welcome welcome to my workshop I'm Jim today I'd like to show you how to make a high voltage DC power supply from an old microwave oven high voltage means six thousand volts this is what a microwave oven looks like after it's been stripped down to get here simply remove all of the screws from the bottom in the back then pull apart the pieces that make up the shell this oven is now upside down with the turntable motor here on the top this metal shell is the oven itself this is the magnetron which generates the microwave radiation and the metal waveguide which conducts the Rays into the cooking chamber this is the high voltage transformer which is the most valuable component for us we can probably use this high voltage capacitor as well there's a resistor wired in parallel with this capacitor for safety reasons but it's enclosed inside the capacitors shell it's not this component this component is the high voltage diode which is wired as closely as possible to one of the capacitors terminals this white plastic cylinder is on the lead from the transformer to the capacitor it's a fusible link it's an inline fuse designed for high voltage low current applications this is the front panel with the push-button switches removed here's the LED display there's a small speaker on the back of the front panel this is the door jam which still has one of the interlock switches mounted as well as the mechanical latch which locks the door closed this oven has a separate low voltage power supply here on the back of the front panel to run the fan and to run the turntable motor this device which looks like a transistor is one of two thermal cutout switches there will almost certainly be a schematic diagram pasted to the inside of one of the side panels take a look at it you might find something interesting I discovered that this oven has a noise filter mounted right down here where the power cord enters the case it's a self-contained little circuit mounted on its own printed circuit board we can probably use it in our power supply I'm now going to disconnect the components I'm interested in before cutting any wires it's a good idea to make a thorough record of the connections this is a schematic diagram of the high voltage side of the oven circuit the high voltage circuit in your oven will look exactly like this even if the details on the low voltage side are different the primary coil in the transformer takes 120 volt alternating current from the wiring in your house these two wires located down here supply the primary current since the primary current is an alternating current it doesn't matter which way around these two wires are connected there are two coils on the secondary side one of them generates the high voltage we can identify the high voltage wire from this coil by tracing it back from the capacitor through the fusible link and to the transformer it's important to note that there's only one wire coming out of the high-voltage coil the other end of the high-voltage coil is connected to the Transformers frame which is bolted directly onto the ovens chassis this is a very important characteristic of the ovens circuit it uses the chassis as the return path for the high voltage current the high voltage coil powers two branches which are wired in parallel one branch is the high voltage diode over here the other branch is the main power flow through the magnetron itself the magnetron also has a heater or cathode like tubes in the olden days a current is forced to flow through a filament heating it up to the point where it glows red electrons then boil off from the hot filament it's their circular acceleration through the resonant chambers in the magnetron which throw off the microwave radiation in any event this filament needs its own current which is provided by the second coil on the secondary side of the transformer these are the two wires which carry the filament current notice that they're on opposite sides of this transformer notice also that one of the filament wires is connected to the high-voltage lead coming from the capacitor the first thing we need to do is to find out what voltage the transformer actually produces I've wired up a temporary circuit to test the transformer you can see that I've kept the noise reduction circuit hopefully it'll reduce noise as its name suggests but it also has a line voltage fuse already mounted on its printed circuit board which would be a convenient component to keep in the finished power supply I've also kept the AC line cord I've soldered a heavy-duty toggle switch into the hot wire of the AC cord the hot wire is the black one the white one is the neutral this toggle switch will be the on/off switch in the finished power supply ground connections are important as much for your own safety as for anything else this bolt is the central grounding point one of the green wires that's connected to the bolt comes from the AC cord the other green wire comes from the noise reduction circuit also connected here is one end of a 10k resistor over which I'll measure the voltage it's a standard quarter watt resistor even though it seems pretty small compared to the other components I've cut off and taped the two leads coming from the transformer which previously powered the filament of the magnetron they're no longer connected to anything and won't do anything in the finished power supply I'll now describe the loop which is powered by the main coil of the high voltage transformer I've kept the fusible link for the time being note that I've removed its white cylindrical case the fusible link is simply an inline fuse holder with a spring-loaded fuse inside I'll be using three resistors in series for the test this is a 1 mega ohm resistor it's rated for 3 watts of power this is a second 1 megohm resistor also rated for 3 watts and I've already referred to the third resistor which is this 10k one I'm going to use an oscilloscope to view the waveform the leads to the we'll be attached at the top of the 10k resistor and it ground at the two points marked by the white arrows this is the oscilloscopes display of the voltage over the 10k resistor it's a sine wave with a bit of distortion at the zero crossings the peak to peak distance on the screen is 3.1 divisions since the vertical scale of the scope is set to 10 volts per division the peak to peak voltage is 3.1 divisions times 10 volts per division or 31 volts in total the horizontal axis of the scope measures time and the distance across the screen corresponding to one period of the wave is about 3.2 divisions the Scopes horizontal scale is set to 5 milliseconds so the period of the wave is 3.2 divisions times 5 milliseconds per division or 16 milliseconds as always the corresponding frequency is 1 divided by the period which in this case is 1 divided by 16 milliseconds or 62 Hertz I'm sure that a more careful reading of the horizontal distance would confirm that the truline frequency is 60 Hertz and not 62 this is a schematic diagram of the test circuit it's important to understand what the circuit does so that we can interpret the oscilloscope readings from a scope we know that the voltage drop over the 10k resistor is 31 volts peak-to-peak because we know the voltage and resistance we can use Ohm's law to calculate the current flowing through this resistor as always the current is equal to the voltage divided by the resistance in this case 31 volts divided by 10,000 ohms is 3 point 1 milliamp s-- to be precise we'd say that it's 3 point 1 milliamp peak to peak this same current flows through all the three resistors because they're connected in series the total resistance is 1 megohm plus a second megohm plus 10k that's a total of two million ten thousand ohms we can use Ohm's law second time to calculate the voltage drop over the total series resistance this voltage drop is highlighted in blue voltage equals current times resistance for this calculation the current is three point one milli amps which is equivalent to zero point zero zero three one and piers and the resistance is two million ten thousand ohms the product is six thousand 230 volts peak-to-peak since the secondary coil of the transformer is wired in parallel with the total series resistance the voltage drop over the coil is also six thousand two hundred and thirty volts peak-to-peak so why is the high voltage capacitor from the microwave oven rated at only twenty-one hundred volts this is a picture of the capacitor from the oven I've circled the voltage rating given on the capacitors label it's quite clear it says twenty-one hundred volts from our test we know that the voltage generated by the secondary coil in the transformer is this blue waveform it is a frequency of 60 Hertz and swings between plus and minus three thousand 115 volts note that 3115 is one half of the six thousand two hundred and thirty volts peak-to-peak which we just calculated since six thousand two hundred and thirty is the peak-to-peak voltage then the peak voltage is one half of that or three thousand one hundred fifteen volts imagine that we rectify the alternating current meaning that we somehow flipped negative voltages into their equivalent positive voltages we could always use a couple of diodes to accomplish that after rectification the waveform is a series of half period sinusoidal bumps the voltage varies but it is always in the same direction in that respect it's like a direct current on the other hand it certainly isn't a constant voltage this type of waveform has its own name it's called rect fied AC suppose that we apply this rectified AC to a resistor any resistor suppose we then compare the result of what happens when we apply a constant 3000 115 volts to the same resistor because the instantaneous voltage of the rectified waveform is almost always below the peak voltage the resistor will dissipate less power than it will in the constant voltage case that means the resistor won't convert as much electrical power into heat it follows that the resistor won't get as hot alternating currents and voltages are compared to direct currents and voltages based on their ability to produce power this is the statement of that equality an alternating voltage with a peak value of V peak has the same effectiveness as a constant direct current voltage 70.7% is high the subscript RMS is a code that tells a person that this voltage is the DC equivalent of some sinusoidal alternating voltage for our transformer the peak voltage is 3000 115 volts so the DC equivalent is 0.7 Oh seven times 3100 fifteen or twenty two hundred volts the RMS subscript reminds us that this is a DC equivalent voltage I've marked this RMS voltage on the graph with the right circuitry we could cause this transformer to do the same amount of work as a 2200 volt battery I'm showing that picture of the capacitor once again if we look at the label carefully we can see that it doesn't say 20 100 volts it actually says 20 100 volts ac those two extra letters ac have the same meaning as the RMS subscript I mentioned before how do we reconcile between the 2200 volts RMS we calculated with the lower 20 100 volt RMS rating of the capacitor the fact that our experimental apparatus was rough-and-ready could explain the difference a more subtle reason is the fact that we tested the transformer with very little load the 2 1 mega ohm resistors are drawing a total of about 4 watts of power but when the microwave oven was cooking this transformer was producing almost a thousand watts as a transformer is loaded up there are inefficiencies which become more important and which decrease the voltage available to an external load it's likely that the two voltages we have 2100 volts RMS and 2200 volts RMS represent the transformer being operated under two different sets of conditions for design purposes as we proceed further I'll assume that the peak voltage from the secondary is 3000 volts the DC equivalent of that is zero point seven or seven times 3000 which is within one percent of 2,100 volts RMS this red sinusoid is the voltage which the power supply in your microwave oven delivers to the magnetron it varies between 0 and minus 6000 volts at a frequency of 60 Hertz there's a good reason why oven designers want the ground reference to be at the top of this waveform but it's a straightforward change to modify the oven circuit to put the ground reference at the bottom the circuit shown on the left is a standard half wave rectifier the one on the right is the circuit in your microwave oven let's just move the ovens capacitor and diode around the loop by 90 degrees that doesn't change the topology in any way but makes it easier for me to compare what goes on in the two circuits in both cases the coil represents the secondary side of the high voltage transformer the red sinusoidal in stone taneous voltage which is generated over the coil I'm going to put the oven circuit to one side for a minute and work through what happens in a typical half wave rectifier these two dashed sinusoids are identical they're intended to make it easier to compare the voltage coming in at the left from the transformer with the voltage drop over the capacitor on the right let's start with the capacitor completely uncharged and with the coil just ready to begin a positive going pulse as the voltage begins it's upward climb the voltage drop over the secondary coil will be in the direction defined by the blue arrow with the top end of the coil at a higher voltage than the bottom end the coil and the capacitor share a common connection at their bottoms and since the capacitor is uncharged the diode will be forward biased and will allow current to flow through into the capacitor the capacitor will charge up in the direction defined by this second blue arrow as it charges up the voltage drop over the capacitor will follow the upward climb of the voltage over the coil this state of affairs will continue as the input voltage climbs up to its peak positive value the diode will continue to be forward biased more charge will flow into the capacitor and it's voltage will continue to climb things will change once the voltage drop over the coil starts back down the right-hand end of the diode will be at a higher potential than the left-hand end so the diode will be reverse biased current cannot flow backwards through the diode so there won't be any current flowing at all the capacitor will remain at the peak voltage to which it was charged as the cycle continues the coil voltage will fall back to zero but there won't be any change in the capacitors voltage nor will there be any change when the coil voltage starts down the negative going pulse during this half cycle the voltage at the bottom end of the coil will be greater than the voltage at the top end the diode becomes increasingly reverse biased as the negative going pulse progresses the reverse bias will reach its maximum at the same time as the negative going pulse reaches its peak at this instant the diode is subjected to a reverse voltage equal to the full peak to peak voltage of the transformer secondary the voltage drop over the capacitor will remain constant while the input voltage climbs back up to zero and even when it starts up the second positive going pulse now I'm gonna stop here to make a change it's not the purpose of a half wave rectifier to sit around watching a charged capacitor the purpose of a half wave rectifier is to provide power at a constant or near constant voltage to some kind of a load what's been missing from the circuit so far is a load which I assume will be a simple resistor powering the load resistor will require that the capacitor use up some of its stored energy so the voltage over the capacitor will decrease instead of remaining constant during the time it takes the input voltage to go through the negative going pulse the voltage over the capacitor will decrease steadily fortunately the capacitor is about to be recharged as the input voltage comes back up to its peak there will be a short period of time during which the diode is forward biased during this period current will flow from the transformer into the capacitor charging it back up to the peak voltage and then the process will repeat the capacitor constantly powers the load resistor but is topped up at the peak of every input cycle the voltage applied over the resistor is not a perfectly constant DC but it is a high-voltage direct-current with some bumps these bumps are called ripple now let's see what the oven circuit does in response to the same sinusoidal input voltage as before we'll start with everything at zero the coil will begin it's positive going pulse the coil and the capacitor share a common connection but this time it's at the top end the diode will be forward biased think of the coil as pushing down on the left-hand side of the diode the voltage at the right-hand side will be pulled down in lockstep so the capacitor will start to charge up as the input voltage climbs up to the first peak the capacitor will follow as the input voltage starts back down the diode will become reverse biased the voltage drop over the capacitor will remain constant at its peak value in fact the capacitors voltage will remain constant as the input voltage Falls to zero as it starts down the first negative going pulse and all the way down to the negative peak it is at this moment that the diode experiences its maximum reverse bias at twice the peak value of the input sinusoid the capacitors voltage will remain constant as the input voltage completes the first negative going pulse and starts up the second positive going pulse once again the intention is that the charged up capacitor supply power to a load resistor instead of remaining constant after the first peak the capacitors voltage will have been steadily decreasing it will now be topped up the cycles will continue with the capacitor topped up at every peak once again the voltage drop over the capacitor and the load resistor is DC direct current with ripple but this is not the voltage applied to a magnetron in a microwave oven that's because a microwave oven uses a trick the magnetron is not connected across the capacitor as you might expect it's connected across the diode therefore the voltage applied to the magnetron is not the relatively constant voltage drop over the capacitor it's the alternating voltage drop over the reverse biased diode the voltage over the magnetron varies through the full peak to peak range of the input from the secondary coil but the charged capacitor offsets this by a constant amount equal to the peak voltage so the voltage over the magnetron never reverses direction this type of circuit is called a half wave doubler it's called a doubler because the voltage over the load acts in one direction only but has a peak value equal to twice the peak voltage of the transformer a better circuit for a high voltage DC power supply is this one it supplies 6,000 volts it uses the transformer the diode the capacitor and the noise reduction circuit from the microwave oven but you will need to add a second diode and a second capacitor of the same or similar type as those from the scrapped oven these aren't expensive and could be robbed from a second oven or purchased from a local appliance repair shop you will also have to add an on/off switch the microwave oven will not have used a toggle switch but will have been activated by relays controlled from the front panel the voltage drop over the secondary side of the transformer is a sinusoid with peaks of plus and minus 3000 volts let's consider the first positive going pulse the top of the secondary coil will be at a higher voltage than the bottom since the bottom of the coil and the bottom of capacitor c1 are directly connected diode d1 will be forward biased current will flow through diode d1 and charge capacitor c1 up to 3000 volts during this rising edge diode d2 will be reverse biased so no current will fall through it capacitor c2 will remain uncharged now consider the first negative going pulse the bottom of the secondary coil will be at a higher voltage than the top since the bottom of the coil and the top of capacitor c2 are directly connected the left-hand end of diode d2 will be at a lower potential than the right hand end diode d2 will be forward biased current will flow out of the bottom of capacitor c2 and it will charge up to 3000 volts this type of circuit is called a full wave doubler it's called full wave because it uses both the positive and negative Peaks from the transformer capacitor c1 is topped up during the positive peaks capacitor c2 is charged up during the negative Peaks this is a look inside the left side of the completed power supply the white arrow points to the transformer the two leads for the magnetrons filament have been cut off and isolated they don't do anything these metal cans are capacitors c1 and capacitors c2 this is the high voltage diode d1 and this is the high voltage diode d2 the power cord from the microwave oven was too short to be practical so I used the power cord from an old computer this is the three prong connector from that computer which is a great way to get the hundred 20 volt AC through the metal enclosure this heavy-duty toggle switch on the front panel switches the hot wire that is the black wire coming from the power cord the 120 volt AC goes to the noise reduction circuit taken from the microwave oven the AC from the noise reduction circuit is delivered to the primary coil of the transformer here this is the view looking into the right side of the power supply from behind this is the transformer again and the unused leads for the magnetrons filament the outside of the three prong connector is just visible on the back of the power supply this red wire is the high-voltage lead from the transformer the other end of the high-voltage coil is connected to the frame of the transformer and is taken off from this lug here the secondary's frame lead is run up to this short red wire which is the connection between the two capacitors the Transformers hot wire is connected to the positive end of diode d1 and to the negative end of diode d2 the negative end of diode d1 is connected to capacitor c1 unfortunately capacitor c2 is too obscured to point out in this image leads from both capacitors run out of the enclosure through grommets inserted into the front panel the leads are kept very short there are only six inches long in order to avoid having high voltage leads flopping around I made a home for them I mounted three bolts to a piece of vector board which is bolted to the front panel but isolated from it the leads can be attached to two of these bolts and secured using wing nuts the third bolt is a spare neither lead has any connection to the ground wire or the neutral wire of the 120 volt AC power cord therefore I didn't use the word ground to describe the low voltage lead I simply labeled it zero volts I'm showing a detail of the end of one of the leads where it's attached to the mounting bolt the lead itself is a piece of number-10 electrical wire it's like the number fourteen electrical wire used in your house just a bit heavier the arrow points to the white insulation I exposed only enough of the copper to make a loop which wraps around the mounting bolt to increase the insulation I added a sleeve of heat shrink tubing and then I added a second sleeve of heat shrink tubing I used a voltmeter and some resistors to calibrate the power supply these two pieces of equipment are a lot more important than a voltmeter particularly when trying out a new piece of high-voltage equipment of course there's the risk of shock the voltage and current which this power supply generates is well inside the fatal range but protection for your eyes is equally important there's a risk that a component will explode and a piece of shrapnel could be at the back of your eye long before you can blink I wired for small resistors in parallel to make a sense resistor each of these resistors is a hundred K so the parallel combination has a resistance of 25 K these are quarter watt resistors so the parallel combination can handle 1 watt in this test and in every test which followed the leads from the voltmeter are connected to the two ends of this sense resistor these blue resistors are the load resistance for this particular test two resistors have been soldered in series each resistor is 1 megohm so the series combination has a resistance of 2 mega ohms these are 3 watt resistors so the series combination can handle 6 watts of power with the voltmeter set to the 250 volt DC range the reading is 75 point one volts this is enough information for us to be able to use Ohm's law to calculate the output voltage from the power supply this is a schematic of the test apparatus the power supply is generating a high voltage as represented by the red arrow whose magnitude we want calculate the voltmeter cannot measure voltages this high so the applied voltage is split between the main two mega ohm resistance and the smaller 25k sense and resistance the voltmeter is connected across the sense resistor and measured seventy five point one volts we can calculate the current flowing through the sense resistor which is represented by the green arrow this current is equal to the voltage divided by the resistance seventy five point one volts divided by twenty five thousand ohms is three point zero zero four milliamps because the internal resistance of the voltmeter is relatively high we can assume that the very same current flows through both the main two mega ohm resistance and the 25k sense resistance I'll round this current to three milliamps I'll now apply Ohm's law a second time this time to the total resistance the voltage is equal to the current multiplied by the total resistance the current is three milliamps or zero point zero zero three amperes and the total resistance is two mega ohms plus twenty five K or two million twenty five thousand ohms multiplication gives six thousand eighty volts this is the voltage generated by the power supply the power supply is not perfect the voltage it generates will decrease as more and more power is drawn from it I reran the tests with different load resistances to see what would happen let me describe one such test for this test I soldered four of the 1 mega ohm blue resistors in parallel their combined parallel resistance is one quarter of one mega ohm or 250 K I used the same 25 k sense resistance as before with the voltmeter set to the 1000 volt DC range the reading was 505 volts the current flowing through the sense resistance is the voltage divided by the resistance 505 volts divided by 25 thousand ohms is twenty point two mil amps the power supply voltage is this current twenty point two milliamps multiplied by the total resistance two hundred seventy five thousand ohms the power supply voltage is therefore five thousand five hundred fifty five volts how much power is the power supply supplying the power is equal to the voltage multiplied by the current 5555 volts multiplied by zero point zero two zero two amperes is a hundred and twelve watts recall that the load resistance in this case was constructed from four one mega ohm resistors soldered in parallel each resistor is dissipating about one quarter of the power generated by the power supply that's 28 watts per resistor unfortunately for them they're rated at only three watts each so they get burning hot very quickly they cannot dissipate this much power for very long I estimate that they would start to smoke in about fifteen seconds i soldered these four resistors in various series and parallel combinations to make different resistance values so I could calibrate the power supply under different loads this is a graph of the power supply voltage which is plotted along the vertical axis as the total load resistance is varied along the horizontal axis we've already looked at the test results and calculations when the total resistance is two million twenty five thousand ohms and when the total resistance is $275,000 I estimate that the voltage will decline to four thousand two hundred volts when the total load resistance is twenty five point two kilo ohms or twenty five thousand two hundred ohms this is a graph of the power supplied by the power supply the power is plotted along the vertical axis and the total load resistance is plotted along the horizontal axis we calculated the power to be one hundred and twelve watts when the total load resistance was 275 K I estimate that the power will peak at 701 when the total load resistance is 25 point 2 K this is a graph of the current from the power supply in milliamps against the total load resistance we ran through the current calculations of three milliamps at two million twenty five thousand ohms and twenty point two milliamps at $275,000 I estimate that at maximum power the power supply will deliver a hundred and sixty-seven milliamps into a twenty five point two K load I would have liked to test this power supply up to maximum power but to do that would require a 700 watt test resistor with a resistance of 25,000 ohms I just don't have enough high resistance high power resistors lying around to build such a beast of a resistor so I had to make an estimate let me explain how I did that the secondary coil of the transformer produces a sinusoidal waveform with a peak voltage of 3000 volts the voltage drop over capacitor c1 is topped up at the positive going Peaks between these Peaks the voltage drop over capacitor c1 decreases as it supplies power to the load resistor capacitor c2 is topped up at the negative going Peaks between those Peaks the voltage drop over capacitor c2 will decay as it supplies power to the load resistor the blue arrow is the voltage drop over the load resistor at any particular instant in time the lead coming out of the front panel which is labeled zero is connected to the capacitor c2 the lead coming out the front panel labeled 6,000 volts is connected the capacitors c1 neither lead is connected to the transformers frame which is that the potential rendered in purple so the voltage drop over the load resistor is this rippled red line this is the voltage drop measured with respect to the lead coming out the front panel which is labeled zero the blue arrow shows the maximum voltage drop over the load resistor which occurs at the time of one of the peaks note that this peak voltage is less than six thousand volts at the instant when one of the capacitors is being topped up to 3000 volts the other capacitor is halfway through its decay cycle let me stop for a moment to clarify something you might be saying that something must be wrong since some of our test results already showed that this power supply can deliver more than six thousand volts that's because the transformer was generating a sinusoid with an amplitude greater than minus three thousand volts to plus three thousand volts under those circumstances the analysis I'm now doing assumes that the transformers peak to peak voltage is exactly six thousand volts now back to the analysis a voltmeter connected over the load resistor will measure the average level of the rippled voltage this average voltage will be about halfway between the maximum and minimum voltage drops over the load resistor as we decrease the load resistance we will draw more power from the power supply the average voltage drop over the load will decrease but there's a practical limit to how far the average voltage can decrease that limit is set by the size of the transformer and the amount of power it's able to deliver my transformer came from a microwave oven that was advertised as a 1,000 watt microwave oven because of the in efficiencies and losses in the transformer in the circuit we will never be able to put 1,000 watts into any load resistance we connect to the power supply I'll assume that the most we will ever get is say 700 watts the voltage drop over the secondary coil will be the plus and minus three thousand volts sine wave we've seen before if this alternating voltage is applied to load resistance R the resistor will dissipate the same amount of power as it would with a DC voltage of 2,100 volts 2,100 volts is the RMS equivalent voltage of an alternating voltage with a peak of 3000 we could ask what is the value of resistance R that will take the transformer to the maximum power level let's start with the power law that power equals voltage times current we'll apply the power law to the load resistance so V is the voltage drop over the resistor and I is the current flowing through it let's not forget Ohm's law for resistors that the current equals the voltage divided by the resistance substituting Ohm's expression for the current I into the power law and then collecting the two terms in the voltage tells us that the power is equal to the voltage squared divided by the resistance this can be rearranged to give the resistance as the square of the voltage divided by the power two thousand one hundred volts squared divided by seven hundred watts is six thousand three hundred ohms our power supply has been designed to deliver voltage at twice this level this won't change the maximum amount of power we are able to get the maximum power will still be 700 watts but we'll get only one half as much current at the higher voltage the resistance which draws 700 watts at a voltage of 4,200 volts is 4,200 squared divided by seven hundred or twenty five point two kills our power supply will be at maximum power when the load resistance is twenty-five thousand two hundred ohms the red curve is the voltage drop over the load resistance the average voltage or RMS voltage will be four thousand two hundred volts whenever capacitors c1 is topped up the voltage drop over the load will increase halfway to the secondary coils peak the same will happen when capacitor c2 is topped up on the alternating Peaks the voltage which is halfway to the peak is nine hundred volts above the average forty two hundred plus nine hundred is fifty one hundred so the maximum voltage drop over the load will be 5100 volts the average which will be about halfway between the maximum and minimum values so we expect the minimum to be nine hundred volts below the average forty two hundred minus nine hundred is thirty three hundred so the minimum voltage drop over the load will be three thousand three hundred volts we can use Ohm's law to calculate the current flowing through the twenty five point two kalo dree sister when the power supplies voltage is forty two hundred volts current equals voltage divided by resistance forty two hundred volts divided by twenty five thousand two hundred ohms is 167 milliamps a quick check using the power loss the resistance will dissipate power equal to the voltage multiplied by the current 4200 volts multiplied by zero point one six seven and Pierce is seven hundred watts good everything seems to be working well so I screwed the cover onto the base note that I built this enclosure with a 3/16 inch gap all around between the base and the cover this allows for some movement of air to carry away heat if you are going to use this power supply under maximum load for an extended period of time you might want to consider providing some active means of cooling that's all for today thanks for watching see you next time you you
