I would have a question regarding to 5-lead unipolar steppers, I just found in a scanner a unipolar stepper motor with 5 wires and it has a nice gearing on it, so I would like to modify to a bipolar stepper. The stepper motor is TYPE 4H4018S2001 12v 0.4A TECO Made in Taiwan. I took a picture and did a little drawing with the numbers on the coil ends. Could you tell me how to modify it to a bipolar stepper : https://dl.dropboxusercontent.com/u/7511244/unipolar_tepper.pdf I cut the plating on the PCB and found which pair of coils could be separated :

wire 7 with wire 3 - ends of two coils;

wire 2 with wire 6 - ends of two coils;

wire 1 with wire 5 - ends of two coils;

wire 8 with wire 4 - ends of two coils;

How should I connect in the right way to get only 4 ends?

  • \$\begingroup\$ A 12v .4 amp motor is high impedance, and thus high inductance, and so basically not worth wasting your time and effort on as it will have low performance for any sane drive voltage. \$\endgroup\$ Commented Jul 5, 2015 at 1:14

4 Answers 4


When I read your post the fist time I missed the fact that you've already cut the traces.

From the photo it appears that Node 5 the common 5th wire is brown. The first inductor pair are red-white wires, and the second are yellow-blue. If you reconnect the trace you cut between red-white(around 10 o-clock on the photo) you will have a bipolar config.

I'm not an expert so I can't tell you if this would work, but it sure looks worth a try.


Your stepper is probably wired like in the circuit below.

  • The • that indicates 'polarity' of the inductor at same direction of current, it defines which end is north, which end is south;
  • Node 5 is common ground.;
  • L3/L4 are paired on the same pole;
  • L1/L2 are paired on the same pole.

When you tie Node 1 to Vcc, Node 5 to GND and leave Node 2 open, then the pole L3 and L4 sit on has a given polarity. However, when you release Node 1 and tie Node 2 to Vcc, the magnetic field will turn around on this pole. The dot of the inductor is on the other side.

This also means that both inductors are in series with each other and their fields will amplify when the center node is released, and only nodes 1 and 2 are used. This would mean that the unipolar stepper would perfectly well work as a bipolar stepper with double voltage rating.

... if only the center tap wasn't connected to the center of the L1/L2. Depending on how you drive the four nodes, you may have an issue with the interconnected node 5.

As @WoutervanOoijen states in the comments, the center tap shouldn't really be a problem, provided that when a coil is de-energized both ends are disconnected. This is usually the case with an H-bridge, but must be carefully checked for the specific type you plan to use.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Are you sure there is an issue? If both coil pairs are energized, both center tabs will be at half the power voltage (hence no current between them). If one is not energized, it is effectively out of the equation. \$\endgroup\$ Commented Feb 12, 2014 at 19:11
  • \$\begingroup\$ @WoutervanOoijen It is an issue if only one end is de-energized. It is important to de-energize both ends at the same time. Maybe I should have written that :o) \$\endgroup\$
    – jippie
    Commented Feb 12, 2014 at 19:14
  • \$\begingroup\$ A standard dual-H-bridge driver for a 2-coil bipolar stepper does so. And if it doesn't for a short time (while switching) the impact will be small: coils don't change their current quickly. I have no experience with this, but my gut feeling is that there won't be a problem. \$\endgroup\$ Commented Feb 12, 2014 at 19:17
  • \$\begingroup\$ @WoutervanOoijen updated my answer. \$\endgroup\$
    – jippie
    Commented Feb 12, 2014 at 19:29
  • \$\begingroup\$ Thank you for your quick answer, but I am a beginner in electronics so I would like to be sure I do it correctly: regarding to my drawing dl.dropboxusercontent.com/u/7511244/unipolar_tepper.pdf I should hook up to the stepper driver the coil wire nr.1 to A1; coil wire nr.3 to B1; nr.6 to A2; nr.8 to B2? and leave all the connections between the coils as they were, thanks \$\endgroup\$ Commented Feb 12, 2014 at 19:45

I saw this question last week and it's been bugging me. But I haven't been near a computer since then (only my phone, which sucks for editing).

I modify stepper motors from uni-polar to bi-polar on a regular basis.

Because you don't have terminal numbers on your coil connections, I'm going to number them according to their position of the numbers on a clock face. Bottom pair of wires comes from the notch at 7:00, next pair of wires comes from notch at 8:00, next pair of wires comes from notch at 10:00, last pair of wires comes from notch at 11:00.

My preference when changing a stepper motor from uni-polar to bi-polar is to wire the coils in parallel if possible. This reduces the inductance (as opposed to a series connection) and also reduces the resistance. The reduced inductance allows faster step speed and the reduced resistance allows higher current operation.

I strongly suspect that each pair of coils is bifilar-wound with 2 different colored wires: tan & red. That is: both wires are wound at the same time. To make the stepper uni-polar, the tan wire from one end is connected to the red wire of the other end and connected to the common point.

I think that one end of phase A is the pair of wires coming from 7:00 and the other end of phase A is the pair of wires coming from 10:00

That would make phase B the pairs of wires coming from 8:00 & 11:00.

The quick way to find out is to gently un-solder the coil wires connecting to the common traces (brown wire) and ensure that all coil wires are now not connecting to each other. A simple continuity test will confirm which wires belong with each other (4 coils, 2 wires each).

If in fact the coil wires are as I suspect, the easiest way to convert this motor to bi-polar is to connect both wires belonging to a phase to the same solder pad going to the Yellow / Blue wires (phase A) and White / Red (phase B).

There wouldn't be any wires going to the common traces on the PCB (Brown wire).


If you hadn't hacked the board, the easiest conversion of this motor would be to cut of the wire connecting the tracks near terminal 5 (9 o'clock), and leave the brown wire disconnected.

Then you could have connected your bipolar driver A to red and white and B to yellow and blue.

You could connect a wire to the spare solder point at 1 o'clock to make a 6 lead stepper motor, once you had cut the connection near terminal 5.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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