# What dictates the current drawn by a magnetron's anode?

Typically AFAIK in a magnetron from a commercial microwave oven, the cathode filament is heated by a 3.3V 10A supply, generating a cloud of electrons by thermionic emission. The anode cavity resonator is then supplied 4000V, at which point the gap is bridged by the electrons (though in a whirl due to magnetic field, creating microwave oscillations as they pass past the cavities etc.) and it starts to draw about 300mA.

What dictates this current to the anode, 300mA?

What is the relationship with the output power? If one were to supply it only half of this for example i.e. 150mA, what would the effect on the microwave power of the oven be? Would it affect it linearly - i.e. one would generate half the microwave power as before? And then heat a cup of water in the oven half as quickly?

What would be the effect of reducing the temperature of the cathode, such that one slashed the number of electrons generated by half, on the current drawn by the anode and the resulting microwave power?

Many thanks!

• If you refer to a normal AC mains operated microwave oven, it actually only emits microwaves on one half of the sine wave where the transformer and capacitor runs in series to reach 4 kV. On the other half of the sine wave the transformer charges the HV capacitor. More info in Bens excellent video: youtube.com/watch?v=I2k2g00onL0 Commented Feb 20, 2023 at 14:40
• @winny thanks for the pointer, though I've watched that video already along with many others. Already aware of what you're saying, the HV capacitor in a microwave (at least the ones I've seen) are rated ~2100Vac, and work with the transformer and the diode to double the voltage to ~4kV on one half the cycle, and charge the cap the other. One way to then reduce the power supplied to the magnetron (as obviously one cannot reduce the voltage, they (commercial microwave magnetrons) only start producing anything at all at around 3900V) is to reduce that capacitance and thereby supply less current. Commented Feb 20, 2023 at 14:53
• @winny My question was then what will the effect of that be, reducing the current, on the microwave power? Commented Feb 20, 2023 at 14:54
• I see. If it's an AC microwave and not a current limited DC HV power supply feeding the magnetron, then the question would be how to limit the power. Commented Feb 20, 2023 at 14:57
• It will be highly non-linear if you use a standard microwave oven since you'll be charging it faster, possibly even to a higher voltage, and you drop below the magnetron "threshold" voltage earlier on the other half-cycle. I would try to simulate it if I where you. Commented Feb 20, 2023 at 15:59

What dictates this current to the anode, 300mA?

and

What is the relationship with the output power?

Once the magnetron starts resonating, it emits an EM wave of a certain power, that power is the supply voltage (4000 volts) multiplied by supply current (300 mA) multiplied by efficiency (maybe 90 %). Hence, for an unspecified fictitious magnetron, it's output power might be 4000 × 0.3 × 0.9 = 1080 watts.

Of course, the efficiency may be less than 90 % for the actual device salvaged.

If one were to supply it only half of this for example i.e. 150mA, what would the effect on the microwave power of the oven be?

Magnetrons are not linear, you may find that reducing the voltage by half would result in only half the current and a quarter of the power output or, you might find that the electrons don't actually reach the anode and the device doesn't produce any output power. However, if the anode is taking current (more than a few mA) you can be assured that some output power is being produced and, that output power will be approximately V × I × efficiency. Here's a V/I/P graph from this magnetron's data sheet: -

As you can see this particular device is quite non-linear in that slight reductions in anode voltage dramatically alter the anode current and, of course, dramatically alter the output power. Your device may be very similar or it may not be very similar.

Would it affect it linearly

Over a small range it would appear to be fairly linear.

What would be the effect of reducing the temperature of the cathode, such that one slashed the number of electrons generated by half, on the current drawn by the anode and the resulting microwave power?

With fewer electrons produced the output power will lower. This will be proportionately reflected in the anode current.

• So you are saying that a microwave where the anode is supplied 1/2 the current, but the cathode heated as normal, will produce exactly half the microwave power and heat up a certain amount of water in the oven 1/2 as quickly (assuming no other heat gain/loss to environment from the water during that time)? Commented Feb 20, 2023 at 15:47
• And furthermore that the same microwave, without limiting the anode current as above, but limiting instead the number of electrons emitted by the filament (by reducing its temperature slightly) to 1/2, will then draw only 1/2 the current as above and output only 1/2 the power, as in the situation above? Commented Feb 20, 2023 at 15:51
• Comment #1 No, I'm saying that to reduce the current by half you might have to reduce the supply voltage by half and, that results in a quarter of the power. I have no idea about the linearity of magnetron power with water heating. I'd imagine that it's fairly linear. Commented Feb 20, 2023 at 15:52
• i.e. in both situations one will have exactly the same output power of the microwave, (i.e. 1/2) despite in the second situation supplying less power to the cathode? Commented Feb 20, 2023 at 15:54
• Comment #2 limiting the number of electrons emitted is, in effect, limiting anode current. If the current is limited to half and the voltage remains the same then power is half. Commented Feb 20, 2023 at 15:54