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Source and Load and the Enhanced Light Bulb method

By on June 22, 2016
light bulb method










Recently in my last repair article about the 750 Watt Power Supply I noticed that a simple Light Bulb like we Electronic Engineers previously used (about 60 to 75 Watt or so), just doesn’t work anymore. The same was the case when I was testing a 42 inch LCD TV in one of my other repairs. Because when we speak about devices that consume 140 Watt or more easily, we have to adjust our Test plan.

This article is about this conclusion, and how I used the following calculations I made to, for instance , protect the 230VAC Mains when also testing High Power Consuming Machines like Washing machines, Tumble Dryers, Magnetrons and Refrigerators. And to prevent components and Copper Layers on Printed Circuit Boards and anything internally shorting from Blowing up by destructive high currents. These Machines dissipate 2000 to more than 3000 Watt easily, even without any destructive shorting currents. And so we need to examine how we approach the protective measurements we need to take care of, like we successfully did before with just a simple Light bulb.

Another friend of mine repairs and sells these Washing machines and other Kitchen equipment in my Hometown, and he was faced by a lot of blowing up Power line Fuses, including short circuited Controller Boards. With lost copper tracks and blown up components as a result.

But first back to my previous article about the very big 750 Watt Power Supply: I had to use about 210 Watt on Light Bulbs in Series with the 750 Watt SMPS unit before my device started to work and before the LCD ATX Tester started to finally show all correct Voltages. 750 Watt is 0.75 Kilo Watt. We continue our calculations: The 750 Watt Power Supply itself needed about 750 Watt/ 230VAC= 3.26 Ampere. And therefore its internal resistance is about R = 230V/ 3.26A = 70.55 Ohm. The Supply started working only with 210 Watt Light Bulbs in series. The Current through 210 Watt of Light Bulbs = 210 Watt/230VAC = 0.91 A. So the internal resistance of the 210 Watt Bulb is about 230VAC/0.91A= 252 Ohm. 70.55 Ohm compares to 252 Ohm as 0.28 : 1.

But as said a Washing Machine still uses a lot more Power. Extrapolated from our previous 750 Watt situation we would need about 2/0.75 x 210 Watt = approximately 560 Watt on Light Bulb Power. To equal our findings. Of course our Power Supply has no Motor like a Washing machine but Light Bulbs are Inductors just like Motors are. Also the Light Bulbs are PTC components. Which means that the current through them is higher when they are cold. (Why their filaments mostly burn through when they are still cold and the switching-ON Current is much Higher). Let’s continue with some electrical theory, see next picture: First picture on the left is the normal DC situation when Current and Voltage are in Phase. Naturally there is no “ j x 2 Pi x F” component in real DC circuits. The next two pictures (ELI and ICE) are the AC situations when we use a Washing machine or PC Power Supply on the AC Power lines. And since a motor is a big Coil and our Light Bulbs also are like a kind of Inductors, the ELI picture would be appropriate.

p=voltage x current

In DC situations we write for Power P = Voltage x Current.

But in AC circuits we use P = Voltage x Current x Cosinus Phi . When Voltage and Current in a AC circuit are in Phase, we say that the power factor is 1, because when Phi = 0 degrees we get for cos 0 degrees => 1 . Which equals our formula back to P = Voltage x Current x 1 = U x I. (like in DC circuits).

Which is how the Electrical Energy Distributors like it. Because Blind Power from large Inductors or large Capacitors takes Energy from the Power Plants but doesn’t cost us any real money. However unstables our Power lines and creates large Blind currents. We can visualize this by drawing Power P, Blind Power S and Reactive Power Q in a Triangle according to the great Pythagoras. The base of our Triangle represents P, and our Reactive Power Q would be the erect perpendicular side of our Triangle. And the oblique Triangle side represents our Blind Power S.

Resulting in these formulas: P = U x I x cos Phi (also = S x cos Phi, because S = U x I),

S = U x I (is in product VA like on Transformers) = P / cos Phi, And finally Q = U x I x sin Phi Continuing we also consider the principle in how we transfer the most Power or Energy into a device, we call a Load, by taking the circuit of a Battery as an example, which is in fact the same for every Source to external Load we use :

power supply1

The EMF is the electrical Force which drives the current through both our internal Battery resistor “r” and through our Load Resistor “R”. When ”E” or our EMF is not high enough there will not be enough electrons moving from A to B. Or it only will be a very low current when the Resistor “R” is too high. Or the internal resistor “r” of the Battery is too high also. And only if the External Resistor R equals the value of r from the Source, the resulting current through our circuit and the voltage over the external Resistor R will both be at their maximum. As would be the product of current and voltage in Watt in our external Load R. Which of course is in Real Power P.

When we know the maximum current that comes out of a Power Source at zero Output Volts, and if we also know the maximum (open) voltage that comes out of our Power Source at zero current, we can draw a straight Line through both measured points. Accordingly we now exactly can predict at what output Load our voltage and our current through it will be at its maximum ! And that would also be the best Power Transfer from our Source to our external Load!

Next picture explains this, but keep in mind that you not always just can short circuit any Power Supply or Source without blowing up something in the end! But I once used this knowledge to test if I could let a Transistor Radio play onto the Telephone a and b Lines. Because I noticed that there still was a voltage of about 50 Volts DC on those Lines after I had ended my Phone subscription. And I could, which felt like a small victory in using my knowledge on electronics at the time!

maximum transfer to load formula

So we know from above circuit that the Power delivered to Resistor R will be at maximum when the Value of R at least equals the value of r of the internal Battery Resistor. Which would be in above given example at R external Load = r internal Power Supply Source = 2.5V / 1A = 2.5 Ohm ! In mathematics we would write the Real, Blind and Reactive Loads in a formula like : Z = R + jX. (again by using a Pythagoras Triangle). When we have an ideal and purely real Resistor we can just write Zr = R.

R represents any Load with voltage and current in Phase. Z is our resulting Reactance or AC resistance. And jX represents our positive or negative imaginary impedances. Positive Imaginary Impedance when our Impedance is Inductive, or a Negative jX when our impedance is mostly Capacitive. A Coil resistance can be written as Zl = j x 2 x PI x F x L, and a Capacitive resistance as Zc = 1 / (j x 2 x PI x F x C). And j is the imaginary part. That is obviously also why they call this Mathematical way of calculating Complex Arithmetic. So the resistance of a Capacitor decreases with increasing Frequency F. And the resistance of a Coil increases when frequency F rises.

Also keep in mind that a Capacitor resists at following Voltage level changes. Whereas a Coil resists to changes in Current. Both are opposite behaving components. Above Battery circuit also explains AC and DC circuits from Line input (100K Ohm) from any Amplifier, or DIN 47 K Ohm (Deutsche Industry Norm EU or the 5-pole plug, as to US ANSI) , to why we need a 75 Ohm Amplifier TV input when we use 75 Ohm Coax cable. Or why we need a 4 Ohm Loudspeaker on a 4 Ohm Amplifier Output. Simply because it generates maximum in- en output transfers. And Maximum results are what it was designed for. Or our TV simply would not recieve any Antenna signal, nor would our Amplifier inputs be able to amplify any of the attached electrical sound sources to the right level of Speaker Output Power. For comparison of these levels we often write these values in dB , because the Bell in DeciBell would be too large to use. A gain of 6 dB would be equal to twice the voltage. +30 dB is factor 1000 times. +20 dB is a factor of 100 times. +30dBW would mean 1000 Watts amplification. Whereas +30dBm would mean a gain of 1000 milliWatts. Also dBW means dB relative to 1Watt, why +3dBW would mean 2 Watt. dBm means dB relative to 1 milliwatt which equals to -30dBW. dBi is the gain of any antenna.

Back to our LightBulb story for very large consuming devices. When we calculate the Internal resistance of a 230VAC 2000 Watt Washing machine, we know that P = Voltage x Current x Cos.Phi.= U^2/R = I^2 xR = 2KW. (for convenient reasons we here say that Cos Phi =1, the actual value probably can be found on the label/sticker of the Washing machine). Also the Power Value on Transformers is given in VA (Volt x Ampère) We also know that the current through the Washing machine = 2000 W / 230 VAC = 8.69565 A.

Which gives us the internal resistance of the Washing machine = 230 VAC/ 8.69565 A = 26.45 Ohm. The Internal Resistance value of the Light bulbs should be at least equal to 26.45 Ohm or less for maximum Power Transfer from Mains to Machine. But in this case I do not think that it is necessary because that would mean 2000 Watt on Light Bulbs in Serie. And from my 750 Watt article destillated that about 0.28 x 2KW would be enough. Which represents 560 Watt on Light Bulbs.

The current through 4 LightBulbs of 105 Watt would equal a value of (105 x 4)/230 VAC=1.826A. Which gives us an internal resistance of 125.95 Ohm. And that represents 26.45 : 125.95 Ohm. Or 1 :4.7618. Which resembles 0.210 : 1. (which was 0.28 : 1 in the 750 Watt Unit Test) But a Power Supply is no Washing machine, and all we need to know is still how to prevent blowingup our Power Circuits and if we have a defect Shorting Device at all?

I made following Circuit for these High current eating Devices that now does work as it was attended for. My friend now finally is able to see looking at about four 105 Watt Light Bulbs (maybe one Light Bulb more still could increase the efficiency of the Test Circuit, but it already does work!) if the Devices are defect without blowing up Fuses or more costly parts anymore.

With a Double Pole Changeover Switch my friend also can disable the Light Bulbs from being active, and the internal Led Light in the Changeover Switch warns him instantly that the protective Light Bulbs are off. Also keep in mind that the AC Mains deviates from frequency (max 10%) as the Voltage does (5%?). And a Washing machine probably also works at a bit lower voltage than the probably average 230V.

You may click on the above picture to enlarge it

Maybe this was a lot of theory for a rather simple solution resulting in previous protective circuit. But it explains why our previous old Light Bulb method needs some enhancing in order to be still as effective at large Power consuming Devices.

Also know that when you see the Light Bulbs modulating, it probably just means that the internal SMPS PWM controlled unit is working when you attached the Bulb(s) in series for instance with a large 42 inch TV. However this doesn’t mean that there still isn’t another smaller Short circuiting Component playing bananas in your Consumer Electronics Device. Like was the case in my 42 inch LCD TV repair article.

I hope it was an interesting story that will help to understand a lot of readers why this Light bulb method is used for so long that it still isn’t replaced by other methods, probably also more expensive, methods. And why we need more Light Bulb Wattage when testing larger Power consuming Devices.

See you all in another story next time!


Albert van Bemmelen, Weert, The Netherlands.

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  1. Parasuraman S

    June 22, 2016 at 11:42 pm

    Oh my God! Limitless is your knowledge and depth in such matters and I think you amuse yourself with a lot of formulas and calculation stuff, that is a 'taboo' for my little brain! I had always scored poor marks in 'maths' (LOL)

    • Albert van Bemmelen

      June 23, 2016 at 2:27 pm

      Hi dear Parasuraman S. I also was bad in Mathematics, and I probably still am. My late Father always knew all his (a-b) x (a+b) Formulas. But I never knew why I would need those Formulas at that time. And I never won any single Checkers game from him either! Nowadays I use the Calculating parts I understand whenever I think it helps solving a problem. And even in this Era of Computers and Android Tablets and so on, Paper and Pencil still are often the quickest way to save and solve many problems. Only writing E-mails, or Letters or making documents are best done on a computer. But those little special Android Apps on Tablets are getting better all the time in help solving many Calculating Problems. (; )

  2. Matsuoka

    June 23, 2016 at 12:52 am

    Dear Albert, thank you for your hard work and detailed explanation. In fact I did apply filament lamp like you to troubleshoot problems of kitchen appliances, such as Inverter Microwave-ovens, IH rice cookers, or IH cookers, with power from 800W up to 2000W, whereas their fuses and IGBT blown, or SMPS shorted. With those a 500W tubular halogen lamp will do it well.
    Furthermore, you may know that inverter aircons use kind of 3-pole stepping motor as compressor. I met many of operation shutdown cases due to protection circuit detected something wrong in compressor driving circuit, or even the compressor was wrong itself. Replacing the suspected compressor with 3x220V500W halogen-lamps connected in star-form helps a lot. This "trick" is also applicable for inverter fridges and washers.
    A small contribution for you all. Best regards.

    • Albert van Bemmelen

      June 23, 2016 at 1:16 pm

      Thank you Matsuoka!for your very much appreciated and very interesting detailed info about using 500W Halogen-lamps in R,S,T multi Phase star-form. I tried to use a 300 Watt Halogen-Lamp in my previous 750 Watt repair article but it only lighted shortly after powering on the 750 PSU and nothing else worked including my ATX LCD Power Supply Tester. So I took the Halogen Bulb for not usable. But maybe just the 2 Halogen-Lamp poles just didn't make decent contact, which is hard to make without a Halogen-Lamp holder? Your experience in this matter invites me to re-test the Halogen-Lightbulb method again next time.

      • Albert van Bemmelen

        June 24, 2016 at 1:08 am

        Thanks, all you guys for all your nice comments.
        Today I also fixed a new Contruction Lamp with a 400W Halogen
        lamp in it , by attaching an outlet power strip and a 230VAC
        input cord to the Light. This way testing any Device is now
        just a matter of simply plugging it in into the power strip.

        It probably is about the same as using four 105 Watt Lightbulbs
        as one 420 Watt Lightbulb in serie. Like used in the article.
        500 Watt tubular Halogen-Lamps will probably even work better
        but I couldn't find them in my local Store. And the reflector
        with the Halogen Lamp holder only allows 400 Watt tubular Lamps.
        But it indeed does work perfectly when tested on a 410 Watt Sirtec
        PSU and a 300 Watt Power Supply. The mentioned 750 Watt Power
        Supply is already placed back in the Big Tower of my friend's son,
        and works like a charm. So I couldn't re-test it with my new
        400 Watt Halogen Test Device I made today. But it probably also
        works fine when attached to the 750 Watt Power Supply.

  3. Albert van Bemmelen

    June 23, 2016 at 1:40 am

    PS: the frequency of the Dutch AC Mains power lines probably differs less than 10% most of the time from our 50Hz. And it is said to be one of the best distributing AC lines in the World. But I read that the frequency is 50Hz between a minimum of -4% and a maximum of +6% which in total is over a range of 10%.

  4. Martin

    June 23, 2016 at 1:53 am

    Hi. Does the light bulb have to be the old incandescent filament type or can you use a Halogen one (or other) as the old type are becoming scarce?

    • Albert van Bemmelen

      June 23, 2016 at 1:33 pm

      Maybe Martin, the Halogen tubular Lamps do work just as well according to the Posts from Matsuoka and others. But only because I had no luck with a 300 watt Halogen Lamp in my previous 750 SMPS repair article I used 210 Watt old standard Light Bulbs (3 of them in parallel). But the Halogen Lamp contacts are hard to connect without the right lamp holder. And they do get very hot instantly which also can be dangerous when the glass splinters, when you for instance would accidentally poor a drop of cold liquid on it. I therefore also tried to use the complete Lamp with the 300 Watt Halogen tubular Bulb and the Triac Dimmer control set to maximum but that didn't work either.

      • Albert van Bemmelen

        June 24, 2016 at 3:36 am

        PS: Martin, see my conclusion in the Post above about the today
        made new Contruction Lamp with a 400W Halogen lamp in it ,
        by attaching an outlet power strip and a 230VAC input cord
        to the Light. This way testing any Device is now
        just a matter of simply plugging it in into the power strip.
        So also a Halogen tubular lamp works fine! A 500 Watt Halogen
        Lamp would probably work even better but couldn't be bought in
        the local Store of my Hometown.

  5. oliver

    June 23, 2016 at 3:50 am

    Hi Albert had same fault ...and I tried ( 2 halogen tube in paraell 250 watts ) that did the trick it was easy to build

    • Albert van Bemmelen

      June 23, 2016 at 1:44 pm

      Thank you Oliver! for your much appreciated and valuable information. I tried the tubular 300 Watt Halogen Lamp in serie in my previous 750 Watt repair article but it just didn't work. so I used 3 standard (2 x 75 Watt and one 60 Watt Lightbulb) instead.
      But maybe it does work also, as you and other readers wrote in these Posts. I will try this again myself as soon I have found some decent Halogen tubular Lamp holders to make a decent test device with it. It could indeed prove very useful!

      • Albert van Bemmelen

        June 24, 2016 at 4:38 am

        I take it Oliver that you used 2 Halogen light Bulbs of 250 Watt each creating in total a 500 Watt Protective Lamp?
        The Largest I could buy in my Hometown were 400 Watt Halogen tubular types. See my today's Post above.

  6. Peter Owens

    June 23, 2016 at 8:30 am

    A tour de force in theory that made my head hurt.
    I don't mean to be picky but where in heck do you buy 105 watt light bulbs, especially the incandescents which are becoming rare.
    Could you not relace with a wire wound resistor & CFL lamp?
    Or small heating element?

    • Albert van Bemmelen

      June 23, 2016 at 1:03 pm

      Hi Peter, I wasn't aware that those 105 Watt Lightbulbs were still sold. But apparantly buying them in The Netherlands at a shops like one called "Action" is no problem. And only because in my previous 750 Watt repair article a 300 Watt Halogen Bulb didn't work we had chosen Lightbulbs which maybe are the last you can buy? But according to above Posts from other readers those bigger Halogen Bulbs also do work?

  7. Yogesh Panchal

    June 23, 2016 at 3:45 pm

    well explained! thanks for sharing .

  8. Albert van Bemmelen

    June 23, 2016 at 6:03 pm

    Thanks Yogesh. Glad you liked it! I could have gone into more
    detail but it would have exceeded the current 6 pages it now took.
    Like why the dB (Decibel) notation for values in Gain and
    Amplification is used. Because our Ears are hearing only
    Logarithmic amplification differences. +6dB would mean 4 times
    gain in Power. Or in other words +6dB means twice the gain in
    Voltage, which is 4 times the gain in Power. ((U x 2)^2 /R)=P.

  9. beh

    June 24, 2016 at 11:40 am

    your articles shows how deep and deeper you can dive in subjects.

    • Albert van Bemmelen

      June 25, 2016 at 12:03 am

      Thanks Beh, I do tend to go in too deep in the matter sometimes. But as
      you know the Base of Knowledge is always wider than the top (;).

  10. Paris Azis

    June 24, 2016 at 8:11 pm

    Hi Albert

    I faced the same weird behaviour when testing a Tagan 550W PC PSU, after I had repaired it. A 100W lamp was glowing periodically. Then I started to add in parallel more 60W incandescent lamps and the proper function occurred at 240W total lamp power.
    Anyway, I think that you went too deeply in this topic, not in vain, but nevertheless without this to be necessary.
    In my humble opinion, given that this test practically exploits the thermistor-like effect of the incandescent lamps, should be used under the rule of using the 60% maximum in lamp power in reference to the nominal power of the device under test. Less than that causes these weird effects, whereas a higher percentage is meaningless in terms of protection of the D.U.T.
    I have seen a technician using a 100W test lamp when testing a 5W SMPS of a TV decoder!! It is obvious that there is no protection in this case. In fact the lamp is protected, with the little PSU being the protecting circuit!! As always, we should use common sense before applying a method for troubleshooting. And above all we must understand the essence of the specific test we intend to apply, before applying it...
    As for the high loads of the KW range, I would prefer either the Variac solution, or some kind of electronic fuse, working in a similar manner with that of the current limit adjustment in our bench PSUs.
    Thank you for the effort you put in this article and for the thorough presentation of it.

    • Albert van Bemmelen

      June 24, 2016 at 11:36 pm

      Thanks Paris, the 60% maximum rule in Lamp Power in reference to
      the DUT seems like a good way to protect a device. But if we could
      easily change the used Lamp Power this reference could change easily.
      After all your example with the Tagan 550 Watt supply that needed 240
      Watt on Light Bulbs, gives a lower figure of 240/550 x 100% = 43.6%.
      And so did my test with the 750 Watt SMPS unit as D.U.T. with 210
      Watt on Lightbulbs, giving another reference of 210/750 x 100% = 28%.
      And indeed the Halogen Lamp method is only to be used in the higher Power
      Dissipating range like the Washingmachines, and Tumble Dryers
      mentioned. The Variac method you mentioned (which is nothing more than an
      Autotransformer were the secundary voltage is -like the middle wiper
      contact from a potentiometer- taken directly and adjustable from the
      only coil which is the primary coil) seems like a very expensive device at higher currents compared to the Light Bulbs. And is I think unsuitable for
      most electronic devices since often lowering the voltage would increase
      the current through the device. I have had a few examples in the past
      were the Fuse in the device blew up because of lowering the VAC input
      voltage. The trouble with using a Lamp as protection in the KiloWatt
      device range is that the electronic components on the Controllerboard
      still can get a much too high short current through them because of the
      lower Lamp serie resistance needed. In this respect I think back on the
      DUOYI DY294 component tester which is a unique device that protects
      any component under test even at high voltage from any destructive
      avalanche current instantly. Which would be a terrific way to protect
      any device under test if it could be translated into a universal Device protector.

      • Paris Azis

        June 26, 2016 at 10:33 pm

        Dear Albert

        Answering your comment in a bottom to top manner in reference to your text, it seems to be obvious that we have different approach about the subject, but in any case this is fully respectable.
        Let me please clarify my view on it. First of all, I don’t trust and therefore I don’t use fancy-cheap testers. When troubleshooting, for static tests, I work permanently with my trusty multimeters set in the “diode” test-position and then I check all the semiconductor devices in this way, both for their functional status and for leakage testing.
        In my humble opinion, being myself a repairs’ technician and not a Quality Assurance clerk, this instrument you propose offers me nothing. It is rather suited for design engineers than repair technicians. I checked the specifications of it and what it can do as well, before expressing my opinion.
        As a repairs technician, I never felt the need to measure the gain of a transistor, or its collector breakdown voltage. All I am interested for is to test the proper function of any semiconductor device and this is easily performed using the “diode” test of the multimeter. Leakages are equally easy detected using the same method. No problem at all.
        On the other hand, in really weird cases, I use a combination of my oscilloscope, set in X-Y mode, along with a self made V/I curve detecting instrument which I built 30 years ago and I can assure you that there is no problem that you cannot identify and solve using this method. None at all! Therefore these fancy instruments have nothing important to offer me.
        As regards your reference about the increase of current when reducing the input voltage which blows the fuses (I have never faced such a case), this must have happen to you by simple coincidence. I support this idea because this increase of current can happen only when testing SMPSes, and even in these cases this phenomenon is both natural to happen (the pulse width widens as the input voltage drops in order to adjust the output to the desired-preset level and therefore the input current increases) and limited as well (because the supervising circuits within the PWM chip will always limit the maximum pulse width in order to prevent pulse overlapping). Your cases, if not coincident, they were pulse overlapping failures, which are the most difficult cases to resolve in push-pull SMPS circuit topologies.
        So in every other case, the Ohms Law is linear and with a fixed load (whatever machine it is) when reducing the input voltage the current cannot increase. This is crystal clear. Therefore the Variac test is absolutely reliable. The cost of buying a variac for using this method depends of course upon its nominal power. But there are, again, different approaches in testing.
        As regards these differences, I really don’t see any good reason to test a washing machine for example, as a total entity, using the lamp test. Of course a relevant variac would cost too much. But why should a technician test the entire unit at once without a “problem break down” procedure? Therefore, applying common sense, one has first to locate the problem and then resolve it.
        Although I am not involved with washing machines’ repairs, I see three major sub-parts to be checked. The water heating resistances (their test can be done by a simple Ohm’s testing both for continuity and leakage, therefore no gigantic variac is necessary), the motor (it can also be tested separately in most cases, or be replaced by lamps which in any case will be of much lower wattage than 1 KW) and finally the electronics, which is the easiest case for the lamp test.
        Even the Matsuoka’s method (thank you Matsuoka for your contribution as well) can be applied with much lower lamp power, I believe, because what we need to identify in this case is that the IC which creates the three phases A.C for the operation of compressor (using the rectified D.C line voltage in its input) works properly. I see no reason to use high load on this test. If an IGBT is shorted out this will be immediately shown by the lighting difference of the three lamps (meaning asymmetrical current between phases).
        Closing here, I didn’t understand your second phrase, the one before the reference of the percentages (and including them). (I already stated that this 60% is a maximum lamp power and both of your percentage references are below that limit, so it seems that we agree on that. So, what exactly did you mean? I didn’t get it).
        Thanks again for the effort you put in the detailed analysis of your article.
        I also apologize for the lengthy text.

        • Albert van Bemmelen

          June 27, 2016 at 2:47 am

          Thanks Paris for your prompt reply. I'll try to answer on your
          thoughtful comments.

          I understand that with fancy-cheap Testers you were referring to
          the DY294 component Tester I mentioned. It may be cheap all right
          but it really delivers a unique testing method at High Voltages up
          to more than 1600 Volt DC that even is capable of testing vulnerable Leds and Zeners without even destroying them in the process!
          And I still do keep using my universal (also cheap Digital LOL)
          multimeter in the Diode/Beep test-position like you do.
          I think however that being able to measure the Break-Down Voltage
          with the DY294 of any Semiconductor, Capacitor or VDR is of great
          importance, because you still can't do this without normally
          blowing up the component under test in the process!
          Neither by using a X/Y curvetracer, an Oscilloscope and so on.
          About using a Variac, Fuses can get blown even on no SMPS circuits
          or SMPS Devices.
          Because primary Fuses in Linear Transformer circuits in for instance Stereo Recievers that use a circuit with selectable AC input Voltage can get blown while decreasing the input Voltage below their standard input value. It happened! Probably because the 110/220V circuit Fuses use different currents.
          I do not want and probably need to defend the Lightbulb method while testing Washing Machines and so on, because it works and is a safe first test to check for High Power shortings without blowing up anything in the meanwhile. Of course afterwards we need to test every single part and circuit internally aswell like you said, but since my Friend recieves hundreds of defect Tumble Dryers and Washing Machines simultaneous he needs to establish with a first quick test what Machines are worth Fixing first !
          I hope you see why the article was written and also why the DY 294 deserves much more appreciation. Because also those PTC and HV
          Capacitors in the High Power Machines often need to be tested on HV Voltages on their Break-Down Value. As also often the Polyswitch
          Fuses (PTC) do need to be replaced in LCD TV Sets. Because their
          value differs often to much in Devices they need to be replaced.

  11. janyves

    June 26, 2016 at 8:23 pm

    Sir.. All you said here is preventing shorts while troubleshooting machines. Well all I do is testing with a limited source of current flow(fuses,breakers) I'll love if this method of yours can be illustrated for me. All I use incandescent bulbs for is transformer short circuit testing. You showed a wonderful theory .....

    • Albert van Bemmelen

      June 27, 2016 at 2:02 am

      Preventing live short currents by using incandescent light bulbs is no
      theory because you are already using this method while testing
      Transformers circuits. So that needs no further illustration.
      But because the standard Lightbulb method does not work as well under
      all circumstances we need to increase the Lamp power in Watt used
      matching to the higher Power consuming Devices we will be testing.
      Like when we are testing a Washing machine. I didn't show any wonderful
      (but thank you! LOL) theory, but I merely explained why and when it
      works as it does in practise.

  12. bassbuddy

    June 27, 2016 at 1:56 am

    Albert, After reading your article I have some comments and additions I would like to share:
    To make reading easier, every comment is preceded by a quote from your article between slashes /…/

    1./Light Bulbs are Inductors just like Motors are/:
    This is not true. A light bulb is nothing but a (PTC) resistor.The very tiny induction formed by windings in the filament can be ignored (at 50 or 60 Hz). If what you write were true an incandescent light bulb (that is where we are talking about) would have a bad cos Phi and would be unacceptable for use on the mains. They may have one thing in common, a high inrush current.

    2. /First picture on the left is the normal DC situation when Current and Voltage are in Phase/:
    DC and phase? With DC no such thing as phase exists! The described case is an AC source loaded with a purely resistive (as is mentioned below it) load.

    3. /But in AC circuits we use P = Voltage x Current x Cosinus Phi/
    That is right, as long as the mains load consists of a combination of passive components, resistors, inductors and capacitors.
    It is good to know that nowadays the FORM of the load current is also of great importance. In many cases the shape of the load current shows little resemblance to that of the mains voltage. Large peak load currents as drawn by, for instance, SMPS-s that directly rectify the mains (and most do) have an in-phase portion of the current that is relatively low, so the average dissipated power is low, and at the same time they generate a lot of harmonic currents in the mains causing a lot of disturbance.

    4. /Blind Power from large Inductors or large Capacitors takes Energy from the Power Plants/
    Blind power takes NO energy from the Power plants (this is the reason why you do not have to pay for it), but large CURRENTS. In case the total load of a power plant would have a bad cos Phi large out-of phase currents would have to be supplied by the power generators (with risk of damaging them, and transported by the mains wiring and would require thicker wiring to reduce transport losses. This shows that the cos Phi is a very important parameter. Big inductions should be compensated by switching capacitors in parallel.

    5. /…know the maximum current that comes out of a Power Source at zero Output Volts/:
    That's the shorting current. Why not say so?

    6. /But I once used this knowledge to test if I could let a Transistor Radio play onto the Telephone a and b Lines. Because I noticed that there still was a voltage of about 50 Volts DC on those Lines after I had ended my Phone subscription. And I could, which felt like a small victory in using my knowledge on electronics at the time!/:

    What is the relevance of this? It only proves that the phone line in this situation was effectively protected against damage or disturbance in case irresponsible people try to use the line for purposes it is not meant for such as "stealing" power from it.

    7. / So we know from above circuit that the Power delivered to Resistor R will be at maximum when the Value of R AT LEAST equals the value of r of the internal Battery Resistor. Which would be in above given example at R external Load = r internal Power Supply Source = 2.5V / 1A = 2.5 Ohm !/

    7b./In mathematics we would write the Real, Blind and Reactive Loads in a formula like : Z = R + jX. (again by using a Pythagoras Triangle). When we have an ideal and purely real Resistor we can just write Zr = R./

    These two things have nothing to do with each other.
    The first deals with the optimum load of a voltage source no matter DC or AC.
    The latter with the phase of the load current in comparison with an AC voltage source. The only common factor is the Pythagorian triangle, like (to use another example) the maximum working height on a ladder of a given lenghth placed against a wall in a given angle.
    It is important to know that the optimum power transfer is achieved when the load resistor has EXACTLY the same value as the internal resistance of the source.

    8. /Also keep in mind that a Capacitor resists at following Voltage level changes. Whereas a Coil resists to changes in Current./:

    To say more clearly what you mean and respecting the passive nature of the components:
    A capacitor voltage CHANGE can only be achieved by a current flowing through it.
    A inductor current CHANGE can only be achieved by a voltage put across it.
    It also makes it easy to understand the nature of the load current phase shift when these components are connected to an AC source. In a capacitor the phase of the current will lead and in an inductor the phase of the current will lag.

    9./ Above Battery circuit also explains AC and DC circuits from Line input (100K Ohm) from any Amplifier, or DIN 47 K Ohm (Deutsche Industry Norm EU or the 5-pole plug, as to US ANSI) , to why we need a 75 Ohm Amplifier TV input when we use 75 Ohm Coax cable. Or why we need a 4 Ohm Loudspeaker on a 4 Ohm Amplifier Output./

    This listing suggests that all these systems are intended for an optimum power transfer in a system with a given system impedance. Only the one concerning the 75 Ohm Impedance of TV Coax systems is (partly) true.

    a. Line input (100K Ohm) from any Amplifier: The associated line outputs are of much lower impedance. A high line imput impedance combined with a low source impedance guarantees that the line output signals do not deteriorate by means of the load impedance with respect to distortion and frequency spectrum.

    b. The same goes for the DIN 47 kOhm inputs. The only exception is the Phono input. Here the input impedance must be 47kOhm. Phono pickup cartridges are quite sensitive to variations in load impedance, even too long connecting cables may distort the signal. The cartridges are optimally designed for a 47kOhm load to give the best performance. But again this has nothing to do with optimal power transfer.

    c. The loudspeaker impedance for a given amplifier is only important because of the current and power limitations of the power transistors used in the final stage. An amplifier with a 4 Ohm internal resistance in its output would have very poor performance (for instance damping factor). Besides this the same amount of power as delivered to the loudspeakers would have to be “cooled away” inside it.

    d. I will explain why I stated that the 75 Ohm Impedance of TV Coax systems is (partly) true. It is true for as far optimum power transfer is concerned.
    The TV cable system has a characteristic impedance of 75 Ohms. This must be respected over the whole system. Deviating from these 75 Ohms results in an unequal distribution of power (standing waves) and reflections in the cable system which result in distorted TV images (like ghost images). Mismatching of impedances also result in strong deterioration of the screening quality of cables. Cables will start radiating and disturbing signals can enter the system. For modern systems where also more and more data are transferred over the lines impedance mismatch can easily result in disastrous data loss. These factors are much more important than the optimum transfer of power.
    A few dB’s loss of the signal is not a problem to your TV set except for very weak signals.

    10. /dBi is the gain of any antenna/

    The factor dBi is the antenna gain in relation to an isotropic antenna.
    In short an isotropic antenna is a VIRTUAL, so a non-existing antenna with 100% efficiency that radiates equally in all directions, it sometimes is referred to as a point source.

    In electronics numerous relative factors are used.
    One example, the dBm is used in audio, related to a voltage of 0.775V. This is the voltage that would generate 1 mW in a 600 Ohms load. This does not mean that it is used particularly in a 600 Ohms system. Apart from telephone systems it seldom is. The dBm is also used in RF, for example in 50 or 75 Ohm systems where the reference voltage would be approx. 0.224 or 0.274 V.

    11. /When we calculate the Internal resistance of a 230VAC 2000 Watt Washing machine, we know that P = Voltage x Current x Cos.Phi.= U^2/R = I^2 xR = 2KW. (for convenient reasons we here say that Cos Phi =1/

    The part of a washing machine consuming most (almost all) power is the water heating element. Once the water has reached its desired temperature the load on the mains is strongly reduced and consists of only motors, pumps, the controller etc.
    The heating element is a pure resistor so during the heating phase the cos Phi is actually very close to 1.
    No assumptions necessary….

    12. /Which gives us the internal resistance of the Washing machine = 230 VAC/ 8.69565 A = 26.45 Ohm. The Internal Resistance value of the Light bulbs should be at least equal to 26.45 Ohm or less for maximum Power Transfer from Mains to Machine.?

    What??? Is this 26.45 Ohms supposed to function as the internal resistance of the mains or so?
    Maximum power transfer from the Mains to the machine is in the case without a serial resistor, namely 100%.
    The power loss achieved with a 26.45 Ohms resistor is 75%. The mains is loaded with twice the resistance of the washing machine (the heating element), resulting in a reduction to 50% of the normal power. These 50% again are divided equally between the resistor and the machine so only 25% of the normal power goes to the washing machine! What is the scientific support for this?

    13. /Also keep in mind that the AC Mains deviates from frequency (max 10%) as the Voltage does (5%?). And a Washing machine probably also works at a bit lower voltage than the probably average 230V.?
    First I wonder what may be the influence of a 10% deviation of the mains frequency? Or 5% of the mains voltage if you call 75% less “a bit”?
    Probably… a bit….probably average… How inaccurate can you be? This strongly smells like guessing.

    14. /And from my 750 Watt article destillated that about 0.28 x 2KW would be enough./
    Well, and saying so what is all this “theory” good for? You could have left out all of it and make this decision in the first place. But for what reason? What is the common factor between the power supply and a washing machine that justifies this decision?

    Further remarks:

    1. The article contains quite some wrong, irrelevant and misinterpreted information and inconsistancies.
    2. In the whole article phase shift is not relevant
    3. The same goes for the optimum power transfer theory.
    4. Washing machines: The amount of supplied power you choose is in fact related to the power dissipated by the heating element. The testing of a washing machine using this method still leaves the possibility to supply a large amount of power to it, enough to damage the electronics (the controller with its power supply, the part that will probably interest us most). It seems much more effective to me to temporarily disconnect the heating element to be able to use a light bulb of much lower power for concentrating on testing the electronics.
    5. Generally you may say that: The lamp is used to detect a shortcut in the tested equipment, at the same time protecting it. The power of the serial lamp should in each case be related to the power normally used by the equipment under test as also stated by Mr. Paris Aziz. Maybe a good idea is to use as serial resistor an array of lamps of different powers, each with a switch in series. Start with the lowest value and add more lamps by means of switches.
    6. Halogen lamps are also incandescent lamps. The difference is that the filament works on a higher temperature made possible by using a halogen, which results in whiter light and a longer life (for the lamp). Halogen lamps can be used in the same manner as normal incandescent lamps.
    7. The mains frequency in the Netherlands in very accurate. The percentage stated is the max momentary deviation. The average is so good that an electric clock that uses the mains as clock frequency never needs to be adjusted,apart from compensation for daylight saving time.
    8. A triac dimmer cannot be used to reduce power because it switches the mains on and off in a 100 (or 120 Hz) pace. If you reduce the power to 50% the first half of every half period is switched off leaving the same peak value. Direct rectifying of the mains (as the majority of computer power supplies do) will result in the same voltage (not considering the load) and very awkward current shapes. Variacs are better but very expensive and must themselves be protected against short circuit

    • Albert van Bemmelen

      June 27, 2016 at 4:59 pm

      Thank you for your comments bassbuddy.

      I will answer on your 14 points of additions if I may starting with point 1.

      1. Lightbulbs are no inductors but PTC resistors at 50/60 Hz. True but they still are coils with as you say very tiny inductions. Because they are no real Resistors or Capacitors either. That is why I also wrote “kind of inductors”. But I get your picture.
      2. In DC there is no Phase. Of course. Also true is that the effective (Veff) voltage of any AC voltage generates the same amount of heat in a resistor as a DC value of that voltage does.
      3. The Cos Phi is of importance when the ELi or iCE influence needs to be corrected like you say. Of course. I never said otherwise.
      4. Blind Power takes no Energy from Power Plants but large Currents which would require thicker transport wires. That may be true but still those large Currents also generate more Power Loss in the Wiring. And Energy is Power X Time. So it is still a bit of the same in result to the transport of Energy.
      5. Why not just say Shorting Current instead of Maximum Current at Zero output Volts. How is that not the same thing?
      6. Relevance of using the a and b Telephone lines to state the in point 5 given example is that it is working and proves the theory in practise! Inclining that I was shorting the telephone lines as being a irresponsible person is not relevant because I used a lower Resistor value in preventing completely shorting both lines.
      7. R = r in and Z = R + jX have nothing to do with each other you say. But they do because as you wrote the Phytagorian Triangle is the Common factor!
      8. In a Capacitor the I phase will lead the V phase in a AC circuit. Yes this already also was shown in the iCE diagram.
      9. Only the TV 75 Ohm optimum Power Transfer is (partly) true. And 100 Kohm inputs have nothing to do with their outputs. Of course, but that doesn’t mean my explanation with the Battery circuit as simple example is wrong.
      10. dBm voltage references may differ. I know.
      11. P = U x I x Cos Phi. No assumption necessary because a Washing Machine is purely resistance. Point taken, but a Washing Machine was just an example of a high Power consuming Device.
      12. Of course the 26.45 Ohm serie resistance of Light Bulbs normally is not there. But that’s why I also said : At least 26.45 Ohm or less. The scientific support as you say is that I really needed at least 0.28 x 750 Watt on Light Bulbs before my 750 Watt Power Supply started working. This may be seem unimportant but it was a given fact!
      13. I never claimed in my article that the Power Loss with a serie resistor of at least 26.45 Ohm was neglectable! Neither did I say that 75% less Power was! Of course as a result it only means 25% Power to the washing Machine left! Your futher remarks on how inaccurate I can be, probably ends in how inaccurate you say I am? Some people always agree to disagree whatever they hear. I wrote that the frequency of the 50 Hz is between a minimum of –4% and a maximum of +6% which is in total over a range of 10%. So I wasn’t guessing or making up anything.
      14. I never wrote about any 0.28 x 2 KW rule! My article explained the why’s and when’s with real measured facts.
      Thank you bassbuddy for your sincere additions to my article in contrast to those people who keep pressing those dislike buttons without contributing any positive vibes. I appreciate it!

  13. Albert van Bemmelen

    November 4, 2016 at 12:59 am

    In addition following post can be very Helpful:

    This simple addition is an explanation on how you can easily see if a device is shorted or not.
    The lights remain switched on for this test in series as usual but now connected with an additional voltmeter to the outlet where the device under test is connected.
    When the Lights are full burning there is a short circuit in the device, and when there is no short circuit but a high current flows they burn even so but something less strong.
    Because it remains difficult to see how bright the lights are on, and whether there has been-or-not a short circuit is an conclusion you can do with this simple extension. With which we now also can see whether there is a short circuit on a (digital) Multi / voltmeter. (in AC mode to with a max voltage selection higher than 230V AC of course).
    Here are the Worked out examples where no one should have any problem with:
    Assume that we use 400 watt bulbs in series (we never stop using Light Bulbs as Short Circuit protection on a device under test) => R lamps U ^ 2/400 = 132.25 Ohms. (Always constant).
    U = 230 VAC always is always Voltage ^ 2 = 52900.

    Example 1 with good Washing Machine with P = 3000 watts.
    P washing machine = U x I = (U ^ 2) / R = (I ^ 2) x R and
    R = 52900/3000 Watt = 17.63333 Ohm.
    U over Washing Machine is about (R washing machine / (R washing machine + R lamps)) x 230 = 27,049V AC.
    In short circuit, this voltage will be much less than 27.05 volts !! And you may NOT directly connect the device without the lamps to the outlet.
    Example 2 with microwave of 1500 watts.
    Microwave R = U ^ 2/1500 watts = 52900/1500 = 35.2666 Ohm.
    A functioning microwave of 1500 watts = (35.2666 / (35.2666 + 132.25)) x 230 = 48.42 V AC.
    Example 3 with a coffee maker of 1000 Watts.
    Coffeemaker R = U ^ 2/1000 watts = 52900/1000 watts = 52.9 ohms.
    You have a good Coffeemaker = (52.9 / (52.9 + 132.25)) x 230 = 65,71V AC.
    ETC. ETC.
    In other words connect a voltmeter simultaneously over the device under test in parallel, and you can just read each on the Voltmeter Display if it is a short circuit or not.
    And the lamps in series protect against blowing any fuses!
    So if you have a device of 3000 watts and the voltage across it is much less than the calculated 27V, then there is a short-circuit!
    You can therefore advance for each device determine which voltage must be present if the device is good.



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