General Pic Question

[richb]

Hi

Before I start I must point out that I come from a CS background and electronics is very new to me. Hopefully this question isn't too stupid, but I have google’d, and the results haven’t satisfied me.

I'm looking at using a PIC16F84 chip to activate an opto-isolator to switch an external device, don't want to use a relay.

My question is, when I set for example port RA0 to Ouput and I send '1' the port is driven high, does this mean i'm outputting the 'VDD' voltage on the RA0 line. Same for when I send '0', am I just grounding the Pin.

If someone could explain the Input/Output concept or point me to some information it would be greatly appreciated.

Thanks

Rich

 

[eng1]

When the output of the port is high, the voltage is not exactly Vdd; it is
Vdd-0.7 V (min). When the output is low, the voltage is 0.6 (max). This parameters are named Voh and Vol in the data sheet.

 

[richb]

ok that makes sence.

What i am confused about is what is going on in this diagram then:-

https://hansmi.ch/download/hardware/wrap-powercontrol/schema.png

I assume then that P0 and P1 are both set to input and when set to '0' are grounding (pulling down) the LED's and optoisolator. Am i correct in saying this ?

 

[Nigel Goodwin]

That isn't a PIC, so it's not a very good example - but assuming it was, P0 and P1 would be set as outputs, set high to turn the LED's (and opto-couplers) OFF, and set low to turn then both ON.

 

[richb]

Thanks.

So I assume then that for this circuit attached, set HIGH would turn the LED's and opto-couplers on and set LOW would turn them off ?

I'm just checking just encase it doesn't work like this

Sorry for the visio diagram, really need some other software.

Thanks

 

[Mike - K8LH]

That should work with a couple qualifiers;

(1) You probably should have current limiting resistors in series with each LED and optocoupler.
(2) PIC I/O pins can sink a little more current than they can source and I suspect that's one of the reasons 'sink' switching to ground is used more often than 'source' switching.

Good luck. Mike

 

[rockin_rick]

Yes, but you need those current limiting resistors so that you do not exceed the PIC's max current out per pin/port spec and so you don't blow the LEDs.

FWIW - A lot of circuits work 'backwards' and use their I/O to ground the circuit (sink) rather than use their I/O to provide the +V (source) because a lot of devices can sink more current than they can source. That PCF8574, for example, can only source (at max) 300uA, while it can sink (typ) 25mA.

Rick

 

[rockin_rick]

Oops, Mike beat me to it while I was digging in the PCF datasheet!

Rick

 

[eng1]

If you use the PIC16F84, as you said in your first message, then you can't use a 3.3V power supply. You need 5 V.

 

[Nigel Goodwin]

If you use the PIC16F84, as you said in your first message, then you can't use a 3.3V power supply. You need 5 V.

Nobody should be looking at using a 16F84 in the 21st century!, it's long time replacement the 16F628 is available in an LF version!.