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- Protege5 2003
Do not install that!!!. My description was correct, there is something wrong with your circuit. The way the IC works the output FET is either (effectively) a dead short or an open circuit. (There are actually two FETS, but they are wired in parallel in the circuit, so I will just use the singular.) When the FET is open (pin 2 floating or high, so no current through the LED) both sides of R2 should be at full battery voltage because there is no current through it, so pin 6 (and pin 4) should be at battery voltage. When the FET is shorted current flows through the resistor, and the voltage across the FET should be very close to zero (and across the resistor ~= battery voltage). That isn't what you are seeing. 0.78V is a very suspicious voltage, that's the voltage drop across a diode. Hmm, maybe you have the pins numbered down the wrong side (mirror image)? Let's see, that could happen if you put the circuit IC on the wrong side of the board, but that wouldn't match the picture and would be immediately obvious. Ah, or you are counting pins from the bottom instead of from the top. To be clear, the numbering is as viewed down on the IC, so that you can see the little tab on one end. Just like in the diagram in the data sheet here:
http://www.alliedelec.com/Images/Products/Datasheets/BM/CLARE/252-0001.PDF
A further test to verify. If the circuit is working then when the + of the battery is on the red wire, and the - of the battery is on the green wire (blower side of L/Y wire, attached to pin 2) then there should be about .75 volts between pins 1 and 2. But you saw 4.46V. Which might be possible if the FET is passing a little current, and yes, it would be if you have it backwards with (effectively R1 and R2 reversed). If the LED is in series with a 10K resistor it will pass 1mA which isn't enough to turn the LED on all the way, but is enough to turn it on a little, so the FET is somewhere between on and off. Then the FET and its series resistor will act like a voltage divider, which is what you are seeing.
In summary, it sounds like the IC is wired backwards somehow. Please examine my pictures carefully and compare it to your device. In this picture, the little white dot in the upper left corner of the IC is pin 1:
http://www.mazdas247.com/forum/attachment.php?attachmentid=184016&d=1305648971
Also double check the part number on your IC, just in case Clare happens to make an IC with a similar part number but reversed pins.
If something is wired in backwards it would be better to redo the soldering on the passive components than on the IC. Wires and resistors take heat a lot better than silicon. If you don't already have and know how to use a solder sucker or braid, then one of these:
http://www.radioshack.com/product/index.jsp?productId=2062731
will make removing resistors and wires a snap.
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