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So I bought a couple of switching power supplies for a project of mine and I am currently recalculating my actual power usage out of my wall socket to figure out how to distribute power. I have made sure that my power supplies will never exceed 90% of its capacity, but the normal operating will be around 40-50%.

I tried to calculate the values and compare them with the ratings of my supplies and they are way more off that I anticipated. Example:

I have a 5V power supply that can deliver 20 amps for DC output. My AC input rating for this supply says: 230V (170-264V) 1.5A. However my calculations for 240V is 0.416A. So the theoretical and stated power usage is 1 amp off. I figured there would be some waste but not that much.

So:

  • Are my calculations way off?
  • Are the producer of the power supply just overly careful?
  • Or does the power supply really waste the much energy in the conversion?
    • Does the power supply have a poor power factor ?
    • Is the part load efficiency poor ?
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    \$\begingroup\$ I don't see any power calculations. \$\endgroup\$ – Eugene Sh. Oct 11 '16 at 18:28
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    \$\begingroup\$ It will NOT actually consume 1.5A - that figure makes allowances for minimum spec voltage, and is heavily sandbagged, safety margined on top of that. Its use is to determine the fuse rating and (e.g.) how many of them you can power from a 13A circuit. To get a feel for actual wasted power (without tools), compare its heat with other appliances ... you can feel the heat rising from a 60W bulb for example. \$\endgroup\$ – Brian Drummond Oct 11 '16 at 18:34
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    \$\begingroup\$ Also, it might consume more power during only part of the mains cycle (resulting in a poor power factor; heavy units shouldn't do this, but no guarantees for no-name Ebay PSUs). The PSU might also draw more power on startup; the maximum current figure should take that into account. \$\endgroup\$ – marcelm Oct 11 '16 at 18:44
  • \$\begingroup\$ @marcelm Ahh, the power on start up. I actually forgot about that. My power supplies are from a decent manufacture and bought from a supplier, NOT Ebay. I dont buy essential hardware from an unknown source. \$\endgroup\$ – evilfish Oct 11 '16 at 19:12
  • \$\begingroup\$ Will measuring the power usage on load be a good indication to what the actual limit is? I know i have to take into account startup power and all that plus have a buffer, but is that a way to go? \$\endgroup\$ – evilfish Oct 11 '16 at 22:31
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this supply says: 230V (170-264V) 1.5A. However my calculations for 240V is 0.416A.

You are comparing two different things. The supply spec gives you the absolute maximum current it will ever draw under any conditions of output load, input voltage, temperature, phase of the moon, and what you had for breakfast.

You are calculating actual current for a specific case based on output power. You are also doing it at significantly higher input voltage than the minimum. Since this is a switching supply, the highest input current will occur at the minimum input voltage. That alone accounts for a 40% reduction in input current.

Of course these power supplies aren't 100% efficient. Look at the datasheet to see how much loss there is at full output power and the worst case input voltage. That could account for another 15% reduction or so. Then there is usually a significant cover your butt margin in specifying input current. The manufacturer doesn't want to get caught with the supply ever drawing more than they said it would. Since most customer aren't going to care that the supply might draw 1.5 A instead of 1 A, the manufacturer gives the more conservative worst case value.

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You can't calculate AC power by multiplying effective voltage and effective current as put on the label. The voltage is there so you can pick the right device for an input voltage and the current is there so you don't overload the circuit, power strips etc.

On the label, there may be two addional values, first is ...VA, which is multiplied effective voltage and effective current, and can be used to determine whether a circuit or power strip would be overloaded (similar to the effective current value), and second is power factor = 0.5 (sometimes cos phi = ...) or similar, which is an additional factor to be considered to get from that VA value to the real power consumed by a device.

That factor is determined by taking the wave form of the current and phase shift between voltage and current into account.

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