Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: tubenit on April 13, 2010, 08:16:42 am
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Guys, I am having some trouble understanding the tube data sheets for a 6L6GC or a 5881. I did look at them but I'm still unclear.
In a single ended amp at 270v on the plates of a 6L6 with a 470 ohm resistor, what would I expect the current ma to be?
Thanks, Tubenit
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I would expect the current to be quite low- around 20-40 ma, possibly lower. When the plate voltage is that low I would probably expect a cathode resistor between 150 and 250 ohms if you wanted to achieve the 70-80 ma I see on the datasheets. 470r would probably be a good fit at higher plate voltages though.
jamie
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There's a couple of extra things to know first...
Screen voltage, rather than plate voltage, is the determining factor when you take about pentodes or beam power tubes. So the screen voltage and bias voltage has to be stated, and between those you can determine plate current.
I'm assuming you were looking at a data sheet like this on from G.E. for the 6L6GC (http://www.mif.pg.gda.pl/homepages/frank/sheets/093/6/6L6GC.pdf).
There is a condition listed for "Push-Pull Class A Amplifier, Two Tubes". In this case, all voltages remain the same, currents are doubled (2 tubes instead of 1). Some sheets will list driving voltage as grid-to-grid, which is the value measured from push grid to pull grid, and you will see it comes out as double the value of the bias voltage. This particular sheet just has the driving voltage for a single grid.
So continuing with this particular sheet, there is a condition listed under that heading with 270v plate and 270v screen, with -17.5v bias. So turn the question on its head, and ask "what cathode resistor must be used to get the data sheet condition?"
Of course, both plate and screen current flow through the cathode resistor, so add the zero-signal plate and screen currents. 134mA + 11mA = 145mA; but that is for 2 tubes, so divide by 2 (unless you want to know the value of a shared cathode resistor). 145mA / 2 = 72.5mA. Bias was stated as -17.5v, so divide that by the current to find the resistance needed (according to ohm's law). 17.5v / .0725A = 241 ohms. Round up to 250 ohms, which is a standard part.
The odd part about this sheet (and a lot of 6L6 variant sheets) is that I think you can find the exact same numbers on the sheet for a metal 6L6 from the late 30's. Plate dissipation, even neglecting that the voltage across the cathode resistor will not factor in, is ~18w, so you're not getting everything out of your 30w 6L6GC.
If you were drawing a loadline, that line would be high on the left side of the curves and low on the right (loadlines for tubes loaded by transformers are a little different from tubes loaded by a resistor, but slope the same way). However, if you would like to see what effect a cathode resistor will have, you can plot a line starting at the 0v gridline and 0mA plate current, with the line continuing up and to the right. You will simply be plotting the current times the resistance of the cathode resistor, and the resulting voltage will be the grid voltage. The problem with doing this with a pentode/beam power tube is that you have to find out what the screen current is separately, add it to the plate current, then multiply by the resistance to find grid voltage.
That, and the fact that plate curves are given with a fixed value of screen voltage, usually 250v, and plate current will be higher if you raise the screen voltage.
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> with a 470 ohm resistor, what would I expect the current
Why do you need to know?
Picking bias resistor values is a last step, not a first step.
Pick your voltages. Pick your current. THEN determine a happy cathode resistor. 100, 220, 470, 2K resistors of the same wattage are all the SAME price. And so cheap you can buy a 10-pack of 100r 5W resistors, string them in series, experiment until you hit your design current.
About 45mA, which at 270V gives 12W which seems a waste of a 19W-30W 6L6.
The GE data for 250V SE says -14V and 72+5= 77mA. 14V/77mA= 182 ohms. That gives 18W in the 6L6. If you use original 6L6 at 270V, you should go to 220 ohms. Using modern 6L6, you could leave it at 180 ohms, get 21W dissipation.