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[DPRG] fairchild QRB1133/QRB1134

Subject: [DPRG] fairchild QRB1133/QRB1134
From: Dale Wheat dale at dalewheat.com
Date: Fri Sep 5 13:53:00 CDT 2003

Chris,

> >  It will vary from one diode drop from your
> > positive voltage to ground, with more illumination driving the voltage
> > lower.
>
> You lost me here. It sounds like you are saying the output voltage from
the
> collector will only vary by a diode voltage drop or 0.5 to 0.8  volts.

No, the output will vary from your positive supply less one diode drop, to
ground (ideally).  For example, if you are using 6V, then the output would
be from ~5V (6V - ~1V) to ground.

> Are you also saying that the more light detected, the less output voltage?


In this configuration, yes.  The more light that hits the phototransistor,
the more it conducts.  In this case, it means the more it conducts to
ground, bringing down (overcoming) the voltage supplied by the pullup
resistor.

> Instead of adding a comparator, could I send the output of each collector
to
> an anolog imput of for instance a PIC chip?  Then I could work out my
> hysteresis and threshold detection in code?


Sure, if you want to mess with it in code.  You can even do some fancy
stuff, including running minimums and maximums, self-calibration, etc.  If
all you're aiming for is a digital output (1 or 0, light or dark, line or no
line), then it's a lot easier (for me, at least) to have a very simple
signal processing circuit up front, and let the processor deal with ones and
zeros as God intended.

> If I have all my logic circuitry regulated or seperated from the motor
> supply, why would I need another regulator for these sensors?


For one, you don't want to load down your logic power supply driving LEDs,
if you don't have to.  For another, the regulator is, in this case,
regulating current, not voltage.  The benefit is that your LEDs don't dim
beyond a useable level if your voltage sags.  If you have plenty of
regulated logic power to spare, then you can use a simple current limiting
resistor (240 ohms if you are using 6V and want 20mA of LED drive current).
This will, of course, cost you battery time.

Thanks,

Dale Wheat
http://dalewheat.com
(972) 486-1317
(800) 330-1915, access code 00


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