G2 power testing and conduction angle

[Andy Le Blanc]

second that, with the issue showing up for all the tubes you've tried it very
well could be something else...

[Ian444]

Azat, for what its worth, I've only been working with valve amps for a few years, my 2 cents, the screen grid is rated for max wattage, e.g. the screen on a JJ EL34 is rated max 8W. If you can measure the RMS voltage across the screen grid resistor, you can then calculate the power the screen grid is consuming. I have not found or heard of any "correct" way to do it, so that is how I do it. This is one of those situations where a signal generator (or computer generated/played sine wave) and a dummy load is useful, so you can give that amp a constant load and measure the B+ and screen supply under medium or full power, and measure screen dissipation at different output power levels. With a sine wave input, the screen consumes max current when the valve conducts heavily on each cycle, i.e. when the anode is at a much lower voltage than the screen. I built my first (and only) guitar amp from scratch about a year ago, with 2 x EL34 and a 3K6 OPT. At hard clipping with B+ sitting at 430V and screen supply at 366V (and 1K screen grid resistors) the screens are dissipating a fair bit more than 8W, something around 12W RMS or more from memory, some alarming figure as a newb. I never saw the screen grids glow and still use this amp. At idle it has a B+ of 463V and screen supply of 458V. Just threw this post in as food for thought. Hope it helps.

[azatplayer]

Thanks Ian. hey where are you? Melbourne? Im in the east side.
This bastard has given me grief like no other build, the glowing grids are a concern tho the amp sounds pretty great. Like Colossal said, could be the price for tone

[pdf64]

The screen current is a combination of dc and ac. My Fluke 189 can be set to measure the rms of that type of voltage, I'm not sure how other meters would cope.
Suggest to try measuring the Vdc then the Vac across the screen resistor, then working out the 2 currents, measure the Screen to cathode Vdc and Vac under the same conditions, then summing the dc and ac powers.
For a max square wave output conditions, expect big numbers.

[azatplayer]

Hey Andy heres the power supply and screen closeup of that amp.
[IMG:673:1277]http://img254.imageshack.us/img254/5981/screensk50.jpg[/img]

Uploaded with ImageShack.us

This old Aussie amp used a 6V6 as a screen regulator http://www.ozvalveamps.org/maxim.htm
Pretty cool.

[Ian444]

Azat, I'm in Briso, would be nice to catch up with you one day. I thought you were in WA?
pdf64, I'm using a Fluke 123 scopemeter, next time I run an amp up, I'll compare that with a Fluke 73 and a rms-reading $40 cheapie DMM. I simply measure the AC voltage drop across the screen grid resistor, and use Ohm's law. If the AC voltage reading is true rms, then I can't see why the DC component would have anything to do with the reading, its just an AC reading with a constant DC superimposed on it? When you look at the waveform on a scope, its just flat (at the screen grid supply DC voltage) except with sine wave half cycle dips, next time I run up an amp, I'll post the waveform if anyone is interested, a scope shot makes it easy to see what is going on.

[Merlinb]

To find the power dissipated in a pure resistor we need to use RMS values:
P(av) = V(rms)^2/R
or
P(av) = I(rms)^2*R

But when measuring the power supplied/consumed by something that is non-linear this won't work. Instead we really do have to plot the instantaneous voltage and current, multiply them together, and then integrate the result. No one wants to do this.

Fortunately, if either the voltage or current is pure DC, then we only need to measure the average (not RMS!) of the other quantity to find the power. In this case the voltage at the screen grid is more-or-less clean DC. OK, there is some modulation of this voltage, but compared to the 300V-or-so of DC, the AC component is very small and won't hurt the calculations much. The power dissipated in the screen is then:
P(av) = V(dc) * I(av)

In other words:
1: Measure the DC screen voltage.
2: Measure the average voltage across the screen resistor, then divide by the resistance to find the average current. (Or just measure the average current directly).
3: Multply the DC voltage by the average current to get the screen power.

If you try and do the same calculation with the RMS current then you will overestimate screen power by as much as 2x.

[Andy Le Blanc]

Here's a simple pentode regulator, It comes out of a Grommes 260-A

Its installed just after the opt center tap, in place of a power resistor or choke,
supplying the rest of the power rail and the screens directly with no screen grid resistors.

This one uses a 6l6, but is probably very similar to that maxim

[pdf64]

Merlin, do you think that the Fluke 189 (Vac + dc) setting is leading me astray then?
The result (as in screen power diss) tends to be about 30% higher using that setting rather that straight Vdc.

But when measuring the power supplied/consumed by something that is non-linear this won't work

Can you unpack that a little, or is it opening a can of worms (best waiting for the power amp book)?
In my ignorance I assumed that rms was 'heating power' and therefore was most appropriate to a heating type of measurement.
Pete.

[Merlinb]

Merlin, do you think that the Fluke 189 (Vac + dc) setting is leading me astray then?
The result (as in screen power diss) tends to be about 30% higher using that setting rather that straight Vdc.

Pete, I'm not sure what you have been trying to do- I only see one post by you related to measurement?

In my ignorance I assumed that rms was 'heating power' and therefore was most appropriate to a heating type of measurement.
Pete.

RMS is heating power in a resistive (ohmic) element only. OK, the screen grid current and voltage are in phase so it is resistive in that sense, but it is a non-linear resistance.

More generally, instantaneous power is the product of instantaneous voltage and instantaneous current:
p = vi

To get the average power (which is what we usually want to know) you must integrate this over a full cycle, or at least over a 'long period' if you're handling quirky waveforms.
P(av) = time integral of vi

However, if v or i is pure DC then it is just a fixed number, and there is no need to integrate a fixed number (constant)- just take it outside the integral. For example, if v is fixed DC:
P(av) = v(dc) * time integral of i

But the integral of i is just the average value of i!
P(av) = v(dc) * i(av)