Phase Inverter - How?

[gearhead]

Hmmm... The bypas cap does 2 things it acts as a filter just like in a power supply and helps to keep a constant voltage on the cathode. This keeps the cathode at pretty much the same voltage positive in respect to the grid. This in effect makes the grid negative in respect to the cathode or provides a "negative Bias" for the grid. It also provides an ac path to ground so it does have some frequency componet. If you change the 22uf electrolytic to a .68 you will notice a change in the tone of the amp stage.

Didn't mention the frequency component, but that makes sense. The electron flow from cathode to plate goes thru that cap.

We usually only talk about the bias as being the voltage on the grid while no signal is present on the grid.

My bad. Thanks! When I say changing bias, now know I should be saying grid to cathode voltage.

From your question I think you really want to discuss what happens when the cap isn't there. One thing to remember is that a small signal on the grid controls a large signal through the tube. That's why we get gain out of the triode. We are using a small signal to control a large change in current. You can think of the triode as a variable resistor. Its resistance between cathode and plate changes depending on the signal applied to the grid. In other words as the grid becomes more pos (Less negative) more current flows from cathode to plate. When more current flows from cathode to plate you also get more current flowing through the cathode resistor and plate resistor since they are in series with the cathode and plate.

Is it really a variable resistor, or is that an analogy? As you said, as the grid becomes more positive, that decreases the voltage difference between the grid and cathode; becomes less "negative". My understanding is that means the grid will not inhibit the flow of (negative) electrons from cathode to plate as much as before; that's why the current flow goes up. Or does the cathode-plate resistance also change?

The more current flowing through the cathode resistor the more positive the cathode becomes in respect to the grid and you loose a little gain. As the signal becomes more negative on the grid the less current flows from cathode to plate and the voltage drop from cathode to plate becomes greater.

Hmm, that last part is a different viewpoint than what PRR outlined before. Or maybe my mis-interpretation of what he said.

If you have a drop in cathode to plate current, the cathode and plate resistor currents decrease too. If you decrease the current thru the plate resistor, the voltage across that resistor goes down. With the B+ stuck at a (pretty much) set voltage, that means the the B+ voltage minus the now-lower plate resistor voltage is -Higher- than before the current drop. This plate-plate resistor junction, where the signal for the next stage comes from, is at a higher voltage because the current decreased. This is the main reason there is an inversion. The tube doesn't invert; the tube circuit does.

[mhuss]

Ok, my turn to try.

Is it really a variable resistor, or is that an analogy? As you said, as the grid becomes more positive, that decreases the voltage difference between the grid and cathode; becomes less "negative". My understanding is that means the grid will not inhibit the flow of (negative) electrons from cathode to plate as much as before; that's why the current flow goes up. Or does the cathode-plate resistance also change?

Yes, it really is a variable resistance from cathode to plate. It's just that the grid-cathode voltage is controlling this resistance (vs. a wiper on a resistance element).

If the grid and cathode are equal, R is minimum, and current is maximum.

As the grid becomes more negative with respect to the cathode, it inhibits the cathode to plate electron flow, so the effective cathode-to-plate resistance goes up and current goes down.

Above a certain amount of grid-cathode potential (a few volts), the cathode-to-plate resistance gets very high, and current stops flowing.

The more current flowing through the cathode resistor the more positive the cathode becomes in respect to the grid and you loose a little gain. As the signal becomes more negative on the grid the less current flows from cathode to plate and the voltage drop from cathode to plate becomes greater.

Hmm, that last part is a different viewpoint than what PRR outlined before. Or maybe my mis-interpretation of what he said.

I think Dana is trying to explain the negative feedback implicit in a non-bypassed cathode resistor.

If you had the cathode grounded, and a fixed negative voltage on the grid biasing it (like is often done with power tubes), there's no NFB, and the amount of cathode-to-plate resistance change per volt of grid bias is fixed mainly by the tube characteristics (maximum gain).

When you add a cathode resistor to "self bias" the stage, there is an idle current flowing through the tube and the cathode resistor. If the grid goes a little positive (decreasing the bias and the cathode-to-plate resistance), more current flows, and the voltage across the fixed cathode resistor increases slightly, therefore the voltage on the cathode increases slightly. This second effect increases the bias again, decreasing the cathode-to-plate current. This second effect is the NFB, as Ohm's law works to counteract the original change in grid-cathode voltage. It's not 100% NFB, just a fraction, so there is still a change in cathode-to-plate resistance even with the NFB, it's just less than it would be without the cathode resistor.

If you have a drop in cathode to plate current, the cathode and plate resistor currents decrease too. If you decrease the current thru the plate resistor, the voltage across that resistor goes down. With the B+ stuck at a (pretty much) set voltage, that means the the B+ voltage minus the now-lower plate resistor voltage is -Higher- than before the current drop. This plate-plate resistor junction, where the signal for the next stage comes from, is at a higher voltage because the current decreased. This is the main reason there is an inversion. The tube doesn't invert; the tube circuit does.

You're right in that the circuit matters -- this is why, e.g., common cathode inverts and common plate does not.

I like (for the common cathode case) to picture the tube cathode-to-plate as a variable resistance in series with the fixed plate resistance.

- A positive-going signal decreases the bias and decreases the cathode-to-plate resistance.
- The total series resistance decreases so current increases.
- More current flowing through the fixed plate resistor means more voltage is dropped across it, so the voltage at the resistor-plate node decreases w.r.t. ground.
Net: positive going grid, negative going plate.

- A negative-going signal increases the bias and increases the cathode-to-plate resistance.
- The total series resistance increases so current decreases.
- Less current flowing through the fixed plate resistor means less voltage is dropped across it, so the voltage at the resistor-plate node increases w.r.t. ground.
Net: negative going grid, positive going plate.

--mark

[gearhead]

Think I got it now. Actually more than I had originally intended

Thanks to one and all who have been patient and contibuted to this thread. I promise my next one will be less geeky, lol.

[jjman]

I have often stared at the LT PI trying to understand how it makes the grid of the “2nd” triode negative as the “1st” is going positive (and vice versa.) And the answer is that it doesn’t.

Rather, the (coupled) cathodes going positive with the “1st” grid makes the “2nd” triode’s current reduced since it’s grid looks relatively more negative to the coupled cathode. The “2nd” grid is not being modulated at all, right?

Revelation at last. That leaves only the oscillator (in Trem) as the remaining “how” mystery for me in guitar amps. A great intro to this forum.

Thanx to gearhead for asking!

[UR12]

I finally got some time to add to my reply. Like Mark said I was trying to show why the bias isn't really changing and how the "NFB efffect" (never really thought of it as NFB) was being applied to the grid. I should have also stated that the varying voltage on the cathode resistor is very small compared to the bias voltage. We only have a few milliamps flowing through a preamp triode section. Those few ma of current doesn't make a huge change in voltage across say a 1k cathode resistor. We would be talking millivolts of change compared to a couple of volts of bias and maybe up to 4 v p-p signal, so the effect is small. While we are discussing it there is a hugh difference in the size of the plate resistor in comparison to the cathode resistor. If you use 100k and 1k for example the plate resistor is 100 times that of the cathode. This is proportional to the amount of gain the stage will develope. The LTPI is producing gain. If you look at a concertina or a cathodyne PI you have the exact same size plate and cathode resistor and one phase is taken from the cathode and the other from the plate. Since both resistors are the same size you get no gain.

PRR brought up the diferential amplifier stage, this is exactly what the Marshall 18watt PI is, with inputs from two different channels going to the grid's of each side of the PI 180 degrees out of phase with each other.

I have read that adding distortion to the bottom 1/2 of a waveform will accentuate even harmonics and the reverse if you distort the top. I have been experimenting with an amp that has a phase inverter at the front of the amp with additional triode stages between the PI and the power tubes. I have seperate tone and drive controls for the top and bottom sides of the signal before it gets to the power tubes. I'm still playing with it and it's presently on the back burner but it has produced some interesting results and been a fun project so far. I'll have to get back on it this winter.

Glad you were able to get it all sorted out gearhead. When you look at all of the theory you will find that there is only so many ways that you can hook up a tube and the component values to make it work are all usually in the same range for a reason. I think it would be very cool for someone to design an amp with a neg rail to supply bias to not only the PI but all of the preamp tubes. 5v negative supply might just be the trick. That would be a cool experiment also!