Hi, all this programmer is extremely cheap but it use LVP which means while the microcontroller is operation one I/O port is unavailable but who cares LOL!, I'm just experimenting :)



IT needs a 74LS04 (which I think is a hex inverter), a 7805 voltage regulator some resistors and a 7V-20V power supply and a 1N4001 diode, an LED etc. I have most of the parts just need the hex inverters and I should be OK :)

So I thought I'd post here so that everyone else here who is is crazy interested in digital logic and microprocessors/microcontrollers can experiment along with me :)

LEt me know what ya think

(hehe, I snitched it from MIT) funny my sister goes to Ryerson University in Toronto, Canada and the class sites are password protected, why wasn't this one?
Posted on 2003-04-14 21:04:35 by x86asm
Not positive, but i think you need a 13V programming voltage as well.. I made my own programmer for the 16F84A and it required such (it has the same DIP layout, which leaves me suspect this).

:NaN:
Posted on 2003-04-14 21:14:00 by NaN

By the way, any cheap options for programming the 18F458, 18F258 et simila?
Posted on 2003-04-15 03:34:29 by Maverick
There are may programmers archived on the PIClist.
Posted on 2003-04-15 06:33:52 by Ernie

Not positive, but i think you need a 13V programming voltage as well.. I made my own programmer for the 16F84A and it required such (it has the same DIP layout, which leaves me suspect this).

:NaN:


I just took a look at the 16F84A's datasheet it doesn't have low programming, I believe the 16F628 is a improvement over the 16F84A? They added a Low Voltage Programming mode to the PIC16F628.

Hey NaN, just before you leave this thread can you answer two simple q's please? Thanks :)

1. I noticed that the PIC16F628 has both a VDD and VSS pins, so does this mean that the PIC can supply power to outside sources? (Max Current it can output is about 300mA)?

2. I also noticed that people place resistors in near the ground, I used to think that once the electricity has gotten close to the ground the designer can forget about it and not do anything to the signal. But I need a little help and understand the danger of not resisting the power before it goes to ground?
Posted on 2003-04-15 06:46:13 by x86asm

Hi Ernie, you wrote:
There are may programmers archived on the PIClist.
Thanks, dude.



Hi x86asm, you wrote:
I believe the 16F628 is a improvement over the 16F84A
Exact.
Posted on 2003-04-15 08:17:13 by Maverick


1. I noticed that the PIC16F628 has both a VDD and VSS pins, so does this mean that the PIC can supply power to outside sources? (Max Current it can output is about 300mA)?

2. I also noticed that people place resistors in near the ground, I used to think that once the electricity has gotten close to the ground the designer can forget about it and not do anything to the signal. But I need a little help and understand the danger of not resisting the power before it goes to ground?


1. VSS is the "Source" and VDD is the "Drain". This is terminology of CMOS and FET technology (so its safe to say there is no TTL inside this chip ;) ). VSS is where you connect your +5V power source, and VDD will then be where you connect your 0V ground connection. The current ratings is to tell you how much current that single pin (and associated internals) will tolerate. If your using inputs on your chip, then your "SINKING" current through this input. The VDD rating comes in to play because you have to SINK these currents to ground. As a designer you should make sure you dont SINK too much current through the chip and prescribed by the VDD spec. The same thing applies to VSS and sourcing output voltages.

2. Im not sure exactly what your getting at, but it sounds to me like its a PULL DOWN resistor. The point here is to ensure the pin sees 0V if there ther is no voltage left. If the power is cut, then the pin will not be left floating, it will detect ground potential, via the resistor, and drain any charge it has through it (causeing internal switching to happen and probably place the CHIP in a safe power down).

Pull up resistors do the excact opposite. If there is no 0V signal on the line, then the pin is pulled up to 5V through the resistor.

Both scenarios imply very little current is drawn through the resistor (as most CMOS applications dont).


Hope this helps..
:NaN:
Posted on 2003-04-15 15:39:50 by NaN

VSS is where you connect your +5V power source, and VDD will then be where you connect your 0V ground connection.
It's the other way around -- see schematic above. The main reason I can remember this is that I'm always looking in the specs for open drain outputs, which are the MOS(FET) equivalents of open collector outputs.
Posted on 2003-04-15 16:28:42 by tenkey
Hey guys you know the 74LS04, I couldnt find it on a samples site so I got the next closest device a 74LS05, its exactly the same thing but with open collector outputs, the datasheet says you need a pull-up resistor for the thing to operate properly so I put inbetween the +5V source and the outputs a resistor?
Posted on 2003-04-15 17:54:53 by x86asm
dang it, these 74LS05's are so damn small, gonna be a pain to work with...
Posted on 2003-04-16 14:48:35 by x86asm
Yes, you need a resistor between power and output. You'll need a separate resistor for each output. So with five gates, you'll need five resistors.
Posted on 2003-04-16 16:06:01 by tenkey
tenkey,

Thanks for catching me here... I got ahead of myself with out thinking ~ Consulting is dulling my sensing is suspect.

x86asm,

I used the same chip on my programmer. They are excellent for scenarios where you want to convert +5 volt logic to any other volt level logic. I used it with 13V and a resistor on the collector on one invertor, and 5V and a resistor on all outputs on the chip. On the input size of the invertors, the CPU provided +5 Volt logic to control all of the invertors.

You can not do this with a 74LS04.

:NaN:
Posted on 2003-04-16 17:56:47 by NaN