Diode help... HELP!

From: Chuck McManis <cmcmanis_at_mcmanis.com>
Date: Tue Feb 22 20:36:14 2000

At 07:15 PM 2/22/00 -0500, Bill wrote:
> > Chuck McManis gave me a circuit diagram...
>
>Hmmm... Here's what I was trying:
>
>[*] for connection
>
>[D] for diode like this: -*->|-*
> |
>

<circuit deleted>

Your circuit is nearly equivalent.

I'm guessing that you don't have the pins pulled high to Vcc through a
small enough resistor. This is what I think is happening.

When the input pins are floating, there is enough leakage to pull them
high, but when they are grounded directly they go low. The back side of a
diode "looks" like an open circuit to the floating pin.

You can test my theory as follows.

Connect pin 1 to a diode then to ground through a switch.

Read back the pin while changing the switch. What do you get?

(given your previous experience it will probably read as a logic '1' both ways)

Now pull the pin high by connecting a 4.7K resistor between the pin and Vcc
behind the diode so you have this:

         Vcc
         |
         R (4.7K)
         |
Pin ----+-----|>|----/ ---Ground

Now repeat the experiment. If I'm right you'll see the switch this time.

If I'm wrong you should look for some Shottky diodes :-)

--Chuck
Received on Tue Feb 22 2000 - 20:36:14 GMT