I think this is getting done kind of backwards. I believe the correct way to
figure efficiency is power output minus drive power that is fed thru,
divided by plate input power (plate current times plate volts).
2200 x .375 = 825 watts plate input power.
Drive power = 20 watts.
Output power = 450 watts.
Without subtracting the feed thru drive power the efficiency would look like
450/825 = 54.5%.
Subtracting the feed thru drive power it would be 430/825 = 52.1%.
But as I said before 100% of the drive power does not get fed thru as part
of it is consumed driving the grid in the tube. A rough estimate of consumed
drive power is around 30%. So in this case we would be left with about 14
watts of feed thru power.
So efficiency would then be (450 - 14) 436/825 = 52.85%.
This doesn't look like much of a difference but when you get into higher
drive power it makes a bigger difference in the overall efficiency
calculation.
In a grounded grid amplifier the driver is in series with the amplifier tube
so part of the drive power is coupled thru the amplifier tube.
73
Gary K4FMX
> IIRC For efficiency it's pretty straight forward it's power out divided
> by power in (not drive) IOW in this case for GG it's Plate current Ip
> (0.375 * multiplied by plate voltage. I don't remember what you are
> using for voltage, but as n example if the plate is 2200 then
> 0.375A*2000V=750 Watts input power plus 20 watts of drive for 770 watts
> input or 450/770 = 58% overall efficiency for the amp. The tube
> efficiency would be 450/750 or 60%
>
> 73
>
> Roger (K8RI)
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