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VRM how power supply ( VRM_PGD ) How to work for Let me try to explain that first the motto of this circuit is to generate VRM power good ( VRM_PGD ) Q92 is a double NPN simple transistor as the collector pin count. 6 Connect 3vsb and transmitter pin numbers. A ground pin and base number .2 enable signal is taken as VRM_PGD_R , and pin 6 of the voltage is 0 V, which means that shutdown circuit transistor and the second base pins. 5 is also connected to the pin number. 6 Therefore, pin 5 also becomes 0 volts and the emitter pin count is 0.4 ground, but the base has no enable voltage. Therefore, the transistor is turned on and the collector pins are numbered. 3 First resistor connected to the voltage divider R1101 ( 4.7k ), secondResistor R1115 ( 100k ) is connected to vcc3 ( 3.3v ). Therefore, as a voltage divider rule, 3.15V will have to go through VRM_PGD . Now let's look at Q90 security guard what under normal circumstances, N -channel MOSFET 's gate is held at a low level SLP_S3_CTRL # signal, VRM_PGD can easily pass through, but if there is a problem anywhere in the circuit , SLP_S3_CTRL # will change goes high, Q90 becomes active and the drain to the source and polar VRM_PGD not the current . file:///C:/Users/Lenovo/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg
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