Calculating power dissipation

[petercapo]

Looked through the archives but could not find what I was looking for. I'd like to know the proper way to figure power dissipation for EL84 plates in the Stereo 35. I've noted the instructions for biasing the tubes in the diytube manual, as follows:

"With an insulated adjustment tool and a DMM (at low voltage setting), adjust bias pots clockwise (up) or counter-clockwise (down) to get the voltage at each TP (i.e. the test point at R19,R22,R25,R28) to be 350mV. This means the cathode current is 35mA. Plate current is about 30mA and the screen is 5mA."

So, given that P=EI, exactly which figures for voltage and current should be used to calculate power dissipation at the plate? Should we:

- Subtract the cathode voltage from the plate voltage and consider the result to be the voltage "across" the plate alone, or is there also a voltage drop across the screen to be subtracted, as well?

- Use 35mA or 30mA as the current through the plate?

Thanks for any help with this.

Peter

[TomMcNally]

You might find this site helpful ...

http://www.webervst.com/tubes1/calcbias.htm

[EWBrown]

Plate sissipation ios the product of the plate to cathode voltage and the plate current. Screen dissipation is tthe product of the SG to cathode voltage and current (usually just a couple of mA), and is generally specified at a couple of watts maximum.. These two power dissipation figures are two separate entities.

The (35 ma) cathode current is the sum of the plate and the screen grid curernts. Across 400 ohms this would measure as 14VDC. This can be between 12 and 16VDC, depending on the tube and the total resistance of the adjustable cathode biasing "network" from ground to cathode.

In the typical ST35 / SCA-35 scenario, the actual idling (no input signal) plate to cathode voltage would be 340 to 350V, factoring in the voltage drops across the OPT primary winding (5-6 VDC approx) and the cathode resistor (13-15VDC approx) .

With 30 mA actual plate current, this would produce a plate dissipation between 10.2 and 10.5 watts Add in an "extreme" maximum of 5 mA for the screen, and that would be 1.70 to 1,75 Watts screen dissipation.

In "real life" situations, the plate current would likely be 33-34 mA and the screen current, 1-2 mA

HTH

/ed B

[petercapo]

"In the typical ST35 / SCA-35 scenario, the actual idling (no input signal) plate to cathode voltage would be 340 to 350V, factoring in the voltage drops across the OPT primary winding (5-6 VDC approx) and the cathode resistor (13-15VDC approx) ."

So, let's say plate to ground is 370VDC and there's 14VDC across the cathode resistor. I can see that you'd need to subtract the 14VDC from the 370VDC, which would give you 356VDC. Not clear on the part about "…factoring in the voltage drops across the OPT primary winding…" Can you say a bit more about this, please? Is this something that can be measured just by putting the meter probes across the OPT primary? And, once measured, it would need to be deducted from the 356?

Thank you.
Peter

[EWBrown]

In the above description, I'm working from the B+ voltage to ground, the plate voltage will be several volts lower:

The output transformer primary winding has a DC resistance of its own, anywhere from under 100 ohms to over 200 ohms, epending on the OPT being used. So just think of it as behaving like a big wirewound resistor, at least for DC measurement consideratrions. This is the same, with either both push-pull and single-ended circuits.

Note, that in most OPTs, the DC resistance of the two plate connections of the primary, to the center tap will be about 10% difference, this is due to the layer-winding technique generally used, and each layer of wire and the insulating material will add up to create an increasing circumference, thusly a longer length of wire, even though the actual number of turns is identical on each side of the center tap.

The one exception to this will be in a bifilar-wound transformer, and these tend to be a lot more expensive "gold-plated" high end items.

So, as an example, the measured B+ votage at the OPT center tap (usually a red wire) is 370VDC, the winding from plate to CT is 200 ohms DC resistance, the cathode resistor is 400 ohms, and the DC voltage across the cathode resistor is 14 VDC, which indicates a combined plate and screen current of 35 mA.

In the SCA-35 or ST-35, with the ultralinear screen grid taps on the primary, the combined plate and screen currents will be the same 35 mA cathode current - this is all a series connected circuit, for our calculations. The screen current is generally quite low, so we can just let the plate take all the "credit" as far as current consumption, in order to keep the mathematics somewhat less complicated. Just think of the EL84 as behaving like a hefty power resistor between the cathode and plate, for static DC measurment situations. The plate dissipation is the product of the voltage drop and current between the cathode and plate

So, in our theoretical amplifier circuit, we have a voltage drop of 7VDC through the primary winding, and 14V across the catrhode resistor, for a total of 21 VDC, which is then subtracted from the 370VDC B+ voltage, which will be 349 VDC. Multiply that by 35 mA and the plate (and screen) dissipation is 12.215 Watts. OK, we're not doing brain surgery here, so we can dispose a couple of the less significant decimal places

12.2 Watts is more than close enough, or even 12 Watts - with tube circuitry, 5% tolerance is more than "close enough" precision.

I hope that this is somewhat clearer than mud

/ed B

[petercapo]

So, from the perspective of Kirchoff's voltage law, the closed loop is from the OPT CT, across 1/2 the OPT primary, then the plate, and finally the cathode resistor, correct? I guess we wouldn't need to include the screen in this loop in order to reckon the voltage at the plate to be used for calculating power dissipation?

In any case, I think you are saying that I could take my meter and put it directly across the cathode and the plate for the voltage figure. And, if I wanted to be picky, I suppose I could desolder a wire and put the meter in series with the plate and measure the current. Then, multiply this current and the voltage measurement together for the plate dissipation, correct?

Many thanks for the basic circuit analysis assistance!
Peter