Capacitor buck full-wave rectification power supply circuit - Power Circuit - Circuit Diagram

Probe current voltage pin 420*4450 head diameter 5.0 over current current and voltage pin

MOS power IC full range

1.5PF SOT23-6 Low Capacitance 5V Breakdown Voltage 6V SRV05

Shunuo varistor / ESD varistor / product complete / Sunlord first-class agent

SMD aluminum electrolytic capacitor

1) Selection of step-down capacitor C1:

a. C1 capacity selection:

The capacitance value depends on the load current. After the load current I is determined, it can be obtained: C1≥1/2лfU

In the formula, the U value of the AC power source is calculated to be minus 10%, that is: I=300mA, U=220V*(-10%)=198V, f=50HZ,

C1≥0.3(2*3.14156*50*198)=4.82uF)

The capacitance value can only be large, not small. In this example, the capacitance C1 is 5uF.

b. Selection of withstand voltage values:

To consider the case where the power supply is 10% positive, as in this example, the mains supply, C1 should choose a 250V AC metal film capacitor.

c. Selection of resistance to transient inrush current:

The internal resistance of the metal film capacitor is very low, allowing the capacitor to have a large current flowing when the energy is being processed. The magnitude of this current depends on the capacitance value and its du/dt value. This value is determined by the structure of the capacitor, the metal film. The type of the lead is determined by the way the line is drawn.

The du/dt value is related to the withstand voltage value of the capacitor. The higher the withstand voltage, the larger the du/dt value, and the du/dt value of different manufacturers also has a big difference, such as the metal film capacitor with a withstand voltage of 250 VAC and a capacitance value of 5 uF. The du/dt value is generally chosen between 3-30.

In this example: C1=5uF, the value of du/dt is 3, then the instantaneous impulse current value of C1 is:
I=Cdu/dt=5*3=15(A)

2) Selection of current limiting resistor R1:

First find the capacitive reactance of C1: Xc=1/2лfC=1/(2*3.1416*50*0.000005)=636.36Ω

Then complex impedance: |Z|=638.3Ω (R1 takes 47Ω)

Find the effective value of current: I=U/|Z|=220/638.3344.7mA

The effective voltage value actually resisted by the resistor: UR=344.7mA*47Ω =16.2V

Find the actual power of the resistor: PR=16.2V*344.7mA=5.58W (R1 selects wirewound resistor, power is 7.5W)

3) Regulated shunt circuit:

Zener tube ZD1 and T1 tube EB junction, R3 constitutes a voltage regulator circuit, T1, R2 form a shunt circuit.

ZD1 selects 11.3V voltage regulator tube; R3 resistance value takes 180Ω1/6W; T1 tube responds to load current change, load current can change within 0-300mA, T1 selects 2W PNP tube, current amplification factor ≥200; R2 When the load current is small, the power of a part of the T1 tube is shared, and R2 is 30Ω/3W.