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A Simple Circuit For Measuring Mains Power Consumption
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.
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.
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.
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.
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…
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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