William Chops Westfield <billw@CISCO.COM> wrote:
> > In such cases it can be useful to connect the PIC output to the base of
> > an NPN transistor, emitter to ground via a resistor, and LED between the
> > unregulated supply and collector.
>
> This WILL cause the transistor to be operating in it's linear region,

Not necessarily. Yes for a bare LED, but maybe not for a bulb.

> otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, which
> could easilly be greater than Vout from the PIC, right?  This shouldn't be
> a problem for typical LED currents, but it does mean that for higher
> current loads (light bulbs?) you'll probably end up needing to go back to
> base and collector resistors instead of the emitter resistor (or risk
> blowing up your transistor due to excessive power dissipation.)

A light bulb will drop much more voltage than a LED, and in fact may drop
more than the available Vunreg - Vpic.

Consider a bulb rated at 50 mA at 12V, fed from Olin's 20V supply, with a
5V PIC controlling the transistor. If the emitter resistor is set for 50 mA,
then the transistor will be in current limiting mode, with 12V across the
bulb and 3.7V across the transistor, which will be dissipating 185 mW.

However, if the unregulated voltage is less than 17V, then the transistor
WILL be saturated, and dissipating very little power.

Furthermore, this type of drive circuit provides a "soft start" for the
bulb by limiting the cold-inrush current. This should extend its life.

With an LED, you can always add a series resistor at the collector of the
transistor, which won't change the LED current (assuming it isn't too big).
It'll merely shift some of the power dissipation away from the transistor.
This retains all of the other benefits of using a current source and the
unregulated supply.

-- Dave Tweed

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