Hi Russell, The company I work for decided to change their name, they needed "rebranding" and with that my email address just changed. The List server is now accepting my mails again, as I have changed the email address on my account accordingly. Now back to the Boost Converter, Sorry if my question annoyed you, I should have looked into the inductor a bit more carefully, it serves me right. The fact that the current on the inductor can not change instantaneously was mentioned in several of the sites, but I kind of ignored it, but obviously it is important. Any info on this would be appreciated. All the sites I found mention this but do not give details of why this is so. I am looking mainly at continuous mode, which it seemed like a better way to do things as di on the inductor could be made smaller hence reducing the peak inductor current. You seem to suggest otherwise. I am not specifically designing a converter for an application, I am more looking into the design of the boost converter as research. I thought I knew how it all worked but obviously not. This is the converter I am looking into at the moment as an example: Vout = 48V Iout = 1.3A Vout_ripple = 0.5V Vin = 12V F = 10KHz Continuous mode di inductor = 0.4Iin At this point I have not selected inductor, switching device or diode. I am looking at it in a theoretical way, hence I welcome some practical advice on this. This is how I looked at it: - Calculated duty cycle based on the Vout Required V Vin available. - Calculated Power at the output using V*I, at this point I ignored switching or any other losses. - Based on P required I calculated Iin. Again ignoring losses. - calculated Iinductor peak = Iin + 1/2di - calculated L based on di, Vin and F Best Regards Luis Best Regards Luis -----Original Message----- From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf Of Russell Sent: 05 December 2009 06:00 To: PICList Subject: [EE]:: Boost Converter Luis didn't yet manage to say (but may when/if the list server accepts his mail) - > Hi Guys, > I am looking at boost converter design at the moment and I came across a > lot of useful and interesting information on the web, but with that I > found a few discrepancies that I would like some advice with. > On one of the sites I found it says that the forward diode should be > dimensioned to be able to handle at least current equal to the peak > current on the inductor, I can not see why, as the diode should never > see that current. Am I missing something here? What value of max current > would you use for this diode normally on your designs? > Best Regards > Luis Your question shows that you do not understand how inductors "work". When attempting to build boost or other converters it is useful (at least) to understand the characteristics of the main components, and the inductor is among the 'mainest' of them all. "The law of natural cussedness" / how it happens / conservation of various things tells us the the current in an inductor can never vary as a step function - it's rate of change can change but, at a point in time the current an infinitely small period before and after is the same. (This is not true for a capacitor or a resistor. With a capacitor you can step change the current but not the voltage. With a resistor you can (and must) change both.) SO if the inductor is carrying a current I just before the controlling "switch" (MOSFET or whatever) is turned off then it will be carrying the same current just after the switch is turned off. The current that was flowing through the switch now MUST find somewhere else to flow, and the diode is the path of choice. In a perfect design the peak inductor current WILL thus flow in the diode. Note that this is the PEAK diode current, not the mean current. Mean current will be smaller and MAY be much smaller depending on various aspects of the design. In a discontinuous converter Idiodemean is ~<= Idiode_peak/2 x Vin / Vo when the output inductor "stands on the Vin pedestal" which always happens with simple switching* and a single winding inductor, and ~<= Idiode_peak/2 x Idiode_peak/2 x Vin / (Vout + Vin) [[ OOMH, BOTE, YMMV, E&OE ... but you get the general idea.]] I won't start in on the continuous case, but its worse :-). * I say "simple switching", as use of eg synchronous rectification allows you to relocate the inductor topologically between the input and output cycles so eg it is ground referenced during output. This is not relevant when a simple diode rectifier is used and a simgle inductor "stands on the Vin pedestal" during output. If you tell us the specifics of the application it may help put things in perspective. Small converters are usually far more forgiving than large ones. As a rule of thumb, the diode will on average carry Iout average. In most cases a diode will happily handle peak currents several time it average. Note that for higher power systems the "average" diode current does not tell the whole story due to higher Vdiode at higher currents than the average current which occur over only part of the cycle. So mean diode dissipation will be higher than if it carried the mean current as DC. In higher frequency systems there will also be losses due to reverse conduction while charge carriers are 'swept out' and other reasonably second order effects. Russell McMahon Applied Technology ltd New Zealand. -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist