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I'm having a little trouble in figuring out how much power is absorbed by my 5v smartphone charger. I'm curious as to how to account the energy loss due to stepdown.

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  • \$\begingroup\$ Measure the power entering the charger from the wall and the power leaving the charger to your phone. The difference of the two is the power lost due to the step down. Then multiply by time to get your energy. \$\endgroup\$
    – vini_i
    Dec 6 '15 at 16:17
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Simple maths: Energy Loss = Input ac energy - Output DC energy.

How to measure input ac energy:

Use something like this: Meter

Plug your mobile charger to this and let it run for an hour or so. It will give you a reading of total energy consumed.

How to measure output DC energy:

You can use a similar power meter if available or use an arduino or some other micro-controller like this:

dc energy

Take readings every second or so and go on summing up to get total energy used on the DC side. This arrangement is required because current and voltage might change over time. Also, you might have to change 1 Ohm resistor according to current in the circuit.

Once you have input and output energy values, you can calculate the wastage.

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  • \$\begingroup\$ 1 Ohm is way too much. Current could easily be 1.5A. \$\endgroup\$
    – mkeith
    Dec 6 '15 at 20:49
  • \$\begingroup\$ Yes, depending upon the smartphone model. However too small resistor value will give you a small dynamic range and precise measurement might be difficult. Even 1 ohm at 1.5A will give 0V - 1.5V. Another way out of this is to use the max output voltage as Aref but ideally it has to come from some other source which is constant over time. \$\endgroup\$ Dec 8 '15 at 5:04
  • \$\begingroup\$ Well, not really. The USB charger will almost certainly not continue to consume 1.5A if the voltage drops to 3.5V (which is also not enough to charge the battery using a buck or Linear charger). So the presence of the resistor will affect the operation of the circuit. I am sorry, but 1 Ohm is way too much resistance to insert in the circuit. Even if the current is only 500 mA. I mean, that is 250 mW in your sense resistor. (and over 2W at 1.5A.). \$\endgroup\$
    – mkeith
    Dec 8 '15 at 5:58
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You can safely think it is zero :)

Well, here is the serious answer. Your charger usually have a little shield giving typical consumption characteristics, but they are usually just maximum ratings and won't help.

If your charger do not overheat, then the loss is very small and you need very precise measurements to calculate them.

You usually know the input and output voltages, but you do not know the currents. You can measure currents, but as the input current is alternating, it is still not enough unless you know the phase shift. So you'll need an oscilloscope and small resistor in a line. Compare input voltage with the voltage on resistor (which will have meaning of current) to determine the shift...

The main trouble is that if the loss is less than 5% for example, all your measurements should have better accuracy than 5% of course.

Of course there is another way. Enclose the charger in the insulated box, put some ice there and the tip of the thermometer. Wait until some ice is melted and temperature falls down to zero. Now turn on the charger. Few hours later measure how many ice have melted due to charger heat - that will give you direct numbers of how many energy was dissipated in the box.

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  • \$\begingroup\$ "if the loss is less than 5%"... don't worry too much about that, it's over 10% even in very efficient chargers. \$\endgroup\$
    – Ben Voigt
    Dec 6 '15 at 21:15

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