buck circuit
The buck circuit is a DC step-down circuit, and we'll show you how to step down a 12V DC voltage to a 5V DC voltage.
1, buck circuit topology: 12V ----->5V
2、Blood Pressure Reduction Principle
a. The switch is closed and the current goes
Positive terminal of the power supply ---->switch ---->inductor ----->(capacitor and load) ----->negative terminal of the power supply
Here there is no current through the diode at this time because the diode is in single phase conduction
One should note that the current at this point will also charge the capacitor
The most important thing here is the inductor, since the current on the inductor cannot be changed abruptly, all the current flowing through the inductor is slowly increasing, through Ohm's law: voltage = resistance * current, so the load voltage is slowly increasing.
When the voltage of the load exceeds 5V, at this time we will disconnect the switch
b. Switch disconnected, current analyzed
Disconnect the switch, the current flowing through the inductor will suddenly become smaller, due to the characteristics of the inductor, the current flowing through the inductor can not be changed suddenly, so at this time the inductor is left-negative and right-positive. At the moment when the voltage on the inductor changes from left-negative to left-negative to right-positive, there is no current in the circuit, so at this moment, the capacitor that supplies power to the load discharges to supply power to the load.
Positive of inductor ----> (load, capacitor) ----> diode ----> negative of inductor
Inductor discharge moment, the current is very large, although the magnetic energy of the inductor to convert electrical energy, magnetic energy is getting smaller and smaller, electrical energy is getting smaller and smaller, the current is getting smaller and smaller, through the Ohm's law: Voltage = Resistance * Current, so the load voltage is slowly decreasing, when the reduction to a certain extent, we close the switch again. The moment of closure, the polarity of the inductor will change, this change of a moment, and by the capacitor to the load power supply
So the voltage change of the load during the whole cycle is like the following, the RMS value is around 5V
How do we control the output voltage is 5V, here we need to introduce another concept, duty cycle
Duty cycle = output voltage/input voltage
Duty cycle = 5/12, that is 5/12 of the time the switch is closed and 7/12 of the time the switch is open. The time here is very short, measured in subtleties, and it won't work with a normal switch, we usually use MOS tubes, IGBTs, etc. instead
boost circuit
1、boost current topology: 5V---->12V
boost circuit is a DC boost voltage circuit, we give you the following, how to boost the 5V DC voltage into 12V DC voltage
2、Boosting principle
a. The switch is closed and the current goes
After the switch is closed, it is equivalent to a short circuit, so the right half is equivalent to a short-circuit state, at which time the current flows through the inductor, forming a left-positive and right-negative power supply, and electrical energy is converted to magnetic energy
b. Switch disconnected, current traveling
Switch off, because the impedance is greater than when the switch is closed, so the current flowing through the inductor will be smaller and smaller, the inductor in order to prevent the current from becoming smaller, it will form a left-negative and right-positive voltage, at this time and the power supply is connected in series, so that the final output voltage is greater than 5V, and here is the core of the boost current boost!
c. The switch is closed again to discharge the capacitor supplying the load, although the capacitor is discharged, the voltage will become smaller and smaller, when less than 12V, immediately disconnect the
d, the final realization of the boost, the final output voltage and what has to do with it? Duty cycle again :D
Vout = Vin/(1-D), it can be concluded that if the voltage is to be boosted to 12V, the duty cycle D = 7/12, so the switch closure time stands at 7/12 and the switch disconnection time at 5/12
The final realization of the output voltage fluctuates up and down 12V, the effective value is 12V
