Make me some code..

[NF9Ω]

I can try. No timer use

DEFINE OSC 4
CMCON = 7
ADCON1 = 7
output PORTB.0
input PORTA.4
b0 var word
b0 = 0
HIGH PORTB.0
pause 200

main:   if PORTA.4 = 1 then enter
        goto main

enter:        if porta.4 = 1 then enter
            
tmecnt:    if PORTA.4 = 1 then extime
              b0 = b0+1         
               pause 100   ' Least count of time 100 ms
              goto tmecnt

extime:     if PORTA.4 = 1 then extime

ledon:  b0 = b0 * 100
        LOW PORTB.0
        pause b0
        HIGH PORTB.0
        b0 = 0
        goto main
        end

If code works, can i give my PayPal................ hehe

[Melanie]

My challenge was to write a piece of code to make the LED in that schematic light... I can't see your code turning that LED on... never mind any timers or buttons, just simply illuminate the LED as wired in that schematic...

As someone who likes to learn new things, it's worth it to me to know how to accomplish this in software (PICBasic, Proton, Assembler, C, Fortran, Cobol, Algol, whatever, I'm good for all flavours)...

For those that want to take up the challenge... I'll save you some time and tell you that the piece of code below WON'T work...

High PortB.0

and neither will...

Low PortB.0

So... a US$10 challenge to the first person who lights the LED (a bog-standard, common or garden nothing special cheapo average LED) in that original schematic (which must remain unchanged) as shown in Post #1 above... and posts his or her working code in this thread...

[Rob]

My challenge was to write a piece of code to make the LED in that schematic light... I can't see your code turning that LED on... never mind any timers or buttons, just simply illuminate the LED as wired in that schematic...

As someone who likes to learn new things, it's worth it to me to know how to accomplish this in software (PICBasic, Proton, Assembler, C, Fortran, Cobol, Algol, whatever, I'm good for all flavours)...

So... a US$10 challenge to the first person who lights the LED (a bog-standard, common or garden nothing special cheapo average LED) in that original schematic (which must remain unchanged) as shown in Post #1 above... and posts his or her working code in this thread...

It's a great challenge but I think it's worth more than $10! I'll add another $10 to the pot for the person who achieves it ;-). Now it's worth $20!

[NF9Ω]

just simply illuminate the LED as wired in that schematic...

Oh. BobK is right.
LED will never illuminate according to circuit.

[Acetronics2]

Hi, Everyone ...

I Win !!!

Just Reverse supply voltage and the Led will light ...

Yesssss ... the Port Inputs clamp diodes Help !!!

Thank you very much, Mel ... I'm waiting for my Bucks !!!

Nobody told the Pic might stay Alive !!!

Alain

[mackrackit]

xobx,

The demand "Make me some code" as you have seen will not get you anywhere.

If you really want to learn, study, write some code and then ASK for help.

Here is someplace to start http://www.picbasic.co.uk/forum/show...ghlight=timers

All I had to do was search the forum for "timers".

[Melanie]

Good try Alain, but the 1N4004 across the input shorts-out the supply...

Game is still open... (excludes supplying the +5v reverse-way round from a high-current source which smokes-out the 1N4004 first, and then the input capacitor, then destroys the PIC and finally lights the LED (which has a panel legend above it saying "now you're completely screwed!").

[mvs_sarma]

Melanie is CORRECT

the circuit is designed wrongly. Forget the micro. Even by connecting the ground at anode of LED it would not light. the LED woud have been wired otherway round. this configuration is like a protection diode across a relay coil driven by a transistor. Perhaps XOBX has to improve on his / her funda of harware design.

YES, it will light, if by using the port pin RB0, we switch a higher voltage than +6.4V to the anode of LED.

[Wilson]

Reminds me of an Industrial Electronics exam I sat for back in the eighties. The circuit was a pretty basic push-pull amp in which we had to write an essay what happens once power was applied.
The schematic had the diodes on the bridge rectifier for the power supply connected wrong.
I just wrote that 2 diodes across the 15VAC supply would blow up. I was the only one that got it right.

Wilson

[Acetronics2]

Hi, Mel

You've wrote it ...

It had to make some smoke, BUT IT WORKS !!!

Do you have some Scottish elders, to modify the rules AFTER the beginning of the Game ??? ...

Bad looser, I Wonder, hmmmm ???

LOL

Alain

PS: Note, I often saw ( some years ago ) LEDs drawn in a circle ... but drawn "reverse" connected.
This could simplify the game ...

[T.Jackson]

lol - Short of reinventing the diode, that's impossible.

[Kamikaze47]

Game is still open...

How about: 'accidentally' cut the wire directly above the "16F628" in the circuit diagram, then PORTA.4=0, then PORTB.0=1, and press the button. LED=lit!

[peterdeco1]

I think you just made $20.

[SteveB]

So... a US$10 challenge to the first person who lights the LED (a bog-standard, common or garden nothing special cheapo average LED) in that original schematic (which must remain unchanged) as shown in Post #1 above... and posts his or her working code in this thread...

How about: 'accidentally' cut the wire directly above the "16F628" in the circuit diagram, then PORTA.4=0, then PORTB.0=1, and press the button. LED=lit!

I don't think this qualifies.

[T.Jackson]

Melanie's challenge proposal had me scratching my head for a few moments, initially thinking that it was indeed possible, after all she is an engineer and doesn't appear to joke much. Because, when someone sets such a low "wager" (only $10 in this case) - it's normally well within reach. In other words it's something that's actually possible, the low wager is chosen based on high risk analysis, the person placing the bet is in fear that there's a good chance that someone might actually win it. So to minimize their loss, in the well foreseen event that someone wins, a low wager is set in place.

If a much higher wager was on offer, I wouldn't have spared a seconds thought. Instead using most of what I know about LEDs to very quickly dismiss it as impossible (or at least for me) I personally feel that the only way you'll get that reversed biased LED to switch on is with additional circuitry, i.e: charge pump (DC to DC converter) to raise the anode ~1.8V above cathode (assuming it's a red LED or around 2V if it's green)

Trent Jackson

[Meriachee]

So that's it? Besides the fact that significant modification, if not outright destruction would be required to get the money out of Melanie, it was an interesting discussion.

[Kamikaze47]

My solution wasn't "significant modification"... It was 1 cut of 1 wire

[keithdoxey]

My solution wasn't "significant modification"... It was 1 cut of 1 wire

Hmmmmm.... sounds like a quote from a bomb disposal technician

However, cutting one wire changes the circuit diagram so no money for you.!!!

PS. If you are a bomb disposal technician you might want to change your name to instill more confidence LOL