SupeR-NovA Posted November 17, 2005 Report Share Posted November 17, 2005 Hi all!I'm designing a switching bench PSU around the UC3482 controller.this PSU will be adjustable between 1v-30v and currnet limiting will be from 30mA to 5A.my first question is, how can current limitting be better implemented? one way is using the 'ADJUSTABLE BUFFERED REDUCTION OF CLAMP LEVELWITH SOFTSTART' as described in figure 19 of the SG3842 datasheet.another way is using a 0.1ohm 5W resistor in series with the output and sense the voltage drop across it.the last one i can think of is a current sensing transformer, but i'm not to familiar with this solution so i would need help about it...so which way is better? Quote Link to comment Share on other sites More sharing options...
Shahriar Posted November 19, 2005 Report Share Posted November 19, 2005 Dear SupeR-NovA I was planning to build the same circuit.you asked: "how much power should i draw from the supply when there is no load connected? "You should draw current as much as the Controller doesn't turn off. because Controller Is supplying from an auxilary winding. the problem that I think you have missed is that it is impossible to build an adjustable 1-30V adjustable SMPS while the controller is supplying from an Auxilary winding. because imagine the output is 2V the the auxilary should be i.e. 10V. now, if you inrease output from 2V to 30V then the auxilary will increase to 150V. (BLOW UP). you should use a seperate PS for controller then put a Resistor at output just for discharging output cap when you decrease the output voltage.you asked:" can i connect the 340v rail ground to the output ground or will this ruin the isolation?"There is a standard for SMPS called "VDE". it says that when the output is higher than 42V the output must be Isolated. For your project it is upto you. If you think it is difficult for you to decrease the voltage down to 1V with using Opto coupler and TL431 (which is impossible if using TL431) then you can forget Isolation! but if you are going to sell your project, Isolate one is better! ;)about current limiting, I think both of them are good but I think transformer one would not be very accurate. I am not sure about this. it is just a guess! I have seen in some schematic that they use the Output inductor as a transformer to determine Over current. There is a document on the web which it has good information for current limiting. it is 6.3Mb :o . Search for it. its name is "Current Sensing Solutions for Power Supply Designers"By the way, The output ripple of SMPS is much higher than a linear one at least 30mV. this is why I Decided to use a LDO at the output of the SMPS then Keep the SMPS output 0.2V Higher than the output. In this case the Power dissipation would be 1W at 5A. Some more benefits of this was: 1- the output ripple was very Very Good.2- Accurate Current limiting was possible with resistor without increasing the output resistance of the PS.3- I could decrease the output down to 0V. Because with using a TL431 and Optocoupler, it is impossible to decrease the output voltage lower than 3.7V. althought at lower voltages than 3.7V Power dissipation would Increase. but it is not a matter.HTH - Shahriar Quote Link to comment Share on other sites More sharing options...
ante Posted November 20, 2005 Report Share Posted November 20, 2005 Hi Guys,First I like to say: don’t omit the isolation between the input and the output, it might be fatal. Which circuit is the starting point for this project, is there a schematic? It would be easier to follow your thoughts if we all could see what you are working on! ;) Quote Link to comment Share on other sites More sharing options...
SupeR-NovA Posted November 25, 2005 Author Report Share Posted November 25, 2005 Hi all, sorry for not replying... have been away for some time..I am using a dual 12v seperate switching supply. one of the outputs is for the UC3843 and fan. the other is for secondary side supply.So does the usage of the seperate supply mean that i can reduce the output down to 0v and not use a resistor in parallel to draw minimum power from the supply?this will sound a bit stupid.... if a core is saturated, will turning power of discharge it, or the core is lost for good? ???it's hard to comeby an LDO with 5A output capabillity... so i'll ommit it and use a secondary output inductor...a few things i'm not certain of...1)is the EER42 transformer a good choice for this design? and how much current can i draw from it safly at 100Khz?2)is the output inducter 300uH 6A value OK? 3)are the output filter capacitors 470uF 50V values OK?4)is R3 current sesnsing resistor (0.33ohm) at current sensing pin of UC3843 value correct?i'm almost finished just a few thouchups and that's it!!! Quote Link to comment Share on other sites More sharing options...
Shahriar Posted November 26, 2005 Report Share Posted November 26, 2005 Dear SupeR-NovA About 330ohm Res. For a Flyback PS, and Vi(min)=190V, Your calculation is right.About the output caps, although there are some calculation for them, but As I have Seen in many designs, if you put bigger Value, it would be better. But about the inductor, as you Yourself Know, You should use the proper Value. I attach a file whichbelongs to a complete 250W Forward design, but because the complete file was too big, I extract just the page you need. (the complete document name is "250 Watt off-line forward Converter". There are lots of Such documents on the web.I think your design is perfect. Build it and test it. It will work fine.by the way, at any circumistances, you need a Resistor at the output to discharg the output cap. because when you decreas the voltage, the cap remain charged and if you connect a low voltage device, then ...HTH - Shahriar250W-Forward.pdf Quote Link to comment Share on other sites More sharing options...
indulis Posted November 28, 2005 Report Share Posted November 28, 2005 Just a few general comments about the switching power supply….The 100KHz 3842 design looks to be current mode forward converter. As such, by definition it doesn’t have current limit. It has power limiting… the output current will continue to increase as the output voltage collapses, thus holding the power constant. Also, the LM339 shown in the secondary will not work!!! You can’t use an open collector comparator as an op-amp!! C7 should be in parallel with C8. You might need snubbers on the secondary rectifier and flywheel diodes. For the min output inductor value don’t forget about 1/2*L* I^2*F. It will be VERY hard to compensate this feedback loop!!!!!! The larger you make the output capacitor, the more it will pull back the crossover frequency (very bad transient response). I don’t see any slope compensation anywhere. You will find that it is VERY hard to make most switching power supplies run with zero load and not loose regulation!!!!What’s the core material for the EER42? What’s you max/min duty cycle? Quote Link to comment Share on other sites More sharing options...
audioguru Posted November 28, 2005 Report Share Posted November 28, 2005 the LM339 Quote Link to comment Share on other sites More sharing options...
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