Post by thread:Re[4]: Clarifying PIC control of DC-DC converter

Scott Dattalo wrote: >Again, in my experience with switching power supplies of this type, >you want to have very hard MOSFET on/off transitions. If you don't, >you will have a less-than-efficient design. This extra power loss >winds up mostly in the MOSFET. I'm not trying to knock your apparent knowledge in this area, but I assumed you were implying that the PIC's port pin was inadequate to drive a MOSFET with a large capacitance. As for size, I picked an IRLL014 device out of the IRF Hexfet data book with the following characteristics: Vdss = 60V Rdson = 0.2ohm Id = 2.7A Ciss = 400pF Package = SOT223 (small indeed) Under ambient temperatures this device will dissipate 2W into a 1" sq. area of the PCB. IRF has a table of turn-on and turn-off times for standard CMOS and TTL gate drives for logic level MOSFETS in thier APP note AN971. I picked a worst case approximation for the above device compared to a device with similar gate and miller capacitance. I rounded up both times to the nearest 100nS. ton = 200nS toff = 100nS I assumed a boost converter topology for simplicity and convenience: Vin = 10V Vout = 16V Fsw = 20KHZ L = 100uH Energy of the inductor storage cycle: Ipk = 2A ton = 20uS Rdson = 0.2Ohm /20uS | 2 Els = | (Ipk/ton * t) Rdson dt = 5.33uJ | /0 Energy of the turn-on was approximated with a linear transition from 10V to 0V with a load current on the inductor as an initial condition which is approximated as a constant for the turn on switch time. This should give a VERY converative estimate. Iload = 0.8A Von = 10V tson = 200nS /200nS | Eson = | ( Von - (Von/tson)*t )(Iload) dt = 0.8uJ | /0 Energy of the turn-off was approximated with a linear transition of Voltage from 0 to 16V and current from Ipk (2A) to 0A during the turn-off. Von = 16V Ipk = 2A tsoff = 100nS /100nS | Esoff = | (Von/tsoff * t)(Ipk - (Ipk/tsoff)*t) dt = 0.53uJ | /0 Total Enery per cycle times the frequency gives the power: Pdiss = (5.33uJ + 0.8uJ + 0.53uJ) * 20KHZ = 133mW No heat sink required. No special gate drive. Mike

At 09:19 AM 5/17/96 -0400, you wrote: This is straying pretty far from a straight PIC discussion, but I felt a few more comments may help those trying to interface FETS to PICs sucessfully {Quote hidden}> >Scott Dattalo wrote: > > >>Again, in my experience with switching power supplies of this type, >>you want to have very hard MOSFET on/off transitions. If you don't, >>you will have a less-than-efficient design. This extra power loss >>winds up mostly in the MOSFET. > >I'm not trying to knock your apparent knowledge in this area, Neither am I,as by your email address, you have a better chance than many of having some, but The basic problem I saw with your approach was the assumption about drive capability. A PIC is NOT going to produce the rise and fall times you gave. The reason you need high drive capability to drive a FET is to quickly charge/discharge the miller capacitance when the drain is falling. Without the ability to quickly transition out of that region your power dissipation can get out of hand. That is one of the things that dedicated hardware will get you. Much can be done with the PIC output, but if you are looking for a high switching frequency to allow better response/smaller inductor, you will have real problems with a PIC In the example you gave, the power dissipation is 65% higher by my calculation but still doesn't need a heat sink. However someone trying to the mythical max of 2W dissipation in the transistor will find themselves in for an unplesant surprise based on your assumptions. > >IRF has a table of turn-on and turn-off times for standard CMOS >and TTL gate drives for logic level MOSFETS in thier APP note AN971. >I picked a worst case approximation for the above device >compared to a device with similar gate and miller >capacitance. I rounded up both times to the nearest 100nS. > > ton = 200nS > toff = 100nS Try using the gate charge graphs in the IR datasheet, adjusted for the voltage and current conditions you are using. Capacitance during switching varies widely as the miller effect multiplies the gate-drain capacitance, and that capacitance changes during the transition. It is more accurate to use gate charge and the equations I= C*dV/dt Q = C*V Using Voh/Ioh values from the 16C74 datasheet to get output resistances, I calculated ton = 1000ns toff = 350ns This changes the total power dissipation from 133mW to 220 mW. Trying to switch extra current will quickly cause problems with these switching times. I hope this discussion helps some folks, Jonathan ************************************************************************* Jonathan King * Unitrode Corp. * spam_OUTkingTakeThisOuTuicc.com Phone 603 424-2410 * 7 Continental Boulevard * http://www.unitrode.com FAX 603 424-3460 * Merrimack, NH 03054-0399 * * * *************************************************************************

Thanks Jonathan, you saved me a little time. At the risk of being flamed, I might add that the thermal resistance of the device Mike quoted is 15 degrees C/watt for junction to case and 42 degrees C/Watt from junction to ambient. I'm not exactly sure of the thermal resitance of the 1 inch square copper foil heat sink. However, I'll buy a six pack for the first person that can keep their pinky on that puppy and sing one verse of "Row-Row Your Boat" while the converter supplies 15 watts! Well, I'm not being exactly fair. The junction is rated for 150 degrees Centigrade, while your finger is rated for only about 70-80 degrees. But you get the point, it will be _very_ hot if you do not use an appropriate driver. Scott