Source and Load and the Enhanced 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.
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 :
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!
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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