Subject:
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Re: Custom Firmware, IR Problems, and Dead RCXs (long)
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Newsgroups:
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lugnet.robotics
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Date:
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Tue, 29 May 2001 02:59:49 GMT
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Viewed:
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684 times
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> Start with the tower. If you use a 9V battery, the IR circuit is
> basically two IREDs, a limit resistor, and an NPN switching transistor.
> Assuming Vf of 1.2V for the IREDs, and a Vcesat of .2V for the transistor,
> you end up with 6.4V across the limit resistor.
>
> In near mode (which has a range of a few feet) the limit resistor is
> 1K, which results in a DC current of about 6.4mA - no problem even in
> straight 100% duty cycle.
>
> In far mode, the limit resistor is 5.6Ohms, which lets 1.14A through the
> IREDs, which are (usualy) rated at 100mA constinuous. Even accounting
> for a 50% duty cycle on the bits due to the inversion of every byte
> and another 50% due to the carrier, you still have about 280mA through the
> IRED on average in the very worst case!
For what it's worth, it's not actually that bad. I don't know the
specs on the NPN transistor, but I suspect 1.14A would kill it almost
instantaneously. One important factor you've left out of this equation
is the 9V battery itself. We often mistakenly assume these are ideal
voltage sources, but the aren't. They have a finite, and fairly high
(not to mention highly variable, depending on freshness, temperature
and battery type) internal resistance. That internal resistance rises
as the battery wears down, but no matter what the value at any given
time, it appears in series with the IRLED's. 9V batteries are not
designed to be high current sources. They have a fairly high internal
resistance (which is why you don't see them used in flashlights - the
filament of the bulb represents a fairly low resistance. Add that to
the resistance of the battery and a significant proportion of the
energy gets dissipated in the battery, not the bulb. This also applies
here. I've tried 11 9V batteries that I have in my cupboard, all of
which I consider to be in "mid-life". They all read as strong on a
battery tester, but that's a relative measurement. An actual
measurement of the output currents ("dead short" through one of my
ammeters) shows them all providing between 1.0mA and 1.6mA. Like our
often idealized view of batteries, it's easy to take an idealized view
of an ammeter as well - that is to say, it's easy to assume it is
actually measuring the current through the load when in fact the
ammeter becomes an additional load itself and the current the original
load actually sees without the ammeter present is actually higher. All
of this depends on the internal resistance of the ammeter itself.
Normally (as with the internal resistance of a battery) we ignore the
internal resistance of the meter since for most applications it is
only a small fraction of the load resistance. However in the case of
trying to measure the internal resistance of a battery by shorting out
the terminals through the meter, the only two sources of resistance
are the internal values of the battery and the meter. In order to
calculate one, you must know the other. For this I turn to my meter's
manual which, as with all important documents, is nowhere to be found
now that I actually need it. If anyone has a meter with a manual, I
would invite you to short out a 9V battery to measure its maximum
output current and thereby be able to calculate the batteries internal
resistance. If you try this, start with the highest current range and
work your way down to something that provides as near to a full-scale
reading as possible. Note that in an ammeter, the internal resistance
changes with the range setting so you will need to get the internal
resistance value from the manual *after* you've selected the proper
range. I look forward to finding out what some of the values are. I'd
like to see some values for carbon, alkaline, rechargable alkaline and
NiCad battery types if anyone can do it.
I strongly suspect that we'll find that the internal resistance of a
typical 9V battery is considerably higher than the 5.6ohm current
limiting resistor used in the high IR setting of the tower. This would
mean that a large proportion of the energy is being dissipated in the
battery and hence the actual current through the LED's isn't anywhere
near 1A.
It should also be noted that your typical wall transformer has an
internal resistance that is considerably smaller than that of a
typical 9V battery. If this is used instead of a 9V battery, less
energy will be dissipated in the source and more will wind up in the
load, which could account for why people who use 9V transformers are
reporting burn-outs.
Matthias Jetleb
(Electronics Technologist - majoring in communications, mind you, but
still.....)
VA3-MWJ
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Message is in Reply To:
 | | RE: Custom Firmware, IR Problems, and Dead RCXs (long)
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| (...) You are right, but the RCX and the tower are right at the hairy edge. The additional 50R impedance will leave the near mode essentially unchanged and I'll sacrifice on the high range. Start with the tower. If you use a 9V battery, the IR (...) (23 years ago, 28-May-01, to lugnet.robotics)
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