Could you please tell me the bootstrap principle of IR2104 in the following circuit diagram (please write in detail, thank you)

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Could you please tell me the bootstrap principle of IR2104 in the following circuit diagram (please write in detail, thank you) [Copy link]

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In another similar post, several netizens have suggested that the original poster should read the working principle to understand it, and then talk about this "bootstrap" circuit. The original poster's tone here is "I don't know how to do this question" please write a detailed introduction. Okay, let me show you how to understand it. First, let's get the manual. IR2104 is a high-voltage driver chip that drives a half-bridge MOSFET. Vb, Vs are the power supply for the high-voltage end; Ho is the high-voltage drive output; COM is the power supply for the low-voltage drive, Lo is the low-voltage drive output; Vss is the power supply for the digital circuit.

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The upper and lower bridges of the half-bridge drive circuit are turned on alternately. Note that when the lower bridge is turned on and the upper bridge is turned off, the potential of the Vs pin is the saturation voltage of the lower bridge switch tube. At this time, according to the original poster's circuit, you should know which is the upper bridge and the lower bridge. This saturation voltage is basically close to the low voltage. At this time, Vcc charges the bootstrap capacitor through the bootstrap diode to make it close to Vcc. When the lower bridge is turned off, the voltage at the Vs end increases, and the two ends of the capacitor cannot change suddenly, so the Vb end is the sum of the Vs and Vcc voltages. The Vb and Vs voltages are still close to Vcc. When the lower bridge is turned on, the bootstrap capacitor acts as a floating voltage to drive the lower bridge, and the losses of the bootstrap capacitor and the lower bridge switch tube during the on-time period are compensated. This is the basic principle of bootstrapping.