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newbie question, Ohm's law and power supply calculation

Ohyes

Sep 7, 2022
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Hi,

I've done my research but I don't have any friend or teacher I can ask to double check my reasoning.
I'm posting this here hoping someone could let me know if what I've done makes sense or if I should review some aspects of it.

I'm trying to make very basic calculations in order to understand the type of power supply I need for my project.

My project is composed of a worm gear motor (12V 3W), a heat sink fan (12V 100W) and an LED (18V 10W).

Overall I believe the Voltage I need is 18 Volts (as each piece will be connected in parallel so don't add up all together like they would if they would be in serie).

In order to find the power in Watts I've divided 113 Watts (all watts added together) by 18 (the highest voltage required in my project).
Overall this is giving me 6.3 Amps.

I came across a fragmented power supply providing 18 Volts and 200 Watts.

I believe such a power supply is providing 11.1 Amps (reached this result by dividing 200/18) and would be suitable for my project.

Am I going in the right direction?

Thanks for your patience.
 

Harald Kapp

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Overall I believe the Voltage I need is 18 Volts
Why do you think so? 18 V will seriously overload the motor and the fan.
You will need separate 12 V and 18 V. Since power drawn from the 18 V rail is comparatively small (10 W vs. 103 W from 12 V), a suitable setup would use a 12 V power supply plus a 12 V to 18 V step-up converter.

In order to find the power in Watts I've divided 113 Watts (all watts added together) by 18
This were correct if you could use a single 18 V supply, but you can't, see above.
Using a step-up converter as suggested above you will have some additional losses in the converter. Let us assume a conservative figure for the efficiency of the converter of 75 %, meaning that for every 1 W of output power you need 1.33 W of input power (modern regulators can be much better with efficiencies > 90 %, the exact figure will depend on the model you use). The total output power on the 12 V rail is then Ptot=10 W + 100 W + 13.3 W = 123.3 W.
This is equivalent to Itot=123.3 W / 12 V = 10.275 A.
One never designs a system at its limits. including a safety margin of at least 10 % that means you need a 12 V supply capable of >135 W or equiv. > 11 A

I believe such a power supply is providing 11.1 Amps (reached this result by dividing 200/18) and would be suitable for my project.
You do not need to calculate in both Amps and Watts. Both are equivalent acc. to P = I × V. Therefore dong your math in eiother Amps or Watts is sufficient.
 

Martaine2005

May 12, 2015
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Yes, you are correct. Thank you for doing a little homework first.

Read @Harald Kapp answer.
I presumed you know the max voltage required. Then lower the voltage to 12V with either regulator.

Martin
 

Harald Kapp

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Then lower the voltage to 12V with either regulator.
If you want to go this way, then the calculation is as follows:
103 W from 12 V stepped-down from 18 V at 75 % efficiency is 137.33 W from 18 V. Add 10 W for the LED and total power from 18 V is 147.33 W.
Add 10 % safety margin and the 18 V power supply should be rated at > 162 W.

Compare the figures:
12 V : 135 W
18 V : 162 W
Clearly the 12 V power supply with a step-up for the 18 V LED is more efficient than a 18 V supply with a step-down for the motor and the fan. This is due to the comparatively high losses in the 103 W step-down converter whereas the step-up converter has much less losses.
 

Martaine2005

May 12, 2015
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You are absolutely correct @Harald Kapp . I was replying to a power supply @18V - 200W. But my math is terrible and I never think about the most efficient way. So thank you for all your calculations.

Martin
 

Bluejets

Oct 5, 2014
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100w heat sink fan and 3w motor....sure that's not the other way around..??
 
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