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I have a 48V stepper running off a 240W supply, max current of the drive is ~4A. I am looking to add an 80mm cooling fan to this circuit. I know 48V fans exist, but 12V are more widely available and easily replaced. In my research, most 12V fans <30dBa draw less than 0.2A of current at max RPM, so I have enough current even with the stepper at max load.

Could I simply use a stepdown converter off the 48V rail to power this fan?

Or should I split from my 120VAC power entry module to a separate 12V transformer (which seems like overkill for one measly fan)?

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  • \$\begingroup\$ Easiest is what you said yourself, get a 48 V fan. Second easiest is to step down the voltage, resistor or DC/DC step down (buck). \$\endgroup\$
    – winny
    Jun 29, 2021 at 20:15
  • \$\begingroup\$ Have you got any 12 V mains adapters to hand that aren't doing anything else at the moment? You could use it to power a fan-fail alarm at the same time. \$\endgroup\$ Jun 29, 2021 at 20:25
  • \$\begingroup\$ Amazon has switching regulators selling for a few dollars that will do 48 to 12v. Depending on how reliable this needs to be I might spend a little more and get something name brand, but either way these things are not very expensive if you just need a quick fix and don't want to buy 48v fans. \$\endgroup\$ Jun 29, 2021 at 21:22
  • \$\begingroup\$ Or four small 12 V fans in series! \$\endgroup\$
    – winny
    Jun 29, 2021 at 21:50

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Not a big fan (as it were) of adding 7W+ of heat by linear dropping or regulation where you are trying to get rid of it. There are also a few other 'potential' issues considering the voltage drop and irregular current draw.

You could add an inexpensive (about 8 USD in MOQ 1) NSD05-48S12 DC-DC converter to create a 12V supply.

enter image description here

These happen to be isolated DC-DC rather than simple buck topology but it does no harm and tends to make it less likely it would fail with high voltage output.

They're 80% or thereabouts efficient, so a 200mA 12V fan should draw around 60mA from the 48V supply.

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  • \$\begingroup\$ Thank you for the detailed response. Wasn't familiar with isolated converters, for $8 I will give it a shot. Not too worried about the 83% efficiency at such low current draw. \$\endgroup\$
    – boggle
    Jun 30, 2021 at 12:44
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I asked a similar question and this is what I used:

Use a circuit like the one below, use either a regulator or a "7805" DC DC equivalent (but the regulator will burn up heat). Make sure the current required for the fan is low (like less than 100mA), I supposed you could parallel circuits. The transistor takes some of the voltage drop away from the regulator (or DC DC converter). The circuit is cheap and you may already have the parts around.

enter image description here
Source: Need a buck circuit to go from 48V down to 12V-9V less than 30mA (or half the supply input for a regulator)

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Easy, cheap & reliable way? Use a dropping resistor. Yes it's old school but it'll work well enough for your purpose if you don't mind wasting a little power.

Here's how -- Simple application of ohms law.

You know (for example) the fan draws 0.2A @ 12V

That means you want a resistor to drop (48-12)=36V and limit current to 0.2A

Using ohms law: E=IR

36 = 0.2 * R

R=180 ohms

It has to drop some fairly decent power...

P = V * I = 36 * 0.2 = 7.2W

So a 10W 180 ohm power resistor, in series with the fan, will work.
Place the resistor in the fans airflow and it'll never even really get hot

You can find suitable resistors almost anywhere. i.e. Ebay has dozens, $5 shipped to your door.

You would, of course, re-run the math above based on the fan you actually select.

Doesn't have to be EXACT either. Plenty of headroom for tolerance. Fan just will run a little faster or slower if you don't "nail it".

Now I'll get some thumbs-down comments because this is an engineering forum and for whatever reason, alotta guys get their pigtails all curled up if a solution doesn't have a CPU or require some exotic circuit concept....

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    \$\begingroup\$ You may destroy the driver inside the fan from the initial 48 V applied to it before it starts to consume current and hence the resistor starts dropping voltage. \$\endgroup\$
    – winny
    Jun 29, 2021 at 20:17
  • \$\begingroup\$ True. To be certain, 'driverless' fan might be in order (i.e. just a standard DC brushed motor... skip the fancy control circuit) Or if it's an issue, a couple zener diodes could easily do the same trick w/o that side effect. \$\endgroup\$
    – Kyle B
    Jun 29, 2021 at 20:21
  • \$\begingroup\$ Thanks, I figured it could be this simple. Been a while since I actually had to specify a resistor that had heat transfer implications and didn't want to guess. Planned on just a basic 2 wire fan, not repurposing a PWM fan I have lying around. As you mentioned, RPM isn't crucial, so long as I'm >80% RPM for decent CFM. \$\endgroup\$
    – boggle
    Jun 29, 2021 at 20:41
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    \$\begingroup\$ Then all you need is another cooling fan to keep your resistor cool, so it doesn't melt the things around it... \$\endgroup\$
    – Simon B
    Jun 29, 2021 at 21:14
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    \$\begingroup\$ I think the thumbs down are more likely to be because the point here is to cool down the system, but you want to add 7W of heat to it... \$\endgroup\$
    – BeB00
    Jun 29, 2021 at 21:42

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