I have a 5V 10A 50WATT DC step-down converter: http://www.powerstream.com/dc-24V-5V.htm

I'm just not understanding how to size the power supply needed to drive this converter at max power.

I have an iGo 15-24V 6.5A 100W Max (130W peak) powersupply. Is that going to be sufficient to drive the step-down converter when it's under max load itself?

Am I understanding correctly, that if I have a 50WATT load, it is always going to need a minimum of 50WATT of power regardless of what voltage is driving it? I understand there will be a loss of efficiency in the step-down, so what's a rule of thumb (10%-100%) for gauging the necessary capacity on the supply.

Or am I missing something fundamental?

  • \$\begingroup\$ what is the efficiency of this 50W converter? \$\endgroup\$ – hassan789 Jan 27 '14 at 6:15
  • \$\begingroup\$ @ZacWolf what you want to say with your last paragraph? it is always going to need a minimum of 50WATT of power regardless of what voltage is driving it? Who is it? The DC-DC converter or the load of it? Load can be the DC-DC converter too. Your question is what power the DC-DC converter is going to need relative to the input voltage? \$\endgroup\$ – Diego C Nascimento Jan 28 '14 at 17:36
  • \$\begingroup\$ It = "50 WATT load" (whatever that might be). I couldn't remember if Power ever decreased in the volt/current bucking formulas. I now understand that because of efficiency, as you and Passerby explain, the power demand can increase (because of the additional DC-DC conversion loss), but the power demands on the power supply (by the buck/load) would never be less, just because the power supply was providing higher voltage/amperage. \$\endgroup\$ – ZacWolf Jan 28 '14 at 21:39
  • \$\begingroup\$ Again is not so clear. Anyway, the efficiency curve will vary according to the input voltage and the load the the DC-DC converter. Depending on design it can be more efficiently at lower input voltages, or the inverse. Also it can have the efficiency reduced with less load for the DC-DC converter. So if the load is not the 9-10 Amps specified, the efficiency could drop or increase. But this does not mean the required power will be 50W. If the DC-DC converter load is like 1A, the required power will be less than 50W, only if their efficiency at this load is so small, at probably is not the case \$\endgroup\$ – Diego C Nascimento Jan 29 '14 at 17:03
  • \$\begingroup\$ If you have a Load actively drawing 50Watts on the stepped-down side of the buck then the supply side is never going to be less than 50watts, just because the supply side makes more Volts/Amps available to the buck. That was my question in a nutshell, because I wasn't understanding how the Ohm's Law equation worked. \$\endgroup\$ – ZacWolf Jan 29 '14 at 20:43


Power In * Efficiency = Power Out

Since we know Power Out (50W) and Efficiency (~84% in decimal 0.84), we can rearrange this.

Power Out / Efficiency = Power In

50W / .84 = Power In

50W / .84 = 59.53W

So if Efficiency and Power Out are fixed, you only need 59.53 Watts in. In a perfect circuit. At 18v minimum, that is 3.33 Amps, and all three (Power, Voltage, Current) are well within the provided iGo's supply specs. Hope you have a 18v~24v tip for the iGo, unless you have one with a voltage selection switch.

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  • \$\begingroup\$ Excellent, this is exactly what I needed. I now understand the rule-of-thumb I was referring to was a function of efficiency, so this formula captures that perfectly. THANKS! \$\endgroup\$ – ZacWolf Jan 27 '14 at 14:56
  • \$\begingroup\$ Upon further reading, I did have one addon question. When you say, "At 18v minimum", I'm not exactly sure I understand. The minimum voltage on the iGo is 15V and the 6.5A "rating" of the iGo is for the 15V (and after doing some reading). That should still be enough, event at 15V correct? I'm not going to be using the tips, I'm disassembling the iGo and creating a new case for all the components. [Milled out of a solid block of aluminum so heat dissipation is planned for). \$\endgroup\$ – ZacWolf Jan 27 '14 at 17:57
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    \$\begingroup\$ @ZacWolf the step down converter has a 18v minimum input (35v max). \$\endgroup\$ – Passerby Jan 28 '14 at 0:57
  • \$\begingroup\$ Of course, my bad. I was focusing on the supply. \$\endgroup\$ – ZacWolf Jan 28 '14 at 15:29

Is that going to be sufficient to drive the step-down converter when it's under max load itself?

According to your DC step-down converter the specified efficiency is 84% at 9-10 amps output.

So if its output is 50W and at a 84% efficiency

$$\eta = Efficiency$$

$$\eta = \frac{Power\ Output}{Power\ Input}$$

$$Power\ input = \frac{Power\ output}{\eta}$$

$$~59,52 = \frac{50}{0.84}$$

So your requirement for the power supply, with the converter operating at the full load is ~60W. So it probably satisfies the needs with 100W continuous power. At 18V input it should be able to supply a current of 60 / 18 = 3.33A. Well within your supply capabilities.

Be in mind that's simple calcs based on resistive loads and does not take into account a bunch of factors that can change power requirements a lot.

Efficiency formulas from http://formulas.tutorvista.com/physics/efficiency-formula.html

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  • \$\begingroup\$ @ZacWolf thanks, I posted the efficiency formula, solved for you, and also the question about the power, before the answer you marked an answer, just curios where my question has a fault for you. \$\endgroup\$ – Diego C Nascimento Jan 27 '14 at 16:11
  • \$\begingroup\$ In my listing @Passerby was first to answer. Possibly because you edited your post later? I also felt his answer was more succinct and to the point. I appreciated your answer for educational purposes, thus the Thank You. \$\endgroup\$ – ZacWolf Jan 27 '14 at 18:09
  • \$\begingroup\$ @ZacWolf no my answer was first, and you can see by the edits that all the formulas are present at the first post, so I don't added the formulas later. The response where is not only to you but all that have similar questions, so it should be educative too. About the minimum 18V is because your DC-DC converter has a minimum input voltage of 18V, that's in the product sheet. \$\endgroup\$ – Diego C Nascimento Jan 27 '14 at 18:54
  • \$\begingroup\$ On the 18V, gotcha. Regarding the order, am I missing something? When I order Answers by oldest, @Passerby answer is at the top of my list... \$\endgroup\$ – ZacWolf Jan 27 '14 at 19:43
  • \$\begingroup\$ @DiegoCNascimento I have to agree with ZacWolf about your answer. The first bit about P=V*I seemed unrelated to the question because you left out the word "minimum", your formulas mixed up energy and power, and you put the result of the calculation on the left instead of the right of the equal sign. Having said that, the same thing happens to all of us from time to time. \$\endgroup\$ – Joe Hass Jan 28 '14 at 0:45

It sounds like your supply is sufficient for the converter. I am guessing the specs are calculated including efficiency, although you need to check the datasheet/manual to confirm this.
Either way, 18V * 6.5A = 117W, so it would need to be less than 50% efficient to be a problem. A rule of thumb is around 80% efficient for switching supplies. As long as you don't over volt with your supply (which it sounds like you cannot) then you should be okay to give it a try.

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