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In this answer, the inrush current for a laser printer is listed as follows:

Inrush Current: (Duration: significantly < 1 second)    
  Model A  (120V): 23 A peak (20 deg C, from cold start) 
  Model AB (240V): 40 A peak (20 deg C, from cold start)

The question posed by Dan Neely in the comments piqued my interest and makes me question my understanding of electrical theory. I would have expected the current draw to be the same, or possibly even half as much, for the higher-voltage model. I should note that I'm basing this assumption on past experience in building out racks in a data center, where we could typically put more 240V servers in a rack than 120V since their current draw was significantly smaller.

So, please school me: why does the 240V model have nearly twice the inrush current as the 120V model?

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    \$\begingroup\$ I don't have time for a real answer, but maybe at time zero the printer's behavior can be modeled using Ohm's law? If the impedance of printer is Z at the moment it's turned on, the current would be I=U/Z. If we double the voltage, the current will double too and will stabilize at lower level once the the PSU actually gets time to start working. \$\endgroup\$ – AndrejaKo Aug 27 '12 at 16:17
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The operating current should be inversely proportional to the voltage as you expected, but inrush is a different issue. The inrush current probably comes from charging up the reservoir caps immediately after the full wave bridge in power supply. Those capacitors are fixed, but get charged up proportional to the line voltage. The current eventually drawn from them will be inversely proportional to the line voltage, but the inrush only sees the capacitance immediately connected to the power line. More voltage on the same size cap means more Coulombs, which means more current over the same short amount of time.

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  • \$\begingroup\$ Thank you, Olin. I suspected that there was something specific to the inrush scenario, but didn't realize that the power supply might have specific circuits to handle that inrush outside of the normal circuitry of the PSU. With that in mind, it makes sense that more push equals more flow. \$\endgroup\$ – Justin ᚅᚔᚈᚄᚒᚔ Aug 27 '12 at 16:30
  • \$\begingroup\$ So the inrush current will be 2x as big but the inrush period should last 1/2 as long? \$\endgroup\$ – KutuluMike Aug 27 '12 at 20:05
  • \$\begingroup\$ @Michael: I don't see where you got the 1/2 as long from. The inrush for the higher voltage will probably last a little longer, and in either case will have a lot to do with where in the power line cycle the unit was turned on or plugged in. \$\endgroup\$ – Olin Lathrop Aug 28 '12 at 11:56
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Inrush generally flows through a resistive device such as a thermistor, and will charge up any capacitors on the mains-side of any power supplies in the device. This is why a larger input voltage means a larger inrush current - the capacitors are like a short-circuit at the moment the power is applied, so it's only the thermistor and the mains voltage that define the maximum inrush current.

The thermistor limits the current charging the capacitors to a safe level, preventing component damage and nuisance trips (blown breakers, fuses, etc.). The thermistor can also help limit the short-circuit current in the event of a shorted capacitor.

Inrush is usually characterized with a programmable AC source or a triac circuit that applies the mains voltage starting at the one of 90 degree peaks, to maximize the peak voltage (and the inrush current).

Cold start refers to the unit being off for sufficient time to ensure that all of the capacitors in the unit are discharged to their absolute minimums. (usually 24 hours).

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