2007-07-13 00:47:33 · 4 answers · asked by aakash s 1 in Science & Mathematics ➔ Engineering
First check to see if you have any shorts between any 2 pins with power off, if so, is problem. If not, then turn on power, black voltmeter probe to common which is the left pin of a TO-220-type package, right pin (facing flat face) of a TO-92-type package, and red to pin next to it, which is the input. The input must be at least about -14V, cannot be more than -35V. If good, move red probe to remaining pin, should measure -12V. If it is less, feel if it is hot (caution-can be hot enough to burn). If it is hot, you may have a shorted output caused by another component. LM79xx and LM78xx-series regulators have built-in thermal and overcurrent protection, and will limit output to a safe (but hot) level if the load is too heavy. See
http://www.national.com/mpf/LM/LM79M12.html
2007-07-13 02:44:38 · answer #1 · answered by Gary H 6 · 1⤊ 0⤋
No. You cannot check for proper operation of an IC regulator using the ohms function. You must power it up. You need to know: 1. The manufacturer's part number of the IC regulator, 2. What pins the input, output, and control are on, and, 3. The full-load regulated output voltage of the IC. You take it from here.
2016-03-19 06:25:09 · answer #2 · answered by Anonymous · 0⤊ 0⤋
7912 Voltage Regulator
2016-12-17 12:11:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Well, if you mean a cold check, the best thing is to check the input to ground with your multimeter in 'Diode ' mode, if there is no short cct or open cct (it should show as a diode) that is a good IC. Usually it won't get a problem at the output side.
2007-07-14 06:06:35 · answer #4 · answered by Holmes 3 · 0⤊ 0⤋