MrHeckles Posted October 26, 2005 Report Share Posted October 26, 2005 Hi guys. I have made a circuit that uses a UC3843 in order to run a 12v lamp and be able to dim the lamp.I can get an the voltage on PIN 1 of the chip (the ouput of its error amplifier) to vary between 0.8 and 7.26 volts (which I believe should be enough to vary the output duty cycle between 0 and about 96%). However the output on PIN 6 is only varying between 0 and 0.13 volts (not the so called 12v for a high output).My question is are there any common sort of reasons I can look for as to why this output is so low? Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted October 26, 2005 Report Share Posted October 26, 2005 Please post the schematic, and there are much easier ways of doing this. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 26, 2005 Author Report Share Posted October 26, 2005 Here we go...I know there are easier ways, but isnt the idea to try and learn things you dont know?Anways - ideas about why the output is so low? Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 27, 2005 Report Share Posted October 27, 2005 Hi MrHeckles,An N-channel Mosfet needs its gate 10V more positive than its source to fully turn-on. Therefore since your light bulb the source's load then it gets only a couple of volts max.If you rewire the circuit so the light bulb is the drain's load, then the Mosfet will work correctly. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 27, 2005 Author Report Share Posted October 27, 2005 Right, I have moved the lamp so it is between the +ve rail and the drain of the mosfet.However, I am still only getting an output of about 0.09 volts from PIN 6 of the UC3843 IC.I dont understand why the output is so low. It should be around 12v. Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted October 27, 2005 Report Share Posted October 27, 2005 What you're doing is fine but there are many easier and cheaper ways of doing this.Using a dual op-amp IC, eg. TL082, CA3240, or reduce the frequency and use the LM358 or LM1458, you could connect the output to +V with a 1k pullup and try a LM393 dual comparator.A single 555, the CMOS 7555 is even better.Or even a 555 and an op-amp, the op-amp can be a uA741, CA3130 or connect the output to +V with a 1k pull up resistor and use the LM311.A CMOS IC.There are many threads that deal with this subject here, these (in my opinion) are probably the most helpful.http://www.electronics-lab.com/forum/index.php?topic=2988.0http://www.electronics-lab.com/forum/index.php?topic=2373.0 Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 27, 2005 Report Share Posted October 27, 2005 Hi MrHeckles,Your lamp will have a very low resistance when cold and prematurely limit the output due to the current sensing at pin 3. Therefore I don't think that this current-sensing circuit will work with an incandescent lamp as its load. :'( Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 28, 2005 Author Report Share Posted October 28, 2005 This circuit was a test to see if I could get the UC3843 to do a variable duty cycle. It was going to drive a mosfet halfbridge driver for a motor.I hooked up a 12v motor (with a diode) instead of a lamp, but the output from the chip was still low.I do not understand why the low resistance would prematurely limit the output due to the sensing at pin 3. Could you please explain why.How can I get around these issues and get the thing to output the correct voltage to turn on the mosfet? Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 28, 2005 Report Share Posted October 28, 2005 Your problem with the circuit is caused by sensing the load current by pin 3.An incandescent lamp draws at least 10 times its normal operating current when cold.A motor draws a huge current when started and when loaded.At first I thought of just connecting a fixed voltage of about 2.5V to pin 3 which might work. Then I saw this test circuit on the datasheet which might work. Both circuits won't need the 0.2 ohm resistor in series with the load. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 28, 2005 Author Report Share Posted October 28, 2005 I did see this when I was figureing out how to do the variable duty cycle. Hence you can see the input to the error amp is the same on my circuit. But doesnt this mean that there wont actually be any current sensing of the load?I thought that I could choose the series current sense resistor to provide 1v to pin 3 at the maximum amperage I wanted to allow and then adjust the output from the error amplifyer to determine the clamp level in the comparator.I tried bypassing the sense resistor all together on my circuit (i.e 0 volts at pin 3) and it did not work. What am I supposed to be after here in order to get the circuit to run, a high voltage on pin 3 or a low voltage? Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 28, 2005 Report Share Posted October 28, 2005 Your 0.2 ohm current sense resistor limits the load current to only 5A. Your Mosfet is rated for a lot more current so if you can keep it cool enough then reduce the current sensing resistor way down so you can drive high-startup-current light bulbs and motors.You can use a piece of wire as a low-value current sense resistor. ;D Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 29, 2005 Author Report Share Posted October 29, 2005 I tried using a piece of wire. It still wouldnt get the lamp (or the motor) to start up. I am also getting a reading of 0v at pin 3, and the mosfet starts heating up.If there was still too much resistance for the piece of wire (sense resistor) why is there no voltage on pin 3. If it was limiting, wouldnt there be 1v or more on pin 3?If I detach pin 3, then it sits there with 1.8 - 2.3 v depending on the error amp output and nothing goes.Is there any way to make this thing start up and limit properly (like a buffer start for high current things)? Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 29, 2005 Report Share Posted October 29, 2005 Hi MrFreckles, ;DI've never seen your IC but from the limited info on its datasheet, it should be giving 0% to about 50% (adjusted by your pot) duty-cycle pulses to the Mosfet at about 40kHz, with a max of 5A to the load with your 0.2 ohm current-sense resistor. The transistor and stuff allow it to reach 100% duty cycle.With pin 3 grounded, the current output should be unlimited so I think your IC is busted. :'( Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted October 30, 2005 Author Report Share Posted October 30, 2005 When I connected pin 3 via the red line it fired up no worries. Of course, no way to vary the output.When I connect pin 3 at the blue circle, it doesnt fire up at all (unless I put a resistor in series with the lamp) even with R7 being just a piece of wire.I figure the IC works fine, but I dont seem to be able to get a low enough resistance between the blue circle connection point and ground. Is there a better place for the pin 3 connection and sense resistor or some way of making the whole system a little less sensitive to such small changes in the sense resistor? Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 30, 2005 Report Share Posted October 30, 2005 I dont seem to be able to get a low enough resistance between the blue circle connection point and ground. Is there a better place for the pin 3 connection and sense resistor or some way of making the whole system a little less sensitive to such small changes in the sense resistor?The datasheet mentions having a "star" ground where all things are grounded at one point, as shown on the schematic I posted. Quote Link to comment Share on other sites More sharing options...
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