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Registered Member #3610
Joined: Thu Jan 13 2011, 03:29AM
Location: Seattle, WA
Posts: 506
That's a good point. Honestly I've never played with a magnetron outside of a microwave oven so I can't say how they behave under different conditions. Most microwaves do cycle them off and on at a fairly slow rate to control power. This is the easiest way because one transformer provides both filament and HV, and a simple relay is all it takes. "Inverter" microwaves I assume use PWM to vary the power level, but I suppose it just pulses at a high rate without altering the peak.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Ash: From an electrical point of view a magnetron looks like a zener diode. It hardly conducts at all until the voltage gets high enough to overcome the electron-bending effect of the magnetic field, then it springs into life all of a sudden: a small increase in voltage causes a large increase in current and RF power. Too much voltage and it goes crazy, for essentially the same reason that a recorder or tin whistle shrieks horribly if you blow it too hard.
Varying the current is the easiest way to go. But because all practical HV power supplies have a fairly high output impedance (that's what those shunts in the MOT are for) varying the voltage amounts to pretty much the same thing.
It's a common newbie fallacy to believe that you can control voltage and current independently. You can only control one of them: the load decides the other. When the load is a resistor, the rule that it uses to decide is Ohm's Law.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve McConner wrote ...
Ash: From an electrical point of view a magnetron looks like a zener diode. It hardly conducts at all until the voltage gets high enough to overcome the electron-bending effect of the magnetic field, then it springs into life all of a sudden:
Yes, I'd assumed that was the case, but you explained it better than I did.
Steve McConner wrote ...
a small increase in voltage causes a large increase in current and RF power. Too much voltage and it goes crazy, for essentially the same reason that a recorder or tin whistle shrieks horribly if you blow it too hard.
Varying the current is the easiest way to go. But because all practical HV power supplies have a fairly high output impedance (that's what those shunts in the MOT are for) varying the voltage amounts to pretty much the same thing.
Would a 250 watt variac be suitable for this? or would I risk burning it out even if I put a smaller capacitor on the output of the secondary? (I'm not that familiar with variacs and I only have one, so I don't really want to learn the hard way)
Steve McConner wrote ...
It's a common newbie fallacy to believe that you can control voltage and current independently. You can only control one of them: the load decides the other. When the load is a resistor, the rule that it uses to decide is Ohm's Law.
As I understand it, using a smaller capacitor in the doubler will reduce the maximum current (same theory as Cockroft Walton multipliers), however, especially if using a smoothing cap, if the load wants to draw more current than is available, this will result in 'voltage sag', presumably, this will find a point of equilibrium, as the voltage will rise until current flows. I can see that this will be a bit of a balancing act, though.
EDIT: I think the issue here is basically whether or not I'm correct in assuming that a smaller capacitor in the doubler will effectively 'throttle' the system? (when used in conjunction with a variac on the primary of the transformer)
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Well, yes, it will "throttle" it.
A variac is just a load of wire. It can be protected by a fuse in series with the sliding contact. Even a slow-blow type will blow before the variac burns. The fuse rating should just be the variac current rating: they're rated in amps, not watts.
You are right with the thing about sag and a point of equilibrium. It's only a fiddly balancing act if you're trying to calculate this operating point or visualise where it might be. The actual circuit always finds the point effortlessly if it exists. If no stable operating point exists, the circuit is an oscillator.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve McConner wrote ...
Well, yes, it will "throttle" it.
A variac is just a load of wire. It can be protected by a fuse in series with the sliding contact. Even a slow-blow type will blow before the variac burns. The fuse rating should just be the variac current rating: they're rated in amps, not watts.
You are, as usual, correct, Steve. 240V, 50/400Hz, 2A. I picked it up to throttle a 250 watt diffusion pump, in conjuction with an MKS butterfly valved pressure control system, that's why I had the figure of 250 watts in my head. With a 2 amp fuse between the output from the variac (not the common terminal) and the MOT will this be suitable?
Steve McConner wrote ...
You are right with the thing about sag and a point of equilibrium. It's only a fiddly balancing act if you're trying to calculate this operating point or visualise where it might be. The actual circuit always finds the point effortlessly if it exists. If no stable operating point exists, the circuit is an oscillator.
I meant it's a 'balancing act' to get the point of equilibrium at the optimum point, for optimum results, ie to adjust the point of equilibrium to where you want it for optimum results. any oscillation will presumably be due to a 'too small' smoothing capacitor.
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