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| General Electronics Chat This forum is for general chat about electronics, eg: Dont know what a part does? Dont know how to read a circuit? Want to get an opinion? |
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| If you have a spare pin on your PIC that you can output a square wave from you can easy generate your own 9v auxiliary supply to drive the gates from, as in the attached picture. If you use shottkey diodes like a BAT43 for example you can come very close to almost doubling your voltage. From this 9v supply you can now power gate drivers (NPN-PNP pair) and still switch these drivers from the PIC. This will now allow you to get the gate on-voltage above 5v to reduce the Rds-on. You can also use a dedicated mosfet driver http://www.intersil.com/data/fn/fn7113.pdf The yellow lines represent the GND reference for each channel on the scope. As you can see the base of the wave on chan B is elevated by about 4.5v above GND All made possible by the principle of DC-restoration | |
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| Since i will be using this H-bridge my power supply would be 12V. This would also mean that my Vds is 12V. Looking at the connection, my Vgs is also 12V(correct me if im wrong, MY H-BRIDGE diagram is located below). The mosfet i bought is the BUZ71A. Plotting the load line, my MAXIMUM current my motor can draw would be around 7A. Therefore i set my Y-AXIS Id as 7A. Since my power supply is 12V, i set my X-AXIS,Vds at 8V(the X-AXIS ends at 8V Vds, there is no 12V. hope im right). Looking at the load line, my Vds is less than 1V and current around 6A. P=VI=6W. This seem abit weird. Although the calculation shows that the mosfet is efficient, the power dissipation seems awfully low . OK, here is the part that im confused with. Vds= Rds(on) x Id. With my Vgs at 12V, i should be able to get a good Rds reading. Looking at the Rds Vs. Id curve, i should be able to get 0.1ohm as my Rds. Therefore, with actual current i obtain from the load line which is 6A, Vds= Rds x Id = 0.1 x6 = 0.6V. So, WHY ARE THERE 2 METHODS TO FIND Vds??? 1 from actual load and 1 from Vds=Rdsx Id. Are they actually the same thing?! By the way, pls tell me whether my load line analysis and calculation is correct. I MUST GET THIS RIGHT!!! Thank you very much!!!!! | |
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| Hi, THEONE mind explaining how your 9V_supply.GIF works? i dont get it. Looks really interesting | |
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| Look at the 2 diagrams, the input Hi and Lo levels are generated by the switches in the diagram (I had to increase the cap values as I can't change over the switches that fast in the simulation to prevent some discharge of the caps) The PIC will of course do that at 10KHz OK, assume the input side of C1 is low (STEP1). C1 will now charge to (5v-1d) and C2 to (5v-2d) where d is the voltage drop over the diode. Now when the input of C1 is forced up to the +5v on the rising edge of the input (STEP 2), this now effectively places the -ve side of C1 on top of the 5v supply. Because the +ve side is now above the 5v line, D1 will now be reverse biased and a voltage of (C1v + 5v) will now be transferred to C2 through D2. The voltage will make the level on C2 to rise up to about 2 times the supply voltage. In this process C1 will discharge, ready for the cycle to repeat again. It is of course important that the input signal can supply enough current to charge-discharge the input cap quickly and that C1 and C2 are large enough to prevent too much drop in the output voltage as the charge on C2 leaks away into the load connected to it. | |
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| Quote:
Ptot(DC) = [T(junction-max) - T(mounting base)]/[Rthermal(junction -> mounting base)] Where Ptot is the power dissipation you may have at DC operation in the device, taking into account the other parameters in the formula. Rthermal(j-mb) you will get under the thermal data and is usually in Kelvin/Watt, which states how much the junction temp of the device will increase for every watt dissipated in the device. Typical T(junction-max) will be 175 deg C. So it is clear from the formula that the cooler you keep your devices on a heatsink, the more power you can handle without melting the junction material. The graph of normalized on-resistance vs temp is used to see the multiplication factor to use with Rds-on at higher temp's. Remember I said measurements are taken at 25 deg C. If you look at the graph, you will notice that the multiplication factor at 25 deg C is 1.0 At a heatsink temp of 60 deg C, your device will only be able to dissipate about 30W :!: Another thing, Rds-on quoted in the quick reference data often refer to pulsed operation with a junction temp at 25 deg C. Always make sure at what temp the data is quoted as these figures can be misleading and rarely obtainable in real applications. It is very dangerous to read only parts of datasheets (graphs), without taking all the factors into account! | ||
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| Do you mind explaining how the SOA curve works?? It does not even show the Vgs and saturation line. And it seems that all the other curves in this datasheet are drawn with the pulse duration of 80us and duty cycle 0.5 . There are no DC lines. What should i do? LAstly, when people plan to buy and use the MOSFETs, do they look at EACH graph one by one ?? sounds really tedious (not that im complaining). Forgot to add, does the MOSFET gate require current? It seems that no one ever mentions about the gate current. For BJTs, base current is really important. | |
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