# LM388 (or LM317) Torque (current) limiting motor driver

I am designing a small handheld electronic screwdriver for a university project. The motor in this screwdriver requires a torque limiter to stop it stripping screws out of plastic parts and i am currently looking to use an LM338 to do this, hoping to use as few components as possible to keep size and complexity down (atleast for V1).

I was going to do it similar to the specification sheet as seen here:

However, this clearly has some issues! 1, my motor can pull 1.45A of current and my small 100ohm potentiometers can not and 2, this also creates a varing potential divider between the LM388 and the motor, slowing the motor down. I just want the motor to be torque limited, not speed limited.

This lead me to a design like this:

However I have not been able to get this working using R1 - 1k ohm, R2 10ohm. The 1km was calulated as shown here:

The 10 ohm is just to prevent shorts and therefore small.

Would anyone be able to give me a hand who has experience with this?

The only other info should probably mention is the motor, which is this:

https://www.servocity.com/81-rpm-mini-econ-gear-motor/

and has a static resistance of 9ohms (I think).

The datasheet for the LM388 is here:

• Tip: "1.45 A of power" is not correct. It's "a current of 1.45 A" (just as we don't say "10 m of distance", it would be "a distance of 10 m"). Show your calculations for output voltage with R1 = 1k (small 'k' for kilo, 'K' for kelvin) and R2 = 10 ohms. Details will catch you out! – Transistor Feb 16 at 22:47
• Appologies! I shall fix them now! For the R1 calc. I did a potential divider with the potentiometer, to get a voltage drop of 1.25 V across R1. 1.25V as this is what the LM338 regulates to on its adjustment pin (which i now realise is un-marked, appologies again). This gave me 860 ohms and I used a 1kohm to check if it was correct. I now see that perhaps this is too far out of bounds? Here is an image of my maths: imgur.com/a/NEoaHBj – Dan Timmerman Feb 16 at 23:02
• Upload the image into the question to make it easy for your readers. I'm off to bed (Ireland) so someone else can take up your question. – Transistor Feb 16 at 23:05
• No worries, appreciate it all the same! – Dan Timmerman Feb 16 at 23:08
• That looks much better. Thanks for the advice. – Dan Timmerman Feb 16 at 23:11

I don't see any way your second circuit would limit the current. Without a current shunt or some other sensor it has no way to know what current is even being drawn.

The first thing to note is that this is a linear regulator, so this circuit is going to be very inefficient.

The second note is that because the LM338 requires a very large 1.25V feedback voltage, the current shunt is going to waste a lot of power at all times.

That being said, here's a topology that might work:

simulate this circuit – Schematic created using CircuitLab

Important facts from the datasheet:

For current regulation applications, a single resistor whose resistance value is 1.25V/IO and power rating is greater than 1.25V^2/R must be used.

The device OUTPUT pin sources current necessary to make OUTPUT pin 1.25V greater than ADJUST terminal to provide output regulation.

The the low end of the current range would be determined by the series resistor. 2.5R for example would result in a 500mA limit. The resistor divider R1 can reduce this feedback voltage to increase the current. R3 ensures that there's an upper limit. In this case approximately 3A.

One major downside is that R2 is going to need to dissipate a lot of power at the high current setting, up to 20W in this example!

Another topology that might work is this one:

simulate this circuit

The switches represent a rotary switch where only one can be closed at a time (or a similar arrangement. Effectively turning the switch changes the shunt resistor and therefore the maximum current. It doesn't provide infinite adjustment, but it's simple and it wastes much less power at high current settings then the other solution.

Note that these diagrams don't include caps or protection diodes and such which may be needed.

• Thanks Drew, this has helped a lot. I think the thing that I didnt process is what you said here: Without a current shunt or some other sensor it has no way to know what current is even being drawn. Of course, the adjustment is required to know the current flowing, and therefore must be inline with the load. That makes a lot of sense. May I ask what the difference is between the lower topology using a rotary switch and using a potentiometer? Thank you for your advice. I really appreciate it. – Dan Timmerman Feb 17 at 11:09
• The upper topology uses a current shunt that produces 1.25V at minimum current. At max current it produces 7.5V, which is then reduced by a voltage divider. Obviously 7.5V is a huge drop. The lower topology changes the shunt at each setting so that it's always producing a max of 1.25V. – Drew Feb 17 at 19:24

Drew circuits are perfectly fine; the real issue is using a linear regulator to drive a brushed motor; it's quite inefficient.

If you look just a little you'll see that there are many DC motor driver with current limiting as a feature (they usually do cycle-by-cycle PWM for that), and usually even inversion (for a screwdriver it's useful). Use an evaluation board for your prototype (they are sold exactly for that)

• Agreed, if it were my project I would not use a linear regulator. – Drew Feb 19 at 1:41