12AU7 with 33k anode resistors and the choke is 30h, 600 ohm DCR.
With a sinewave at the input, the PI outputs are balanced but look like a square wave with a sawtooth on top
Any thoughts on what might cause this?
Help request diagnosing PI
Moderators: pompeiisneaks, Colossal
Help request diagnosing PI
I'm playing around with a "choke tail" PI. Basically a LTP but instead of a tail and cathode resistor it uses a choke.
12AU7 with 33k anode resistors and the choke is 30h, 600 ohm DCR.
With a sinewave at the input, the PI outputs are balanced but look like a square wave with a sawtooth on top
Any thoughts on what might cause this?
12AU7 with 33k anode resistors and the choke is 30h, 600 ohm DCR.
With a sinewave at the input, the PI outputs are balanced but look like a square wave with a sawtooth on top
Any thoughts on what might cause this?
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Steve
Re: Help request diagnosing PI
The short answer is inductors and resistors aren't interchangeable. Why on earth would you expect your "playing around" to work correctly? You've misbiased the tube and substituted a part that has a non-linear frequency response into a place where a linear frequency response is needed for proper functioning.
https://www.electronics-tutorials.ws/in ... uctor.html
http://www.valvewizard.co.uk/acltp.html
That being said... at a glance the output looks like it might sound interesting. Have you plugged in and played through it yet?
https://www.electronics-tutorials.ws/in ... uctor.html
http://www.valvewizard.co.uk/acltp.html
That being said... at a glance the output looks like it might sound interesting. Have you plugged in and played through it yet?
Re: Help request diagnosing PI
The DCR of the choke is 600R as previouly stated, seems to be biased ok, a little over -6vdc on the grids:
This is the Boozehound Labs circuit I borrowed from:
His description:
What we need is a device that has relatively low DC resistance so that it won't drop a lot of voltage, but high AC impedance so that it will provide a substantial load for the cathodes of the phase inverter, and give us the linearity we want.
The choke replaces the cathode resistor, to allow the use of a high current 12AU7 as a phase inverter/driver. This high(er) current driver allows more current to be delivered to the grids of the output tubes, extending the sweet spot well into the region where grid current flows. Phase inverters tend to have neat names: cathodyne, long tail pair, concertina. I nominate this phase inverter be the "choke tail" phase inverter.
The phase inverter no longer looks like a long tail pair because the increased AC impedance of the choke makes the "tail" unnnecessary. The bias voltage is generated by the DC drop across the choke, but the large AC impedance typically generated across the cathode bias resistor plus the large tail resistor is generated across the choke as well.
Here's the circuit I have it in, pardon the pencil sketch...
What we need is a device that has relatively low DC resistance so that it won't drop a lot of voltage, but high AC impedance so that it will provide a substantial load for the cathodes of the phase inverter, and give us the linearity we want.
The choke replaces the cathode resistor, to allow the use of a high current 12AU7 as a phase inverter/driver. This high(er) current driver allows more current to be delivered to the grids of the output tubes, extending the sweet spot well into the region where grid current flows. Phase inverters tend to have neat names: cathodyne, long tail pair, concertina. I nominate this phase inverter be the "choke tail" phase inverter.
The phase inverter no longer looks like a long tail pair because the increased AC impedance of the choke makes the "tail" unnnecessary. The bias voltage is generated by the DC drop across the choke, but the large AC impedance typically generated across the cathode bias resistor plus the large tail resistor is generated across the choke as well.
Here's the circuit I have it in, pardon the pencil sketch...
You do not have the required permissions to view the files attached to this post.
Steve
Re: Help request diagnosing PI
What is the amplitude of the input signal?With a sinewave at the input, the PI outputs are balanced but look like a square wave with a sawtooth on top
Re: Help request diagnosing PI
I've been playing around with the B+ feeding the PI. When I took that scope shot, the B+ was at 230vdc and the bias at about -4vdc. The input to the pi grids was about 4vac. At 2vac input, the output of the PI looks normal.
So is this the result of pulling grid current on the 12AU7 PI tube? Is this kind of like "nipple" distortion on a cathodyne?
If I put the B+ back up to 350vdc, do you suspect it will normalize? Add a grid stopper to the PI?
So is this the result of pulling grid current on the 12AU7 PI tube? Is this kind of like "nipple" distortion on a cathodyne?
If I put the B+ back up to 350vdc, do you suspect it will normalize? Add a grid stopper to the PI?
Steve
Re: Help request diagnosing PI
I think you are just overdriving the PI.The input to the pi grids was about 4vac. At 2vac input, the output of the PI looks normal.
Re: Help request diagnosing PI
Thanks Sluckey. The waveform didn't really look familiar - was expecting it to just square off 
Steve
Re: Help request diagnosing PI
I originally shot from the hip about misbiasing, and missed because I've never seen an inverter exactly like this before. My gut reaction was that it was misbehaving similarly to the Marshall 20 watt amps (1917, 2019, 2022, 2061, that family) whose inverter is the closest match to this that I'm familiar with. I still think their misbehavior is similar, but not at DC; 600Ω looks to be a totally appropriate value at DC after doing a double take.
I'm willing to admit in advance that my assessment could totally fall apart under scrutiny, because this is a weird circuit and I could be missing something that's obvious to someone else.
At 80 Hz the tail inductor's impedance is 16kΩ, an octave higher it's 31kΩ, on the high E it's 62kΩ. I'm not ready to accept that these disparate values are going to "give us the linearity we want" yet. The way I understand it, the higher impedance at the cathode at higher frequencies is going to rotate the load line counterclockwise. That part's straightforward enough, I hope. But it also simultaneously changes the percentage of the rail voltage at the output cap on the anode similar to increasing the value of the "bottom" resistor of a voltage divider. Maybe someone can convince me to change my mind about how this is working, but this is where I'm at right now. This non-linear frequency response leads to some more interesting stuff though.
When the grid starts clipping like Sluckey diagnosed, the waveform stops being the sine wave frequency and instantly becomes a higher frequency signal, which immediately behaves differently at the cathode because the tail inductor's impedance isn't linear. This is my explanation for why clipping a sine wave doesn't give it a nice square top like you'd expect.
This is where things get really difficult to predict, especially for me because I'm not especially good with math. Voltage leads current in an inductive circuit. So as the AC voltage is doing it's happy little dance through the tube and across the inductor, the current through the inductor is lagging behind and not quite keeping up with the voltage swings. Since the "bias voltage" is based on the current flowing through the inductor "right now" the "DC" voltage that the AC signal is riding on top of isn't quite as stable as it would be with a resistor tail.
My brain is getting tired now. Feel free to shoot whatever holes in this theory you can find.
Also, what does it sound like? I'm intrigued.
I'm willing to admit in advance that my assessment could totally fall apart under scrutiny, because this is a weird circuit and I could be missing something that's obvious to someone else.
At 80 Hz the tail inductor's impedance is 16kΩ, an octave higher it's 31kΩ, on the high E it's 62kΩ. I'm not ready to accept that these disparate values are going to "give us the linearity we want" yet. The way I understand it, the higher impedance at the cathode at higher frequencies is going to rotate the load line counterclockwise. That part's straightforward enough, I hope. But it also simultaneously changes the percentage of the rail voltage at the output cap on the anode similar to increasing the value of the "bottom" resistor of a voltage divider. Maybe someone can convince me to change my mind about how this is working, but this is where I'm at right now. This non-linear frequency response leads to some more interesting stuff though.
When the grid starts clipping like Sluckey diagnosed, the waveform stops being the sine wave frequency and instantly becomes a higher frequency signal, which immediately behaves differently at the cathode because the tail inductor's impedance isn't linear. This is my explanation for why clipping a sine wave doesn't give it a nice square top like you'd expect.
This is where things get really difficult to predict, especially for me because I'm not especially good with math. Voltage leads current in an inductive circuit. So as the AC voltage is doing it's happy little dance through the tube and across the inductor, the current through the inductor is lagging behind and not quite keeping up with the voltage swings. Since the "bias voltage" is based on the current flowing through the inductor "right now" the "DC" voltage that the AC signal is riding on top of isn't quite as stable as it would be with a resistor tail.
My brain is getting tired now. Feel free to shoot whatever holes in this theory you can find.
Also, what does it sound like? I'm intrigued.