FETs 1. You know how to make a NOT gate and a NAND gate using CMOS technology. Now design an NOR gate. A NOR gate’s output is low if any of its two inputs is high. In any other case it is high. Tip: a two input NOR gate can be built using only four transistors. Can you also design a three input NOR gate? 2. Use Farnell, RS, Digikey or any other supplier database to find a good MOSFET that will turn on and off a load. When the load is connected it draws 1A of current. The MOSFET is connected to an Arduino development board. Arduino can provide a 0/5V control signal. Be sure that you choose the transistor that can handle the 42V when the transistor is off. We need a MOSFET that will be easily solderable so you should look for a through-‐hole part. We don’t want a heatsink so you need to be sure that you will not overheat the transistor junction. You need to find out how much power your chosen MOSFET can safely dissipate without the heatsink. 3. In some situations we need to limit the speed of MOSFET switching. How would you do it? Is it possible to e.g. slow down turn-‐on without affecting turn-‐on? 4. An IXTN36N50 MOSFET is used in a converter as an active rectifier switch. This means MOSFET’s body diode is used to do the rectification. Whenever the diode is conducting MOSFET can be turned on so that current can flow through the channel instead of body diode. Look at the electrical characteristics of MOSFET’s structure and its body diode and see if you can figure out whether current will flow through the MOSFET’s channel or though its body diode at 5A, 15A and 25A. 5. Driving a power MOSFET very often requires an additional gate driver circuit. This is because a microprocessor can only give 0/5V or 0/3.3V signals are not good enough to achieve low RDSon values in the MOSFET. This circuit is typically connected to an external 15 or 20V power supply. It can also supply very high gate currents (few amps) to speed up MOSFET switching. A very common problem in power electronics is supply of power to the high-‐side gate driver (driving QH). Imagine the DC bus voltage is 600V. When QL is off the emitter of the top transistor is emitter of the top transistor is at 600V potential. When QL is on the emitter of QH is at 0V. This means that the power supply of the Qh gate driver has to be isolated from the main circuit. Dbootstrap QH QL There is a workaround called a bootstrapping (red circuit). An additional diode is added. Explain how this works and what are the limitations of this method. 6. In the circuit below is it possible to replace BJT transistor with a MOSFET?