[AMRadio] Ferroresonant transformer revisited |
Jim Candela
jcandela at prodigy.net
Sun May 7 13:32:22 EDT 2006
Rick asks: Jim, correct me on my math if I am heading the wrong direction. Since I don't have a true RMS voltmeter can't I take the square root of the power dissipated in a resistor across the output to find the actual RMS voltage? My best answer: E^2=PR, so E = sq root (PR) Yes, but how do you measure P? Another idea: If you have an RF ammeter, this will measure rms current regardless of ac waveform or even dc, then I=E/R so E = IR. Pick a known resistor that will work with the voltage and ammeter you have, and do the calculation. The ammeter must be the thermoucouple type however to get the correct answer. One drawback of the "quasi-sine-square" wave from a Ferro transformer is that certain power supplies are designed to charge the capacitors to the peak value of the AC voltage, and the designers expect the peak to be 1.414 times the RMS value. With a square wave the peak and RMS are the same, and with a Ferro you might have a value of 1.2 or so. Certain equipment like a SSB linear with a full wave voltage doubler P/S will be way low on peak power when powered from a Ferro versus normal sine wave power. Some of the power inverters on the market that transorm say 12 volts DC to 115 vac do so with a peak voltage of about 150 volts to the load, and dead time either side of the 60 HZ half wave pulse to in effect maintain a workable RMS at 115, and yet peak to about 150 to satisfy capacitor input power supply needs. This can be done quite efficently, and is pretty clever. Regards, Jim JKO
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