The bootstrap circuit, also called the boost circuit, uses a bootstrap boost diode, a bootstrap boost capacitor, and other electronic components to superimpose the capacitor discharge voltage and the power supply voltage, thereby increasing the voltage. Some circuits have elevated voltages that can reach several times the supply voltage.
Boost circuit principleTo give a simple example: there is a 12V circuit, there is a FET in the circuit that needs 15V drive voltage. How does this voltage come out? Just use bootstrap. Usually a capacitor and a diode are used. The capacitor stores the charge. The diode prevents the current from being reversed. When the frequency is high, the voltage of the bootstrap circuit is the voltage input to the circuit plus the voltage on the capacitor, which acts as a booster.
The bootstrap circuit is just a name given in practice, and there is no such concept in theory. The bootstrap circuit is mainly used in the class A and B single-power complementary symmetrical circuits. The class A and B single-supply 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. So use the bootstrap circuit to boost.
Commonly used bootstrap circuit (taken from fairchild, instruction manual AN-6076 "Design and Use Guidelines for Bootstrap Circuits for High-Voltage Gate Driver ICs")
Switching DC boost circuit (so-called boost or step-up circuit) principle
The boost converter, or step-up converter, is a switching DC boost circuit that can have an output voltage that is higher than the input voltage. The basic circuit diagram is shown in Figure 1.
Assume that the switch (triode or mos tube) has been disconnected for a long time, all components are in an ideal state, and the capacitor voltage is equal to the input voltage. The following is divided into two parts of charging and discharging to illustrate this circuit.
Charging processDuring the charging process, the switch is closed (the transistor is turned on), the equivalent circuit is shown in Figure 2, and the switch (triode) is replaced by a wire. At this time, the input voltage flows through the inductor. The diode prevents the capacitor from discharging to ground. Since the input is direct current, the current on the inductor increases linearly at a rate that is related to the inductor size. As the inductor current increases, some energy is stored in the inductor.
As shown in the figure, this is the equivalent circuit when the switch is turned off (triode cut-off). When the switch is turned off (triode cut-off), the current flowing through the inductor does not immediately become zero due to the current holding characteristic of the inductor, but slowly becomes zero when the charging is completed. The original circuit has been disconnected, so the inductor can only be discharged through the new circuit, that is, the inductor starts to charge the capacitor, and the voltage across the capacitor rises. At this time, the voltage is already higher than the input voltage. The boost is complete.
The boosting process is an energy transfer process of an inductor. When charging, the inductor absorbs energy, and the inductor emits energy when discharged. If the capacitance is large enough, a continuous current can be maintained at the output during the discharge. If this on-off process is repeated, a voltage higher than the input voltage can be obtained across the capacitor.
Simple 1.5V liter 9V power supply circuit diagramThe 1.5V liter 9V power supply circuit diagram is shown in the drawing. This circuit is a batch oscillation boost circuit. BG1 and L1, L2, C1, etc. constitute an oscillator. BG1 is an oscillating tube and operates in a switching state. L1 and C1 are oscillation feedback elements. L2 is an oscillating energy storage winding. For convenience, the circuit also designed an automatic electronic switch composed of BG3. When the base of BG3 has no load, there is no base current, BG3, BG2, BG1 are all cut off, the whole circuit stops working, and no power is consumed. Therefore, this circuit does not require a separate power switch.
When the two points of A and B are connected to the load, BG3 is turned on, and BG2 is also turned on. The load supplies the base current to BG1, and BG1 is turned on. The energy flows in from the power supply and is stored in L2. At this time, the collector voltage of BG1 is very low, D1 is cut off, and the load is supplied by the residual voltage of C2. When BG1 is turned off, the current in L2 cannot be abruptly changed, and it will produce a higher counter-electromotive force, which is rectified by D1 and output. When the output voltage is higher than the regulation value of D2, the b and e junctions of BG2 tend to be reversed, and the base current of BG1 will decrease, forcing the oscillation to weaken, and the output voltage will also drop to automatically output the voltage. Control is near the regulation value of D2.
Component selection and production debugging:
BG1 selects NPN type silicon tubes with reduced saturation voltage, such as 9013, 8050, etc., and requires ICM"300mA, β"200. BG2 can use PNP silicon tubes such as 9012 and 9015, and BG3 uses NPN type tubes such as 9014. The smaller the penetration current is, the better. L1 and L2 are wound on a 高频8MM high-frequency magnetic ring (detached from an old electronic ballast or energy-saving lamp) with a MM0.1MM enameled wire. L1 is 6åŒL2 is 36åŒ.
I use this circuit to supply power to the DT890A digital multimeter. The measured working current is: buzzer and capacitor 20uF, 2uF block is below 45mA, other gears are below 25mA. When the battery voltage drops to 0.9V, in addition to the buzzer block that consumes a large current, the capacitors 20uF and 2uF block have a power shortage display, and the remaining gears have no power shortage display. The circuit is simple to manufacture, stable in performance, economical and practical. No need to debug, as long as the wiring is correct, it can work normally.
Digital Multimeters If a 1.5V battery is used to replace a 9V stacked battery with a boost, it is usually necessary to install a separate power switch. Inconvenience to production and use. The circuit described in this article controls the start or stop by detecting the presence or absence of the digital multimeter's operating current. Therefore, as long as the power line is connected to the output of the booster circuit, the digital multimeter power switch can be used.
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