2 LEDs and one pin left

[cncmachineguy]

Referring to the attachment 2 posts above,

Am I just missing something? by my calcs, assuming 2V drop for the LED's the current would be 3/10000 = .3mA.

Will this be enough to light the LED? By my calcs, I would think 150-200 ohms would be more suitable. Any help to clear this up for me?

[Darrel Taylor]

I don't think the attachment 2 posts above will do any good with 10k resistors.

I had actually started making a flash demo for this when it first started, but with so many options it got too complicated.
I really think the correct theory is in the link from post #13 though. The link is also in the flash below ... under the title. (batsocks.co.uk/ ....)

I only made it through simulating +5V. Things are clickable in the image below ... try it.
At +5V the resistor is 100 ohms (with 2 diodes).
But if your situation is different, perhaps I can add to it. ...

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Oh, it says I attached the .dsn file ... I didn't, cause it wasn't done.
But if you have Proteus, I will make it work for all conditions ...

[cncmachineguy]

Thanks Darrel. That is an awesome gadget. Is there any way to get you to replace d2 and d3 with variable resistors (or fixed if need be) and bypass the 200 ohm on the pin? This way it will be closer to the other circuit. I would love to test it (and yes, I am trying to use your sim instead of my breadboard) I think a more approiate value would be 0-1000 ohms. If not, its all good.

[Darrel Taylor]

Well, the problem with that ... is that the variable resistors would only control current.
But the correct operation of the LED's depends on the voltage.

If the total forward voltage of the LED's is less than VDD, both LED's will be ON no matter what the resistor values are. The only difference will be their brightness.

Given a specific VDD voltage, LED color (V forward) and foward current, the correct configuration can be found.

And to add 2 variable resistors to the flash demo, would increase the frames from ~50 to more than 500,000. I'm affraid that will take a few years to create, and you probably won't wait the 2 weeks to download it.
I suppose it could be done more efficiently.

But again, if you have Proteus, it's a "Piece of Cake".

[cncmachineguy]

thanks darrel, once again, it seems a case of I know not what I ask. Don't worry about the demo then. As for the actual circuit, I will have to mull over why the 2 resistors in series with the led's doesn't create a voltage divider.

I would just LOVE to have proteus. I tried looking it up 1 time, seems it was in the 4 digits, just a little much for me

[Darrel Taylor]

Actually, I'd rather know your conditions and do the math.
What is it you need to do with your LED's?
VDD, LED Forward Voltage / Forward Current?
See if the theory works?

... As for the actual circuit, I will have to mull over why the 2 resistors in series with the led's doesn't create a voltage divider.

Because the voltages across the resistors are changed by the voltage drops that are controlled by the LED's.

I would just LOVE to haave proteus. I tried looking it up 1 time, seems it was in the triple digits, just a little much for me

That's what I thought too at first.
But they have a $250 "starter version" that gives you a 12F, 16F and an 18F.
Of course it comes with all the other logic chips, analog stuff and test instruments.

Even if it's not the exact chip you target, the "Family" is usually close enough to know your program has a chance.

Beware:
Once you get used to it ...
You probably will spend $1000 for the full PIC version.

[cncmachineguy]

The LED's will be board level indicators, showing drive enables, disabled, or in a fault
red is 2Vf @ 10-20ma green is 2.2Vf @ 10-20ma

Vcc 5V

Now I could use 3.3Vsupply for the LED's But was trying not to add that.

$250 I could deal with, its the beware I am listing too!