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Microwave Oven Transformer

  1. Jul 23, 2009 #1
    If one had a microwave oven transformer (MOT) from a microwave that was said to be 1200W, and the transformer itself had an output of 2300V, would that mean that the current would be:

    Power = Current * Voltage

    1200W = Current * 2300V

    Current = 0.52A

    The reason I ask is because: I'm looking to power said MOT by a Variac that is said to be 0-140VAC at 6A. I'm trying to verify that I won't overload the current of the Variac with said MOT.

    Also, I'm going to rectify the output of the MOT with a diode bridge consisting of four diodes said to be 9kV and 500mA. I understand that at 9kV and 0.5A, the power would be 4.5kW -- way higher that the output of the MOT. But, does the maximum current of the diodes increase as the voltage decreases? In other words, even though they say 500mA, could they sustain higher current at lower voltages, such as 500V?

    I'm sure to many of you these are laughable questions. :shy: Any help, though, would be much appreciated! Thanks in advance.
     
    Last edited: Jul 23, 2009
  2. jcsd
  3. Jul 23, 2009 #2
    MOTs are dangerous to play around for reasons stated, 0.5 amps at 2000 volts is lethal.

    the primary winding is usually pulsed by DC or the positive side of AC only, and the secondary of many MOTs is tied with magnetic core to serve as a ground, if you feed AC directly to primary then your whole MOT becomes hot, meaning if you touch your transformer you can be killed.

    As for current rating on the secondary, check the wire gauge, and look it up in a table.

    you can make a nice welder with a MOT, but I strongly advise against using it for high voltage
     
  4. Jul 23, 2009 #3

    vk6kro

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    Probably the power supply in your microwave oven is the most dangerous item in your house. So, it is probably necessary to warn you of this. Even one such transformer is able and willing to kill you.

    Diodes are rated for current according to the heat they can dissipate. High voltage diodes are usually a stack of lower voltage diodes in one package and their dissipation depends on the voltage drop across the diode stack times the current flowing.

    It does not depend on the transformer voltage.

    Discarded microwaves often have a perfectly good diode in them that is used as a half wave rectifier carrying the full current of the magnetron. These are well worth salvaging if you get the chance.

    The current in your calculation would be 1 amp

    Power = Current * Voltage

    2300W = Current * 2300V

    Current = 1 A

    However, the primary current would be more like 20 amps (2300 watts / 120 volts) in normal operation so you may find the fuses in your variac blowing if you try to use that sort of power from the transformer.
    Also these transformers are relatively low inductance in the primary so they draw a large leakage current even with no load on the secondary.
     
  5. Jul 23, 2009 #4
    Vk6kro, sorry, the 2300W was a typo. I meant 1200W.

    Also, just for you guy's reassurance, I won't even think about powering this thing up before I have it all securely fastened into an enclosure, probably made of thick plastic, with only the control knob to the Variac, an I/O switch and a momentary "power/fire" button on the panel of said enclosure. The output of this whole power supply device will be run through 15kV 18AWG wire.

    Vk6kro, you said that "Diodes are rated for current according to the heat they can dissipate... and their dissipation depends on the voltage drop across the diode stack times the current flowing... It does not depend on the transformer voltage." But, wouldn't the voltage across the diode depend on the voltage coming out of the MOT?

    Just FYI, http://cgi.ebay.com/9KVolt-Microwav..._trkparms=65:15|66:4|39:1|293:1|294:200#shId" that I'm currently interested in.

    So, 1200W/120VAC would give 10A of current in the primary of the MOT (in which case I would find a Variac capable of such a current.) But, would the MOT would pull 1200W of power in its primary through the full range of 0-120VAC as adjustable by the Variac? This concept is what really has me lost/confused. :confused:

    Similarly, would the output of the MOT, which depends on the output of the Variac, remain at 1200W through the full range of voltages possible by the MOT's output? For example, if I had the Varaic turned down low, the MOT would consequently output a lower voltage, say, 500V. Would the current that is coming out of the MOT be 1200W/500V = 2.4A? Again, trying to make several well-known equations, which all seem dependant on each other, balance out in my head is very confusing.

    Sorry guys, I wish I could be more participative on these forums versus constantly inquisitive. Thanks for your time.
     
    Last edited by a moderator: Apr 24, 2017
  6. Jul 24, 2009 #5
    At the risk of repeating other posts, your analysis of the rated current (0.52 A) is correct. (I V = 1200 watts, or more exactly, 1200 volt-amps). This essentially is the current (IR drop) limit of the transformer windings. If you exceed this current, the copper windings overheat.
    I believe this transformer is safe to power with your Variac, but you might want to limit the input voltage to 120 volts, because this limits the interturn voltage of the windings, AND limits the peak magnetization of the core. If you go to 140 volts, you may push the magnetic core into saturation. Does the transformer have any other secondaries that may need to be included in the 1200 w (actually volt-amp) rating?

    The diodes have two separate ratings: 0.5 amps, and PIV = 9 kV. Do not exceed either.

    I know a guy who was knocked unconsious by touching the high voltage in a microwave oven. Be extremely careful, and stand back from your project whenever the transformer is powered. If you have capacitors, be sure they are discharged before touching anything.
     
  7. Jul 24, 2009 #6

    vk6kro

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    We have to give these warnings because there are lots of people reading these wise words and we don't want someone frying themselves! So please excuse the warnings if you already know how dangerous a microwave transformer is.

    The 1200 watts is actually the microwave power available for cooking or heating. The actual power the transformer uses is shown on the back of the the oven. It could be something like 2000 watts.

    The current a transformer draws on its primary depends on the load on the secondary. If there is no load on the secondary, the primary will not draw much current. BUT it will draw some current. This is called the leakage current.

    The diode. This has two problems.
    First, it has to stop current flowing in the reverse direction. All diodes do this, but some high voltage ones can stand more reverse voltage than others. This is called the peak inverse voltage of the diode.
    Secondly, when the diode is conducting, it has a small voltage across it (usually 0.6 volts, but in this case it will be higher) and a current flowing through it. It is only this small voltage that matters for heating, because when the big voltage occurs, there is no current flowing.
    Can you see that these two voltages happen at different times?
    Both these ratings are important because diodes can fail if one of them is exceeded.
     
    Last edited by a moderator: Apr 24, 2017
  8. Jul 24, 2009 #7
    So, with the Variac turned up to 120V, wouldn't the primary on the MOT pull significantly more (1200VA/120V=10A) current than the 6A of the Variac?

    Worst case scenario: the load is a short-circuit (maybe I want to make a couple sparks when I'm not charging caps.) So, would it be advisable to put capacitor or inductor between the Variac and MOT primary to limit current, because based on what you said, a short-circuit would probably instantly blow the Variac's fuse, right?
     
  9. Jul 24, 2009 #8

    vk6kro

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    So, with the Variac turned up to 120V, wouldn't the primary on the MOT pull significantly more (1200VA/120V=10A) current than the 6A of the Variac?

    No. If there is no load on the secondary of the transformer, there won't be much current in the primary. The current might only be an AMP or so.

    Worst case scenario: the load is a short-circuit (maybe I want to make a couple sparks when I'm not charging caps.) So, would it be advisable to put capacitor or inductor between the Variac and MOT primary to limit current, because based on what you said, a short-circuit would probably instantly blow the Variac's fuse, right?

    Sparking will possibly cause the primary current to go very high and may even melt the fine wire of the secondary. So, yes, you need to limit the current somehow.
     
  10. Jul 24, 2009 #9
    Wouldn't the load be, in one scenario, the charging capacitor?
     
  11. Jul 24, 2009 #10

    vk6kro

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    Yes, but the capacitor is only 1uF in a microwave and charging happens very quickly.
    You might see it briefly on a meter when you apply the capacitor, but it would probably happen too quickly to blow the variac fuse.

    It might, though, so try it at a lower voltage first.

    In fact, I would suggest you never feed more than about 20 volts into the primary. That would still give you 400 volts.
     
  12. Jul 24, 2009 #11
    I see. That's about all the voltage I'll truly need. But, I'm actually interested in charging a 400VDC 12,000uF capacitor. Considering such a large capacitance, I was just thinking that it would be a high load, hence pulling a high current through the primary of the MOT.
     
  13. Jul 24, 2009 #12

    vk6kro

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    Yes, it might be a problem. You could limit the current with a 24 volt lamp bulb or two 12 volt ones in series with the primary. Wind the Variac up gradually from zero when you first try it.

    Don't leave the charged capacitors around unattended. Use them or discharge them. Even 400 volts is dangerous.

    A good advice used to be to keep one hand behind your back so you couldn't get across high voltages.
     
  14. Jul 25, 2009 #13
    If I want to limit the current coming in to my circuit to 5A (I do,) I'll need 24Ohms:
    120V/24Ohms = 5A​


    So to get 24Ohms of inductive reactance, I'll need 0.063 Henries:
    24Ohms = 6.28 * 60Hertz * 0.063Henries​
    That's 63,000uH of inductance -- seems a little uncommon.


    To get 24Ohms of capacitive reactance, I need:
    24Ohms = 1 / (6.28 * 60Hertz * 110.58uF​
    That's 110.58uF -- seems much more available.

    Can most electrolytic capacitors be used for AC as well as DC?
     
  15. Jul 25, 2009 #14

    negitron

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    68 mH is a standard value. Probaby close enough.


    No, but you can put two in series, with like polarity together. Assuming each has an equal value, C, the resulting capacitance will be C/2.
     
  16. Jul 25, 2009 #15
    Hmm, after doing some searching, most inductors of this high of an inductance seem really low current. I'd imagine that to get one that is 68mH and can handle 5A would be massive (and expensive.)


    On the other hand, however, I found a http://mouser.com/Search/ProductDetail.aspx?qs=sGAEpiMZZMsCnlYck6hSqLhQjFWSHEDhL0WgC2TwEi4%3d" [Broken]. So, I could connect two of those in series with like polarity and use it as a 110uF AC capacitor limiting current to 5A? :-)
     
    Last edited by a moderator: May 4, 2017
  17. Jul 25, 2009 #16

    vk6kro

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    If you want to limit the voltage to 400 volts or so, you can do the calculations with 20 volts as the input.

    That means the suggestion above of using 24 volt lamp bulbs would be OK. Probably a lot easier than using reactive limiting. These have the added advantage that you can see if there is current flowing because the lamp will glow.
     
  18. Jul 25, 2009 #17
    I see, that's fantastic. So a 24VAC LED lamp would work? So, once I dial the Variac to 24V, would the lamp limit the current to whatever it's rated for, regardless of the load?
     
  19. Jul 25, 2009 #18

    negitron

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    Right. Think of it this way: the absolute maximum current the bulb will draw at a given voltage occurs when the bulb is alone in the circuit; anything you add in series with that can't make it draw more.
     
  20. Jul 25, 2009 #19
    LEDs are not suitable for AC circuits. Use only filament-type lamps. Also, if you use capacitors to limit the AC current, use only motor starting capacitors, preferably ones rated for continuous use. If you put capacitors in series, make sure the two capacitors are equal capacitance. Also put resistors across each capacitor to equalize the voltage drop across each capacitor. It is much better to put AC capacitors with adequate voltage rating in parallel.
     
  21. Jul 25, 2009 #20

    negitron

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    Not true; many of the products our company makes use LEDs as indicators on AC circuits. In addition to the usual current-limiting resistor, if you also add a series diode with a PIV rating greater than the peak voltage of the AC source, your LED will be protected from reverse bias breakdown. This is necessary, since most LEDs have a low PIV on the order of 5 volts or so.

    EDIT: And now I see he meant to use the LED lamp as his current-limiter on the MOT circuit. I missed that. As Bob says, that's not going to work at all.
     
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