I'm working with a 12V DC linear actuator and only have a 24V 2A power supply, so I tried lowering the voltage using power resistors (2Ω – 100W).
In simulation (Proteus), the system worked fine: I connected a 2Ω resistor in series and then a combination with another in parallel, but I didn't get the expected results. Then I made mixed connections, one in series and one in parallel, and the voltage to the actuator dropped to about 12.3V. When I tested it in real life, following the connections, the parallel resistors heated up because they were shorting out.
My question is, is it feasible to use these resistors to lower the voltage to the actuator?
The idea is to only use this high-power resistor; a step-down resistor or a transistor isn't valid.
The answer is yes and no. Yes it can be done, but very unwise as you will just waste a lot of power in the form of heat.
That heat is power being wasted. NOT shorting out, resistors are a series device, if you short them they will be cold but the short may burn up.
It sounds like you haven't passed 101 yet.
If you're using two of these resistors in a voltage divider configuration with an inductive load at the center tap, then the power dissipated may be as high as 288 W (when the current has reached steady state, which will short out one of the resistors, sending 12 A through the other 2-Ω resistor)*.
More importantly, if the load draws any current, then the voltage divider center voltage will not be constant, so you will only get 12 V when your actuator first starts up (zero current).
*Assuming negligible armature resistance, which may not be the case.
Here are the details I currently have for the linear actuator:
Voltage rating: 12 V DC
Max current under load: ~2.0 A (measured with multimeter during extension)
Stroke length: 100 mm
Speed: 10 mm/s (so full travel takes ~10 seconds)
Rated force: ~100 N
Resistance (calculated): ~6 Ω (based on V = IR)
As for usage:
Duty cycle: Around 10–15 seconds of movement per minute (not continuous).
It's used to extend and retract briefly, not constant actuation.
The idea was to power it from a 24 V DC supply using resistors to drop the voltage to around 12 V, but I’m experiencing serious heating in the power resistors, even though I simulated the configuration successfully.
You measured the current to the motor as 2 amps at 12 volts, which you multiply to give you the watts, 24 watts. That same current goes through your resistors while they drop 12 volts from the 24 volt supply. How many watts are your resistors rated for? Obviously less that 24 watts.. You need wire wound resistors rated at more the 24 watts! Try 50 watts mounted so air can circulate.
Thanks for your input you're right that using power resistors isn't the most efficient or practical solution for stepping down voltage. That’s exactly what I’m confirming through testing.
The simulation did work (as expected under ideal conditions), and I’m fully aware the heat is just wasted power, not a short circuit. I never implied the resistor was shorted — I only mentioned the excessive heating, which made me revisit the idea of using a buck converter.
I’m not looking for an "electrical purity test" — I’m exploring practical constraints, verifying assumptions with real measurements, and asking for insight from others who’ve worked with linear actuators and DC loads.
Also, passing "101" is about knowing why something works and when not to use it, which is exactly what this post is about.
You'd be better off using an ohm-meter to measure the resistance with no power applied to the actuator.
By my estimate, the current drawn through the resistors under load will be 7 A and 5 A, respectively. The first resistor would be dissipating 98 W of power, which is awfully close to the 100-W rating.
@deavisss are you using heat sinks for your power resistors?
I had assumed the OP used a single resistor to drop 12 volts, I did not understand and still do not understand why he made a voltage divider using two resistors. One in series is sufficient!
Try applying ohms law to the resistor calculations and you will see the problem. Also, a voltage divider is called a voltage divider for a reason, also apply ohms law to that.
We are dealing with electric motors here and they don't care if there is an over voltage for a short time. Incandescent light bulbs are/were used in series to test motors with various name plate voltages.
Of course there is! In principle, you are converting half of the supplied power into heat, rather than useful work.
12V linear actuators briefly draw the full stall current (7 to 10 Amperes) every time they start moving, so it is best to use a 12V power supply capable of handling more than that.
You can pick up 12V power bricks for next to nothing at thrift shops and computer recycling operations.
You might want to consider the effects of a loaded potential divider. The rule of thumb is that you need ten times the current down the potential divider than you are going to draw out of it.
As others have said this is a really bad idea, as the voltage output get will change as the load changes. And the load will change as the actuator turns on and off.
Here's another reason why a voltage divider will not work:
If your power supply is limited to 2 A, and your load draws 2A, then the second resistor in the voltage divider may not draw any current, meaning that the only plausible arrangement is to put a 6-Ω resistor in series with the actuator (so that the actuator itself becomes the second half of the voltage divider). The problems with this is that due to inductance effects, back-EMF, and the (presumably) low armature resistance, the voltage across the actuator could vary anywhere from 24 V down to a few volts or less (the lower bound will depend on the magnitude of the armature resistance). Also, under start-up or stall conditions (or under any load in excess of the rated load), the power supply will experience a current overload.
So it seems that your ideas for how to power this actuator are fundamentally flawed, and that it would be best to reconsider your approach.
