kuDo wrote: ↑Tue Nov 03, 2020 10:44 pm
The signal from the tone stack is around 110v? Wow that higher than I thought.
That is indeed way to hot to pass through to the effects. Is this setup how you personally have build and used it?
The signal needs to be attenuated a lot. Doesn’t that mean that the biggest part of the pot is unusable?
The signal needs to go down to input signal lvel right? What is that, about 150mV?
So the impedance will be around 1M with the pot in place, that’s oke, because there was already a pot in place. The master volume. Seems logical.
But if I attenuate, the recovery stage needs to be able to get the signal back to 110 volts again.
The 110Vp is the worst case and it is at 82Hz and it is the negative side of the signal only. You are more likely to get 80Vp with the usual settings.
I have built and used loops with this topology, but the values may have been slightly different.
If there is a small signal coming out of the tone stack and you want the same level at the effects send, then you will use the entire sweep of the send pot.
I want the signal to be as low as 100mVp at the effects send.
This loop isn't unity gain, but the signal from the recovery stage is capable of way more than what is needed to overdrive the output stage of the amp.
kuDo wrote: ↑Tue Nov 03, 2020 11:06 pm
Am I right when I say that you can’t see the effect of the feedback within the CF? As in, when you put signal in, the graph says it will amplify, but I know the signal won’t get stronger than the amplitude that gets in due to the feedback.
The graph says that the CF has the same gain as a common cathode gain stage with the same components. This is true. The difference is that the grid to ground voltage has to be way higher for a CF in order to get the same grid to cathode voltage.
Okay, so one other thing. Just to see if I understand correctly.
So how can I use the Load line I have drawn to see if the stage has enough to feed the loop?
The AC loadline has been created with the parallel value of the cathode resistance, pot resistance and input resistance of the load. This now gives the possibility to see what the max voltage swing can be in relation to the input signal on the grid.
1.
For example:
a signal of 8,5 volt is presented at the grid, The voltage swing on the output of the cathode according to the AC loadline will then be 30V.
This being the first stage in this CF should give just a little les then the input signal so about 8 volts. This confuses me.
2.
When I read that the output is 30V with the given load. How much of that signal is now going to the effects pedal input? How can I calculate that?
If it should be clear already I apologize, but I really would like to know.
Lookin forward, I also created a load line for a Effects loop that Merlin has on his webiste with 12ax7 tubes.
In this graph I have also drawn a AC load line for the parallel kathode resistance with a load of 10k.
If I see this graph is seems to me that this schematic is not really suitable for the job. Because the voltage swing is really tiny.
Merlin has this schematic on his website so I think I am doing something wrong here. Clearly he knows his Valves
Any thought on that?
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kuDo wrote: ↑Wed Nov 04, 2020 12:00 pm
Lookin forward, I also created a load line for a Effects loop that Merlin has on his webiste with 12ax7 tubes.
In this graph I have also drawn a AC load line for the parallel kathode resistance with a load of 10k.
If I see this graph is seems to me that this schematic is not really suitable for the job. Because the voltage swing is really tiny.
Merlin has this schematic on his website so I think I am doing something wrong here. Clearly he knows his Valves
Any thought on that?
The DC load line uses the sum of the DC resistance at the plate and the DC resistance at the cathode. For Merlin's 12AX7 loop, the plate has a 100K resistor and the cathode has a 1K resistor, so the total DC resistance for the load line is 101K.
The AC load line uses the sum of the AC resistance at the plate and the AC resistance at the cathode. There is no load at the plate, so the AC and DC resistances are both 100K. The AC resistance at the cathode is 0.83K, so the total AC resistance for the load line is 100.83K. The difference between these two load lines is less than the thickness of the line used on the graph.
Merlin's ECC83 loop is well designed and will work great for you. R9, R10, and R11 won't be necessary for your application.
I found a loop that is similar to this on the Valve Wizard site. Whereabouts on the site did you find this particular one?
The AC load line uses the sum of the AC resistance at the plate and the AC resistance at the cathode. There is no load at the plate, so the AC and DC resistances are both 100K. The AC resistance at the cathode is 0.83K, so the total AC resistance for the load line is 100.83K. The difference between these two load lines is less than the thickness of the line used on the graph.
Normally I take the voltage of the Anode that is 50 or 100 volts lower than the bias point.
That voltage I then divide by the Cathode resistance and I then know where to set a point from where I can draw a line crossing the bias point.
That than is the AC load line which rotates around the bias point.
You now add both the load impedance and the Anode Resistance. How are you now going to draw the load line?
The AC load line uses the sum of the AC resistance at the plate and the AC resistance at the cathode. There is no load at the plate, so the AC and DC resistances are both 100K. The AC resistance at the cathode is 0.83K, so the total AC resistance for the load line is 100.83K. The difference between these two load lines is less than the thickness of the line used on the graph.
Normally I take the voltage of the Anode that is 50 or 100 volts lower than the bias point.
That voltage I then divide by the Cathode resistance and I then know where to set a point from where I can draw a line crossing the bias point.
That than is the AC load line which rotates around the bias point.
You now add both the load impedance and the Anode Resistance. How are you now going to draw the load line?
Thanks for leading me to the schematic.
Load lines are always the sum of what is connected to the anode and what is connected to the cathode. Take the case where a 1.5K cathode resistor is fully bypassed by a 22uF capacitor and the plate is connected to the power supply by a 100K resistor. When you go to draw load lines for this, you don't draw a dead vertical AC load line just because the AC cathode resistance is zero. You have to add up everything that is connected to the anode and cathode that conducts AC for the AC load line. The 100K resistor connected to the plate conducts AC, so it must be included in the AC load line. If the anode isn't AC coupled to another resistor, then the entire 100K is the AC resistance for the anode.
Look at Valve Wizard's cathodyne load lines. He added the DC resistances for the anode and cathode for the DC load line. He has 100K resistors that are AC coupled to the 47K resistors for an AC resistance of 32K at the anode and 32K at the cathode. He added the AC resistances to get 64K for his AC load line.
Not sure if this thread is of any use to someone. But to conclude it for myself.
I think I learned a thing or two here, for which a am really grateful.
For the amp that I am going to build I will go for the 12au7 variation I started this thread with.
I took another look at it with my new found knowledge. And did some calculation just to see what is going on when it's working.
Kathode bias line
Ik = 4 / 2k2 = 1,8mA
Ik = 10 / 2k2 = 4,5mA
A = -µ Ra / (Ra + ra) = -17 100k / (100k + 7k7) = 15,78
Zload = Ra || Rload = 100k || 1M = 90.909 Ohm
Zac = Ra || ra = 100k || 7k7 = 7.149 Ohm
38.461 : 90.909
90.909 / (7.149 + 90.909) = 0,92
A after impedance bridging
15,78 * 0,92 = 14,5
inverse amplifacation factor = 1 / 14,5 = 0,069
Divider in front of send stage = 0,069 x 1.075k = 74.175 Ohm
in series with a 1M Ohm resistor. To create amplification of 1 throughout the loop.
Zin = Rg / (1 - A) = 470k / (1 - 0,94) = 7,8 M Ohm
f = 1 / 2 pie Rg Cin = 1 / (2 pie 7,8M 1nF) = 2Hz
Cin = 1nF
Zload = 22k || 10k = 6k8 Ohm
f = 1 / 2 pie Zload Cin = 1 / (2 pie 6k8 470nF) = 4,97Hz
Csend-out = 470nF
f = 1 / 2 pie Ra Cout = 1 / (2 pie 100k 47nF) = 33Hz
Cout = 47nF
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