0
\$\begingroup\$

I have two stepper motors (ripped from old scanners) and two L298N ICs. I have to connect these motors to the STM32 MCU, and I'm going to use this scheme:

Schematic

I want to make one board for two motors just to save the space, the question is: Is it safe to connect L298Ns directly? I mean, do I need to use some diodes or something?

is it possible to just do this

If it's possible, do I need to double Vsupply (e.g. 24V instead of 12V)?

\$\endgroup\$
2
\$\begingroup\$

It's fine wiring the two circuits in parallel from a 12V supply. I'd probably put a zero ohm link in series with each group in case one went badly wrong - I could then take the link out and see that one was still working and the other was broken. Maybe even consider a small resistor so you can measure the two individual currents to both circuits - something like 0.1 ohms.

On the negative side, the L298 isn't very good for this sort of thing - it will drop (on each transistor) about 1.3V at 1A and if your stepper motor coil needs the full-juice from the 12V, be prepared to be disappointed with results. However, if you only need (say) 250mA to drive the coils then it should be OK. Here's the data anyway: -

enter image description here

What this means is that from a 12V supply the stepper motor coil will not see 12V but something more like 9.4V at 1A load. If this is a continual drive the chip will dissipate 2.6W per channel.

There are better chips around that use MOSFETs and drop a fraction of the voltage.

EDIT - added info on a better chip

The DRV8841 is a better choice for what I believe you might need. It's dual H bridge therefore will drive one stepper motor and works from over 8 volts up to over 40. It's got PWM circuits that can regulate the current too (that can of course be disabled). The on resistance of each transistor is typically 0.25 ohms at 1A - a full H bridge would dissipate 0.5 watts and you'd lose 250mV per transistor meaning that instead of about 9.5V being applied to the coil, it's be more like 11.5 volt. Also, because the power transistors are FETs, the parasitic diode in nearly all FETs, saves you having to uses the external diodes in your circuit.

\$\endgroup\$
  • \$\begingroup\$ If this is a continual drive the chip will dissipate 2.6W per channel. You have 4 channels so that's getting on 11 watts to get rid of. It has two H-bridges, each one with 2 transistors active at any time. So the total power dissipation would be 2.6 W * 2 = 5.2 W, not 11 W. \$\endgroup\$ – m.Alin Mar 19 '14 at 16:05
  • \$\begingroup\$ @m.Alin Thanks for letting me know. I forgot that both coils are never activated together. \$\endgroup\$ – Andy aka Mar 19 '14 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.