How much negative is allowed on collector BJT

Actually, without losses, it'll swing to the full Vcc voltage *above* Vcc, and than back to ground again.

OTOH, should it somehow want to swing below ground, and you were using a MOSFET, the body diode would conduct.

If you were using a BJT, the transistor would go into inverted mode, where the C pin acts as an emitter, the base is pulled negative and depending on your circuit, has a base current, which causes a current amplified by the inverted-mode beta, to flow from ground (the E pin now acting as a collector) into the inductor.

In most classic BJTs inverted-mode beta is quite low, but for low-saturation-voltage BJTs, like those made by Zetex, inverted-mode beta is often quite high.

--
 Thanks, 
    - Win

please explain why it doesn't head for infinity...without losses...

Class E, not C -- the inductor is "overcharged".

If you put a diode in series with the transistor, it looks like this,

Clearly, it's quasi-resonant where the resonant part is free-wheeling, then stretched out in time during the charging step. :)

This.

If C drops below GND, it at least pulls B down (because B-C junction gets forward biased). If there is B drive at the same time (which would usually be the case in a class E amp), the emitter pulls both of them up (because it's acting like a collector), but only as strong as its inverted hFE can muster.

This is what makes MOSFETs great for class E: any reactive current is clamped by the body diode. To do the same with BJTs, you need to add the diode yourself. (IGBTs too -- co-packs are standard fodder for this reason. Back in the days of BJTs and CRTs, they were common in horizontal output BJTs, too. I've seen a few solenoid driver darlingtons this way, too.)

Yeah, ZTX651, PBSS303NX (the Nexperia equivalent) and such have amazingly good inverted hFE. Very good switches indeed!

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: https://www.seventransistorlabs.com/

In a traditional class C amplifier, it also acts as a mixer, where supply voltage is in equilibrium with output amplitude. This is true when the negative peak of the waveform (where the transistor is on) is just touching GND.

That is, suppose you drove extra power into the tank: as long as the transistor is switching synchronously with the waveform, it acts as a synchronous rectifier, and will draw power rather than deliver it.

No real amplifier/oscillator is quite so ideal, so there's some factor in there due to conduction angle and impedance of the resonant tank. Further factors due to real losses, collector impedance and so on.

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: https://www.seventransistorlabs.com/

It was stated to be resonant, so the voltage rises, current being routed from the inductor to the capacitor, until the inductor current drops to zero. That will happen 1/2 cycle later, at 2 Vcc. Now the current flows into the inductor until the capacitor is again at zero volts after one cycle.

--
 Thanks, 
    - Win

Correct, that's how it behaves.

I have a circuit connected to the collector node, which offsets it down, so the collector is pulled lower than emitter at intervals

Cheers

Klaus

Ok, so it is the classic VEB max of about 5V that comes into play here also

Thank you for the clarification

Cheers

Klaus

Your explanation makes sense qualitatively, but I don't see that it explains the voltage quantitatively. When the transistor is conducting it pulls the collector to ground and current through the inductor rises from the power supply voltage on it. It continues to build current and field until the transistor releases the collector or practicalities intervene like saturation. At that point the field starts to collapse and current continues to flow charging up the capacitor. The voltage on the collector and capacitor will continue to rise until the current in the inductor is zero. What in this process limits the voltage on the collector/capacitor to

2 Vcc?

The voltage on the capacitor will be determined by the energy put into it as

1/2 * C * V^2. The energy in the coil is determined by the time integral of the current put into it. Longer charging time in the inductor will give a higher voltage when the circuit resonates.

--

Rick C 

Viewed the eclipse at Wintercrest Farms, 
on the centerline of totality since 1998