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This 24V 150W power supply below is marked by the following ratings on its surface:

enter image description here

I use it for 200-240VAC and it is not clear to me how to interpret these values.

The AC input range is given as 200-240VAC.

The max AC current drawn is given as 1.8A.

The output voltage is 24VDC.

The max output current is given as 6.25A.

The cos phi is given as 0.45.

So I try to write an equation between the input and output real power as follows:

Pin × cos phi = Pout

For max input real power:

Pin × cos phi = 240 × 1.8 × 0.45 = 194.4W

For min input real power:

Pin × cos phi = 200× 1.8 × 0.45 = 162W

For output real power:

Pout = 24 × 6.25 = 150W

So Pin is always greater than Pout. Does that mean it is because of other power losses?

More importantly I want to ask something different:

Let's say The power supply output is always kept at its max that is to say at 24V and 6.25A. Now if we vary the AC line voltage between 200V up to 240V (by keeping the load current at 6.25A), would the AC input current remain as 1.8A? If so, something is odd to me that I cannot explain to myself. Maybe my approach of interpreting this information is wrong?

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  • \$\begingroup\$ You calculated the input power varies from 162 to 194 W as the input voltage increases from 200 to 240 V. The maximum output power is 150 W. Why do you say that the output power is greater than the input power? You also assumed that the input current does not change with input voltage which is probably not true. \$\endgroup\$
    – Barry
    Sep 23, 2019 at 12:31
  • \$\begingroup\$ Sorry I wrote wrong I corrected now Pin is greater than Pout. \$\endgroup\$
    – GNZ
    Sep 23, 2019 at 12:42

2 Answers 2

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Let's say The power supply output is always kept at its max that is to say at 24V and 6.25A. Now if we vary the AC line voltage between 200V up to 240V (by keeping the load current at 6.25A), would the AC input current remain as 1.8A?

No: to see why it is so, let's remember that the apparent input power is related to the load power by the following formula $$ A_{in}=\frac{P_{out}}{\eta \cos\varphi}\label{1}\tag{1} $$ where

  • \$\eta\$ is the efficiency of the power supply
  • \$\cos\varphi\$ is the power factor.

Now, assuming that \$\eta, \cos\varphi\$ are approximatively constant respect to the output power \$P_{out}\$, and remembering that $$ A_{in}=V_{\mathrm{AC_{in}}}\cdot I_{\mathrm{AC_{in}}}, $$ by substituting this equation in formula \eqref{1}, we get $$ I_{\mathrm{AC_{in}}}\simeq \frac{P_{out}}{V_{\mathrm{AC_{in}}}\eta \cos\varphi }\label{2}\tag{2} $$ therefore, keeping the output power constant and lowering the input voltage makes the input current rising.

Few observations

  • The nominal input current (\$1.8\mathrm{A}\$ for the range considered in the question) for a given range usually refers to the maximum value measured at the minimum input voltage (if the datasheet is properly written). This is due to the fact that this value also serves as a design parameter for the upstream part of the circuit (overload protection devices like fuses and other kind of breakers), so the designer should rely on it.
  • The assumptions \$\eta\simeq\mathrm{const.}\$ and \$\cos\varphi\simeq\mathrm{const.}\$ are only approximately true, thus you can expect noticeable variations in the real behavior \$I_{\mathrm{AC_{in}}}=f(V_{\mathrm{AC_{in}}})\$ respect to the one predicted by formula \eqref{2}. However, ceteris paribus, you can always expect an increase of the input current when lowering the input voltage.
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    \$\begingroup\$ Your assumption of approximately constant η is only valid for modern switching supplies, it should be noted. η is a function of input and output characteristics to the point that it's not even a useful metric for a linear supply. Of course, linear supplies are extremely rare in modern electronics, but I feel it's still worth mentioning. \$\endgroup\$
    – Hearth
    Sep 23, 2019 at 13:17
  • \$\begingroup\$ Thanks for the very clear answer. Just one last question. You mentioned and demonstrated at the lowest ACin voltage the current becomes max and it is used as a design parameter for fx overload protection. I understand that and makes sense. This power supply in question can opearte at down to 100V and can draw 3A current. Does that mean depending on the voltage(230V vs 110V at different countries) we should consider different fuse sizings? \$\endgroup\$
    – GNZ
    Sep 23, 2019 at 13:36
  • \$\begingroup\$ @Genzo, Yes: this is a dual range input power supply, therefore if you should chose different fuse sizing for each range. The info on the input current provided by the producer is aimed at this purpose. \$\endgroup\$ Sep 23, 2019 at 13:41
  • \$\begingroup\$ @Hearth, nice observation. I also did not mentioned the fact that, even in modern switching power supply, my assumption \$\eta\simeq\mathrm{const.}\$ is nearly true only at least above the 20% percent of the nominal load. If the power supply is loaded with less of 10% of its nominal load, its efficiency falls down quickly. \$\endgroup\$ Sep 23, 2019 at 13:43
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You over-complicated things.

Let's take the 240V example:

Max output it can sustain is 24Vx6.25A = 150W - this complies with its spec. On the input side, you have a worst case scenario when you have 200V and you get 360W out of that.

Given that any low quality rated PSU has an efficiency of 75-77%, that is well within specs. Note that quality PSUs get quite easy over 90% (the ones marked Gold, Platinum, etc). Considering such a low efficiency you're still left with 270W to work with while you will not be able to drain more than 150W safely due to the output specs limitation.

So there is absolutely no way you can exceed the input current limitation even if you keep the PSU at max load. If the power voltage varies - let's say it increases from 200 to 240 suddenly - you will get a current variation (in the current example it will be lower).

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  • \$\begingroup\$ 360W or 360VA? Why dont you take account cos phi? \$\endgroup\$
    – GNZ
    Sep 23, 2019 at 12:46
  • \$\begingroup\$ For the sake of simplified calculation. Things would be a lot easier if the PSU would have it's efficiency rating stated. It's quite simpler to take power factor into consideration. And I'm not sure why you calculated like that in the 1st place. \$\endgroup\$
    – Overmind
    Sep 23, 2019 at 12:48
  • \$\begingroup\$ Note: the officially stated working efficiency of this PSU is up to 90.5%. \$\endgroup\$
    – Overmind
    Sep 23, 2019 at 12:57

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