# Reduce 36 vdc to 12 vdc

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I have a 12vdc 130mA no load motor that I want to use for an application. My only power source is 36vdc. I assume the motor will burn up if I run it on 36vdc. Is there a simple way to reduce the voltage to 12vdc?

Thanks Jim

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Series resistor sized to drop the appropriate current, of a power capacity sufficient to handle the I-squared-R power, maybe on a heat sink?

Else you're looking at a DC-DC converter.

GWE

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Assuming you are not concerned with efficiency and can afford to waste twice the power the motor is drawing:

The "simplest" way would be to use a series resistor of the correct value and power dissipation rating to drop 24 volts. That will only work if the motor were very lightly loaded, because as the load on the motor increases, the current drawn would rise which would cause additional voltage drop in the resistor, greatly limiting the motor's speed and power.

The "next simplest" would be to use a series 24 volt Zener diode with a power rating adequate to handle the full load motor current. (Zener power rating in Watts = motor's full load current * 24). These are not hard to find through electronic parts distributors.

But, whenever I have a "one off" electronic component requirement these days and don't want to run into minimum order charges from an electronic parts distributor I look on eBay. Here's something that looks like it could do the job for you:

If you put string 4 of those diodes in series with the motor you'd drop pretty close to 22.4 volts, and with their 5 watt rating the motor could pull almost 900 milliamps without overdissipating the Zeners. Your motor probably won't mind eating 13.6 volts instead of an exact 12 volts.

If you bought the 25 of those Zeners for \$5 on eBay, and only have one motor yo're worrying about, you'd have 21 of them left as spares in case something stalls the motor and the first set of Zeners creates that "magic blue smoke". An appropriate sized fuse in series with the motor could also help protect those Zeners (and the motor) from overcurrent.

HTH,

Jeff

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The simplest way is to turn 24V @ 0.13A into heat with a resistor. Ohm's Law says the resistance value should be 185 ohms, and its power rating should be at least 3.2 W.

Now that works for "no load" only. Under load, your motor will draw more current, the voltage loss across this resistor will go up proportionally to the current, and the power dissipated as heat will increase as the square of the current.

If this motor will see widely varying loads, then you probably need a DC-DC converter because it will automatically adjust to keep the output at a steady 12V. Take a look at astrodyne.com for some info and prices. (The model ASD10-48S12 is probably what you need)

Another possibility is to buy a surplus power rheostat. A 250 ohm, 10 W unit should do the trick. You will have to adjust it manually to keep the motor happy.

jim n judy wrote:

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Buy a three pin voltage regulator from Radio Shack, 130ma is not much. Since 36V is much higher than 12V, be sure that the regulator is derated. You need to keep the regulator from overheating, since most of the power will get burned up in the regulator. 130ma times 24 volts is

3.1 watts, so a heat sink will probably be needed on the voltage regulator to help dissipate at least 3 watts. The best way to avoid problems is to make sure that the regulator and heat sink are larger than needed.

You might need some kind of overvoltage protector on the voltage regulator for the back EMF generated by the motor when the power is removed, depending on the switching. If you put the switch between the regulator and the motor that will protect the regulator.

Richard

jim n judy wrote:

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The cheapest way is to find a 185 ohm resistor < at least 2 watt > theoreticly 20 watt bulb in series may work but it's just a calculation based on a hot filament ... of course a cold filament would have less resistance but how much at that load is trial and error. You would be wise to put a .25 amp fuse in there before trying anything. This is just a calculation ... so do at your own risk. Go to radio shack and but a 12 volt 1 amp regulator ... but if your not a tech type your screwed.

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What you suggest will work, but in my experience the 3-terminal regulators all have various issues when driving inductive loads like a motor. The answer is to surround them with additional components to protect them from damage/shutdown. So much for simplicity.

We're > Buy a three pin voltage regulator from Radio Shack, 130ma is not much.

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BFI method:

TIP121 80V NPN darlington 0.60 at mouser (mount on big electrically isolated aluminum heat sink- 3W dissipation)

1N5243B 13V 0.5W zener 0.05 " 2.7K 1W MOF resistor 0.13 " MF-R020 200mA resettable fuse 0.51 " ------ 1.29

Best regards, Spehro Pefhany

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Look at the LM317 three terminal regulator from National semi, it only cost a couple of bucks and with 2 resistors can adjust from 1.2 to

40vdc ,current limit with heat sink is 2.2 amps. There are several application notes from Nat. Al

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This is a goofy idea, but I think it might work. Put a cam on the motor shaft (or any driven part of your setup), which closes a switch for 1/3 of a revolution. Wire the motor through this switch. You might have to start it manually, but once started, the motor will be getting 36 volt pulses, with a

33% duty cycle.
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None of the common 3-terminal regulators have high enough max input voltage to be really safe off a 36V unregulated supply. LM7812's are rated for only 35V absolute max.

Of the common ones, the LM317 is sorta okay because it should only see

24V and it's rated for 40V absolute max, but if the output is shorted or if the LRA exceeds the current limit it will see the full input voltage (36V nominal (+ ripple + regulation?), which is awfully close to 40V). Things get worse if the motor is plugged, of course, and a diode from the output to ground would help with that.

Best regards, Spehro Pefhany

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Kinda the equivalent of a (non-posi-track) differential? (stall one motor and the other one goes like crazy)

Best regards, Spehro Pefhany

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Modular dc-dc converters are cheap and

I did a quick Google search on 36 VDC-12VDC linear converters and \$300 bucks ain't cheap. A DC motor has to reach speed before it settles down to .130 ma It seems a current regulator is the best way to go .. however ..he did say "simple"

I still say if he series the motor with a 15-20 watt bulb it the best and cheapest way to go

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Or -- a switching regulator could do the task, as long as there was no need for isolation.

I'm not sure whether you can get switching regulators as pre-built modules, however.

Good Luck, DoN.

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You can find "wall wart" power supplies that will do 12V @ 130MA. An old computer power supply will give you a very stable 12V. HTH

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That depends. On the older ones, quite often it requires a certain minimum load on the 5V output before it starts giving reliable power on the 12V output. The 5V switching regulator is the master one, from which power to derive everything else comes.

If you have problems -- hand a low-valued resistor on the 5v line -- something like about 2.5 Ohms if the minimum load from the 5V is listed as 2A.

Enjoy, DoN.

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Or an 1157. ;-)

Best regards, Spehro Pefhany

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