Hi Steve!
The formula can be worked several ways, which is what makes it so neat!
R = V*V / (2*P) or
P = V*V / (2*R) or
V peak = SQRT (2*P*R) Same thing!
The point is that the output matching network is presenting a certain
load to the transistor. *If* the transistor is being driven hard enough
(which the 38 special is), the transistor will swing from almost zero to
2*V ****at the impedance the output matching network presents to
it****.
If it is not driven hard enough, you will not see the full zero to 2*V
voltage swing. But whatever voltage swing occurs will be across the load
the matching network presents.
In the 38 special, we see 5 watts out for a 12v supply. Therefore, the
matching network is transforming 50 ohms at the antenna port to a load at
the transistor of
R = V*V / (2*P) = 12*12 / (2*5) = 14.4 ohms.
Now that the matching network is known to transform 50 ohms at the
antenna to 14.4 ohm at the transistor, with 24 volts applied, the power
at the transistor is:
P = V*V / (2*R) = 24*24 / (2*14.4) = 20w
The matching network of the 38 special will transform this to 50 ohms to
produce:
V peak = SQRT(2*R*P) = SQRT(2*50*20) = 44.7v peak which is close to the
45v peak that I measured. The net effect of the matching network is to
raise the voltage from the 24v at the transistor to 45v (peak) across a
50 ohm load.
Likewise, with 30v at the transistor, P = 30*30 / (2*14.4) = 31.25w, or:
SQRT(2*50*31.25) = 55.9v peak across 50 ohms at the antenna
Again, no matching network change required at all! Just change the
voltage to change the power. 12v gives 5w, 24v gives 20w, 30v gives
31.25w.... No changes required (except for a larger heat sink!)
Remember that the current drain goes up also. Double the voltage to 24v,
and the current also doubles. Twice the voltage, twice the current, same
resistance, gives four times the power (20w). Likewise, 2.5 times the
voltage (30v), requires 2.5 times the current as well, giving 6.25
(2.5*2.5) times the power. If 5w requires about 0.7 A, 24v requires 1.4A
(20w out), and 30v requires about 1.75 A (31w out).
I would not recommend going over 30v. The IRF510/511 is only rated to a
maximum 20w of heat dissipation, given an ideal 25 degree C case
temperature. Assuming 60% efficiency, during key down, the transistor
would be dissipating right at the maximum rating of 20w of heat
dissipation. This is probably ok for cw, since the duty cycle is low.
The time averaged heat dissipated will be 10w or less.
I guess you would not want to try to run RTTY on a 38 special at 30v!
Like I said, I have verified 20w operation at 24v and I will try for 30w
at 30v later this weekend.
Neat stuff! Neat little toy (hi!) radio!
- Dan Tayloe, N7VE(ex-KK7BD), Phoenix, AZ, QRPL # 696, Az ScQRPions
>One thing though, your formula isn't quite right. It's Vc(Vc) / 2 Po
>to give the impedance of the output transistor. If you crunch the
>numbers with a 13.8 volt Vcc and with 24 Vcc for different output
>power from 5 to 40 watts, then divide those numbers into 50 ohms and
>take the square root to get the actual turns ratio for matching
>impedances, you will find the (normalized) turns ratio varies about 1
>to 4 with 13.8 volts and 1 to 3 with 24 volts, with the higher turns
>ratio needed for the higher power levels.
>
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