Alt-BEAM Archive

Message #04745



To: beam@sgiblab.sgi.com
From: "Ben Hitchcock" beh01@uow.edu.au
Date: Tue, 22 Jun 1999 09:33:18 +0000
Subject: [alt-beam] Re: Feedback explained?



----------
>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?
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.

That's my understanding, at least, and there are probably flaws in it - feel
free to point them out.

Ben


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