While up in Rotterdam I started thinking about Class C Amps and the standard formula used to calculate power out and load resistance: Rl=(Vcc-Ve)^2/2Po. I understand why this formula works for Class A amps: The Vcc-Ve term describes the maximum voltage you can get at the output. The rest of the formula is just a version of P=IE and P=E^2/R. The 2 in the denominator converts peak to average. The books tell us that this same formula applies to Class C amps. How could that be? I wondered. Doesn't the output of a Class C amp look (pre-filter) like a series of pulses at the operating freq? Wouldn't that require a somewhat different formula?The answer came from SSDRA and LTSpice. SSDRA page 25 explains "If we assume that the collector voltage varies from zero to twice the Vcc level while delivering the desired output power, the load needed at the collector is given by the familiar relation Rl=Vcc^2/2Po." (Emphasis added.) The voltage at the collector is being pulled down nearly to zero as the voltage at the base goes positive and the transistor conducts. You can see this in the waveform in the LTSpice screenshot above. Then, when the input voltage dips below about .6 volts, the transistor goes into cutoff and stops conducting. At this point the energy stored in the inductor in collector circuit is dumped onto the collector, raising the voltage there to about twice Vcc. That the ugly spike you see at the top. Wow, you can really see from this the need for output filtering.
As I was exploring this issue, I cam across an old LTSpice VideoCast from December 2006. See below.
BTW: These are the kinds of questions explored in the book "SolderSmoke -- A Global Adventure in Radio Electronics." I'm hearing that delivery is very fast, especially in the UK.
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