When I look at, let's say an iPad AC adapter, the input/output rating on the charger states:

Input: 100-240V 0.45A (AC)
Output: 5.1V 2.1A (DC)

I know that the input and output ratings are maximum. The AC voltage in my country is 230V. Through a simple calculation, I can deduce the following (correct me if I am wrong):

Input power: 103.5W
Output Power: 10.71W

Now, my question is this: what is the true power drawn from the wall socket? Is it 103.5W or 10.71W? If it is 103.5W, then I presume the iPad adapter is 10% efficient?

  • \$\begingroup\$ Skaty, keep in mind that voltage and current go hand in hand. P=V*I Double the voltage, and you can halve the current, while keeping the same power. The adaptor will pull less current at 230v than it will at 120v. That's before we get into actual draw and efficiency and everything else that was answered below. \$\endgroup\$ – Passerby Mar 15 '14 at 14:32
  • \$\begingroup\$ I chose that answer, even though it's irrelevant to my question, as it explains why the Input and Output ratings on an adapter is as such. @Passerby I know. Let's assume 100V and 0.45A (based on ratings), why is it 45W? Where is the 30W++ gone to? I know AC is not a constant power draw, assuming the current is peak, based on RMS values, 0.31A and 100V (since wall sockets are R.M.S. values), it's 31W, which is far from 10.71W that is output. Andy answered it thoroughly, even though it does not answer it directly. \$\endgroup\$ – Skaty Mar 15 '14 at 14:32
  • \$\begingroup\$ That's fine. You are allowed to choose whichever answer you want to accept if you feel it answers your question. EE.SE is all about sharing knowledge, not about collecting points :D My comment in caps is to the people who voted to close your question simply because your question mentioned an ipad. You ask a valid electrical engineering question. \$\endgroup\$ – Passerby Mar 15 '14 at 14:35
  • \$\begingroup\$ @Passerby Oh, I didn't realise that. I stated iPad charger since that is what I had beside me. It can be anything and my main concern is not the iPad charger but a project that I would like to implement, involving various adapters. \$\endgroup\$ – Skaty Mar 15 '14 at 14:40
  • \$\begingroup\$ Exactly. That's my point. But some people on this site will vote to close any question with the slightest mention of a consumer product in it. Really frustrating. \$\endgroup\$ – Passerby Mar 15 '14 at 15:25

Undoubtedly the charger you have is a switching regulator. I say this because that is what all modern chargers appear to be and the input voltage range is wide enough to make this assumption valid. It's not a big charger - roughly 10 watt output means it is small in my book and more than likely it will be based around the following: -

  • Raw AC voltage is rectified to DC (peak will be about 338V DC on 240 Vac input and about 140V DC on a 100 Vac input)
  • This gets smoothed by a capacitor - probably in region of 220 uF (rated at 450V)
  • A switching circuit will convert this high dc voltage to 5.1 Vdc

As a thought experiment, if you connected the AC input of the charger to a 140V DC supply and the charger's output to a load resistor that took 2.1 Adc (10.71 watt load) and assumed the power conversion efficiency in the charger was 80%, you would expect to see about 14 watts taken from the input DC supply of 140V. This means a current of about 100mA.

Input power = 140 Vdc x 0.1 Adc = 14 watt

Output power = 5.1 Vdc x 2.1 Adc = 10.7 watt

So, when you connect it to an AC supply of 100V AC RMS, why could a current of 0.45 A flow? To understand this you have to recognize that this device's AC input current (as measured through an RMS measuring ammeter) is not representative of real power into the device. Unlike DC circuits (where it would be representative of real power), AC circuits like this can draw very non-linear (non sinusoidal) currents whose RMS value could be quite high compared to the "useful" current.

This means you can't make the assumption that input power into the device is Vac x Iac. The device has (more than likely) a bridge rectifier and smoothing capacitor and the current drawn will almost be like a spike of a few milliseconds every 10ms (50Hz supply). This, without doing the maths could mean that the input RMS current is twice the "useful" current: -

enter image description here

This would take your "useful" and needed input current of 100 mA to 200 mA - a bit closer to the 450 mA stated.

Also to consider is the inrush current - the manufacturer's rating of 0.45A may include some measure of inrush current into this figure but we don't really know.

Remember also that the rating will be most valid when the input AC voltage is at 100 VAC and not when the voltage is much higher (as per your calculation).

  • \$\begingroup\$ Good explanation on the spike in an AC-to-DC rectifier that is causing the 0.45 A current. So, from what I can infer, the power drawn would just be based on the rated efficiency (based on guidelines), right? \$\endgroup\$ – Skaty Mar 15 '14 at 14:31
  • \$\begingroup\$ The RMS of the current is probably about 200mA at the lowest rated input AC voltage. The spike could easily be over 1A but it is short duration. Power drawn will be based on efficiency of about 80%. \$\endgroup\$ – Andy aka Mar 15 '14 at 16:08

As others have explained, the input and output current and voltage markings on the supply have little or nothing to do with the efficiency. If you want to know the efficiency, however, there is typically a marking that is relevant.

If you look carefully at the iPad adapter, you should see a Roman numeral (mine is beside the CE marking, and is a 'V' or equivalent 5 in Arabic numerals). That is a code for the Energy Star Version 2.0 Level 5 efficiency rating.

(photo modified from Apple web page) enter image description here

As a 10W (rating on the label) supply, it has a guaranteed minimum efficiency of \$0.0626 \cdot ln(10) +0.662\$ or 80.6%, where ln(10) is the natural logarithm of the label rating in watts.

As you might expect from Apple, that is a very good efficiency rating and exceeds the mandatory minimum in places like California where one is imposed.

  • \$\begingroup\$ Thanks for the tip, was looking for an efficiency gauge. :) \$\endgroup\$ – Skaty Mar 15 '14 at 14:16
  • \$\begingroup\$ Nice find! Never knew that \$\endgroup\$ – Passerby Mar 15 '14 at 14:26
  • 1
    \$\begingroup\$ Added a calculator for applying the correct formula to the newer standards. \$\endgroup\$ – MBer Jun 27 '16 at 22:54

Whereas the output value is the actual nominal continuous output, the input value is the connection value or fuse value, i.e. you can attach 20 of these devices simultaneously to a wall outlet rated at 2000W, or, when (significantly) more than that current flows in due to for instance an electrical fault within the device, the fuse will trip. It has nothing to do with efficiency; this is seldomly noted or even tested. The AC connection value is just a safety requirement.

  • \$\begingroup\$ What do you mean by attach 20 of these devices simultaneously to a wall outlet rated at 2000W? \$\endgroup\$ – Skaty Mar 15 '14 at 14:16
  • \$\begingroup\$ @Skaty: In most of the world, wall outlets are rated at some guaranteed continuous capacity. Here it is 16A at 230V. Traditionally most AC powered devices have a connection rating, i.e. the power that they are rated to use. This is used to give the end user a guideline figure of how many devices he can attach to that outlet. So if you have an outlet capable of delivering 2000W and your individual devices have a connection value of 100W, you can safely attach at most 20. \$\endgroup\$ – user36129 Mar 15 '14 at 18:33

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