Re: Value of Zener diodes

Jon - thanks again.

The drive I am using is not a chopper type unipolar one. The power
supply is is current limited to the rated current limit of the stepper
being used (the circuit is from the National Semiconductor LM317
datasheet, though it omits a few important details), and the voltage
is set at 24 to 30 volts. Although a high voltage will be evident at
the start of a coil charging, by the time full current is supplied for
a step, the voltage will be the much lower (as a rule) rated coil
voltage. I haven't actually looked at the voltage with a scope, but
that is the theory.If all steppers I would use were low voltage
perhaps the zener would be unnecessary? Would a zener take care of
spikes which might be relatively high (I guess you refer to these)?
Some steppers I have are specified at 12 or 24 volts, so a zener would
definitely make sense with them, I guess.

Again the current limited supply implies an average voltage
somewhat above the rated coil voltage, but when the driver switches
off current to a coil, I think the voltage at the tap is at the
recommended coil voltage, so in many cases the power rating of the R's
would be lower than your calculation suggests.

Please continue to correct my misunderstandings, I appreciate your
patience and willingness to educate.

Grant

--- In CAD_CAM_EDM_DRO@yahoogroups.com, Jon Elson <elson@p...> wrote:
>
>
> grantfair2001 wrote:
> >
> >When you say "so a 40 V Zener diode could be used. but, this is
> >pretty dangerous", what is the danger?
> >
> >
> Hmmm, maybe I was miscalculating!
> The center tap of a chopper-type unipolar drive is at the DC supply
> voltage. When one transistor turns on, grounding that end, the other
> end goes up to 2 X the DC supply voltage by simple transformer
> action, plus a little more, due to the collapsing of the leakage flux.
> If you limited this peak voltage with a 40 V Zener going back to the
> DC supply, that would cause the DC supply + 40 V to be felt by the
> transistor. If your DC supply was 30 V (as you mentioned), + the 40 V
> Zener, that would clip the voltage to 70 V, which sounds reasonable.
>
> (My error was I was adding 40 V to the 60 V transformer-action peak
> at the off side, and that was EQUAL to the transistor's rating.
But, that
> is not how the circuit works. It is the Zener plus the supply voltage.)
>
> >Also, when you say the resister/diode combination require the R to be
> >based on the current involved, how is this calculated? Is this just to
> >ensure the R's ability to dissipate the power, or is there more?
> >
> >
> No. The value of the resistor is determined by R=E/I. So, you
figure how
> much voltage you can tolerate above the DC supply (again, 40 V is a good
> number) and your motor current (was it 3 A?) and R=40/3 = 13.3 Ohms.
> So, you'd use a 12 Ohm resistor. The power rating depends on the step
> rate, and the fact that the resistor could never see a duty cycle
greater
> than 50 % anyway (unless you combined currents from both ends with
> the standard diodes that are necessary anyway). Assuming you used
> 4 resistors, and the steps were coming fast and furious, so the current
> hardly has time to die away, P=I^2*R (/2 for the duty cycle) so you
> get 3^2*12/2 = 54 Watts. Actually, a 25 Watt resistor will probably
> handle it ok, due to the exponential decay of the current. Hmm, 4
> 25 Watt resistors! Note, also, that the Zener diodes would have
> to dissipate a similar amount of heat! Are you starting to see why
> nobody makes serious motor drives unipolar? You start out thinking
> you will save money with only HALF the transistors, but then all these
> uncontrolled inductive pulses start REALLY complicating the drive
> when you try to tame the windings.
>
> Jon