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Re: [TowerTalk] SteppIR problem (K7NJ)(K4SAV)

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Subject: Re: [TowerTalk] SteppIR problem (K7NJ)(K4SAV)
From: "R. Kline" <>
Date: Mon, 11 Jun 2007 22:19:16 -0600
List-post: <>
R. Kline wrote:

>There has been mention that the source is a current source, and depending
>upon the length of the control cable, the current would increase, etc.
>couldn't be correct, since the load (motor), cable resistance, and source
>are all in series - so the current would be the same everywhere.  What may
>be happening is that the voltage increases in an attempt to keep the
>constant.  Perhaps the supply "senses" the voltage output and shuts down if
>it exceeds a predetermined maximum voltage.
>riki, K7NJ

Apparently you didn't read my explanation, in answer to WC1M's question
about how this works.  Here it is again.

An interesting observation.  One can easily make the mistake of saying
that the current to the motor is regulated so the supply current doesn't
change due to the extra resistance, because the motor current doesn't
change, but this is not exactly true, as you are alluding to.

Think of the load as an RL network.  The pulses delivered to this load
are both width and frequency modulated, in other words they change to
provide a fixed amount of average current to the motor.  So for an
increased resistance the pulse width has to be wider than that for a
lower resistance.  This is because the rise time to this network will be
slower.  Looking back at the power supply this requires more average
current from the supply because of these wider pulses.  (The rep rate of
these pulses also changes but that's just another detail.)  However the
peak current required from the supply is the same in both cases, because
the regulator IC shuts the current off when it reaches a preset value.

It is this max peak value that is independent of the amount of
resistance, not the average current from the supply.

That does raise the question of what the power supply's capability is,
in terms of peak and average current capability.  It is possible that
the power supply filter cap can deliver the high peaks but that the
supply itself cannot sustain the higher average current required.

Jerry, K4SAV


I've tried reading your message over and over, but just can't follow it.  I
get the feeling that you know what you're talking about, and I would like to
understand what you're saying. 

Let's return to basics.  Without getting involved in a lot of mathematics,
the equivalent graphical definition of average current for one period of the
waveform would be:
(1)  Graph the current vs time
(2)  Determine the area bounded by the current graph (step 1)
(3)  Divide the area (step 2) by the elapsed time from the beginning to the
end of the waveform.

It follows from the above that if the amplitude of the current pulse
increases, the average current will remain unchanged if the length of time
of the current pulse is decreased.  Conversely, if the amplitude of the
current pulse decreases, the average current will not change if the length
of time of the current pulse increases.

Please, then, explain the following points to clarify your previous
(a) Why is the rise time "...of the network..." slower?

(b) Why does shutting off the current result result in the SAME REQUIRED
peak current ?

(c) Assuming the same average current, why are longer current pulses

I get the impression that you've put in place many of the pieces to the
puzzle, but the logical connections between them are somewhat fuzzy.


riki, K7NJ




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