VEX 393 Motor Question

Hello all,

Just as a heads up for those reading, my post doesn’t have to do with a robot I’m constructing for a VEX competition. It’s instead a question I have regarding a VEX component.

I am pursuing a science fair project for high school, which involves a VEX 393 motor. I have a question regarding whether or not I will be able to use it for my experiment.

Just for context, here’s the rough outline of my experiment:

Question: How Does Load Resistance Impact the Power Output of a DC Motor?

Independent Variable: The load resistance of the circuit.

Dependent Variable: The time it will take for a winch to lift a weighted string over a certain distance.

1. In my experiment, I will have a fixed resistor of a certain resistance in series with the winch system that is powered by a single DC motor. This winch system is connected to a weight of a specific mass by a string.
2. Using Ohm’s Law as well as the resistance value of the resistor I am using, I will be able to calculate the theoretical value of power that the motor will exert.
3. I will turn on the circuit and measure with a stopwatch how long it will take the winch to lift the weighted mass. Using the equation W=Fd, I will calculate the Work that was performed by the winch. Then, using the equation P=W/t, I will calculate the experimental value of power that the motor had exerted.
4. I will perform step 3 (or more) times to reduce error.
5. Then, I will replace the resistor I used with one of greater resistance, and repeat steps 2-4.

(I also attached a PDF of my proposal in case you’d like to read that):
James Lee - Science Fair 2020_2021 - Initial Project Idea.pdf (184.3 KB)

All components will be placed in series with a voltage source that emits a constant voltage, regardless of the load resistance.

The problem I have is from my engineering teacher. To quote him:
“The motors are variable speed, but I’m not sure if they’re varied based on current. If so, the project will work great. If they’re variable based on voltage, it might either A) have enough power to lift it and perform the same function every time or B) not have enough power to lift it.”

So, my question is, does anybody know if VEX 393 motors are varied based on current? I figured this would be the best possible place to figure something like this out.

Due to COVID, I don’t have access to the robotics lab, and I can’t figure this out on my own. I will have to borrow parts from my school and do the project from my house. So, if I find out that my motor won’t work after I had already brought it home, I will be out of luck.


In normal use, that is, controlling the motor using a Cortex and the MC-29 controller, the speed control is by PWM (Pulse Width Modulation), which would be a whole different project. However, the 393 motor is a standard DC moter, which can be operated, and have it’s speed adjusted, by simple voltage control. So if you can use a variable DC power supply, I think you’ll be able to do your experiments.

“Bonus points” if you study on PWM and make a comparison between a PWM controlled motor and the same motor running on a variable DC power supply.


One thing to keep in mind is that you’ll want to keep your current below 1.5A or so otherwise you risk tripping the internal thermal PTC , which heats up and drastically increases resistance (thus lowering current) when subjected to currents higher than 1.5A for varying periods of time. See this thread for more information: Motor torque-speed curves - REV2


Good point. As a Non-competition, Non-VEX use of the motor for an experiment, you might want to bypass the PTC for your research, but be careful that you don’t stall the motor for too long during your experiments. Also, give the motor plenty of cool-down time between each experimental run.


It’s possible that I am simply misunderstanding the experiment you intend to conduct, but I get the impression that you may be misunderstanding some electrical principles. Let me explain.

You provide this circuit schematic:

This is not wrong by any means, but do you know what a simple DC motor can be modeled by? Nothing more than a resistor:

Note, though, that the equivalent resistance R_M of the motor will vary based on the motor’s mechanical load and potentially also based on other factors like voltage V_M across the motor (not to be confused with V_S).

In your step 2, you say you will calculate the theoretical power P_M the motor will draw, using nothing more than Ohm’s Law (V=IR) and P = IV. I recommend you try and work out the algebra for that, and see what you find :slightly_smiling_face:

Don’t forget:

  • Both of these relations only work for individual circuit components, not complete circuits.
  • It is possible to simplify circuits (i.e. combine circuit components) using the rules of series and parallel resistors.

Expand to see the algebra worked out (try it yourself first though!) and conclusion

(Sorry that MathJax doesn’t work inside these Hide Details blocks)

Remember the goal here is to find P_M in terms of known quantities only, i.e. V_S and R:


You can see that P_M is stuck dependent on R_M — an unknown (and even potentially variable) quantity. So there is no way to theoretically determine P_M in your setup; it can only be found experimentally.

If anyone wants it, here is the source code to generate my images used in this post:
Circuit symbols came from here:


Wow, everything here is so helpful! It’s true, I’m not the most experienced in electricity. As a Junior in high school, the only class I’ve ever taken on electricity was a dual enrollment class at our local county college. I am enrolled in the PLTW curriculum in my high school and I will be taking Digital Electronics from this February to June. Despite my lack of expertise I’ve always had an interest in electricity, so I’ve chosen this particular topic as a means of learning more about it myself. I’m open to any advice you may have.

Regarding not being able to measure R_M – I looked it up on Google, and it states that the resistance of a VEX 393 motor is 1.5 Ohms. Would it be possible for me to measure the resistance of the motor myself with an ohmmeter? Also, how could the resistance of the motor change? I’m guessing this has something to do with inductance, but I’m not too sure.

Also, just out of curiosity, how/where did you create that schematic? I think it looks really clean and I’m interested in being able to replicate it myself for future circuits.


So, as long as the voltage stays the same, the power in the motor will increase as the current in the motor increases, correct? (P=VA)

Regarding the resistance of the motor, and this is a very simplified version of how a motor works, at the same voltage, with the more load you put on a motor (that is to say, the harder it’s working) the lower the resistance is.

From the 393 info sheet from VEX, at 7.2V there are two values listed, the free current and stall current. Free current is the amount of current drawn when there is no load on the motor (it’s spinning freely), stall current is the amount of current drawn when the motor is unable to move because it’ can’t overcome whatever is opposing its movement (it has stalled). Those two values are 0.37 for free current and 4.8 amps for stall current respectively.

Using Ohm’s law, at 7.2V on the motor, the effective resistance when free spinning is about 20 ohms, and the effective resistance when stalled is 1.5 ohms, the value you found.

I get that it’s counterintuitive that resistance is lower when something is, you know, resisting movement of the motor. I tried to come up with a metaphor or something to explain it and came up blank there, sorry.

Now, your teacher had asked if speed of the VEX motors are determined by current or voltage. Again, to drastically simplify, it’s driven by voltage. This means if you have too much resistance from your resistor, you’re gonna have a hard time with the motor moving at all.

The other thing you’re going to be concerned about is the amount of power dissipated by the resistor you chose. Basically, resistors convert energy into heat. The resistors you buy will have a rating in watts that say how much heat they can safely dissipate before they burn up or general bad stuff happens. If you buy just random resistors from Amazon, odds are they’re either going to be such high resistance they’re gonna stop your motor working at all, or they’re gonna be rated for too low of power and they’ll burn up. Neither is good.

I don’t know if your plan is locked in stone yet, but I think a better experiment might be to measure current versus load. So you start with your pulley with nothing and see what kind of current it draws. Then you add a small weight and see how much it draws, and so on. Since you know voltage and current, you can calculate power, and you can then compare input power (how much power is going to the motor) versus output power (calculated using the work/time formula you said earlier).

Generally, I don’t think adding a resistor in front of a 393 motor is going to give you particularly good/interesting results and it has some pitfalls that, quite frankly, you don’t know enough about electronics to realize. I hope that doesn’t come across as rude, I certainly also didn’t know those pitfalls back when I was in high school either.


You are on the right track here. R_M is NOT the motor coil resistance, which you can measure with an ohmmeter (and should be 1.5 Ω). R_M is not a real resistance but rather just a stand-in representation for V_M/I.

Inductance and magnetism do play a part if you want to really analyze the theory and get a fully theoretical P_M. However, you can still do a lot with just Ohm’s Law as long as you are mindful of quantities you can only determine experimentally.

Very good questions!


Well…yes… but you don’t get to control the current. One of the simplest, but hardest, concepts in electronics to explain is that Ohm’s law rules everything. If you hold the voltage constant, the current will be based on the “resistance” that the motor has…but for motors it really isn’t resistance (at least not exactly like a resistor) because we’re dealing with a inductive (a magnetic coil of wire) based system.

Back to the motor: it will draw more and more current as the load increases on the motor, to the theoretical point of “infinite current” when it stalls, but physically that’s not possible.

(trying to keep it on the pre-college/EE level)…


I use a combination of Inkscape and LaTeX. It’s the only solution (only free solution, anyway) I have found that offers the flexibility I need for effectively modeling more complex circuits.

You can get the source SVG and LaTeX files I used from the attached Zip file at the bottom of my first post. In Inkscape, File > Save a Copy > PDF format > Omit text in PDF and create LaTeX file. Then include the exported LaTeX file in a LaTeX document as shown (you can certainly make it nicer, but I was only doing it to take screenshots).

There are free online circuit designers that you can use to still make pretty nice diagrams, but a lot more easily. They will definitely serve your purposes just fine for this experiment :slightly_smiling_face:


No dude, you’re not at all being rude. The more I look into this, the more I see the many ways this experiment could fail. I realize that what I’m trying to do is too advanced for me and I should aim for something a little simpler.

I’m submitting my proposal by midnight. I really like your idea. I have a homework assignment to do, but afterward I will draw up a new procedure with your idea. Would you mind looking it over for me at about 7 pm?

Yeah, sure. 7 PM what time zone?

1 Like

For those interested in more details on the 393 and (now discontinued) 269 motors, we dug into this in depth several years ago. The information is spread over several topics and a bit disjointed due to the nature of the forum, but a good place to start might be here which includes links to some of the earlier topics.


Oh, sorry, didn’t see this. 7 PM ET, so 15 minutes from now.

Ok, so this is what I’m thinking:

  1. Using a breadboard and jumper wires to make electrical connections, connect a winch that is powered by a single DC motor in series with a voltage source and an ammeter.
  2. With no mass attached to the end of the winch, turn on the voltage source and record both the current displayed by the ammeter and how long it takes for the winch to roll up a designated length of string.
  3. Reverse the polarity of the motor on the breadboard to unwind the winch and bring it back to starting position.
  4. Perform steps 2-3 19 more times (a total of 20) to reduce error.
  5. Tie an x gram weight onto the end of the string, and repeat steps 2-4 so it has been tested 20 times.
  6. Repeat steps 2-5 so no mass, an x gram mass, y gram mass, and z gram mass have all been tested 20 times each.

Also, quick question: So you’re saying that increasing the mass the winch has to lift will change the current travelling through the motor?

Breadboard is probably fine, but technically it’s not really advised to use a breadboard with anything high current like this. If possible, just use alligator clips or something to connect stuff.

Personally I’d say go with more data points and fewer runs at each point, doing it 20 times is just excessive, and you’ll get a clearer idea of the curve with more data points for different weights. 3 runs is probably fine, I’m lazy and would probably just do one.

Make sure your weight range represents a good range of value from no load to nearing the stall torque (1.67 Nm with the default gear ratio) of the motor to get the most comprehensive data, Not sure if you know the physics regarding torque or not.

And yes, as you increase the amount of power the motor has to use (in this case, increasing the weight it’s lifting), current increases assuming you keep voltage constant. This continues to go up until you reach stall torque, at which point the motors stop being able to move, and the current is at its highest.

To use a very imperfect metaphor, you try to push a light box it doesn’t take much effort (current). The box gets heavier, you’re working harder. Eventually it’s so heavy you’re pushing with all the strength your muscles have and the box still isn’t moving. The box could be 500 pounds or 5 million, either way at this point you’re using the same amount of effort and getting equally nowhere. Don’t read too much into this metaphor in trying to understand the relationship between current resistance and voltage, it’s not great.


No, it’s cool man, I get the picture lol.

Also, I apologize for asking you so many questions. I just have one more and I should be good.

Unfortunately no, I’m not too well versed in torque. I’m currently taking AP Physics 1, and we’re on Unit 3 which is Circular Motion and Gravitation. Torque and Rotational Motion doesn’t come up until Unit 7.

I looked up another post on vex forum regarding torque. To quote someone else,

"Torque is measured in inch-pounds. 1 inch-pound is the amount of torque required to lift a 1 pound weight that is 1 inch from the axis of rotation of the motor output. So if you had a motor that supplied a torque of 1 inch-pound, and you put a wheel on it with a radius of 1 inch, the motor would be “pushing” through the wheel with a force of 1 pound.

The maximum torque of a 269 is 8.6 inch-pounds and the maximum torque of a 393 is 13.5 inch-pounds."

If my winch is lifting something a foot or so away, I’m not sure how this would affect the torque.

I found a hanging mass set on Amazon. I can add up to 10 g masses onto it, meaning I can have 10 data points (10 g to 100 g).


Do you think these are still a bit too close together to see any noticeable difference? If so, do you think I should go for something like 100 g - 1 kg? I’m not sure what the limit is here until I reach stall torque.

OK, basic physics lesson. If you’re already in physics, some of this should be review.

F=MA. F is force, M is mass of the thing you’re lifting in kg, A is acceleration due to gravity. We’re gonna say acceleration is 10 m/s^2 because it’s actually 9.8 and 10 is easier.

So if you have a 100g mass, 0.1kg * 10 m/s^2 gives you a force of 1N.

So you a motor and let’s say the stall torque is 1 Nm. The units there are Newton meters. What that basically means is that 1 = Newtons of force * meters from the center of rotation (motor shaft). I.e you have a pulley 1 meter in radius on a motor. So, if you try to pull up your 100 gram weight on a motor with 1 Nm of torque with a pulley with a wheel a meter in radius or greater, it will stall. Any less than that and it’ll be able to pull it up.

Alternately, if you have a a 200g weight, now your radius needs to less than 0.5 meters, because you’ve doubled the number of Newtons of force you need, so you have to halve the number of meters.

Now, if your pulley has a 10ft rope or a 5 inch rope hanging down, doesn’t matter (we’re ignoring the weight of the rope), all that matters is the radius of the pulley wheels, and the weight of the thing being lifted.

Did that make sense?


Also, inch pounds and Newton meters are different units that measure the same thing, torque. Except inch pounds uses imperial which is awful never ever use imperial. (Newtons and pounds are both measures of force, inches and meters are both measures of distance)