View Full Version : Cheap Unipolar LPT Driver, High Amperes

03-24-2004, 01:12 PM
Ive drawn the schematics for a cheap unipolar driver but capable of driving high ampere steppers, because I got some 1'3V/3'9A steppers and there is no real schematic able to drive that amount of current. I've built and tested with Master5 at a P266 with w98, an old PC-AT PSU and:

- Surplus unipolar Nema23 1V3/3A9 unknown torque
- Surplus unipolar Nema34 1V2/4A8 unknown torque

and variable satisfaction ;) the nema23 has good 'tactile' torque and moderate speed, nema34 very good 'tactile' torque at low rpm.

Credits to the idea to www.cnconabudget.com
and this thread: http://www.cnczone.com/forums/showthread.php?s=&threadid=2854

Next steps will be triying to add some current control IC to avoid RBallast.

Now, for the schematic:
- Wired for LPT pin2 Direction, LPT pin 3 Step, repeat for other axis (pin3&4 for Y, pin 5&6 for Z) or rewire for other pins as needed
- Wired for outputs always enabled
- Use Steper Calc (credits to Eric Van Andel) for Value of Rballast

03-25-2004, 08:18 PM
What voltage, and amps does it support?

03-26-2004, 01:23 AM
From IRFZ44N datasheet
Voltage: 55V (VDSS)
Current: 47A (ID)

03-26-2004, 06:01 AM
Originally posted by abasir
From IRFZ44N datasheet
Voltage: 55V (VDSS)
Current: 47A (ID)

Instead IRFZ44N I've used IRLZ44N, wich is logic level Mosfet (Vgs=5V) in order to assure full ON when switching. Apart of that, characteristics are the same.

But do not expect to reach 47A :cool: because the T0220 package is totally unable to support that level of current. I'm testing it and at 5A/5V the mosfets are cold, and at 10A/5V they are ligthly warm (no heatspreader, no forced air)

These are the safe operating areas (ID/VD/Time) from the datasheet:

04-01-2004, 09:21 PM
Hey, do you have a better quality picture for schematic?.

04-12-2004, 03:29 PM
The clamp diodes look like they are in the wrong place. Consider the stepper winding as a auto transfomer when one side has voltage across it the other will be shorted by the diode it will still work to degree buts it's not best. the clamp diode should be accross the Mosfet and the opersite side will clamp the side which has just turned off. This transformer action also means that the voltage rating of the mosfet needs to be twice the power supply voltage.

Mariss Freimanis
04-12-2004, 09:09 PM
Couple of design tips:

1) Return the diode cathodes to the +Supply voltage, not across the coils as you have it now. Otherwise you will have excruciatingly slow current decay on "winding current off" time. That will severely limit max speed.

2) If you return the cathodes to where they should go, the peak drain voltage on the MOSFETs will be twice the supply voltage. The MOSFETs you have picked will come apart at 27.5V theoretically, probably at 24V if shoot-thru currents due to diode reverse recovery time are taken into account. Use fast recovery rectifiers (<100nS Trr), a current rating equal to the motor current and a rated voltage at least equal to the supply.

3) Don't pick a fragile MOSFET. The "on" resistance may look real sexy on the data sheet but the number that lets you know how rugged the device is the "Single Pulse Avalanche Energy" rating. The more milli-Joules the better.

4) Use RC snubbers across the ballast resistors. MOSFETs are real fast. Too fast in fact. The rate at which they turn-on, turn-off makes a wire wound resistor's inductance significant. A 200MHz scope would show truly breathtaking voltage spikes you never knew were there. Snub them.

5) Do not breadboard or wire-wrap a MOSFET power circuit. Use a printed-circuit board with big islands of copper pour in the high current sections. Use +Supply to ground capacitors. Reason? Wire is not wire at these speeds; it is an inductor (1nH per inch more or less).


04-15-2004, 05:12 AM
wow. i wish someone would rework that schematic using Mariss's suggestions. the output would probably be suitable for any uniploar indexer circuit - a great output stage i would bet.

04-15-2004, 12:20 PM
Mariss, tachus, you simply mean the shematic below? And, if possible, could you explain a little bit more about the snubbers? I understand the concept and the utility but I cannot figure how to wire it.

Thanks in advance


Mariss Freimanis
04-15-2004, 01:04 PM
First off, the diode does absolutely nothing as drawn. Move the anode to the MOSFET drain, the cathode to the + terminal of the power supply. Inductors (windings) do not "like" current thru them to change abruptly. They will generate what ever voltage is necessary to keep it flowing until the stored energy (J = L * I squared / 2) is exhausted. This is the way a spark plug coil works; you don't want that on your MOSFET drain.

When the diode is moved the way I suggest, the winding current circulates thru the coil in the original direction, thru the diode (anode to cathode), thru the limit resistor and back into the coil again. This completes the circuit.

The peak voltage on the MOSFET on "turn-off" will be 1 diode drop above the supply voltage. The peak voltage in the circuit will be at the coil to resistor node. That will be twice the supply voltage (assuming a zero-Ohm coil).

On the otherhand, the circuit as drawn will have its peak voltage on the MOSFET drain and will be limited only by the MOSFET's Vds (drain to source breakdown voltage). This is not a good thing.

A snubber is a resistor in series with a capacitor. It is meant to dissipate short voltage spikes. This combo is placed drain to source on the MOSFET. Pick the resistor value to carry an Amp or so. Pick the capacitor so that the RC time constant equals the width of the pulses you want to snub.

Example: 24VDC supply, 2A, 5uS pulse. That should be plenty.

R = 24 / 2 = 12 Ohms
T = RC so C = T / R C= 5uS / 12 Ohms = 0.41666 uF Use a .47uF cap in series with the 12 Ohm resistor.


04-15-2004, 01:21 PM
Originally posted by Mariss Freimani

When the diode is moved the way I suggest, the winding current circulates thru the coil in the original direction, thru the diode (anode to cathode), thru the limit resistor and back into the coil again. This completes the circuit.

Yeah yeah yeah oh yeah, that's the kind of things I love to read here, because is explained a way that is so simply that is almost beautiful. Btw, that's the reason I love electronics too ;)

I've attached a new schematic with the hope to help to the others that were in my own situation.


PS: Mariss, if you travel some day to Spain, drop a PM before taking the plane and you have assured an invitation to a cofee -as a minimum ;)- at the arrival

04-15-2004, 02:00 PM
i love the fact that Mr. Freimani not only designs these things for a living, but also is willing to share his hard-won klnolwedge so freely.. i cant begin to say thank you.

04-15-2004, 02:30 PM
Ahm Vac, when you travel to Spain as said in other thread.. the coffee invitation applies to you of course ;)


Mariss Freimanis
04-15-2004, 03:03 PM
Thanks for the nice words but I screwed up my previous post due to a caffine-starved brain. I saw a single coil and that is what I answered to. That is not the case for a unipolar 6-wire situation. Its been a long time since I have looked at unipolar drive circuitry; I use full-bridge topology because it gets around a lot of problems.

A step motor center-tapped winding is best thought of as a center-tapped transformer because it is one.

This means if one end-wire goes to ground (MOSFET "on") and the center tap goes to the +Supply, the other end-wire will have twice the +Supply voltage on it while the current was changing in the coil. Once the current stopped changing (about 3 L/R time constants), the voltage would decay to below +Supply (+V - I*R). The diode connection I suggested would over-heat the diode and rob the motor of power during that time.

So, basically lose the diode altogether and keep the snubber. Alternately, replace the diode as drawn in the original circuit with a zener. Then you don't need the snubber. The tricky part is the zener voltage MUST be more than twice your +Supply voltage but less than the MOSFET rated breakdown voltage.

The above is the unpleasant consequence of a unipolar drive. Only 4 MOSFETs are required but circuit voltages exceed twice the supply voltage. A full-bridge drive requires 8 MOSFETs but all circuit voltages are equal to or less than the supply voltage.

Everything else being equal, MOSFET "on" resistance goes up with the square of the drain to source (Vds) rated voltage. This means a 200V rated MOSFET has 4 times the die area as a 100V MOSFET having the same "on" resistance (Rds). Since price theoretically tracks the chip size, the 200V puppy costs 4-times as much.

Mariss Freimanis

P.S. My name is too long by one letter. CNCzone whacked off the trailing "s" in my name. I'm not Italian :-)

04-15-2004, 03:33 PM
Ok, so forget the circuits, go back to Urgundiz's version but with well sized zeners, and all happy ;)

And I maintain the coffee, because is not related to your citizenship but your way of being, and because I've seen you are at South California before the offering ;)


04-17-2004, 10:49 PM
I thought it was better to have a zener
and a schottky D-S (rather than depend on the body diode?)

published rds-on figures I suspect are never approached in real world situations

Melnais Balzams - thats kinda like a coffee...


Mariss Freimanis
04-18-2004, 12:18 PM
What is the point? Neither are ever going to conduct.


09-02-2007, 01:52 AM
Alright I've read through all these posts and what not. The question or request rather. Could someone piece together what changes should be made to enhance the circuit and draw up a new schematic with a parts list. I have 3 Keling KL23H286-20-8B. They are over 400 oz. in holding torque nema 23 and I bought a driver off of hobbycnc.com but it can't nearly push them at full speed and rightfully so. The driver I have is only rated at up to 42V at 3 amps per motor(3 axis). My motors say they draw 8 amps in unipolar. I have a router all built but It won't go any faster than 30 ipm. Just cuz of the motors not getting the current they need. All the drivers out there are rated low as far as amperage goes. And the ones that are rated high are redicously priced. Well as far as I'm concerned. Well thanks for any help.


09-02-2007, 09:44 AM
It is OK to re-invent the wheel as long as you learn from it. Take a look at this two threads, the cheap unipolar drive design is already done and tested:
http://www.cnczone.com/forums/showthread.php?t=25361 and http://www.cnczone.com/forums/showthread.php?t=38873



06-23-2009, 09:45 AM
Hi, I have been reading this thread. I would like to experiment using this snubber/steering diode configuration. I am curious as to what the current limiting resistor should be (RBALLAST). You mentioned it should be a 1A but what value is adequate. Thanks for the valuable information.