Stepper motor query

Hey guys just wanted to pick your brains over a little project im working on...my first venture into stepper motors

I have a KT42JM06 Japan Servo stepper motor (24v 1.5a 2-PHASE) Its a very precise motor with a 0.6deg step and weight damper fitted so its perfect for the job...

Im trying to set it up as a focusing motor, now i setup a simple parallel port driver using 2 x ULN2003AN driver IC's in parallel which is all ok but the motor is 3 phase and going over the specs of the motor shows 6 excitation states with 1 pole being disconnected at each state. Anyone know any work around for this or can it be done with high and low states still, if it is required the only real solution i can see is 3 solid state relays running from another 3 data ports to disconnect those outputs in sequence..

The closest information i could find was someone else trying to setup a similar series and he was trying to use:

U V W
1 + - -
2 + + -
4 - + -
5 - + +
6 - - +
7 + - +

cheers

 

You don't need relays, just use a tristate buffer like the 74HC125 to produce a high-low-floating tristate signal.

 

Personally I would use a stepper motor controller IC, and the you can just feed it increment/decrement pulses.

PS - you dont neccesarily have to drive the stepper in bipolar mode. Try experimenting with driving one or two coils at a time with a single current direction, and you may find it works acceptably but with a larger step size or even reverse operation. You could tie leads 2, 4 and 6 to ground, and then just cycle through energising 1, 3 and 5.

 

I'll have to try experimenting again that ic is just what I was looking for except I'll have to come up with some sort of driver after it to switch the main loads.

I was going to use a dedicated driver but there arnt many options locally that supported this steppers requirements... The motor is 3 wire 6 excitation states

 

Edit, just read that it is 3 wire (so delta system)... makes it trickier.

For the driver you should be able to have 3 NPN transistors connected to ground and 3 PNP transistors connected to V+, with the three phase wires connected in between. You can then selectively activate each one of the 6 to create a positive or negative connection as required. Logic on the PNP's will be reversed, and dont activate both transistors at the same time.

It actually the same design as a three phase inverter or speed controller - what you have is essentially a three phase sychronous motor with a large detent torque.

Edit2 - if you're looking for a circuit, it's also known as a 3 phase H-Bridge. If you keep the 6 control signals seperate, then the circuit becomes very simple, ableit reliant on correct control to prevent short citcuits.

If you go here and scroll down to "Bipolar Motors and H-Bridges", the first diagram is close - just add another block with switches E and F.

The main block of this circuit here is also very typical. If you replace the MOSFETS with BJT (convenient at those current levels) then make sure you include the protection diodes too like the first circuit.

 

Easiest way, have two sets of three 1.5a switches. One set is connected to +24V and D0,D1,D2 on the parallel port. The other set is connect to 0V and D3,D4,D5 on the parallel port. All switches will be off when 0 is applied from the parallel port data lines.

Then simply write yourself a new sequence table, using this basic idea:
U+ = D0, U- = D3, V+ = D1, V- = D3, W+ = D2, W- = D5,
eg: step 1 is ??010001 where ? = who cares

Use whatever you want for the switches, just remember you need to be able to switch positive and negative with them. I'd use BD139/140's myself.

 

thanks for the help so far guys hows this looking so far dont you mean that half the outputs will be active (inverted) as at 0v on the base of the PNP's will output high?

 

I couldn't remember which was the NPN version

Printer ports default to 0V, which means using PNP transistors would result in damage to the stepper motor.

 

The BD140 is a PNP transistor, yes you are right also i didn't take into account the fact that they will all be active high on all channels before i get to change the input signals

hmmm i guess i might have to look at a mosfet based driver?

 

It shouldnt be a problem. If the port defaults to 0, then the NPN's will be off, and the PNP's will be on. That means all 3 coils will be pulled to a high voltage, but no current will flow.

If need be you can invert the pnp logic by connecting a small NPN to it's base.

 

I realised that after I posted it, but that all assumes there isn't a circuit fault, all it takes is a solder bridge, bad transistor, or miswiring and who knows what the result would be.

Besides, its easier for testing purposes and programming if all transistor outputs default to off. Not to mention what happens if you disconnect the stepper motor without removing power.

The circuit will be fine with all BD139's, but remember to connect a reverse diode from emitter to collector : E -|>|- C so the inductive load's back emf doesn't blow them.

 

Is that possible with a NPN as you have the 24v supply line on the collector, the emitter would have no connection? or am i misreading something

i can see adding some NPN's to invert the output of the PNP's would work though

 

I think realistically he'll need some PNP's in there. If they are all NPN's then he has to get the Base voltage on the positive side switch higher than the load voltage, and that may not be an option.

I think from the data sheet the stepper was 5V, so an mix of NPN's and PNP's should be directly drivable from the parellel port or buffer with only a 5V supply.

 

What a load of rubbish, all transistors would see the same relative voltages on E and C terminals. The reason we usually use NPN + PNP is to simplify the drive circuits in things like amplifiers.

I've got a circuit here that switches the 0v side of a 12v circuit here from a <12v, its not blown up or anything.

I did dig up a data sheet for the BD139, max Emitter to Base voltage is 5V, so minimum base on voltage is 19v (24v-5v), might as well use 24v.

No the stepper is 24V, plus the base current for a BD 139 is 500ma, and printer ports aren't rated past 50ma or something.

 

I would be using the ULN2003an as a pre driver anyway that puts out 500ma @ 24v per channel so that's no issue for the bd139 would it work though for switching the 24v supply to the motor since it would either be 0v or 24v on the base and the emitter would be open circuit?

 

Huh? I'm talking about the high side switch. It's much easier to use a PNP.

Let's look at the high side, assuming we're running things in saturation. For a NPN the Collector would be connected to VCC, then there is the nominal internal C-E voltage drop of 0.2V, then the Emitter is connected to the load, then the other side of the load to ground. The Emitter voltage (and the load) is VCC - 0.2V.

Working back, the nominal B-E voltage drop is 0.7V, so the required base voltage would be VCC - 0.2 + 0.7 = VCC + 0.5V. In a single supply circuit, where does that magic extra 0.5 volts come from?

If instead you supply only VCC to the base, the transistor will not switch on fully and will develop a higher CE voltage, dissipate more power and become hot.

With a PNP on the high side, you only have to lower the base voltage 0.7V below VCC to get it to turn on. Very easy with a pull down.

You dont actually need to apply high B-E voltages. It's a diode junction and unless you try and cram ridicuous amounts of current through there will maintain around 0.7V. When you're in the lower end of the operating envelope, all you need is 0.7V above the Emitter voltage for full saturation, on the condition that you satisfy the current requirements with a current limiting resistor.

the OP said 24V 2-Phase, the data sheet he linked to says 5V 3-Phase. Which one is it?

You dont need 500mA of base current. Hfe is about 25 at the high loads, so for full saturation you only need to apply load current / 25. The data sheet for the stepper said 1A, so you only need 40mA base current.


It turns out that the C-E voltage of the BD139 is 2V at 1A, so it's probably not a good choice anyway. A better choice might be a TIP32/TIP42 which have a similar gain but 6A rating. C-E voltage at 6 amps is 1.5V, should be alot less at 1A. This means less losses, less heat, and better efficiency.

If it is a 5V stepper, MOSFETS would be marginal as you're not guaranteed a good turn with a 5V gate signal. You'd have to buy special logic level MOSFETS. If it is 24V then a MOSFET would be perfect - once again P-Channels on the high side and N-Channels on the low side. You'd need to protect the gates from high voltages though with say a voltage divider circuit.

 

The sticker on the motor says DC24V 1.5A/2 Phase

I believe the spec sheet refers to some other rating as someone else brought it up on a forum with his motor and they said it was wrong.

As its stands i have 24v supply for the load. Now as a start i have the ULN2003AN driver straight after it which will put out 500ma @ 24v across out all outputs. Now i can do away with that or make use of it as a more powerful output and protection for the parallel interface? Only problem is to reduce the current for the NPN's it requires decent power rating resistors...

 

thanks for the help so far guys, i had a run of good luck the other day and managed to source the original driver board with the SLA7611M IC driver for the motor, just gotta work out how to control the inputs on the chip and it should be good to go

http://www.sanken-ele.co.jp/en/prod/semicon/pdf/sla7611me.pdf