0
\$\begingroup\$

I'm currently building my first stepper motor driver, a bipolar stepper, 4.5-48V 1A phase current with 4.5 Ohm phase resistance. I've built the circuit of the L6206 following the "Typical application" in the data sheet and controlling it with an AVR ATmega.

Basically it works fine, but the current is much too high. At 12V my 2.5A power supply switches off and the L6206 gets hot very quick. At 7V the current at stand still is more than 1.5A, running it is at 1A.

So I suppose I have to limit the current, but I am not sure how. The datasheet mentions an integrated PWM current control but I can't find any further information on how to use it. It explains how to the overcurrent protection can be used to roughly regulate the load current what I have tried, it works but the motor then runs choppy and I feel sorry for it.

Another thing I don't understand is why the driver still works if I remove the charge pump?

\$\endgroup\$
1
\$\begingroup\$

I think you may have selected the wrong part for what you are trying to do. Without any extra circuitry, the L6206 will only provide voltage-mode control. If you set the duty cycle to 100%, you will end up with

$${12V \over 4.5\Omega} \approx 2.7A$$

in the active phase.

You can reduce that by reducing the duty cycle, but if you want to have something that will control current for you, an L6207 might be better suited:

AN1762 Section 2

\$\endgroup\$
  • \$\begingroup\$ How can I reduce the duty cycle, can I PWM the enable voltage? \$\endgroup\$ – Torsten Römer Feb 14 '16 at 3:42
  • 1
    \$\begingroup\$ You could, but pulsing the IN input is more common. It changes how the motor current will decay during the off-cycle slightly. \$\endgroup\$ – Daniel Feb 14 '16 at 18:45
  • 1
    \$\begingroup\$ I tried pulsing enable with 1, 4 and 30 kHz, it did not work well and the motor was running very rough. Pulsing IN with 15 kHz seems to work very nice though. But anyway, I think I'll go for a DRV8825 now, it can do a lot more and costs about the same. \$\endgroup\$ – Torsten Römer Feb 15 '16 at 2:57

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.