On Sun, 19 Oct 2003 03:43:32 GMT
wrote:

I am working through an electronics course and hoping that someone can
explain this to me. Ok here goes

VCC
|
|
\
/ R1
\
/
|
|
|-------------------------------------- Device Pin
| |
| |
/ V
/ _ LED1
| SW1 |
| \
| / R2
| \
| /
| |
| |
GND GND

I think part of your problem is a confusion between voltage and
current. Think of an analogy with a water plumbing system: voltage
corresponds to pressure and current corresponds to water flow.

Vcc gives you a voltage (pressure) of 5V, the voltage (pressure) at
ground is zero.

When the switch (valve) is closed (open to the flow of electrons or
water) all the pressure drop occurs across R1, so the output voltage
is zero, since we agreed at the start that GND was zero volts. Under
this condition there is still a flow of electrons (current.) In fact,
the current here is larger than it is when the switch is open.

I suspect that the quandry for you is the fact that the min. voltage
occurs when the current is at a max.

When you open the switch, the current reduces, because the total
resistance from Vcc to ground has increased. Now there is some
resistance (and voltage drop) between the output and ground, so you
get the output voltage you calculated with Kirchoff's voltage law:
3.5V.

In the first case ALL the voltage drop occurred across R1, but in the
second case you have R2, R2, and the LED. Note that the power
disippated in R1 will be much greater with the switch closed, which is
what you'd expect, since it is carrying much more current then.

-
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Jim Adney
Madison, WI 53711 USA
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