Do these resistors on the voltage dividers also control gain?

[frank57]

Bit of a newbie question:
How do the resistors(r32 1M on soldano jet city atomic 16 clone)
and r6 470k(on hiwatt) affect the gain?
Higher value, less gain or more gain into the next stage?
They're voltage dividers with a bright cap right?

[jjasilli]

Yes, but more precisely:  A single resistor cannot be a voltage divider.  A single resistor can only subtract, NOT divide; this subtraction is called "attenuation".  A voltage divider is a circuit, or network, of at least 2 resistors: a series resistor & a shunt resistor.  So, R32 & R33 are a voltage divider.  For more info see www.aikenamps.com > tech info

Personally, I would call the caps bypass caps rather than bright caps.  In the divider circuit, the series resistor blocks highs and maybe mids too.  The bypass cap across the series resistor probably restores the original frequency mix, not just hi's.

[frank57]

I think I'm starting to get it.
So on the Hiwatt,r6 and r7 are the voltage divider?
So to calculate the value of the two resistors acting together  you would use the same formula as for
resistors in parallel?(r6 x r7 /r6+r7)
If you removed r6 and the cap then, the gain would go up?

[tubeswell]

On that hiwatt scheme R6, R7 and R8 (and R4 for lower frequencies and R5 for lower frequencies and the plate resistance of V2B) all form an AC load on V2B. Both R4 and R5 are partially-bypassed (R4 is partly bypassed by C4, which bleeds treble from V2B's output, and R5 is partly bypassed by C5, which boosts mids and higher frequencies in V2B). The lower frequencies will be 'attenuated'.

Properly speaking (as 'seen' by AC), R4 is in series with the plate resistance of V2B, and for lower frequencies, these are also both in series with R5. R7 is in parallel with all of those (to AC).  But R6 and R8 form a further voltage divider which attenuates the (AC) signal appearing at the junction of R7 and R6 by a factor of R8/(R6 + R8). C7 is just bypassing the voltage divider that is R6 and R8. That's my take on it anyhow. Others may add or subtract accordingly

[frank57]

It's fascinating stuff.
So going to 220pf on c4 would roll off treble higher up in frequency ?
One thing I tried was r8 at 47k which really started to make the amp work better
(also plate was 100k).
So probably r6 being reduced would have had a similar effect?
What happens if r7 is reduced as well with r8 at say 47k ?Less gain?

At one point I bypassed v2b by taking out c4 r4 r5 c5 and c6.
I went straight to the r6 470k resistor, but I had no gain to speak of although it did reduce the hum and buzz way down on the amp.(but I put r8 back up to 470k)Would removing r6 or lowering it bring the gain back in such a setup?

[RicharD]

What helps me is to simplify a circuit into it's resistive components only.  (see attachment).
Amps circuits typically are best dissected from output to input so let's work from right to left.  First let's see what V2A is seeing.  The input impedance of a 12AX7 is very high, greater than 10M.  This is in parallel with R8.  Since the difference is so great, we can ignore the internal input impedance and say that Zin of V2A is 470k.  Next we see that r6 & r8 are in series.  They form a voltage divider.  since their 2 values are equal, we know Vout = 1/2Vin.  This divider network only allows 50% of the signal to pass.  We can take it a couple of steps further and see how this affects the loading of V2B.  So R6 & and are in series so their combined resistance is r6 + r8 = 470k + 470k = 940k = Rt.  This is in parallel with R7 so Rl = (R7 * Rt) / (R7 + Rt) = (330k * 940k) / (330k + 940k) = 244k  Now Zsource = Ztube + RK = 1500 + 62500 = 64k.  This tells us that we want our total load not to be less than 64k, preferably about 2x that or greater.  If we assume the DC supply to have a 0 ohm impedance, then R4 is in parallel with Rl so the total load on V2B is (220k*244K) / 220k+244k) = 116k, (which is about 2x Zsource).  Knowing this number and B+ is all you need to plot a load line and finger out the gain of the stage.

R6 & R8 could easily be changed to a pot so as to vary the amount of gain reduction.  As it sits, AV of the R6 & R8 network is 0.5

[JayB]

Not to forget that the cap is bypassing one resistor setting the frequency knee of the circuit. It kind of does two things at once.

[Fresh_Start]

Butter... When looking at the load V2A "sees" (Zin), don't we need to look at R8 (470K) in parallel with the sum of R6 (470K) plus R7 (330K)?  After all, R6 is directly connected to R7 despite the presence of C7.  About 309K by my questionable calculations.

However, does the load vary depending upon frequency?  IOW for frequencies above a specific point determined by the capacitor/resistor network, does R6 stop being a factor?

Confused...

Chip

[tubeswell]

Butter... When looking at the load V2A "sees" (Zin), don't we need to look at R8 (470K) in parallel with the sum of R6 (470K) plus R7 (330K)?  After all, R6 is directly connected to R7 despite the presence of C7.  About 309K by my questionable calculations.

However, does the load vary depending upon frequency?  IOW for frequencies above a specific point determined by the capacitor/resistor network, does R6 stop being a factor?

Confused...

Chip

The output in a thevenin circuit is the knee of R6 and R8, both of which are in parallel to R7 and the unbypassed frequencies in R4, and ra, and the unbypassed frequencies in R5. And I think Buttery needs to link up the B+ side of the V2B plate resistor with the ground return in his diagram, because B+ and ground look one and the same to AC (the B+ is shorted to ground at AC, through the filter caps).

[frank57]

Here are the voltages in the area in question.
It seems like quite a complicated and interesting area.
The tracks are awfully close in that area on the pcb as well.

[RicharD]

There's about 4 different conversations going on in this thread.  First off, Chip is right, I miss calculated Zin of the 2nd stage.  We don't really care about this right now in a DC condition.  As long as the grid is somewhere between 0 and V Rk, we're fine.  With no signal it's at 0 volts.

Next I broke off the DC bias condition for the tube.  If you calculate Ztube, it will not = 62500.  Ztube will change constantly with ac signal.  Rplate is just a published average value used for some calculations.  It is not an absolute.

The next diagram shows ac loading.  The output impedance can roughly be calculated as .5(Rplate+Rk).  So if Rplate is fluctuating with ac, then so is Zout.   Effectively B+ and ground are 0 ohms apart, so the total load the tube sees is RL= R4 || R7 || (R6+R8) = 116k  The more load you put on a tube, the harder it works and the less gain you can get.  Now R6 and R8 are equal and form a divider so the point where the next stage is tapped is at .5 of Vac source.  

OK but what about that bypass cap around R6?  C7 and R8 forms a hi pass network meaning anything above a certain frequency can pass.  Obviously this attenuation curves but you can calculate the -3dB point, or 1/2 power using the equation 1/(2*pi*R*C) = 1/(6.28*470k*470pF) = 721Hz.  This is not drastically affecting our loading on the previous stage but frequencies above 721Hz are being allowed to pass more freely.  R6 is having less affect at these frequencies and they are passing at closer to Vac source.  Hence the name, "Brite Cap."

Now C6 and RL form another hi pass filter.  Using the same equation we see that f-3dB = 1/(6.28*296k*.022uF) = 25Hz.  So if we start at the plate, frequencies below 25 Hz are being attenuated through the 1st networ.  As the signal hits the voltage divider, frequencies below 721Hz are being passed more through R6 and therefore subject to the voltage divider, and frequencies above 721Hz are passing more freely through C7 and suffering less attenuation.  

  

[jjasilli]

There's about 4 different conversations going on in this thread.  Yes, let's look at the original question:  Do these resistors on the voltage dividers also control gain?  Technically the answer is No.  There are 4 arithmetic functions:  addition, subtraction, division & multiplication.  The last 3 apply to the question.  As mentioned above:  1 resistor subtracts voltage.  2 resistors arranged as a voltage divider, divide voltage.  Multiplication = gain.  Tubes multiply voltage to create gain.

The tube's gain is controlled by its characteristics & the supporting circuitry around it:  grid leak resistor; plate resistor & plate voltage; cathode resistor & cathode voltage; cathode bypass cap; and finally grid leak resistor of the following stage (because for AC or signal operation, it's in parallel with the plate resistor).  These things will give the tube, IN CIRCUIT, its multiplication, or gain factor.  If the gain factor is 10, then if you put 1 volt in, you get 10 volts out.  The gain factor operates on the input voltage, but is independent of it.

The voltage divider circuit in front of the gain stage does not CONTROL gain, per se.  Rather it presents a suitable signal voltage to the input of the gain stage so as NOT TO OVERLOAD IT.  Or for guitar amps to allow for desired overdrive, without "unduly" overloading the gain stage.

[RicharD]

Gain typically implies an increase as described but gain can also be stated in fractions or decimals.  A cathode follower typically has a gain of .9 ish.  We usually count on thumbs, round up, and call it unity gain (AV=1).  A voltage divider circuit where the 2 resistors are equal could be said to have a gain of .5  If we changed R6 to a 250k and R8 to a 750k, the network would have a gain of of .75  If we flipped those 2 values, we'd have a gain of .25  But you must take into consideration the hi pass filter formed by C7 and R8.  Above I calculated frequency -3dB to be about 720Hz.  Frequencies above this tend to pass around R6 and have less gain reduction.  The higher the frequency, the closer to AV=1.  Frequencies below 720Hz are blocked by the hi pass filter and forced to pass through R6 and become attenuated.  The lower the frequency, the closer AV comes to .5

[jjasilli]

Good point; I stand corrected.  Guess I had too narrow a concept of "gain" as applying only to active circuits.  It can properly be said that passive circuits, like a voltage divider, have a gain of less than 1. 

[frank57]

Wow. This is a very complicated part of the amp.
Now if I used 220 pf cap there, then we would be allowing signals even higher up to pass
but not below?
One thing I tried was reducing r8 to 47k to get the 3 stages more under control.
Overall the amp had a better sound.
That really killed a lot of the hum and reduced the gain,but you lost a lot of bass and mid.
It would have made more sense then to try different values on r6 and r8 together?
In the stripped down version of the amp with that whole stage stripped down(*v2b out),
you have no gain and the problem on the amp seems to be related to after the master volume.
Perhaps going into the phase splitter.
Turn it up you get hum past 1 oclock.Bigger hum the higher you go after that.

[RicharD]

Yes, changing the cap to a 220pF pushes the F -3DB point up to 1500Hz.  It's a high pass filter, frequencies above the knee are allowed to pass.  Clip lead in a 1 meg pot in place of R6 R8 and dial it to satisfaction.  Pull it, measure it, and replace with the new R6 R8 values.