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Diode as Pull-down

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Iomega

New Member
It's well known to not have floating pins as inputs to a microcontroller, and pull-up or pull-down resistors should be used.

My question is whether a diode attached to ground would accomplish the same thing. I know that a diode with current flowing through it has a ~2V drop across it.

Will it be a robust pull-down in the attached circuit when the switch is open?

Thanks.
 

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Noggin

Member
No. I'd suggest moving R2 to the top of R1 and then use another resistor to pull the pin low. The other problem you have with your circuit is that the switch won't pull the input pin all the way high. It'll pull it up to roughly 4v which will likely be enough for the PIC to register it as being high, but I don't like it.
 

Speakerguy

Active Member
Can you use a pin on Port B and use the Port B internal pull ups instead? And re-work the circuit around that?
 
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be80be

Well-Known Member
It's well known to not have floating pins as inputs to a microcontroller, and pull-up or pull-down resistors should be used.

My question is whether a diode attached to ground would accomplish the same thing. I know that a diode with current flowing through it has a ~2V drop across it.

Will it be a robust pull-down in the attached circuit when the switch is open?

Thanks.
Are you just wanting to use the led to show if you have pressed the switch
I do my switches like this saves power
 

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Diver300

Well-Known Member
Most Helpful Member
My question is whether a diode attached to ground would accomplish the same thing. I know that a diode with current flowing through it has a ~2V drop across it.

Will it be a robust pull-down in the attached circuit when the switch is open?
That circuit will be very poor.

LEDs do not conduct electricity once the voltage gets below 1.2 volts or so. That is just about the threshold voltage for PIC inputs, so it will act as a floating input. Also, LEDS generate voltage when exposed to light, so the circuit's function will depend on ambient light conditions, which will make things even worse.

Put a real pull-down resistor in there.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
It's well known to not have floating pins as inputs to a microcontroller, and pull-up or pull-down resistors should be used.
Actually, no it's not 'well known' - there's really no problems associated with leaving PIC inputs floating.

If you're wanting to actually use it as an input, then it's obviously important to ensure that both high and low states are accurately defined, and pullup resistors are most commonly used for that purpose.
 

Mike - K8LH

Well-Known Member
Iomega,

An LED and its current limiting resistor should work fine as a pull-up or pull-down.

In the circuit below I sample the switches periodically by switching the pins to inputs then I switch them back to outputs to illuminate the LEDs. A switch press does illuminate the LED but in the circuit below each switch press toggles the LED so you can't really tell that the switch is lighting the LED.

Regards, Mike

 

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Russ Hensel

New Member
Actually, no it's not 'well known' - there's really no problems associated with leaving PIC inputs floating.

If you're wanting to actually use it as an input, then it's obviously important to ensure that both high and low states are accurately defined, and pullup resistors are most commonly used for that purpose.

I think it is fairly well know. At least that is what I have been taught. If you want a pin to float it should be an output. Or I may be wrong.
 

Noggin

Member
If it is an output, then its not floating. If it is a floating input, it can draw a few extra milliamps
 

Mr RB

Well-Known Member
Actually, no it's not 'well known' - there's really no problems associated with leaving PIC inputs floating.
...
Really? I thought with the earlier PICs that used TTL digital inputs there was a problem with latchup, and that was part of Microchips decision to start changing all the digital inputs to schmidt triggers in the mid 90's??

I know the PIC schmidt type digital inputs are pretty immune to latchup, you can leave them floating.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
If it is an output, then its not floating. If it is a floating input, it can draw a few extra milliamps
That happened on CMOS 4000 logic chips, it's not a problem with PIC's - there's been much discussion over the years about what option is best.

Personally, and if I think about it, I set unused pins as outputs, but I'm not obsessive about it :D
 

Noggin

Member
Well, I'm not going to argue with you about it, it won't do either of us any good. I will accept what you say as an unsubstantiated fact, but with a grain of salt, and continue setting my pins unused pins to output.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Well, I'm not going to argue with you about it, it won't do either of us any good. I will accept what you say as an unsubstantiated fact, but with a grain of salt, and continue setting my pins unused pins to output.
As do I - but if you ever feel the urge, monitor the current and try different methods.
 

mdorian

Member
Why you don' put a 50k or higher between GND and RD2 to be shore ? The LED will not act as a reliable pull down.
 
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