The two channel amp 5C2+6SJ7 optional grid leak/cathode biased circuit

printer2

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Resistance grid to what? The switch is a on-on switch with the ground in middle terminal.
To ground.

There should only be a couple of volts across the cathode resistor of the 6SJ7.

The switch on the 6SJ7, does it switch back and forth across the width of the chassis or lengthwise? I see the two blue wires, one going to the 470k resistor (which I am guessing is the grid leak resistor) and the 2.2k resistor. I can not see the switch clearly. Does it have three terminals or six terminals? The blue wires are suppose to be on either side of the center terminal that goes to ground.

IMG_9040 (2).JPG
 
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Jerry garrcia

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I can not see the switch clearly. Does it have three terminals or six terminals?
Starting without any power application to the amp.
The switch is a three pole.


It is is not going length wise. Middle terminal to ground. Right terminal connection from turret closer to the socket that the cathode resistor/cap is connected to. The left terminal connected to 470K resistor connected to the grid.
Resistance/Volt values 6SJ7:
Switch towards the speaker
Plate: O.L. / 193 V
Grid: 4.9M / -364 mV
Screen: O.L. / 24.6 V
Cathode: 0 / -0.1 mV

Switch towards back
Plate: O.L. / 196 V
Grid: 398K / 2.4 V
Screen: O.L. / 26.7 V
Cathode: 2.2K / 2.8 V

It’s very quiet in Pentode mode
Regardless of bias switch position and quite high B+
 

Jerry garrcia

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Now all of a sudden the pentode channel went quiet… max volume on guitar and on amp. Ok volume. Like on 2-3 on triode channel. Then quiet. No sparkling or anything. Changed tube but no different. Checked voltages on the pins and was the same. Chinastick and nothing. Something is rotten in the state of Denmark.
Anybody know what to do? Was something wrong from the beginning due to low output volume and now this. Strange.

Edit: bad solder joint at the volume pot. Now sound but still low on pentode channel
 
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printer2

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To ground.

Starting without any power application to the amp.
The switch is a three pole.


It is is not going length wise. Middle terminal to ground. Right terminal connection from turret closer to the socket that the cathode resistor/cap is connected to. The left terminal connected to 470K resistor connected to the grid.
Resistance/Volt values 6SJ7:
Switch towards the speaker
Plate: O.L. / 193 V
Grid: 4.9M / -364 mV
Screen: O.L. / 24.6 V
Cathode: 0 / -0.1 mV

Switch towards back
Plate: O.L. / 196 V
Grid: 398K / 2.4 V
Screen: O.L. / 26.7 V
Cathode: 2.2K / 2.8 V

It’s very quiet in Pentode mode
Regardless of bias switch position and quite high B+
Understand the switch now, never seen a double throw with both terminals on the same side. The screen resistor should be 1M acording to your schematic. This is in between the 470k resistor and the 4.7M. The meter should be able to read it and not give an overload display.
 

Jerry garrcia

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The meter should be able to read it and not give an overload display.
That is correct. Have it in place and connected to the right pin. Pin 6 of the 6sj7. A bad solder joint? But when I measure resistance after the screen cap it’s 0 R. When measuring resistance from the turret that the screen resistor and cap connects to it’s still O.L. Shouldn’t it be that? The 1 M resistor connects to the preamp node up stream
DAECE28D-6E9F-48B5-888A-61B4A8E435A0.jpeg

Could that be the problem that explain the low volume? Also quite high B+ to the preamps. Should it be lowered? Up the voltage dropping resistor?
 
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printer2

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Pin 6 and across the 1M resistor with the amp off and the capacitors discharged. After you get a reading measure the voltage from pin 6 to ground (roughly 30V) and then the voltage across the 1M resistor.
 

printer2

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That is correct. Have it in place and connected to the right pin. Pin 6 of the 6sj7. A bad solder joint? But when I measure resistance after the screen cap it’s 0 R. When measuring resistance from the turret that the screen resistor and cap connects to it’s still O.L. Shouldn’t it be that? The 1 M resistor connects to the preamp node up stream
Pin 6 to 'after the screen cap', what point is after the screen cap? Do you mean pin 6 to the junction of the 22nF screen cap and the 1M screen resistor? From the screen you will get zero ohms.
 

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Pin 6 and across the 1M resistor with the amp off and the capacitors discharged. After you get a reading measure the voltage from pin 6 to ground (roughly 30V) and then the voltage across the 1M resistor.
Pin 6 across the 1M shows 1.06M
Pin 6 to ground shows 26.7V.
Voltage before 1M shows 290V
Voltage after 1M shows 26.8V
Pin 6 to 'after the screen cap', what point is after the screen cap? Do you mean pin 6 to the junction of the 22nF screen cap and the 1M screen resistor? From the screen you will get zero ohms.
just to see if the ground connection was correct. Rest language barrier I suppose.
 

printer2

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Try unsoldering the wire from the 0.1uF coupling capacitor for the 6SJ7 and unsolder the wire going to the 6V6 grid leak resistor. So the coupling cap is connected to the 6V6. See what the volume is like without the mixing resistors, no volume control. Just the 6SJ7 straight into the 6V6 grid stopper.
 

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Try unsoldering the wire from the 0.1uF coupling capacitor for the 6SJ7 and unsolder the wire going to the 6V6 grid leak resistor. So the coupling cap is connected to the 6V6. See what the volume is like without the mixing resistors, no volume control. Just the 6SJ7 straight into the 6V6 grid stopper.
To much work that clouds my amp project. I’m in a big messy organisation that needs some delicate manoeuvring to not create a massive chaos.

Why didn’t I think of this 🤯. So by connecting the coupling cap of the 6sj7 directly to the 1.5K grid stopper of the 6v6, I’ll exclude the two pots, mixing resistors and grid leak resistor.
If it functions well, regarding to output and bias switch, I’ll know that the problem lies in that chain.
Tack!
 

Jerry garrcia

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Connected coupling cap from pentode to pin6 of 6v6. From pin 6 a 1.5K resistor connection to grid, pin 5. Good output volume controllable by volume knob of guitar.
So problem is ether mixing resistor, pot or bad soldering?

A strange thing is that when I plug in guitar in triode input jack there is a tiny output volume from speaker. It shouldn’t be any connection to that one. Volume very low.
81859327-9DB2-4886-B0EE-8E7050614E9E.jpeg
81859327-9DB2-4886-B0EE-8E7050614E9E.jpeg
 

Jerry garrcia

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I will change the pot and reflow the solder to the pin connections first to see if it starts to work. If not mixing resistor?
Also need to put some heat shrink on the volume pot by the bias switch since it’s really close and sometimes touch the metal of the switch.

@printer2 regarding the cathode resistor bypass cap I wonder why all 6SC7 designs have a lower value cap than you suggested? All of the one I found have a cap of 6.6-25 uF. Now I have already put in a 220 uF cap but this question is just for learning.
 

printer2

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I will change the pot and reflow the solder to the pin connections first to see if it starts to work. If not mixing resistor?
Also need to put some heat shrink on the volume pot by the bias switch since it’s really close and sometimes touch the metal of the switch.

@printer2 regarding the cathode resistor bypass cap I wonder why all 6SC7 designs have a lower value cap than you suggested? All of the one I found have a cap of 6.6-25 uF. Now I have already put in a 220 uF cap but this question is just for learning.
What the capacitor does is shunts the AC component around the resistor to ground. Without a capacitor the voltage developed across the resistor changes according to the change in current. Stay with me, I will try to keep this short as possible (you really should understand this anyway).

localfeedbackloadline1.jpg


Here is a load line chart for a 12AX7 with a load line for a 100k plate resistor. The bottom scale is the voltage on the plate, the left hand side the current through the triode, the lines at an angle are the grid bias voltage. To draw the load line of the plate resistor we use what we know of Ohm's law to draw the line. We know that if we go more negative on the grid bias we reduce how much current going through the tube. At some point where the grid bias is enough top stop all current flow the line hits the Va (voltage anode) line. On the other end when the voltage across the cathode to plate is 0V it is like the triode is removed from the circuit. In that case we only have the 100k plate resistor in the circuit. (I am going to shift the blue line up a little to make the math more palatable).

Say the supply voltage is 300V. With a very high resistance there is no current flow (triode is biased off), we are at 0 mA at 300V. With zero resistance with just the 100k resistor we have 300V / 100,000R = 3mA. So on the chart we draw our blue load line using these two points. Now we look at the grid lines and we decide where to set the bias. If we have a 2V peak to peak signal setting the grid bias to 1V gives us the most linear amplifications (least distortion) as the line segments are roughly the same length. The -1.5V - 2.0V segment is a little shorter and will cause a little distortion, so is life.

Which is all great, but what does this have to do with the capacitor? As the voltage goes from -1.0V (our bias point) to 0V and to -2.0V the current changes from 0.65mA to 1.9mA. This change in current causes a change in voltage across our cathode resistor. Just because I have my calculator out I'll find the change in voltage at the cathode. 1.9mA - 0.65mA = 1.25mA Say we are using a 1.5k cathode resistor we can figure out how much the change in current causes the voltage on the cathode to change. 1.25mA x 1.5k = 0.0015 x 1500 = 2.25V. The voltage at the cathode will change 56% of our signal voltage (4.0V Peak-Peak).

So what? Well, the voltage developed at the cathode by the change in current is opposite to the input voltage. It opposes the signal voltage at the grid and causes the gain to be lower. It is called localized NFB. To get the amount of gain that the tube can give use with the components we are using we put a capacitor across the cathode resistor. When we do this the AC change in voltage gets shunted around the cathode resistor and we have the maximum gain.


The larger the capacitor the lower the frequency the capacitor shunts the AC around the cathode resistor. A 22uF capacitor will usually be good for around 20 Hz (depends on the cathode resistor, I won't throw more math at you). If we want less bass going through the triode we reduce the capacitor value, the NFB that develops in the cathode resistor starts to work below where the capacitor becomes less effective. So we may use a 2.2uF or 0.68 uF to reduce the bass, the 0.68uF capacitor letting the NFB to operate higher up in frequency than the 2.2uF or the 22uF.

So back to your original question. Why a 220uF capacitor than a 22uF? Because the cathode is shared in the tube you are using the change in current of the one stage effects the other. Above where the capacitor is effective the two stages can do their thing without one knowing what the other is doing. But the 22uF capacitor is less effective below 20 Hz, at 1-2 Hz it has almost no effect at all. And this is where things get ugly. Because the second stage gets an inverted signal from the first stage plate, the AC signal it develops across the cathode resistor is fed back to the first stage. Basically the two stages together acts as a low frequency oscillator. What you will get is a thing called motorboating.

(Fun fact, just googled mottorboating and got the following - The act of placing one's head between a woman's breasts and making the sound of a motorboat with one's lips whilst moving the head from side to side.) You would get ta thump, thump, thump... at about 0.5 - 2 Hz. The way you stop that is to make the cathode capacitor larger so it is effective even at 1 Hz. So rather than a 22uF capacitor we use a 220 uF capacitor. This should reduce the motorboating to 0.1 Hz or less. This would not be audible but may still cause problems. But other factors reduces the gain down there and it is not a problem.

You know, I probably should have just stuck with the really short answer.

As far as what part of the volume or mixing resistors are the issue, you just need to do the same thing as you have done. Reduce the circuit to its simplest configuration (you did that with going straight to the 6V6) next either add the volume control or the mixing resistors. The add the other section until something messes things up. This is the part where it is hard to give advice in troubleshooting remotely. It can be one device, section, solder joint. Any one of them or more in concert can be the problem.
 
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Jerry garrcia

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The act of placing one's head between a woman's breasts and making the sound of a motorboat with one's lips whilst moving the head from side to side.)
I’ll focus on the most important. Remember the term motorboating from the movie “Wedding Crashers”. Hoped it wasn’t the same 😀.
Thank you for the explanation! Been reading all these books but a post from you covers about five chapters.

Been troubleshooting this evening. At the country house and couldn’t let go of the amp. Told the wife I had to shop some food and drove 200 km back and forward to get the amp.
Think there is something strange in there. The pentode in this circuit has much lower output volume than in the rest of the 6SJ7 circuits I built.
1) Connected it directly on the grid of the 6v6 and nice tone but still lower output than previous builds.
2) Reconnected the bias switch. Made it the wrong way first. Now the switch functions well.
3) connected the coupling cap of the pentode after the mixing resistors and still a bit low output of it but the triode had high output volume.
4) connected pentode coupling cap before it’s mixing resistor and then the triode output volume was lower than before and the same output to low volume of the pentode. The volume is ok for home use but lower than it should. Also strange that the triode lost a lot of output volume.
5) tried a new pot and no difference.
6) both tone circuits works fine.

So the main problems are:
- to little gain from the pentode.
- triode lose output volume when the pentode is connected to its mixing resistor.
- also strange that when the pentode was connected directly to the grid of the 6v6 there was some tiny volume from the triode when guitar was connected to the triode input jack. 🤔

Will need to do some more troubleshooting tomorrow. Have to see if the circuit differs much from previous builds. One thing is that the B+3 is about 40V higher than in builds before.
 

Jerry garrcia

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Did some new measurements. No cord in input jacks.
B+ 362V
B+2 354V
B+3 289V

Resistance/voltage to ground.
Resistance/Volt values 6SJ7:
Bias switch towards the speaker
Plate: 10M / 211 V
Grid: O.L / 5 mV
Screen: 11.8M / 52.5 V
Cathode: 2.2K / 2.4 V

Bias switch towards back
Plate: 10M / 196 V
Grid: O.L / -1.1 V
Screen: 11.8M / 31 V
Cathode: 0R / 0 V

Resistance/Volt values 6SC7:
Plate 1: 10.5M / 175 V
Grid 1: 65K / 0V
Plate 2: 10M / 151 V
Grid 2: 67K / 0.4 V
Cathode: 2.2K / 2.3V

Resistance/Volt values 6V6:
Plate: 10M / 333 V
Grid: 127K / 0V
Screen: 10M / 346 V
Cathode: 272R / 17.5 V

Quite high voltages on preamp section. Increase resistor between B+2/B+3?
 

printer2

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You could lower the voltages if you want to match the other amps. You should be able to calculate how much resistance (more or less) with the posts I gave. The cathode bias/grid leak bias switch seems to be doing its stuff from the difference in voltages with each other, good to see. Have you noticed much difference between the sound of the two modes? I need to get around and trying it but just too busy with the house work. One thing you could do also to get a little more gain is change the 100k plate resistor to 220k.

The drop in gain with the triode and the mixing resistors is because the two 220k resistors are in series and the one goes to ground through the other's volume control and plate (which goes through the filter capacitor to ground). It splits the output of the triode (or pentode, whichever you are using as a reference) across the two mixing resistors and the voltage of the signal is cut in half as compared to just one of the tubes going straight to the 6V6. Just how it works to isolate them from each other so that changing one volume control does not effect the other channel too much. You could also go to 470k resistors, Marshall uses them instead of 220k for a little more gain. The higher value loads down the signal a little less.

I don't know why this amp is performing different than the others. I would do the same preamp measurements on the other amp and compare them with this one. It is also a good idea to have a schematic or layout with the voltages to use at a future date when something changes in the amp. It would be good to do AC measurements through each stage. Maybe make a boost pedal that you can use as a buffer and then output a sine wave (400-1kHz) from a digital source like a wave file. Rather than headphones, plug it into the buffer and then the amp input. The buffer is only to protect the signal source, normally nothing will feed back into the audio source but who wants to chance it?


Ran some numbers, the 2.2k cathode resistor might like a 220k plat better than the 100k. That or trying (and it is easier to do) tack on another resistor in parallel with the 2.2 (with the 100k) and bring the cathode resistor down around 800-1,000 ohms.
 
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Jerry garrcia

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Just how it works to isolate them from each other so that changing one volume control does not effect the other channel too much.
Didn’t really get you there.

Will change the mixing resistors to 470K’s and see the effect.

Also reduce the cathode resistor value to 892 Ohm by adding a 1.5K resistor in parallel with the 2.2K. Progress.

Thought this one would be a walk in the park. Maybe easier to build from known circuits…
 
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Jerry garrcia

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So it’s complete I guess. Changed the mixing resistors to 470K which gave the triode a bit of a boost. When that channel is cranked up it really rocks LOUD with a nice distortion. Below 5 a clean, warm and soft tone.
The pentode is still the ugly duckling. When the volume is at 100% it is about the same volume as the triode on 30%. Changed the cathode resistor value to 890R. Tiny difference but not an enormous effect on the output volume. Checked everything else and changed tubes but minor differences. May have to use that boost pedal in front of that channel, after the AB box.
Irritating.

@printer2 the 470R cathode resistor for the 6v6 was the right call. With the 270R the 6v6 ran at >20W. With the 470R it went down to around 14W.

Checked previous builds. Only difference of the pentode circuit is that in both previous builds the only difference are the caps values. 0.022 uF in them and now a mix of 0.1, 0.047 and 0.022uF

Edit:
It really drives me crazy since as these two preamp tubes (especially with the bias switch that really “tames” the pentode compared when it is in grid leak bias) together with the WG8C speaker, delivers everything one would want from a 5W octal preamp amp. Except for the low gain pentode, so it is impossible to drive it to dirty distortion. In previous builds the pentode produced high gain. As always “Something is rotten in the state of Denmark”.
 
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printer2

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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.

4qkNvvR - Imgur.png

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.
 




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