If you leave the diodes "open", they will have NO effect on the circuit whatever. In this case any given diode "pair" will appear as a set of cathode-to-cathode connected diodes, and there will *never* be any conduction through them. To dissipate the flyback energy it is imperative that the set of common cathodes be connected *somewhere* so that the current can circulate. Otherwise the stepper motor coils will create extemely high voltages when they are turned off. By the way, you don't need a separate resistor for each coil. You can use a single resistor connected between the common cathodes and the Vcc line. Fr. Tom McGahee ---------- > From: Russell McMahon > To: PICLIST@MITVMA.MIT.EDU > Subject: Re: [OT] Driving a Stepper > Date: Monday, August 16, 1999 2:11 AM > > Paul B said > +AD4- I'm not sure what you're saying here? Wouldn't you want to tie it > to > +AD4-a 2xVcc rail like you just described? > > > Tying to 2Vcc+- would be SAFEST. > What I am saying is that if you can tolerate the ringing voltage it has > least effect on operation if you leave the diodes open (or with a very > light load). > > Operate the circuit - look at the diode common point on an oscilloscope - > decide whether the result is tolerable for your driver. Depending on your > switching waveforms, relative on and off times and the amount of leakage > inductance you may see very little ringing here as the other end of the > same winding is clamped to ground. This is what produces the 2:1 > transformer action but also limits the voltage excursion. Be sure that what > you do is OK in all circuit configurations - eg do you ever suddenly turn > it off rather than stepping? (yes, you do :-)). > > > > RM > > > -----Original Message----- > > +AD4-Russell McMahon wrote: > +AD4- > +AD4APg- In one design I drive a small unipolar stepper similar to this one > +AD4APg- with a ULN2803 darlington driver with built in diodes and I leave > the > +AD4APg- common diode point open. In this case the driver is happy. YMMV > :-). > +AD4-