10GV8 Bias point for the K-12G

[Ty_Bower]

Originally Posted by jmillerdoc

What I know for a fact now is that the B+ is 250V and the effective cathode resistor value is 300R. The B+ is 250V at the center tap of the OPT...

That seems too high. The power supply caps C8 and C9 are shown on the schematic as being rated for only 250V. Did you take your voltage reading with the tubes in the circuit? If you unload the power supply, the voltage will be much higher than it would during normal operation.

Also, I think I assumed the g2 was at -20V, why I did this I don't know at the moment...it all seemed to click when I ran through this at that time. Can you help me through this?

The plate of the tube (also known as the anode, or A) will be at a high voltage (hundreds of volts DC). Its voltage will swing along the load line of the tubes as the amplifier operates. At idle, it will be slightly below B+ due to voltage drop across the primary winding of the output transformer. A typical OPT might have 100 ohms DC resistance, so you might drop 4 volts across the winding if the plate is operating at 40 mA. During operation, the plate of the tube can swing much higher than the B+ due to the reactance of the OPT. This can be a difficult concept to appreciate.

In a pentode wired amplifier, the screen of the tube (also known as the second grid, or g2) will be held at a fixed, high voltage potential. The screen provides the accelerating potential which attracts the electrons from the cathode. After they fly through the vacuum towards the screen, most of the electrons zip straight through it and are collected by the plate. It is common to keep the screen at a somewhat lower voltage than B+. In the K-12G schematic this is accomplished by the 470 ohm resistor R18. If you have four output tubes and each tube's screen is drawing 8mA, R18 will drop 15 volts. The screen supply voltage will be that much lower than B+. You should also keep in mind the acceleration potential of the screen needs to be calculated as the voltage from screen to cathode. We should probably be looking at the tube curves where Vg2 is specified equal to 190V~170V. Unlike the plate, the screen voltage of a pentode wired output tube does not (at least, should not) vary during operation.

The control grid of the tube (usually called "the grid", or g1) must be kept at a voltage lower than the cathode. If the potential at the grid becomes greater than (or equal to) the cathode, electrons will cheerfully become drawn to the grid and your tube will pull "grid current". While this can be part of the amplifier's design, I suspect the K-12G is not one of those kinds of amplifiers. The driver tube would need to be capable of handling significant amounts of current, which in this case it cannot. Very small changes in voltage on the control grid will result in large changes in current flow through the tube, as represented by the lines drawn on the curve diagrams.

the values above are from the schematic and the knowlegde that the unrectified trafo voltage is 180-0 through a bridge rectifier, 180x1.4--->250V.

As noted above, the voltage of the unloaded power supply will be quite a bit higher than when the tubes start drawing power. I'd expect your actual working B+ is closer to 210, as previously indicated.

When I now look at the curve if I pick about 50mA (just below max power curve) above the 125V X-axis, the point falls right at the -15V line. If I calculate this out 15/0.05=300....300R is exactly the value (300/2=150 on the schematic) of the original design.

The voltage at the plate of the tube will not be 125V at idle. It will be B+, minus whatever is lost to the DC resistance of the OPT primary (maybe 4 or 5 volts).

The schematic I found for the K-12G shows a 150 ohm resistor for the pair of output tubes. If you know the voltage drop across that resistor is 15 volts, then the pair of tubes must be drawing 100mA. Volts equals current times resistance. 15 volts equals 0.1 amps times 150 ohms.

I can say one thing for sure, the more I work through this stuff the more I begin to understand it (I think!).

Good! Once all this makes sense, you can try to graph the actual working loadline for the push/pull pair, and then your head will be dizzy again!

50mA is much more in line with where this tube should operate, from what I can tell.

At lot depends on the real B+, and it's much easier to figure everything out if you've got a measured cathode voltage. If I assume B+=210V, and assume the screens draw 8mA, then Vg2-k = (B+) - (4*Ig2*R18) - (Vk). We thought the cathode voltage (Vk) was about 15 volts, so Vg2-k = 180V. We've only got curves for Vg2=190V and Vg2=170V, so I'll have to guess at something in the middle. I'll make another wild guess here, and assume the plate current is 42mA. Ik = Ia+Ig2 = 50mA per tube. That's 100mA for the pair, which across the 150 ohm cathode resistor should give Vk = 15V. The voltage from plate to cathode is (B+) - (~4) - (Vk) = 191V. Looking at the plate curve (Vg2=170V), I see Ia~>40mA for Va=190V and Vg1=-15. I'll accept that my guess for plate current agrees with the assumed conditions.

Calculating dissipation, we said that Va-k was 191V and Ia was 42mA. That means the plate is dissipating 8 watts, which right at the datasheet limit. Screen dissipation should be (Vg2-k) * (Ig2) = 180 * 8mA = 1.44 watts. Again, right at the datasheet limit. Personally, I'd still rather see these tubes run a little cooler (unless you like replacing tubes often). I was also looking at the Vg2=170V curves, so my results are going to be a little less conservative than reality.