Hi Sean,

Thanks once again.

I went back to the dtasheets and saw the PWRT problem mentioned by you.
BTW, I also have external reset Circuitry connected to MCLR -- 10K to VDD, 10mfd
to gnd.
Re: Power-up timer.  Logically, shouldn't the timer start timing after proper
voltage is reached ? ie. 2 V or more ? Why do they have it this way ? Any idea ?
At what voltage does the timer start ?
Shall try a few changes with the power supply and line input, and get back to yo
u.

Regards  -- Biswanath Dutta


Sean Breheny wrote:

> Hi again Biswanath,
>
> At 08:22 AM 3/12/99 +0530, you wrote:
> >Thanks Sean --  Answer to your questions --
>
> You are very welcome. I just hope we can solve your problem. I know someone
> here can.
>
> >
> >1) I'm using a prototyping kit which has a rectifier, 1000mfd, 5V
> regulator --
> >and is supposed to take in AC. It has a jack which just fitted a small AC
> >adapter which I happened to have -- it produces 12V DC, no regulation -- so I
> >just plugged that in. Getting about 4.95V
> >
> >There may be a problem here -- with the power supply rise time as you have
> >mentioned.
> >How important is this & why ?
>
> No,it sounds as if your supply is pretty good. I'd just try adding a 10uF
> or so electrolytic cap and a 0.1uF ceramic cap directly between the PIC's
> Vdd and Vss pins. I think your supply should rise quickly and monotonically
> as it is. The problem with the supply not rising quickly and monotonically
> is this: when a PIC first receives power,it holds itself in reset for a
> short period of time (if the Power up start timer is enabled). If the power
> is not up to its full value by then,the PIC can begin running code in an
> unreliable manner or lock up. If the power doesn't rise steadily (i.e. it
> dips) during power up,then the PIC can enter a metastable state (it locks
> up and won't run code until the power is removed and re-applied). I must
> admit that I don't know the particular mechanisms by which this happens in
> a PIC,but I think the basic idea is that a PIC is a large sequential logic
> circuit. In order to work properly, certain registers (such as the PC) must
> get set to their startup values. If the power is not good enough,the
> mechanisms which are supposed to set these registers may not set them,or
> the registers may lose their contents after being set.
>
> >
> >2) I'm using a 2N2222A transistor -- emittor to gnd, 22K to base, 5.6K
> >collector to +5V, collector to Pic input. AC voltage is isolated from the
> mains
> >by a trafo which gives 22V, connected between 22K and gnd.
>
> This could possibly be the problem,although I can't offhand see why it
> shouldn't work O.K.. You should be able to do what you want by simply
> putting about a 220K resistor between the PIC pin and the output of the
> transformer. This will allow the PIC's protection diodes to clip the
> voltage with very minimal current flow.
>
> >
> >3)  470 ohms in series with each LED.
> >
>
> This sounds O.K,it would yield somewhere around 50mA max,less than the rating.
>
> >Hope this helps to understand the problem.
> >
> >Thanks to Mr. & Mrs Paul also for responding.
> >
> >Biswanath Dutta
> >
>
> Try what was suggested above and let us know what happens.
>
> Sean
>
> |
> | Sean Breheny
> | Amateur Radio Callsign: KA3YXM
> | Electrical Engineering Student
> \--------------=----------------
> Save lives, please look at http://www.all.org
> Personal page: http://www.people.cornell.edu/pages/shb7
> mailto:shb7@cornell.edu ICQ #: 3329174