mesamune80

i had thried the above mothod which is proposed by ericgibbs .i found that the voltage go to my uC is about 2.71. is this voltage enough for my uC to detect as high?

ericgibbs

hi,
You have used the method I suggested, BUT you have not used the resistor values I calculated.
You still have a 1K0 to +5V from the OPA ouput and the 10K adjustable is not a good idea.

The 2.7V is only marginal, its not high enough for reliable operation, I would suggest at least 3V.

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Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's site: http://www.winpicprog.co.uk/
Gramo's site: http://www.digital-diy.net/

mesamune80

i see,so u mean i have to use 700 (can i really get this from the market?what i got is 680ohms) and 300 (this can be obtained)
And how could you manage to calculated those value?
can u show the method to as well.From what i know is by using voltage divider but why must 700 and 300? the ratio would be 3/10 X12V =3.6V ?
is this what you mean how about the LED voltage?

ericgibbs

hi,
A 680R and 330R would be fine, if the PIC pin voltage is still a little too low use a 330R in place of the 270R.
The original values were base on about a 10mA current from the +12V supply, this would be enough to light the LED [ I assume its a Red LED.?}

The 3.6V would be acceptable

Do you follow.?

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's site: http://www.winpicprog.co.uk/
Gramo's site: http://www.digital-diy.net/

mesamune80

ya i am using a red and a green LED,so you what you mean is 680 and 330 .And check the voltage output whether it is enough or not if not replace 330 with 270R right?hope there is no problem after this.
ya one question here,what PIC programmer do all of you using?
i need a USB programmer now,and dunno which one to choose >_<"
Any recommendation?

ericgibbs

hi,
No, you replace the 270R with a 330R, the higher value of the bottom resistor the greater the voltage will be at the junction of the two resistors.

Use Ohms law to calculate the voltages on the divider.

EDIT: I use PIC Start Plus.. with MPLAB

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's site: http://www.winpicprog.co.uk/
Gramo's site: http://www.digital-diy.net/

futz

PICkit 2 or Junebug! You won't be sorry. Both are excellent, and inexpensive.

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bananasiong

Or you want the PICkit2 clone which is available locally. The price is reasonable too.
http://www.cytron.com.my/listProduct...sp?cid=81#2965

I think it programs most of the 5 V PICs.

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mesamune80

Thanks for all the replies, i am currently using EPIC Programmer which is using parallel port. It seem to not program sometimes.This is the reason why i need to find a new programmer.i'll take a look at above link and see which one is suitable for me,thanks guy! =)

mesamune80

ericgibbs i had tried your method which is using 680 and 330 it gave me 4.x V for the output to uC but my LED not lit at all. why is the course for this i wonder? not enough current to drive the LED? or what?

mister_e

Did you read through this
http://melabs.com/support/epicfixs.htm

If using XP, do you have the printer polling patch installed?
http://melabs.com/support/patches.htm#xppolling

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ericgibbs

Hi mesamune,
The supply to the LM393 is +12V, I am assuming you are using a RED LED. which should have a forward voltage drop of about 2V.

So that leaves about 10V across the 680R and 330R divider, the current thru the divider will be 10V/1010R = 10mA,
so across the 330R there will 0.01*330R= 3.25V + the LED Vfwd = 5.25V.

The LM393 has an open collector output, so as the output transistor in the LM393 is OFF, it will not sink any current and load the divider.
Also the input impedance[resistance] of the PIC input is a fairly high impedance it will not load the divider.

So ALL the current flowing in the resistors 680R and 330R in series will flow thru the LED, as the current is about 10mA
the LED should glow.!

Could you post the diagram with voltages marked at the top of the 680R, the junction of the 680R and 330R ,
also the voltage across the LED.

You must have a fault elsewhere, are you sure that the PIC pin is set as an Input.?

Lets know.

EDIT: checking your figures that gave you +4V.... I suspect that the LED is blown short circuit.
Assume that the LED is blown s/c... so 12V/1010 = 11.8mA... 11.8mA * 680R = 3.9V. [You measure 4V.!]

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's site: http://www.winpicprog.co.uk/
Gramo's site: http://www.digital-diy.net/

mesamune80

I am using EWB 8.0 to do this circuit simulation,but afer i configure the circuit to what you mentioned earlier it gave me this result:
i am using LM339N for comparator in my circuit (quad type )
Anything wrong with my circuit?

Attached Images

windows comp,jpeg.JPG (125.0 KB, 5 views)

mesamune80

Hi Mister E ,my EPIC keep on promp me "code verify error at 0000" 3FFF should be 118A,continue verifying?
i duuno why suddently become like this i already used it long time ago,no problem. ;(

ericgibbs

hi,
I'll look thru the circuit.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's site: http://www.winpicprog.co.uk/
Gramo's site: http://www.digital-diy.net/