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The 1k capacitor is going to ground because I tried to get the power to 1mA. Which obviously didn't make the chip work as all the power just went through the LED. It is more of a test, but without it 1.4 Amps flow through the chip.
The chip is still programmable, out of the four I've been testing with, all are fully functional. That is, if I write a program to them, I can read it back later.
I don't have a capacitor on, well I tried it with no success, because I don't have anything reasonable close to 100 nF. I thought that it was just to keep the chip from resetting.
Oh, and I didn't mention this, I don't think that the chip is resetting during my flashing program because it starts in the off position. But the LED comes on anyways if I have the 1k resistor, without it, nothing but heat.
This is one of those frustrating problems when I I'm just dying to get my hands on it. I think at least 3,000 miles separates us so that isn't an option.
Firstly, a warning, you should never allow the voltage on any pin get higher than VDD or the chip WILL be permanently damaged. The only exception is during programming but special circuitry protects it in that circumstance. By adding the 1K resistor in the VDD line, you probably dropped the voltage to less than that on the line driving the LED so there is a potential for damage occurring.
The capacitor is necessary I'm afraid, they only cost a few cents and are plentiful. It's needed because the chip needs to have a clean supply between VSS and VDD. As the chip runs, it draws rapidly varying currents, the capacitor acts like a reservoir to stop the current causing voltage changes on the supply.
I'm still baffled by the high current though. Common sense and experience tell me that if it consumed 1.4A, even for a few seconds, the silicon would be history. The photograph isn't clear enough to see but are you sure there are no other components on the board under the tape?
There is another minor problem with your circuit/program. You appear to connect the LED and 490Ω resistor (looks like 390Ω incidentally) from GPIO 0 to the VDD but your program makes GPIO 0 high. This would make the pin go to the same potential as VDD so the LED would not light. Either change the program so GPIO 0 goes low or connect the LED between GPIO 0 and VSS, cathode end of the LED toward VSS.
The photo shows the LED and it's resistor going to the positive side of the battery but you say it goes to VSS which is the negative side. Please don't take this as an insult but you do have the batteries the right way around don't you? The positive end is gold colored has the small 'dimple' and should go to VDD, the flat end is black, negative and goes to VSS.
here is no in-built regulation or current limiting, it is up to the designer to ensure that maximums are not exceeded.
um... so does that mean that the negative side of the battery is considered ground?
I thought that the power flows from the negative to positive thereby making the positive ground.
The positive negative thing always gets me. I will be away from home until tuesday, but I know that if I switch the polarity that it doesn't overheat, but the LED would have been backwards so it may have worked.
Yep, negative side is ground. Side of the LED with the flat on it is the cathode which is the end most negative.
Hopefully the problem is solved. If it is, I take my hat off to Microchip for making a 40c IC that can withstand so much cooking and still come out alive!
Let us know if the problem is solved. You have an audience out here who can sleep at night again.
Well the LED flashes. I tried switching the + and - but I forgot to turn the LED around also.
What I really need is for someone to clear up for me + - ground and current flow. When I try to combine all this things just don't work out right.
What I think I understand now is that - is ground. Electricity flows from ground to the other side meaning a positive current flow. You connect components with their leads opposite, so the + side of a LED to the - side of a battery.
Think of the LED and battery positive ends chasing each other in a loop rather than having to put positive on the + end to make it work. I treat diodes as having a cathode and anode rather than -/+ ends, I think assigning what looks like a voltage is misleading.
I'm glad it's working now. Boy, they built them PICs tough don't they!
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