scott wrote:
>> Well anyway, FWIW, the IC managed to drive the LED just fine. (I 
>> bought special 5V LEDs; apparently normal ones take roughly 3V.)
> 
> "normal" ones work at all sorts of voltages, it even varies from piece 
> to piece quite significantly.  But you never drive an LED directly by 
> applying a fixed voltage, you always drive it by regulating the current 
> to a fixed amount (like 20 mA or whatever).  Adding a series resistor to 
> a raw LED is a quick and crude method of fixing the operating current.

This doesn't make sense to me.

Presumably the resistence of the LED is finite and fixed. How does 
adding another resistor help? There are several schematics in my 
electronics kit that involve LEDs and no resistors at all.

>> You think using a 2-pole switch with the poles connected to each of 
>> the power rails will do it?
> 
> Yup, or just connect a high value resistor (eg 10 or 100K) between the 
> input and the other power rail - so you have your switch arranged like 
> this:
> 
> http://www.seattlerobotics.org/encoder/mar97/basics4.gif
> 
> That way, when the switch is open, the input is connected to 0V, and 
> when it's closed it is connected to V+.

Now, see, I've never been able to comprehend stuff like this. To me, 
this diagram just looks like when you close the switch, all the current 
will flow straight from one rail to the other, shorting out the battery 
and not providing any current at all to the input of the gate.

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