Holy cow! I think I’ll need to bring out my defibrillator, make a bilateral temporal bipolar brain shock to reset it all after this reading. Was flipping through “War and peace” to get in the right mood for a good night sleep. Now it feels like I hung out with Ken Kesey at an illegal rave party. How will I be able to sleep after this?A picture helps. Beside the triode and the pentode symbols I put the sample internal impedance of the tubes, 68k for the triode and 200k for the pentode. At the second stage of the triode we have a pair of 200k resistors. At point A the AC impedance of the circuit has the 100k resistor goes to ground through the filter capacitor. It also has the internal resistance of the triode in series with the 1.5k resistor. The 68k and 1.5k equals 69.5k (the internal impedance of the tube is different for different triode types and can change depending on how much current is going through the tube). This 69.5k resistance is in parallel with the 100k plate resistor. the two parallel resistances combined have a equivalent resistance of 41k.
Say the triode at point A is making a signal voltage of 20V. It has a load of the following resistors, 220k and 220k, 440k. Now we calculated the impedance of the first stage as an equivalent resistance of 41k. This is the source resistance, the 440k is the load resistance. Without the load resistance the first stage is making 20V and if we measure the voltage at point A we will read 20V. But if we connect the 440k to it we now have current flowing through the circuit. The total resistance would be 41k and 440k in series for a total of 481k. We use Ohm's law to calculate the current through the loop. 20V / 481k = 0.041mA (or 0.000041A, it is best to use Amps and resistance values in ohms to calculate with.
Now we take either the 41k or the 440k to calculate one of the voltages across a resistor, I will pick the 440k. 440k x 0.000041 = 18.1V, this is the voltage across point A to ground The original voltage the triode could produce without a load was 20V but since the triode circuit has its own internal resistance and a current going through a resistance causes a voltage drop, 20V - 18.1V = 1.9V is dropped internally in the triode. The lower the resistance of the load resistance the more voltage that gets dropped internally in the triode. If you had a short to ground the only resistance in the circuit would be the equivalent resistance of 41k and Ohm's law says all the voltage will be dropped internally and point A would be at 0V. So with no load the open circuit voltage would be 20V, a short would be 0V and and as the load resistance goes up more of the voltage appears at point A and across the load resistor. This is why we use higher resistances in order to retain as much voltage as possible. When you went from 220k resistors to 470k resistors you end up losing less signal voltage.
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Now the easy part, the output of the first triode goes to the top of the first 220k resistor and the voltage is split between the two resistors. So if we have 18V at the top resistors and since they are the same value we will have 9V at point C.
Now jumping to point E. Just to make it easier we are going to say the pentode is putting out 10V at the top of the pot. And we have the bottom pot at 0%. In this case we have the voltage split equally across the 220k resistors with 5V at point E. If we turn the top pot up to 100% we have that point to ground with our equivalent resistance of 41k. So the pentode circuit will see 220k + 220k + 41k = 481k. At point E we with then have the 10V signal split between the 220k resistor and the 220k + 41k. And using the voltage divider rule we have 10V x (261k / 481k) = 5.4V.
So the mixing resistors act as a voltage divider. If the top pot is set at 50% and I am going to wave my magic wand at the triode resistance equivalent and say it is 0 ohms rather than 41k, just to make the math easier. So the 1M pot at 50% we have two 500k resistances in parallel, which is 250k.
So now with the pentode pot at 100% putting out 10V we have the 10V split across 220k, 220k, 250k. With the triode side of point E we have 10V x ( (220k + 250k) / (220k + 220k + 250) ) = 10V x (470k / 690k) = 6.8V
So in the end we will end up with roughly 5.4 to 6.8V of the 10V signal. Now if we did not have the mixing resistors we could get the maximum voltage from the pentode side when the triode side pot is set to the middle of the resistance range of 500k to ground (remember 50% of rotation of a log pot is not 500k). Now if you turn the triode pot to 0% all the signal from the pentode is shorted to ground and you will have no signal at point E. If you have the triode pot at 100% you will effectively shorted out the signal with the 41k load on the pentode side.
So the mixing resistors is a necessary evil if you just want to use the volume controls and have them operate as you would expect. But if you have the opposite pot turned up or down you would have little signal from the channel you are using and you may think something is wrong. You could turn up the pentode channel to a point and then rotate the triode pot to find the point where you have an equal resistance on either side of the pot and therefor not load down the pentode channel.
You can jumper across the two 220k mixing resistors and see how the signal of the one channel can be loaded down by turning down, or up, the other volume control.
A lot here again, once you see the concept you start seeing how the AC signal is effected by the other parts of the circuit.
The pentode side should be operating as the other amps (if you have the switch on the grid leak bias circuit). Something is causing it not to it seems.
Will need to read it a couple of times more and I guess my patients tomorrow will meet an absent doctor flipping through his old high school physic’s books.
So for the layman:
- tomorrow I’ll will update the schematic with all the changed stuff I’ve done.
- start up the amp and plug a guitar in to the pentode jack. Turn the volume at a 100% of the volume pot of the pentode. Then playing while I’ll adjust the volume pot of the triode to see if I’ll get a increase in output volume of the pentode and try to find the sweet spot?
- then place a jump wire directly from the volume pot, of the chosen channel, to the turret that connects to 470K grid leak resistor of the 6v6. Then do the same as above and repeat it with the other channel (the un-bypassed mixing resistor channel is still connected to the 470K (previous 220K) mixing resistor from the volume pot)?
One thing that struck my mind. One thing that differs from this build than from the previous, that includes a high gain 6SJ7 circuit) is that in this build 1) both channels has their separate tone control and
2) the previous builds with two different channels the both has an extra “gain stage” (if you could call the paraphrase phase inverter a gain stage).
Or didn’t I choose this version since the triode had to little output volume ?
Don’t remember now and will have to check. Ether way in the last two schematics the output volume stuff was that the triode had a higher volume than the pentode. Now it is the opposite….
In this build this schematic is almost correct except for the grid leak resistor for the pentode is 470K, cathode resistor 890R. For the triode there is an added 1M grid leak resistor. For the 6v6 the cathode resistor is 470R and increased the resistors between the filter caps and increased the mixing resistors to 470K ones. Must really update the schematic…
Strange and will now contemplate what you @printer2 really wrote.
And if anybody has any good idea what could be the cause of this and what to make of the tone circuit of the reverb amp I’m all ears. Actually thinking about skipping it since it now mostly make something great a bit less great. Maybe a bias switch instead?