All,
I have a sudden interest in calculating transformer ratios for
solid state power amplifiers. Mostly, because I have a Chinese kit
linear I'd like to interface to my QRP radios.
Not sure I remember how to do this, but here goes:
* Suppose I want to deliver 100W with a power supply of 13.8V. What
turns ratio do I need for the output transformer?
We have two devices in push-pull. Each device delivers 100W, but
only half the time. I'll do it for one device.
I'll assume that my device can swing its output voltage to within a
half volt of ground and Vdd. So its total available swing is 12.8V,
or 12.8V PEAK. Assuming a sine wave, the RMS
voltage is the peak voltage divided by the square root of two,
or about 9.05V.
Here's where I get a little confused. For a single-ended amp, I
would say "Peak-to-peak swing is 12.8V" - peak voltage 12.8 divided by
2, rms = peak divided by squareroot-of-two. But for a push-pull
stage, each transistor delivers one ALTERNATION of the output wave.
So I guess I don't have to put in that divide by two to get from
peak-to-peak to peak.
So - we have a device delivering 9.05V (rms). To deliver 100W,
( for half the time ), current = voltage / power, or 50/9.05, or 5.52
Amps. And the desired load impedance would be R = E/I, or 9.05/5.52,
or
about 1.6 ohms.
For the transformer, we have a center-tapped primary, with 1.6 ohms
impedance on each side of the center tap, or 3.2 ohms total. So our
impedance ratio is 50ohms / 3.2ohms, or 15.6. Let's call it 16.
Turns ratio is the square root of impedance ratio. So that gives us a
turns
ratio of 4.
Does this sound about right?
Forty years ago, I could literally do this stuff in my head...
I was just reading a footnote in my 1976 "Devices and Circuits"
college textbook:
"* Most FETs have power ratings under 1 W, although a few are
available with ratings as high as 10 W."
- Jerry KF6VB