Alt-BEAM Archive
Message #04749
To: beam@sgiblab.sgi.com
From: Bob Shannon bshannon@tiac.net
Date: Mon, 21 Jun 1999 20:01:30 -0400
Subject: [alt-beam] Re: Feedback explained?
Ben Hitchcock wrote:
> ----------
> >From: Bruce Robinson
> >
> >I've come across the feedback concept a number of times while wandering
> >the BEAM landscape, Ian. I've seen a similar suggestion to yours in a
> >few places -- motor slows down, current increases, supply voltage drops,
> >and the cap charges more quickly.
> >
> >After a lot of reflection, I think this kind of feeback is bogus. (Boy,
> >am I asking to get dumped on for that remark!) The thing is, for a
> >74HC14 (and for most basic CMOS chips), the trigger threshold is
> >approximately proportional to the supply voltage. And in one of the
> >Stills/Tilden papers, I came across the formula for the time delay -
> >
> > T = R x C x ln(Vmax / Vth)
> >
> > R = resistance (ohms)
> > C = capacitance (farads)
> > Vmax = maximum voltage (e.g. supply voltage)
> > Vth = threshold voltage
> > T = time (seconds)
> > and finally, ln is the symbol for natural logarithm
> >
> >So if Vth is approximately proportional to Vmax, then ln(Vmax / Vth) is
> >a constant, and the only thing that will affect the delay is R and C.
>
> True, for CONSTANT battery voltages. But what happens if the walker is
> happily walking away, and in the middle of a step. Say the motor voltage is
> 5 volts. Also say that the normal unloaded time constant is, say, 2
> seconds. Also, to make things a bit simpler, let's say that the trigger
> voltage is 2.5 volts, and that the cap charges from zero.
>
> Now let's freezeframe at the point where it started a step one second ago:
> the cap voltage is a bit above (because the charge curve is exponential)
> 1.25 volts. Let's say about 1.4 for the sake of argument. So you still
> have 1 second to go before the inverter will trigger.
>
> Now grab the leg so it can't move. The battery voltage will decrease due to
> the increased load, right?
Not really. Many walkers use regulators, its not the battery voltage that
drops here, its the output of the prior Nv stage, due to its limited ability
to source current.
The battery voltage may remain at 5 volts, but feedback still happens!
The logic level out of the prior Nv may well drop to 2.8 volts however,
depending
on your motor driver.
> Let's say it decreases to 2.8 volts. (That's a bit exaggerated, but it
> serves to illustrate a point.)
> Now, what's half the supply voltage? 2.8 / 2 = 1.4 V. What does this mean
> for our little walker? It means that the inverter will switch STRAIGHT AWAY
> instead of waiting for another second.
>
> So the time constant is decreased.
>
> As far as I understand it, the time constant changes when the battery
> voltage changes, but not when the battery voltage is constant. So if you
> hit an obstacle at the end of a step, you won't be able to see any change at
> all. But if you hit one halfway through, then you will.
This does not fit my observations. The supply may be fixed, yet the output
level going into your Nv can be pulled down by an excessive (per specs) load
from the motor driver.
> That's my understanding, at least, and there are probably flaws in it - feel
> free to point them out.
>
> Ben
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