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[Visitor (112.0.*.*)]answers [Chinese ]Time :2020-01-09
To give a simple example: there is a 12V circuit, and a field effect tube in the circuit needs a driving voltage of 15V. How to get this voltage? It is to use bootstrapping. Usually a capacitor and a diode are used to store the charge. The diode prevents current from being reversed. When the frequency is high, the voltage of the bootstrap circuit is the voltage of the circuit input plus the voltage on the capacitor, which acts as a boost.
The bootstrap circuit is just a name given in practice. There is no such concept in theory. The bootstrap circuit is mainly used in Class A and B single-supply complementary symmetrical circuits. Class A and B single-source complementary symmetrical circuits can theoretically make the output voltage Vo reach half of Vcc, but in actual tests, the output voltage is far less than half of Vcc. An important reason for this is a voltage higher than Vcc. Therefore, a bootstrap circuit is used to boost the voltage.

Common Bootstrap Circuits (from Fairchild, Instruction Manual AN-6076 "Design and Use Guidelines for Bootstrap Circuits for High-Voltage Gate Driver ICs")

(The so-called boost or step-up circuit)
The boost converter, or step-up converter, is a switching DC boost circuit that can have an output voltage higher than the input voltage. The basic circuit diagram is shown in Figure 1.

Assume that the switch (triode or mos tube) has been turned off for a long time, all the components are in an ideal state, and the capacitor voltage is equal to the input voltage. The following sections are divided into two parts, charging and discharging, to illustrate this circuit.

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