At 08:29 PM 2/29/2004 +1300, you wrote:
> >You would have used something like a dump resistor between the ULN2003
>diode common and ground
> > (and not connect the diode common to Vdd).
>
>Good point to note - when driving a multiphase stepper from a ULN2003 the
>common diode connection must *NOT* be connected directly to Vdd.

Ah, yes, I agree, on unipolar steppers. Vdd will be bad.

>This is
>because the stepper acts like a transformer - when you connect one winding
>between Vdd and ground the other winding which is between Vdd and an "off"
>ULN2003 driver is transformer coupled to the on winding and assumes an equal
>and opposite voltage to the "on" winding. As a result the voltage on the
>"off" terminal rises to about twice Vdd !!!!!!!!!!!!! If you connect the
>catch diode terminal to Vdd you are shorting out the off winding via its
>catch diode. At the least this will drastically affect stepper operation
>(and probably make it not work properly or at all) and at worst will destroy
>the catch diode. Thereafter the winding will not be clamped in any way and
>you will have 'trouble". In such cases the diode common needs to be
>connected to something which can deal with the 2 x Vdd output or left to
>float. (If left floating you have to deal with any leakage inductance spikes
>some other way). Several possible energy dissipating terminations are a
>resistor to Vdd, a capacitor (reduces spikes)plus discharge resistor or a
>suitable zener.

Don't forget the discharge resistor or zener as you suggest, because otherwise
the  capacitor voltage will increase, possibly beyond the 50V transistor
rating,
after many cycles.

Here's the schematic of the zener to Vdd.

                                               i
                                               --->
                        .-----------.--------------->|-----+
                        |            )|                    |
                        |            )|       /            |
                        |    +5V  o -' ----|>|-------------+
                        |            )|      /             |
                        |   i2 |     )|                    |
                        '     \ /   -'--------------->|----+
                        |          |                       |
                      |/         |/                        |
                     -|  Q1     -|   Q2                  (com)
                      |>         |>
                        |          |
                        |          |
                       ===        ===
                       GND        GND


Q1 has just turned off, and the coil current i(t) (<= steady state
coil current) is flowing through the zener, raising the collector
of Q1 to 5V+Vz. i2(t) is increasing towards steady-state coil
current.

Since the transistors are rated for 50V, a 35V or so TVS would allow
the inductive (~10mH on a small motor) current to drop faster. The
energy is ~1mJ/pulse.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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