bright/bypass cap parallel with the resistor 1 2CW.

The formula for the voltage at the knee is resistor2/(resistor1 + resistor2).
E.g.; R1 = 90k and R2 = 10 k, the voltage at the knee would be 10k/(90k + 10k) = 0.1 or 10%, so the voltage at the knee would be 10% of the voltage at the input. The AC swing would be similarly knocked down.

(Is that what you were asking?)

[Tubis]

Grid - R - Signal In - R - Ground
A resistor right before the grid is not a part of an interstage voltage divider.
Thats a grid blocking resistor that will limit the grid current, introduce harder clipping and roll of high freq because of the miller capacitance between the plate and the grid. The higher the gain of the stage and the higher the resistance, the more high freq roll off.

[defaced]

Hmm, two different answers. Maybe this is a better question than I thought (I thought I was just missing some 101 level stuff). I attached a picture of an amp I'm eventually going to build that uses the two different ways on two consecutive stages. This amp, and seeing inputs wired each way has prompted this question.

A resistor right before the grid is not a part of an interstage voltage divider.

I guess this is part of what confuses me: in both cases, there is a resistor between the incoming signal and the grid of the next stage. So don't they both work as grid blocking resistor? (assuming it's actually intended for that, mounted on the socket, all that fun stuff)

[Steve Conner]

Yes, they both do. Well, the network on the left can't attenuate, but I think I know what you meant.

The "grid blocking" resistor in the Option 1 network is just the Thevenin equivalent impedance of the network, which will be about 330k, maybe 400 allowing for the plate resistance of the previous tube stage.

If you want to know more, read up on Thevenin equivalent impedance. It is a 101 level concept, if you were doing a degree course in EE.

[defaced]

From a conceptual standpoint I get the Thevenin equivalent (I took intro EE classes in college), but in practice I'm clumsy with the topic because in tube circuits I have no clue what gets stuffed into the proverbial "black box".

Well, the network on the left can't attenuate, but I think I know what you meant.

But that's wired like a volume control, which definitely does attenuate, so.....?

[Steve Conner]

My bad! I meant the network on the right The one on the left is indeed a volume control. It's option 2 that (to a first approximation) doesn't attenuate the signal.

As for what goes inside the black box: The Thevenin impedance of a voltage divider ("volume control") is just the two resistors in parallel. If it's driven from a source that has an impedance of its own, you add that into the top resistor.

Practical example: A 1M volume control set to 50% resistance has an output impedance of (500k in parallel with 500k) = 250k.

So the pot wiper looks like it has a grid stopper resistor of 250k in series. If you drive a 12AX7 grid from this, with its Miller capacitance of 100pF, the resulting low-pass filter is 3dB down at 6.4kHz. Can you see why Fender used bright caps?

Note, this only applies to the 50% setting which is the worst case.

[Joey Voltage]

My bad! I meant the network on the right The one on the left is indeed a volume control. It's option 2 that (to a first approximation) doesn't attenuate the signal.

Taking the loading effect of the grid leak out of the equation, Option 2 still attenuates the signal because the source impedance of the 39K stage still forms a potential divider with the 330K grid leak. The output impedance can actually be higher than you expect from colder biased stages with unbypassed cathodes.

[Enzo]

I could be wrong, but I don;t think you got two different answers. If I read him right, tubis was decribing the case that does not attenuate and explained what it was instead.


A plain old vanilla volume control is a classic example of voltage divider.

Imagine this: 9v battery with 100k resistor across it. That 9v should be dropping along the length of that resistor inside. We can't get in there, so let's make it two 50k in seriesw, then that across the battery. Now we expect 4.5v across each resistor.

Call one end ground for sake of argument, and we have 9v across the whole 100k and then 4.5k across one half of it. And that is what your option 1 is. In your example there is 470k and 1M in series for 1.47M. In the middle of that is the "output" at the 1M point. Whatever comes from V2a will be across that 1.47M resistance, and the grid will see 1M/1.47M of it. In other words signal will drop about a third. 9v of signal would become about 6v.


Now go back to our battery, and we still have two 50k resistors wired together. But this time, connect the battery across just one of the 50k resistors. Refer to one battery end as ground - it makes me happy to do so. If we apply the 9v across the 50k, current flows, and Ohm's Law tells us how much. We now also have the other 50k sticking out into the breeze. We sure wouldn;t expect to find 18v there. But how much current is flowing through that resistor? Not any, zero. How much voltage is dropped when zero current flows? Not any, zero.

The circuit becomes a 50k resistor across the battery, and a 50k resistor sticking out. That resistor, since no current flows through it, will have the same voltage at both ends. If there is 9v at one end, there will be 9v at the other. There is no voltage division. And that is option 2. Your signal comes out of V2b and finds itself across the 330k resistance. No current flows through that 220k, all it does it place the signal potential on the grid. No voltage reduction occurs. SO it is not a voltage divider.


I know in the real world it is not quite that simple, but for purpose of your question, I hope it helps.

[Joey Voltage]

I guess now I'm not too clear on what the question is . whether it is just a general question about voltage dividers, or if he is asking about the particular circuit he posted.

[defaced]

I could be wrong, but I don;t think you got two different answers. If I read him right, tubis was decribing the case that does not attenuate and explained what it was instead.

Gotcha. I completely missed that the first time, but now I see that's what he was saying.

The circuit becomes a 50k resistor across the battery, and a 50k resistor sticking out. That resistor, since no current flows through it, will have the same voltage at both ends. If there is 9v at one end, there will be 9v at the other. There is no voltage division. And that is option 2. Your signal comes out of V2b and finds itself across the 330k resistance. No current flows through that 220k, all it does it place the signal potential on the grid. No voltage reduction occurs. SO it is not a voltage divider.

Ok, that makes sense for the signal. So basically this is how you grid leak bias a triode, and provide grid stopping, without attenuating the incoming signal.

[Joey Voltage]

Ok, that makes sense for the signal. So basically this is how you grid leak bias a triode, and provide grid stopping, without attenuating the incoming signal.

well your not grid leak biasing, you are providing a DC reference for the grid in those examples. and essentially yes to everything else, but you will always attenuate the signal some because the source resistance (output impedence of the driving stage) forms a potential divider with the the grid leak. Most of the time you don't have to worry about it, because the grid leak is usually very big in comparison, if you use a lower load you might have significant attenuation.

[defaced]

Fancy seeing you here. Looks like I'll have to go do some reading and try to analyze the way the stage on the right is working.

[Enzo]

All the added stuff about impedances are true, I was just ignoring everything else except the voltage divider itself, to explain why one is a divider and the other not.

[J M Fahey]

one way is easier to layout than the other, so knowing if there's a big difference between the two would be nice.

Plain vanilla answer: no, they are not the same.
Left option pads gain a little, right one does not, both cut highs .
The math was explained above.
Good luck on your new handmade amp.