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If a power supply says the input power is 100-240V, ~1.5A, I'm assuming that the "1.5A" draw would be at 100V, not at 240V, right? I.e. would the total max watts be 150W regardless of the input voltage?

The context of this is trying to determine the maximum draw for something based on the input power specs on its power supply to ensure a circuit doesn't get overloaded if too much is on it.

I want to make sure to get the max draw correct whether this is in the US at 110V or Europe at 220V.

As a side question, the output rating on this supply is 19V, 3.42A. Does that mean that it's pretty inefficient?

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  • \$\begingroup\$ Efficiency must be seen in context, like ripple, size, off etc. \$\endgroup\$ – PlasmaHH Jul 6 '16 at 17:38
  • \$\begingroup\$ The power supply also has a wattage rating right? Something to think about. Calculate the wattage of the power supply at the input and output from the voltage and current specs. \$\endgroup\$ – laptop2d Jul 6 '16 at 17:53
  • \$\begingroup\$ @laptop2d Nope, just volts and amps \$\endgroup\$ – g491 Jul 6 '16 at 17:56
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    \$\begingroup\$ That figure is a required overestimate of the actual current draw under worst case but continuous operation. Your product will not meet CE if you go over this labeled value but won't be questioned if you are well below it. Solution - estimate really high, add VAT and round upwards. \$\endgroup\$ – winny Jul 6 '16 at 18:02
  • \$\begingroup\$ @winny Do you know what voltage that amp draw is referring to? \$\endgroup\$ – g491 Jul 6 '16 at 19:52
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If a power supply says the input power is 100-240V, ~1.5A, I'm assuming that the "1.5A" draw would be at 100V, not at 240V, right? I.e. would the total max watts be 150W regardless of the input voltage?

Yes, the PSU input power performance will be given by

$$ P_{OUT} = P_{IN} \times efficiency $$

... the output rating on this supply is 19V, 3.42A.

$$ P_{OUT} = VI = 19 \times 3.42 \approx 70~W $$

If the efficiency is 85%, for example, the input power would be $$ P_{IN} = \frac {P_{OUT}}{efficiency} = \frac {70}{0.85} = 82~W $$

At 100 V current would be < 1 A.

The 1.5 A is worst case surge current. If the efficiency was only 50%, for example you would have 70 W dissipated in the power supply and it would be very hot to the touch. The main current should be much less than 1.5 A in normal operation.

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19V x 3.42A is 70W that's less than 75W so

Power factor correction may not be present.

The power supply may consist of a rectifier followed by a capacitor and a DC-DC converter. in that case the RMS amps will vary depending on the supply voltage and frequency.

240V as the higher capacitor charge level ans slower would result in much shorter current spikes, ans thus worse power factor, so it's unlikely that the current would halve if you double the voltage.

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The nameplate current rating is a worst case value. Power supplies are expected to operate at 10% below the minimum voltage on their nameplate. So your power supply would be tested at 90V and full load (maybe even overload, not sure on that). Furthermore there will almost certainly be a fudge-factor involved since drawing less current than specified on the nameplate is acceptable while drawing more current than specified on the nameplate is not.

Unfortunately that doesn't tell you much about how much the power supply will actually draw under normal conditions. You don't know how big a fudge factor the manufacturer applied, you don't know how the efficientcy and power factor will change with load. The current will almost certainly reduce with increasing voltage but exactly how much by is not at all clear.

Reputatble power supply vendors will usually provide you with figures for efficiency (\$\eta\$) and power factor(\$P_f\$) which you can use to calculate input current.

$$I_{in} = \frac{I_{out}\times V_{out}}{\eta\times P_{f}\times V_{in}}$$

Be aware though that these figures are likely to be "typical" rather than maximum. They may also only be specified at specific load conditions requiring you to make assumptions about their values in between.

Another thing to consider is what happens to your system during a brownout. In some countries it's not unheard of for the actual mains voltage to be much lower than the nominal mains voltage. In a system with mostly resistive loads a brownout would mean reduced current flow but on a circuit mostly supplying wide input switchers it will mean increased current flow.

You need to ensure that even under brownout conditions you don't go melting wiring. So you should either size cables to the nameplate current of the devices or put in appropriate overload protection and accept that under brownout conditions that overload protection may be triggered.

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  • \$\begingroup\$ So if the max is 90V * 1.5A or less, then at 220V, what would the max current be? Would it be different than 90/220 * 1.5? Jasen's answer makes it sound like it might be. Thanks \$\endgroup\$ – g491 Jul 9 '16 at 18:36

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