[The Speaker Guy]

Hey hackers, cool site.

I purchased some LaserJet II used steppers, they're on their way. These are supposed to be rated for 1.4A per phase unipolar, 5.2V. If I drove these from 28V until 1.4A, then dropped to 5V, it that efficient?

Background:
I am beginning a CNC machine, smaller scale, likely about 12" x 8" table, gantry style, Most common use will be PCB, but small woodworking is a possibility.

The reason I ask this is the drive of this is an interesting part. I am an EE, and have access to all the requisite tools parts and SMT soldering. So I am disinclined to use an off the shelf IC, and use an analog approach.

The motors have 200 steps per rev, I have bought 3/8 Acme 12tpi, giving a theoretical resolution of 0.41 mils per step.

Thanks
John

[ViperTX]

Hackers.....where??

Well John as a fellow EE....I have no idea what you're asking in the first paragraph.

The reason that you run steppers at a higher then their rated voltage has alot to do with the way each phase is driven....Pulse Width Modulation (PWM)...someone apparently did some empirical studies and determined that it would be safe.....(SOA) to use a voltage up to 20X the rated voltage as long as you were using PWM....

I've never personally studied or tested it....but I would if I was planning on delivering a product to the marketplace...I would want to validate the SWAG....

41 mils....oh....4 ten thousands.....yep...that's correct...

[pminmo]

Couple of the circuits on my website you could use. Chopping to limit the current is the way to go.

[The Speaker Guy]

See attached schematic. This is a quick hack, as I need to know more about the resistance and inductance of the motor.

Circuit operates as follows:

An incoming pulses turns on the low side FET. This same pulse is level shifted to drive the high side FET. Because of D1 blocking, the highest voltage is applied across the load. When the load current hits 2A, the amplified sense signal overides the high side gate signal, turning of the high side FET. The load voltage Vload drops to where it is caught by D1, supply a steady state (slighlty indereasing) current to the load.

Based on the second answer, I think I can do this with a PWM converter as well, avoiding the dual supply requirement.

Thanks

John

[Mariss Freimanis]

1) You have no inductive current return path whe IRF510 turns off. Add a diode, anode to n-channel drain, cathode to your +24V bus.

2) You are using a p-channel as the top MOSFET. A bootstrap gate drive supply would allow you to make it an n-channel as well. P-channel MOSFETs have 4 times the Rds of a same die-size n-channel. They are not true complimentary devices like bipolar junction transistors.

3) I assume you plan to have 4 identical channels of the circuit shown. That makes for 8 MOSFETs, the same number you would need for 2 full-bridges. Consider a dual full-bridge drive, they have no voltages present that exceed the bus voltage by more than +/- 1 diode drop.

4) Take a look at IR half-bridge MOSFET drivers like the IR2104. They make life much easier. If you'd rather "roll your own", please see the attached pdf. It has a working and tested circuit I designed a few years ago.

Mariss

[The Speaker Guy]

Thanks Mariss;

understood. This was to indicate the general topology, and was not a final solution. The FETs are OK, but a 1N4002 is practically useless. <<Sigh>>, it's what comes in the PSPICE eval library.

I am sure I'll have more questions when I actually GET my motors !