AbhimanyuSingh Posted August 23, 2010 Report Share Posted August 23, 2010 This is one of the test circuits given in datasheet of SG3524, with a little modification, mostly insignificant, like changing the value of inductor and using just one BJT instead of 2, as given in datasheet. I am not getting any output there. The expected % volts is not there. The signal out of pin 13 has a duty ratio of almost 0. What could the reason be. What possible mistake have I made. Even if I change the voltage at pin 2 (off course remove this voltage divider and put a rheostat), I dont find any change in duty ratio. Also once it happened, in another occasion that I was getting 2 volts at that point once I switched on the circuit and that voltage remained there even after i switched off the power supply. What could be a reason for this. The circuit is built on a bread-board. So its a little lousy, but there were no loose-wires. I am posting the circuit diagrams. The first file is the circuit i have built and the second file is the test circuit given in datasheet of Texas Instruments. Quote Link to comment Share on other sites More sharing options...
indulis Posted August 23, 2010 Report Share Posted August 23, 2010 These are both buck converters. First, what do you have for a load? You gotta have one, or else it will pulse skip or just shut down once the 500 Quote Link to comment Share on other sites More sharing options...
AbhimanyuSingh Posted August 23, 2010 Author Report Share Posted August 23, 2010 Thank you for your reply and for your concern.There is no load, which I thought would be inappropriate at first. The value of the inductor is 0.150 mH. Ramp voltage, as in normal situation is there on pin 7. Pin 12 and 13 are shorted and there is signal with almost 0 duty ratio on pin12 and 13 shorted. Quote Link to comment Share on other sites More sharing options...
indulis Posted August 24, 2010 Report Share Posted August 24, 2010 Put some minimal load on the output... say 20%. Quote Link to comment Share on other sites More sharing options...
lucy Posted November 18, 2010 Report Share Posted November 18, 2010 what is load? Quote Link to comment Share on other sites More sharing options...
indulis Posted November 18, 2010 Report Share Posted November 18, 2010 "Load" as in draw a load current out of the output. Quote Link to comment Share on other sites More sharing options...
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