Class E amplifier question

Here is a question for all electronics gurus here. Can a class E amplifier be termed efficient - given the shunt capacitor from BJT collector to ground, which is an AC short -- unless the capacior value is very small, such that the capacitive reactance is large. Any flaws in this argument ?

Reply to
amal banerjee
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The capacitor is a part of a tuned circuit, so the simple short circuit idea is moot.

I have a kilowatt RF amplifier, where there is a capacitor from the tube plates to ground and another across the output (Google for a pi-filter).

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-TV
Reply to
Tauno Voipio

A transistor dissipates instantaneous power Vce*Ic, so discharging a charged capacitor heats the device. The trick to get efficiency is to only turn on the transistor when Vce is low, namely pump a lot of current into a resonant load for part of the cycle, as the resonator voltage swings close to ground. The resonant load gets pumped by the full supply voltage times the current, a lot of power, but the transistor dissipation is low, high Ic but low Vce while it's on.

Resonant power supplies aren't called class E, but work the same way. Some are called "zero voltage switching" circuits.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
jlarkin

The PI filter is not essential, i.e., the capacitor to ground(in parallel with the load) is of the low pass L impedance matching sub-circuit, and and can be replaced with a high pass(inductor to ground) L impedance matching sub-circuit, with marginal change in performance characteristics. Ub fact, the L impedance matching sub-circuit can be done away with. I have dirtied my hands a fair bit with SPICE simulations(HSpice, Ngspice) to say this.

Reply to
amal banerjee

Thanks for the explanation. Most treatments of the class E amplifier(including SPICE simulations) use an ideal switch, instead of a real BJT or FET.

Reply to
amal banerjee

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