Fixed bias, phase inverter AB1 PP... examples?

[bluefoxicy]

I've been googling this for a while.. I asked about fixed bias a while back and got some help, but it's still unclear to me and I've been trying to understand how this works since.

I'm told I need a negative power source; I can make one. -5V from a recto tube filament supply, silicon rectified and cleaned up with a good 2200uF capacitor. -5V to a pot plugged into the grid, but... where the hell do I plug +5V? How do I resolve the circuit? Or am I relying on "building a charge" but not on "a current flow" so I can just tape off the +5V (this does not make logical sense to me, this isn't static)?

Can someone illustrate on a stock Valve Jr. V1? Here's a schematic, make up a +5V power source (can be a battery, rectified AC, I don't care; rectified and filtered AC will work right?) and turn this into fixed bias:

http://www.euthymia.org/DIY/VJstock.jpg

Another question I have is on a phase inverter in a push-pull circuit... anyone up for showing me a 12AX7 PI into two EL84 or 6V6 tubes? I think I get the general operation -- center tap OT primary to B+, use one tube for positive signal on the top (arbitrary) tap of the OT primary, the other tube for negative signal on the bottom (arbitrary) tap, bias the power tubes so each clips a little below 0 signal, and send the bottom tube an inverted signal. It's inverting the signal with NO GAIN (mu == 1) that I don't get.

Mu....

[Rob Mercure]

BF,

I'm not sure that I understand what you're trying to accomplish (relay supply?) for the first query so I'm going to pass:

As to the second, go to the "Fender Amp Feild Guide" and look at any of the Black Face Deluxes if you want an example of a 12AX7 PI feeding 6V6s. I won't comment about the tap, bias the power tubes so each clips a little below 0 signal, and send the bottom tube an inverted signal. cuz it isn't necessarily true, nor true in any specific amp I can think of. The clipping point is where the signal pushs the control grid to a non-linear point on the characteristic curves (simplified) - might be the point you refer to and might not. And I don't understand the last question either - if the "top" side of a PP amp is driven by a certain signal level you want the "bottom" side driven by the same level signal 180 degrees out of phase. So you take the "top" signal and invert it without amplifying it - if you amplified it the circuit wouldn't be "balanced" - not amplifying is a gain of 1 - "mu" refers to something totally different but related (the "mu" is the theoretical maximum voltage gain of a triode [rough, quickie definitioin] - but it's dependent on circuit parameters and the circuit rarely approaches the theoretical value. If I remember correctly the 12AX7 in a typical Fender input stage, with a mu or 100, has a voltage gain of around 30-35 or so).

Rob

[tubeswell]

On the bias supply, remember that the voltage on the grid has to be negative with respect to the cathode.

in my example you need about 50V for the bias supply. (I guess you could tap it from one PT high voltage AC winding, but the vaues of your caps would need to be higher, and I think it would have a small impact on the rectifier supply)

When you stick a 1000V diode (1N4007) backwards in series with the PT supply (i.e. with the banded end pointing back to the PT), it will convert the AC in the PT supply into negative DC, and then with a 10+uF (about 150Vmin) electrolytic cap after the diode (that has the +ve end pointing to ground), it will smooth out the negative DC ripple to about -69V. Then you can use a voltage divider to reduce the -ve voltage to what you want (i.e. so it achieves the desired current flow through the output tube(s)) to supply this -ve DC voltage to ('the bottom of') the output tube grid load resistors.

[bluefoxicy]

When you stick a 1000V diode (1N4007) backwards in series with the PT supply (i.e. with the banded end pointing back to the PT), it will convert the AC in the PT supply into negative DC, and then with a 10+uF (about 150Vmin) electrolytic cap after the diode (that has the +ve end pointing to ground), it will smooth out the negative DC ripple to about -69V. Then you can use a voltage divider to reduce the -ve voltage to what you want (i.e. so it achieves the desired current flow through the output tube(s)) to supply this -ve DC voltage to ('the bottom of') the output tube grid load resistors.

I...what???? I'm totally lost. Also I thought I couldn't use the power rail (the high voltage tap powers this) for the bias power supply, i.e. I need a second PT?

[Shea]

I...what???? I'm totally lost. Also I thought I couldn't use the power rail (the high voltage tap powers this) for the bias power supply, i.e. I need a second PT?

You can't use a filament winding because it won't provide enough voltage. Your bias supply should be able to provide at least -20 volts for EL84s, more for other tube types.

The bias supply usually does come from the HT winding. Some power transformers have a tap in the HT winding. The AC voltage at that tap is less than the AC voltage at either end of the HT winding, so when you half-wave rectify the voltage coming out of that tap, you get a DC voltage that is at the right level for your bias supply.

When you don't have a bias tap, there are at least two ways to make a bias supply. One is to put a zener diode, reverse-biased, in series between the center tap of the HT winding and ground. The end of the zener that is connected to the center cap will have a negative voltage with respect to ground that is equal to the zener voltage. That might be a good way to do that for this amp. When you remove the bias resistor and connect the cathode of the EL84 to ground, then the plate voltage (i.e., voltage from plate to cathode) will go up by the amount of the voltage drop that you used to get across the cathode resistor. If you set up your bias supply using this zener method, then the B+ voltage will be reduced by the amoutn of the zener voltage, which will partly counteract the increase in plate voltage caused by removing the cathode resistor.

There are at least two potential drawbacks to that approach: (1) if the zener shorts, the EL84 will have no bias voltage and will burn up pretty quickly, and (2) the bias voltage won't sag when the B+ sags, which might make the responsiveness of the amp feel a little weird. Or, it might not.

The other way to make a bias supply when you don't have a bias tap involves connecting a film/foil capacitor to one leg of the HT winding, with a resistor to ground connected to the other end of the capacitor. AC gets through the capacitor, but because of the capacitor's reactance, the AC voltage you're going to get at the junction of the resistor and capacitor will be less than the full AC voltage at either end of the HT winding. Then you half-wave rectify and filter that reduced AC voltage. You have to do some math to figure out the best values for the cap and resistor to get the AC voltage down to where you want it. I think there are some old Marshalls or Hiwatts that used this method. Start looking up older Marshall and Hiwatt schematics, and look for one where the power transformer does not have a bias tap on the HT winding.

FWIW, I've read that you can't hear a lot of difference between fixed bias and cathode bias with an EL84 anyway. Because they take pretty low bias voltage and have sensitive grids, supposedly, with a cathode-biased EL84 you don't get the squishiness that is usually caused by shifts in the cathode voltage in cathode-biased amps with other types of tubes. I can't vouch for that opinion myself, though.

Shea

[BackwardsBoB]

For an exampe of bias voltage generation, check blueguitar.org, and look at the Pignose G40V in the miscellaneous amp schematics section. This is similar to what tubeswell describes. It capacitively couples the AC off the high voltage windings to generate a voltage through a diode.

When you say that "It's inverting the signal with NO GAIN (mu == 1) that I don't get,"
do you mean that you don't know how to do it, or you don't know why you need to?

It's possible to build a P-P PI stage that simply generates the gate drive for one power tube, and then inverts this signal with unity gain using a second triode circuit, but there are several other more popular ways to do it.

[bluefoxicy]

When you say that "It's inverting the signal with NO GAIN (mu == 1) that I don't get,"
do you mean that you don't know how to do it, or you don't know why you need to?

I dont' understand how the circuit works or how it does it, or other ways of doing it.

Coworker is telling me to go to EE school. *poor*

[tubeswell]

Hi Bluefoxicy

Have you seen this page?

http://www.el34world.com/charts/Biascircuits.htm

I found it very instructive

[Chuck H]

FWIW, I've read that you can't hear a lot of difference between fixed bias and cathode bias with an EL84 anyway. Because they take pretty low bias voltage and have sensitive grids, supposedly, with a cathode-biased EL84 you don't get the squishiness that is usually caused by shifts in the cathode voltage in cathode-biased amps with other types of tubes. I can't vouch for that opinion myself, though.

I sure can. I installed a fixed/cathode bias switch in an EL84 build once. I intentionally set the cathode bias a little hotter than the fixed bias. That way any sag wouldn't result in a percieved volume loss. The result was a switch on the back of the amp that doesn't do anything. At least, it does so little that I have to wonder if any change is just my imagination.

FWIW I think the thread moved a little sideways. Blue was originally asking about fixed bias phase inverters. The amount of voltage needed to do that should be much lower than the bias needed for power tube grids. Blue, check out some of the Mesa schems. They use a fixed bias inverter that works really well. I know for sure it's used on the Subway amps. It requres a -15 volt supply. Even with a voltage doubler circuit the 5v winding is only good for 11 or 12 volts. But this is where the other good info in this thread on bias supplies will apply.

Chuck

[bluefoxicy]

Found it.

RC-30 page 83, Grid Voltage Supply. With alternate power source, negative goes to Grid, positive goes to Filament negative. In AC filaments, positive goes to center tap of two low ohmage (20-50R) resistors connected between the filaments. If it's using the high voltage, just find a more negative tap than the cathode.

I could use a voltage doubler. Maybe. I don't know.