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Given the same load on the output (secondary) of a transformer, the input is mains voltage, and that I don't want to try messing about with the transformer windings at all, how can one control the current supplied to the load, without affecting the voltage?

EDIT: I want to test if my high-voltage load will still work in the way I need if I supply less current to it. I have the dedicated transformer, and don't want to mess about with that. I have been given possible ways to achieve this in a permanent and efficient solution. What I need though firstly is a way to test if it still works with the power limited at all, as if not that would be a waste of money. Hence the best and quickest way to test this is to find a way to limit the current supplied by the dedicated transformer, without changing the voltage it supplies, as the load won't then work.

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Oliver Walters
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  • @tobalt Cheers, adding a bunch of power resistors and heatsinks will be my backup plan then. I have changed the constraints - the input is mains power. Could I do it more efficiently limiting mains current somehow? – Oliver Walters Feb 13 '23 at 11:21
  • What is your actual problem you are trying to solve? Draw a schematic and/or block diagram. – winny Feb 13 '23 at 11:25
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    @OliverWalters if you limit mains current, your power will no longer be constant. EDIT: Oh I see you removed the need for constant input power. Yes now it is essentially the concept of a voltage-regulated power supply, which can be extremely efficient. They will keep output voltage constant and supply the necessary current to the load and take only the necessary input current from mains. – tobalt Feb 13 '23 at 11:29
  • Thanks. Your block diagram shows what everyone expected, you need constant voltage to feed your load. Get yourself an SMPS made for the voltage and current you need. – winny Feb 13 '23 at 11:42
  • Does this answer your question? [Choosing power supply, how to get the voltage and current ratings?](https://electronics.stackexchange.com/questions/34745/choosing-power-supply-how-to-get-the-voltage-and-current-ratings) – winny Feb 13 '23 at 11:42
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    this isn't quite making sense to me. The output voltage will be a function of input voltage, transformer turns ratio, and load resistance. Are you testing the load or the transformer? If you want to reduce the load current, you must either reduce the output voltage or artificially increase load resistance (e.g. by adding resistors). But if you are testing the regulation of the whole thing, the usual way to do that is with a variac - basically adjust the input voltage (simulating mains supply fluctuations). – danmcb Feb 13 '23 at 11:50
  • @winny I need ~4000 volts. Not aware of cheap SMPSs which do this. As I said also, I have ways of implementing this permanently by buying a bunch of stuff and making a dedicated power supply for it, but I'm looking for a way to test if its worth doing that using the transformer which I have for it already. – Oliver Walters Feb 13 '23 at 11:53
  • @danmcb I'm testing the load - which normally receives specific high voltage power from the transformer when supplied by mains. I want to know, using the transformer I have, if I can supply it less current and it still work. – Oliver Walters Feb 13 '23 at 11:56
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    *how can one control the current supplied to the load, without affecting the voltage* <-- if the load is constant (as per this: *Given the same load on the output (secondary) of a transformer*) then you can't because ohm's law prevents that from happening. @OliverWalters – Andy aka Feb 13 '23 at 11:57
  • @Andyaka just a nitpick: the load could be non-linear and so not following Ohm's law, but you are right anyway: given a fixed voltage the load will react absorbing whatever current it needs according to its characteristic curve (unless it's a very very strange load, with a non-monotonic curve and different operating points for the same voltage, but also in this case the operating point is "chosen" by the load and its past history). – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:09
  • Your requirements seem inconsistent and it seems you are a bit confused about the basics. Could you tell us what load are you trying to power and what is the purpose of your test? This may help us to clarifying the problem and giving you some meaningful answers. – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:12
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    BTW, an unrelated note: when you draw block diagrams is customary to draw them with the power/information flow going from left to right. That is power supply to the left, load to the right. This is just a basic convention in EE world. – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:15
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    Please put that into the question body. 4 kV won't come cheap even with a mains transformer, but granted, cheaper than SMPS. – winny Feb 13 '23 at 12:15
  • @LorenzoDonatisupportUkraine the op said this: *Given the same load on the output (secondary) of a transformer* hence, my comment can be seen as an opportunity for the OP to present what he/she actually wants to happen. At the moment it is unclear. – Andy aka Feb 13 '23 at 12:15
  • @Andyaka I perfectly agree with you. I nitpicked just to be sure the OP wasn't further confused by your comment. He seems a newbie and sometimes comments with subtleties are lost on them. – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:18
  • He's been a member for over 5 years @LorenzoDonatisupportUkraine – Andy aka Feb 13 '23 at 12:20
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    SAFETY NOTICE: since from what you say you seem a newbie in the field (sorry if I assumed wrong) I feel compelled to warn you that you are working with dangerous stuff. Depending on the power of your transformer, 4kV can easily kill you if you mishandle something. A few milliamps through your heart is more than enough to stop it and at 4kV it's not difficult to achieve (unless the current output of your transformer is severely limited). High voltages can also easily generate arcs that can ignite things. Please be careful. – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:26
  • @Andyaka Yes, but barely active and who knows his level of expertise (I'm registered on a bunch of SE sites in whose topics I have just a mild interest, but in most of them I'm almost a lurker with almost no expertise). – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 12:29
  • @LorenzoDonatisupportUkraine The load is a magnetron from a commercial microwave. Understand the dangers of messing about with these, as have read other answers on this site. What the magnetron is outputting into (as this isnt the standard comercial microwave either) is another part I need to test and design, and whose characteristic impedance will then vary also. – Oliver Walters Feb 13 '23 at 12:32
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    hmmm. The dangers are not limited to HV shock if you are playing with magnetrons. Just the RF coming out of one of those things can do you serious harm without you even knowing it is happening. Be VERY careful. – danmcb Feb 13 '23 at 12:35
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    honestly, my gut feeling reading the above is "back away slowly". As someone that worked on radar long ago and quite a bit of experience in various forms of electronics, this is NOT a field to be messing unless you REALLY know what you are about. – danmcb Feb 13 '23 at 12:46
  • Is the load DC or AC? – bobflux Feb 13 '23 at 15:15
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    @danmcb Same feeling here. I never worked in that field, but I have a friend who for a period was enlisted in the Italian Military Navy as a EE in the field of EM compatibility and radar systems. He used to joke on how many seagulls were fried during testing and operation of radars installed on military vessels. He also told me how risky it was for the maintenance guys that climbed the masts to repair the antennas: sadly sometimes something happened to go wrong and a poor guy was irradiated badly. – LorenzoDonati4Ukraine-OnStrike Feb 13 '23 at 15:44
  • If your aim is to test some microwave components I have some doubts this is the right approach. I'm not an expert, but remembering from my old university days this is done with a low power microwave source followed by a microwave attenuator to set the actual power delivered to the load (no voltage or current, but EM power and S parameters). A magnetron is something used for high power microwave generation. Any test done with such a thing should be carried out by qualified operators with appropriate safety procedures in a suitable environment. – LorenzoDonati4Ukraine-OnStrike Feb 14 '23 at 09:02
  • Yes as I said I have an understanding of the dangers of microwave radiation, particularly worried about my eyes. Also dust from the beryllium oxide cermaic insulation on the magnetron. And also the high voltages obviously. This is why I will be careful. But equally if people only ever did what they'd already been told how to do, nothing new would ever have been invented. But yes if I end up cooked, or blind or dead then there's only one person to blame! My aim is basically to get it working enough for a specific purpose but using much less power, so its basically to make it safer anyway – Oliver Walters Feb 14 '23 at 12:19

1 Answers1

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Given the same load on the output (secondary) of a transformer, the input is mains voltage, and that I don't want to try messing about with the transformer windings at all, how can one control the current supplied to the load, without affecting the voltage?

For a resistive load, you can't without violating Ohm's Law, I=E/R, to change I you need to change E or R, so with the same load, and the same voltage, the current will be the same. For a non-resistive load, pretty much the same thing but with harder math.

You can limit the current, but that works by reducing the voltage. If you have a 10\$\Omega\$ load and a 100 V source the current will be 10 A. Adding a current limiter to limit it to 5 A will reduce the voltage to 50 V. Adding a 10\$\Omega\$ resistor in series with the load will reduce the current to 5 A, but again the voltage across the load will be 50 V. That is what will happen when you 'supply less current' to a load, the voltage will decrease. Changing the transformer to one that is rated at less current will either overheat the transformer, reduce the voltage to the load, or more likely both.

GodJihyo
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  • Not all loads are resistors, and Ohm's Law is only for resistors, but it does work kinda the same way for every loads. The equation I=E/R is for resistors but almost every load uses a various function of the form I=f(E) so you cannot change I without changing E. – user253751 Feb 13 '23 at 18:37