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First off I'm a beginner at electronics I don't understand much so bear with me.

I've just constructed my first bipolar stepper motor circuit today on a breadboard. I've actually extracted the stepper motor from an old Epson printer. I've used an lm317t adjustable voltage regulator to power the motor driver. Input is a 9v battery and I give 5v as motor power supply to the l293d motor driver. There is a separate microcontroller and its power and ground which is 5v as well.

Now I've used a pot to adjust and multimeter to check the output motor voltage of the lm317 regulator. I have set the pot such that the output is at 5v

Now my question is when the stepper motor runs successfully, the multimeter reading drops from 5v down to 4v. Why does this happen?

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  • \$\begingroup\$ Did you measure the voltage going in to your regulator (battery voltage) with the motor running? \$\endgroup\$ – brhans Apr 27 '16 at 17:18
  • \$\begingroup\$ No but its output is fluctuating between 4 and 5v. I don't have any problems with the circuit. Its fine for a first time project. I just had a question \$\endgroup\$ – user108320 Apr 27 '16 at 17:20
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    \$\begingroup\$ You're missing my point. If the regulator's input voltage is not at least output + dropout, then output will not be what you expect it to be. So measure the input and you'll know if that is the problem or not. \$\endgroup\$ – brhans Apr 27 '16 at 17:23
  • \$\begingroup\$ Yes see I want a 5v output. The input is 9.3v exact from a 9v battery. \$\endgroup\$ – user108320 Apr 27 '16 at 17:25
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    \$\begingroup\$ You didn't measure that 9.3v while the motor was running. As I said before, measure the battery voltage while the motor is running. \$\endgroup\$ – brhans Apr 27 '16 at 17:29
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9V batteries can only source a small amount of current. You are likely exceeding the current that the battery can comfortably source (or, to phrase it differently, you are drawing enough current that the battery's internal resistance causes too large of a voltage drop).

If you measure the battery voltage, you will likely see it dropping below ~8V, which is the drop-out voltage of the LM317 when you are operating it at 5V output. The LM317 may need an input voltage of up to 8V (5V + 3V) to provide a stable 5V output (the recommended Vi-Vo is 3V).

Use several AA batteries in series, or a wall-wart that can supply a decent amount of current.

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  • \$\begingroup\$ But the output of the regulator is going to an h bridge l293d stepper motor driver. Doesn't the driver amplify the current to drive the stepper motor \$\endgroup\$ – user108320 Apr 27 '16 at 17:26
  • \$\begingroup\$ The stepper motor drive sequence is coming from the microcontroller. Isn't the output of the microcontroller pins amplified by the driver ic? \$\endgroup\$ – user108320 Apr 27 '16 at 17:28
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    \$\begingroup\$ You can't pull current out of thin air. The current provided by the motor driver to the motor must go through the regulator. \$\endgroup\$ – uint128_t Apr 27 '16 at 17:28
  • \$\begingroup\$ The stepper motor drives a large amount of current. But isn't the IC providing that current by amplifying the regulator input current \$\endgroup\$ – user108320 Apr 27 '16 at 17:31
  • \$\begingroup\$ Alright, I'll reiterate: P=IV, and power is conserved (you can't break physics). The voltage going into the motor driver from the power supply is (approximately) the same as the voltage driving the motor. Therefore, the current is the same. You cannot pull current from thin air, that current comes from your power supply (in this case, your battery and regulator). \$\endgroup\$ – uint128_t Apr 27 '16 at 17:35

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