Hence, the gate terminal is in a non-conductive state. At the start, no voltage is applied to the gate terminal. The junction J 2 is reverse biased and no current will flow inside the IGBT from the collector to the emitter. Junction J 1 is forward biased due to the V CC. The collector is positive to the emitter.
The voltage source (V CC) is connected across the emitter and the collector. The voltage source (V G) is connected to the gate terminal in a positive direction to the emitter and collector. This is done to ensure that the parasitic thyristor doesn’t latch-up which would cause the latching of the IGBT. The base and emitter terminals are connected through the resistor R b of the NPN transistor. The collector of Q 1 shares the same base as Q 2 and the collector of Q 2 is the same as the base of Q 1. The drift region experiences a resistance which is known as R d and the resistance offered by the p body region is known as R b. Q 1 BJT is a p-n-p transistor and Q 2 is an n-p-n BJT. In this structure, it can be observed that it is an n-channel MOSFET and has two bipolar junction transistors (BJT) which are Q 1 and Q 2. P-channel IGBT shares a similar structure to the N-channel IGBT except the doping is reversed in each layer. The P+ layer at the bottom is the drain or the collector. The n+ layer at the top of the IGBT is the source or the emitter. The other two layers are called the body region (J 2) and the drift (J 1). The injection layer is very integral to the characteristics of the IGBT.
The only difference is that the injection layer is p + and not n + substrate as in Power MOSFET. If you are using a microcontroller output rather than a battery then it won't be able to deliver the current needed to switch the IGBT quickly.IGBT is made of silicon and is very similar in structure to Power MOSFET. According to the IRG4PC30U datasheet, \$V_\$, you might be able to remove M1 and M2 (and the associated resistors) and just use the 5V directly - or maybe not. The emitter will always be a few volts below the gate. With an NPN BJT emitter follower, the emitter will always be about 0.7 volts lower than the base voltage - regardless of the collector voltage. Your IGBT is being used in a way similar an emitter follower would be used with regular NPN bipolar junction transistor (BJT.) Since you seem to have fixed the "burning" problem and now only have a "low output voltage" problem, I think I can help.
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