A Simple Circuit For Measuring Mains Power Consumption

By on December 5, 2016



homemade-wattmeter-circuit

 

 

 

 

 

 

 

 

I am sure that in many cases you would like to confirm if the wattage of the equipment under repair you have on your bench is within normal limits, or, at least, to identify that its power draw after a repair stays in its nominal value or if something is still wrong with it and draws more power.

Of course I know that nowadays one can find in the market cheap digital equipment (unfortunately unavailable during the time I was professionally active) which fulfill this requirement. Nevertheless the circuit shown below is a proposal for you to build it, just because it’s the simplest way to measure power. It works in cooperation with an ordinary digital multimeter which is capable of measuring AC millivolts.

CAUTION: It is always dangerous to work with mains voltages, especially if you happen to be a novice. Unless you master the basics of electricity and you know very well what you are doing, please, do not try to use this jig.




Next to that, it is always wise to use an isolation transformer when testing or repairing various mains connected devices. This will protect you effectively against the deadly “phase” which otherwise will always be present there waiting for the chance to strike you! So, respect your life and stay safe!

This, below, is the circuit. wattmeter-circuit

I came across it at the end of the nineties, surfing in the web… And I used it extensively since then…

The circuit is voltage-specific. This means that if you use it in a different mains voltage the result you will get will be erroneous. It was originally designed for 220V AC nominal mains voltage. Recently I had the time to analyze it in order for me to understand its designer’s thought when developing it and then I modified it for 230V mains voltage. Afterwards I thought to redesign it for 120 V mains voltage users and share the entire entity with all of you.

The circuit produces (theoretically) exactly 1mV/Watt readings on the multimeter’s display. This is the base of its design while its beauty is its simplicity. It is implemented by using only four fixed value resistors, without the need of adjustments after its implementation.

It is obvious that the circuit is a simple application of Ohm’s Law. The two in parallel connected 0,47Ω/5W resistors equal an equivalent resistor of 0,235Ω in value, namely the half of the individual nominal value of the two resistors composing it, but of double their nominal power, which is 10W now. Another characteristic element of the circuit, based on the same fact of being purely resistive, is that it practically measures the current the load draws from mains.

The trick is that the resistive divider, connected across the main shunt resistor which the two power resistors form, “steals” a part of the AC millivolt voltage which represents the value of the current flowing through the load, resulting in direct indication of Watts (when, of course, the load is a pure resistor, like an incandescent lamp). In that aspect you can call it a quasi-power meter, or a pseudo-power meter. Nevertheless the indication you get in your multimeter’s display, using its AC mV range, represents the apparent power your load draws from mains.

You can see it functioning below, indicating in the display of my multimeter the power consumption of that minion 40 Watt lamp I use for tests, in comparison with an ordinary, cheap power consumption meter of our modern time. homemade-wattmeter

If you compare the indication of mV on the multimeter and the respective indication of Watts on the power meter, at the center of its display, you will see that they are the same.

In addition, on the display of the power meter, at its down left side, you can also see the power factor in percentage form, showing 100. This proves that the load under measurement is purely resistive or in other words that the cos φ is 1, that is, there is no phase difference between the voltage and the current within the load and what we see in this case in the multimeter’s display is the real power consumption that this (resistive) load consumes.

This 0,235Ω shunt resistor is connected in series with the load. Depending on the amperage the load draws from mains, a voltage proportional to that current develops across its terminals. Furthermore the two 0,25W resistors of 33K and 2,2K respectively, connected in series forming a resistive voltage divider, are finally connected in parallel to the current measuring resistor. The voltage which represents the power consumption of the D.U.T, is developed across the terminals of the 33K resistor. The terminals of this resistor are the connection points to the digital multimeter’s voltage inputs, while the instrument is set to measure   `.

Let’s analyze the circuit in order to understand how it works.

Suppose that a (resistive) load of 1W at 220V AC is connected to it, as shown above, in order to measure the power it draws.

This load will draw from mains a current of:

I = P / U = 1W / 220V = 4,545mA

This current (applying Ohm’s Law) will develop across the 0,235Ω shunt resistor a voltage of a value:

 

U = I * R = 4, 545mA * 0,235Ω = 1,068 mV

 

This voltage will cause in turn a current flow through the total divider resistance of (33+2,2 =) 35,2K :

 

I = U / R = 1,068mV / 35200Ω = 30,34nA

 

And this current will finally develop a voltage across the 33K resistor:

 

U = I * R = 30, 34nA * 33000Ω = 1,001mV

 

This is the voltage the multimeter reads for 1W consumption.

One could wonder about the loading effect of the 35,2K resistance on the 0,235Ω current shunt. Well, the 35,2K resistor is about 150.000 times bigger than the 0,235Ω one and therefore we expect that it won’t affect its nominal value resulting in unacceptably faulty readings.

Anyway let’s examine mathematically this issue as well.

The total shunt resistance, namely the variation of the 0,235Ω of the shunt resistance due to the effect of the parallel connected resistive divider to it, will be:

 

Rtot = R1 * R2 / R1 + R2, where R1 = 0,235Ω  and R2 = 35,2K

 

Replacing these values to the above formula we get:

 

Rtot = 0,235Ω * 35,2K / 0,235Ω + 35,2K = 0,2349Ω

 

This simply means that the loading effect is negligible and practically does not affect the accuracy of the measurements. One could also wonder what the maximum power that can be measured with this jig is. Let’s examine this question too.

This maximum power to be measured depends on the maximum power the overall shunt resistor can withstand without suffering thermal destruction. So the starting point to assess this power is to take the shunt resistor’s nominal 10W power as a base for the calculations. For safety reasons again, we limit this power to 60%. This means that we will not allow the shunt resistor to consume more than 6 Watts.

Solving the formula: Pmax = I^2 * R for I, we get:

 

I = √Pmax / R = √6W / 0,235Ω = 5,05A

 

This will be our maximum permissible current flow through the shunt resistor. This current will flow through the device under test when it consumes an amount of power of:

 

Pmax = 5,05A * 220V = 1111,6 Watt, or, rounded up, 1100W

 

However, momentarily, just for a single quick measurement only, we can exploit the full nominal power of the shunt. In this case the shunt must withstand a current of:

 

I = √Pmax / R = √10W / 0,235Ω = 6,52A

 

And will measure an amount of power of:

 

Pmax = 6,52A * 220V = 1435W, or, rounded up, 1400W

 

Needless to say that if your multimeter is not auto-ranging and the lower range of your AC V is limited to 200mV, like mine, for measurements related with power consumptions above 200W you should change the range in advance, to the next higher one.

If we apply the same mathematical analysis again, as above, for 230V mains nominal voltage, this circuit, for 1W, will deliver at its output only  958μV instead of 1mV, namely 42μV less than required, always theoretically speaking, which introduces an error of minus 4,2%.

To correct this error, in order to use the jig on 230V AC mains, we need to replace the 2,2K resistor of the resistive divider with another one (theoretically again, as calculated) of 675Ω. Because this value does not exist in the common E12 series of resistors, we can chose the nearest to it of 680Ω which meets the requirement almost perfectly.

In the above photo, showing the power measurement of the 40W lamp, I had already replaced this 2,2K resistor by its 680Ω counterpart.

Verification:

 

This 1W load will draw from 230V mains a current of:

 

I = P / U = 1W / 230V = 4,347mA

 

This current (applying Ohm’s Law) will develop across the 0,235Ω power resistor a voltage having a value:

 

U = I * R = 4, 374mA * 0,235Ω = 1,0217 mV

 

This voltage will cause in turn a current flow through the divider consisting of 33K+680Ω = 33,68KΩ resistance:

 

I = U / R = 1,0217mV / 33680Ω = 30,33nA

 

And this current will finally develop a voltage across the 33K resistor:

 

U = I * R = 30,33nA * 33000Ω = 1mV

 

 

For 120V AC mains, the voltage divider needs to be totally recalculated.

 

Given that the high values of the resistors composing the resistive divider must be kept at that level, we can keep the 33K resistor as it is, for the multimeter connection across it, and then make some other changes.

Because the mains voltage is much lower now, first we need to reduce a little bit the voltage at the terminals of the shunt resistor in order to make the recalculation easier. For this purpose we connect in parallel 3 resistors of 0,47 Ω/5W instead of two of the original design. The shunt now equals 0,156Ω/15W in total.

The 1mV per Watt ratio must remain unchangeable. Therefore with the 33K resistor of the divider unchanged, a new value is required for the second member of the divider (namely for the replacement of the originally 2,2K resistor), calculated as follows:

1W (resistive) load will draw from 120V mains a current of:

 

I = P / U = 1W / 120V = 8,333mA

 

This current (applying Ohm’s Law) will develop across the 0,235Ω shunt resistor a voltage having a value:

 

U = I * R = 8,333mA * 0,156Ω = 1,3mV

 

This voltage will appear across the terminals of the divider. Given that we need the 1mV across the 33K resistor, the current through it will be:

 

I = U / R = 1*10^-3 / 33 * 10^3 = 30,3nA

 

Therefore the rest 0,3mV (1,3mV – 1mV = 0,3mV) will be developed across the new resistor, with the same current flowing through it as well. This means that the resistor must have a value of:

 

R = U / I = 0,3mV / 30,3nA = 9,9KΩ

 

This value does not exist again in the E12 series of resistors. Therefore we must create an equivalent of it from the existing resistors of this series. So, if we connect in parallel two resistors of 22KΩ and 18 KΩ respectively, we will obtain the required value. Indeed:

 

R1 * R2 / R1 + R2 = 22K * 18K / 22K + 18K = 9,9KΩ

 

This means that the second member of the resistive divider, for 120V mains, will be composed of two individual resistors of 22KΩ/0,25W and 18 KΩ/0,25W respectively, connected in parallel. This pair replaces the original 2,2KΩ resistor of the above original schematic for the proper operation of the test jig in this voltage.

Let’s examine once again what is the maximum power we can measure with the modified test jig for 120V AC mains system. This time the shunt resistor’s total power is bigger than that of the original circuit (15W instead of 10W). If the maximum permissible power consumption for the shunt resistor is to be 60% of its 15W nominal rating, namely 9W, applying the Power formula we have:

 

Pmax = I^2 * R,   and solving for I, we get:

 

I = √Pmax / R = √9W / 0,156Ω = 7,59A

 

This will be our maximum permissible current flowing through the shunt resistor. This current will flow through the device under test when it consumes power of a magnitude of:

 

Pmax = 7,59A * 120V = 910,8 Watt, or, rounded up, 900W

 

However, momentarily, just for a single quick measurement only, we can exploit again the full nominal power of the shunt. In this case the shunt must withstand a current of:

 

I = √Pmax / R = √15W / 0,156Ω =  9,8A

 

And will measure an amount of power of:

 

Pmax = 9,8A * 120V = 1176W, or, rounded up, 1100W

 

As I already have mentioned above, this test jig will measure real power (measured in Watts) only in cases where the load is purely resistive. In all the other cases it will give you the apparent power the load draws from mains. Just remember in these cases to read the indication of your multimeter as Volt-Amperes. Not as Watts. This simple circuit will miss the cos φ factor of the loads to be measured and this is what makes the difference.

First of all, in order to test its function and make sure that the results it produces are reasonable and as expected, we can connect an incandescent lamp in series with it and compare the reading displayed on the multimeter with the rated power of the lamp. This is usually printed on its body. If we are very close to it, the best case being to see the reading on the instrument coinciding with the rated power of the lamp, the jig works perfectly.

Now if we connect it to a load presenting impedances (not being purely resistive), its power factor comes into play. In these cases the test jig will ignore the power factor and always give you results related to the apparent power drawn from mains. Therefore without a resistive load connected to it, you should consider these results as apparent power values only. Not as real power measurements whatsoever.

You can see below the comparative indications on the consumption of a non loaded linear transformer. how-to-make-wattmeter

As you see, the multimeter shows 10,63VA (Volt-Amperes now, not Watts) while the power meter shows a load of 6,8Watt with a power factor of 65%.

If we divide the (real power of) 6,8Watt shown on the power meter by the (apparent power of) 10,63VA the multimeter displays, we get a power factor for this specific load of 0,64, which is very close to the relevant indication the power meter displays. This is an additional proof that the jig works properly, although as I mentioned, the purpose of using it should be limited in making indicative or comparative measurements.

I was using this jig in order to check any likely differences in power consumption before and after a repair, especially after replacing critical components, like, say a line output transformer on a CRT TV.

In older times, especially in this case, during the vacuum tube era, I used to always check the cathode current of the line output pentode tube and compare the measured value to that of maximum specified as permissible one, in order to make sure that the current through the primary side of the line output transformer was not exceeding the safety level.

On the other hand I always prefer to measure current by using a known resistor and measure the voltage drop across it, rather than exposing my precious multimeter in relevant risky measurements using it directly as an ampere meter. The danger to damage the instrument, when being dizzy during troubleshooting a difficult malfunction of a device under repair, is always there.

The common mistake (a common result for anyone being dizzy) in similar cases is to forget the plus probe on the Amps input and then try to measure voltage…Bang!!

Good scenario: internal fuse of the instrument blown. Bad scenario: The instrument was unfused…Most likely you’ll need a new instrument…if nothing else followed the…explosion…!! So never forget that. Remove the red probe from the Amps input, immediately after measuring current, and put it in the V input. Just because this is exactly what I am afraid of, I always prefer to measure current as voltage drop on a known resistor instead of making direct current measurements…

Hopefully this simple circuit has drawn your interest and you will make use of it at least for fun, despite the fact that there are readily available cheap gadgets for that purpose in the market nowadays…

paris aziz

This article was prepared for you by Paris Azis from Athens-Greece. He is 59 years old and has more than 30 years’ experience in electronics repairs, both in consumer and industrial electronics. He started as a hobbyist at the age of 12 years and ended his professional carrier as a senior electronics technician.  He has been a specialist in the entire range of consumer electronics repairs (: valve radio and BW TV receivers, transistorized color CRT TV, audio amps, reel and cassette tape recorders, telephone answering and telefax devices, electric irons, MW cooking devices e.t.c) working in his early stages at the official service departments of National-Panasonic first and JVC afterwards, at their premises in Athens.

Then he joined the telecoms industry, working for 20 years as field supporting technician in the sector of DMRs (: Digital Microwave Radio transmission stations), ending his carrier with this subject. Now he is a hobbyist again!




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35 Comments

  1. Anwar Shiekh

    December 5, 2016 at 8:30 am

    I like clever things like this, but with Kill A Watt meter costing $20 it might not make sense; parts and postage might cost that much.

    Likes(3)Dislikes(0)
    • Paris Azis

      December 5, 2016 at 3:58 pm

      Hello Anwar

      Thank you for your positive comment. I already mentioned the relevant cost reference in the text above and obviously I agree. The point is to be able to take this measurement, in an urgent case, while that ordinarily cheap meter is not there...For me it was something like ....survival electronics..

      Best Regards

      Likes(4)Dislikes(0)
  2. Steve

    December 5, 2016 at 9:07 am

    Nice job great article loved it. You always have great articles looking forward to the next one.

    Steve

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    • Paris Azis

      December 5, 2016 at 4:07 pm

      Hey Steve

      Thank you for your kind words. I try to offer the readers anything practically useful as regards electronic repairs, being at the same time interesting for them either as an effective method or an applicable problem solving technique.

      Best Regards

      Likes(0)Dislikes(0)
  3. Anthony

    December 5, 2016 at 9:36 am

    Hi Paris,
    A very thorough and detailed explanation by you as per usual ! Thanks for sharing this and though I probably won't get
    around to making it....at least it's great to know it can be built and that the expense is prohibitive.

    Kind Regards

    Likes(1)Dislikes(0)
    • Paris Azis

      December 5, 2016 at 4:21 pm

      Hey Anthony

      Thank you for your support. Cost is not always a decisive factor. Perhaps in a given case you lose some money, but for sure you earn valuable knowledge for all the similar repair cases you will encounter in the future.
      Old time engineers were working with the renown logarithmic ruler. Modern engineers use calculators or computers. The point is this. How can you trust their results in case you cannot prove that they are correct? What finally makes you an engineer? Worse than that, for us today, those old timers were able to calculate things without the need of electricity. And I find the idea of being autonomous perfect!

      Best Regards

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      • Anthony

        December 6, 2016 at 10:10 am

        Hi Paris, I'm sorry....my sentence should have been "not prohibitive" in relation to the cost of constructing
        your project. I agree with you that modern devices can make our minds lazy when it comes to calculating
        maths and other problems we must deal with. Thank you again for the theory lesson.

        Kind Regards

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        • Paris Azis

          December 7, 2016 at 9:56 pm

          Hey Anthony

          The point is that we always must be in position to evaluate the results of any tool we use for our job. Not to trust them blindly. This is, always in my humble opinion, the difference between a mature technician and anyone else who tries to repair something occasionally.
          When I was working for Panasonic, I remember that after repairing a calculator, the company proposed to us, through their servicing manuals, to perform simple arthmetical tests in order for us to very that the processor we just replaced was performing its job correctly. The results of the numbers entry should be a string of zeros...
          So it was not considered as given that, replacing the processor, everything would be perfect as expected. We should always test after the replacement...
          I not only accepted this idea, but also adopted it applying it in every repair of mine, ever since.
          On the other hand, professionally talking, if this test was to be performed by the user-customer after the return of the appliance to them and its function was not flawless as it should be, the name of the company would be tarnished and of course nobody wanted this to happen.
          Therefore good reputation should always work as an additional counterbalancing factor against low quality, non professional, or irresponsible repairs.
          Irresponsibility is finally the easiest way to lose customers you obtained so hardly...This is what must be permanently kept in mind.

          Best Regards

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  4. Robert Calk

    December 5, 2016 at 10:19 am

    Thanks Paris, another great lesson. I use a Kill A Watt EZ Model: P4460.01. It is easy to use and was pretty cheap too.

    Likes(1)Dislikes(0)
    • Paris Azis

      December 5, 2016 at 4:25 pm

      Yes, Robert, I know that and I cannot deny that it is true. But, please, take a look at my answer to Anthony and see my aspect of seen things. Under this light is this article of mine written.

      Best Regards

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      • Robert Calk

        December 6, 2016 at 1:05 pm

        Hi Paris,
        Yes, I understand and agree with you. It's a good thing to understand how things work. You do a great job!

        Likes(1)Dislikes(0)
  5. PARASURAMAN

    December 5, 2016 at 11:03 am

    Mind boggling mathematics! So, I skipped through to the last, to save myself (LOL)! But good information!

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    • Paris Azis

      December 5, 2016 at 4:34 pm

      Thank you Parasuraman.

      It is simple mathematics. Nothing scary anyway.
      We need this basic knowledge in order to understand how a circuit works. Otherwise the result is as usually happens..."I keep on replacing it, but it burns repeatedly"...
      We fall in a loop without a chance to escape...So we need to know the basics in order to understand the nature of the failure. Something beyond the blind component replacement and just hopping or wishing for the appliance to function properly. Who else is there to help us but our selves?

      Best Regards

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  6. Albert van Bemmelen

    December 5, 2016 at 2:20 pm

    Hi Paris. Your interesting article makes it look like we are being ripped off by the Power Meter Manufacturers. I, like many others already own a couple of these Digital Powermeters at home. But I will have a closer look on your mathematical explanation later on when I've got some time to spare. PS: Like you, I added an interesting Post to the "Source and Load and the Enhanced Light Bulb method" that also costs nothing but makes a great tool to be able to read digitally and exact if there is a short in High (or lower) Power Devices.
    It also comes with some easy understandable equations:
    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. 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! And you can read it digitally simply by using any standard Digital Multimeter.
    You can therefore advance for each device determine which voltage must be present if the device is good. etc.etc.

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    • Paris Azis

      December 5, 2016 at 4:44 pm

      Hello Albert

      In general, about using the lamps as a protective measure against short circuits, I told you and I repeat that again, I agree. I only had some concerns about their use in that way, without knowing that your friend wanted a fast go-no-go test of the incoming appliances.
      But as you see, "the cheap meter wins" in our days. Well, I respect that, but my point was another one. You can figure it out if you see my answers in the comments above.

      Best Regards

      Likes(0)Dislikes(0)
      • Albert van Bemmelen

        December 5, 2016 at 6:43 pm

        Thanks Paris. Indeed what would we do without our cheap meters. In that light of explaining the obvious we both are trying to write a clear and helpful article for the less experienced readers among us. Many experienced Engineers would probably think of it as plain old basic knowledge. And renewing some of the old Mathematical rules never hurted anyone.
        It is indeed like you say valuable knowledge for all the similar repair cases we will encounter in the future. In case of understanding the resulting complex Phasor Diagrams of AC circuits, or any other complex circuit, I now find Tina Designer a very helpfull tool too. It shows Nyquist Diagrams, Smith Diagrams and Polar Diagrams in the Network Analyzer Menu. This great Electronics Program is even capable in transfering any complex function of a drawn circuit in seconds by using the Symbolic Analysis AC Transfer! It is therefore also a great Program for educational purposes! (Like the now old redundant Calculator).

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        • Paris Azis

          December 6, 2016 at 2:59 am

          Agreed Albert. You understood correctly what my point was about this article. And the point is simple: always double-check and never take something as given. So simple, but also so important.
          Thank you for your information about that program, but as a stubborn old timer I am, I prefer to use my old calculator. And I can reassure you that if had the chance I would use that mechanical logarithmic calculator.
          It is a pity that the children of our time cannot perform a simple arithmetic division operation with pencil and paper only...
          Well, I love electronics, but feel that I must survive without a battery or mains supply when they are absent for any reason. I was trained properly for that by the education system we had at that time of my youth. My generation was as well. But the children of our time are not. And this brings me sorrow.
          But this is another long story...not likely to be resolved in a few lines of comments about electronics, although not being completely irrelevant. Anyway...

          Greetings

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  7. Yogesh Panchal

    December 5, 2016 at 3:44 pm

    Paris,
    you have back up of very good testing meters! can you just brief about an isolation transformer for testing the equipment of aground 1000w PC SMPS.

    Likes(1)Dislikes(0)
    • Paris Azis

      December 5, 2016 at 7:54 pm

      Hey Yogesh

      I am not sure I undestood what exactly you mean. If you just need information about the wattage of the isolation transformer needed for testing 1000W PSUs, you should simply use a transformer of higher wattage due to the power factor problem. This is because the transformer will suffer higher currents than the nominal ones. You should choose at least 50% higher wattage for the transformer, just for safety reasons alone.
      Of course if there is a power factor correction circuit included in the PSU under repair and functions properly, the 1000W transformer could withstand the current draw. But we always need to stay safe and keep our equipment safe as well. So, I recommend 1,5kW or even higher transformer.
      Now, if your meaning is a "how to do it" puestion, then the answer is simple. You connect the device under test in the secondary winding and go on with the repair. Keep in mind not to ground any of the secondary terminals. It is a mistake I have seen many times. The secondary must always be floating. If you ground one of the two secondary terminals your isolation is already lost. It is like not using an isolation transformer at all, and you will remain exposed in the danger of being electrocuted accordingly.
      So, only the core of the transformer must be grounded. This is the safety ground, in case that there is a leaking current from mains,i.e. the primary winding, to the core of the transformer or even to its secondary winding. Many people confuse it with the "neutral" ground of the big power distribution transformers, but this is not a similar case. In fact we need both the secondary winding terminals "neutralized" using the transformer. This is the meaning of using it. So, if you accidentally touch anyone of the two terminals of the secondary, nothing happens to you. But you will jump to the ceiling if you touch them both!! Keep that in mind!
      Hopefully I have answered your question. Feel free to ask me anything further.

      Best Regards

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      • Yogesh Panchal

        December 6, 2016 at 12:45 am

        Paris,
        Thanks for the reply. I just want to know Here in India the standard of the Mains electricity is 230V, 50 Hz so what would be the Transformer rating? and how to Build the unit for 1.5Kw pls. guide.

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        • Paris Azis

          December 6, 2016 at 4:35 pm

          O.K Yogesh, I understand now completely your requirement.
          My article about the Farfissa transformer calculations explains fully the procedure. You just need to follow all the steps described therein, starting from the core size for the nominal power you need and proceed in calculating the turns of the windings. All you need is a scientific calculator, or you can use the Excel of your P/C. Anyway if you have any further questions, please ask my email address from Jestine and contact me.

          Best Regards

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  8. Humberto

    December 5, 2016 at 9:57 pm

    Hi Paris, thanks for sharing a Professional Tutorial of this topic. Keep up.

    Likes(1)Dislikes(0)
    • Paris Azis

      December 7, 2016 at 9:17 pm

      Hi Humberto

      Thanks for your support.

      Best Regards

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  9. Gerald

    December 6, 2016 at 10:26 am

    Hi Paris,

    I like this method. Modern (and cheap)watt meters are great and useful but there is nothing like understanding how things work. You feel more in control than just relying on a display...

    I have been looking for a simple way of measuring the power factor and found one HERE

    I tried it and it works.

    Keep sharing this sort of things.
    Cheers,
    GM

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    • Paris Azis

      December 6, 2016 at 4:38 pm

      You'got my point Gerald. Thank you for your support and the cos phi info.

      Best Regards

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  10. Clair Morrill

    December 8, 2016 at 3:23 am

    Paris Azis, if yow want this circuit to last longer space the power resistors an inch (2.5 cm) apart and mount them to a metal heat sink, use mica washers and transistor heat sink compound too. Clair

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    • Paris Azis

      December 9, 2016 at 12:18 am

      Hey Clair

      For the relatively small loads I usually work with, namely ordinary home electronic appliances this is not necessary. On the other hand this test jig is to be used for simple measurements indicating power draw from mains instead of cumulative power consumption ones needing its permanent connection to heavy loads. Anyway thank you for your comment.

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  11. Chris

    December 8, 2016 at 5:50 am

    Hi Azis, I read your article with great interest and I like it.
    I show this to some of my friends kids who are interested in electronic
    stuff and I hope they would be one day a big kids with a great
    knowledge as you are.
    Repairing is not based only on part changing...

    Keep up and I'm waiting for your next article.
    My best regards.

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    • Paris Azis

      December 9, 2016 at 12:42 am

      Thanks a lot Chris.

      Yes, I absolutely agree, electronics repairs is not a simple parts changing procedure alone. The essential requirement on the part of the repairs technician is to understand and have a complete, self-explained notion about the failure mechanism. Incapability to answer this question escorted by self-reassurement that the repair is successfully completed, many times results in repeated failures and relevant returns of the appliance ending in personal humiliation.
      On the contrary, understanding the failure mechanism is what really assures the successful repair and leads the technician to the field of knowledge enhancement and enrichment as well.

      Best Regards

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  12. Hafidh Mahmood

    December 13, 2016 at 2:50 am

    Thanx for sharing Azis. I have always worked on the notion that the key to diagnosis,however,is knowledge of how the circuit is supposed to work rather than the availability of elaborate test equipment or a degree in electronics.Knowing exactly how a circuit is supposed to work is just a few steps from pin-pointing where your fault lies.If you do not have an idea of how circuit is supposed to work then how do you expect to safely work on it!
    By the way, a very good Fluke 8060A you got there.I had one since 1986, gave me good service for more than 12 years until the display failed.

    Cheers

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    • Paris Azis

      December 15, 2016 at 5:01 pm

      Hello Hafidh

      Thanks for your support. I fully agree with your comment.
      As for my Fluke 8060A, this is my second Fluke. I bought it some months ago through eBay. It was a New Old Stock piece. I spent a lot of money for it, perhaps I could buy something much more flexible and modern, but I love this instrument since the early nineties when I used for first time in the company I was working at that time. I already had the industry standard Fluke 87 since 1989 and it works fine up to now. I only replaced the elastomeric connectors for its display which was fading...

      Best Regards

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  13. Lawrence A. Mc Coig

    December 15, 2016 at 12:42 am

    I liked the article. I entered the circuit in Electronics Simulator and am getting 1 volt/ per Watt and not 1 mV/ Watt. Did I miss a measurement somewhere?

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    • Lawrence A. Mc Coig

      December 15, 2016 at 4:29 am

      My mistake. In America, we use period decimal points instead of commas that we use to group thousands. So, instead of 470 ohm resistors it was .47 ohm 5-Watt resistors. Things worked well then.

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      • Paris Azis

        December 15, 2016 at 5:07 pm

        Hi Lawrence

        I am happy to hear this. I only calculated the 120V version of this circuit without testing it. All my tests were done on the 230V version. But mathematics never fail, yes?

        Best Regards

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  14. Ulises Aguilar Pazzani

    February 26, 2017 at 2:13 pm

    MR paris , great circuit

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