Capacitor step-down components should be correctly selected - Database & Sql Blog Articles

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

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

GW MB10F 0.8A 1000V rectifier bridge

The simplest capacitor step-down DC power supply circuit and its equivalent circuit are shown in Figure 1. C1 is a step-down capacitor, generally 0.33~3.3uF. Assuming C1 = 2uF, its capacitive reactance XCL = 1 / (2PI * fC1) = 1592. Since the on-resistance of the rectifier tube is only a few ohms, the dynamic resistance of the voltage regulator tube VS is about 10 ohms, the current limiting resistor R1 and the load resistor RL are generally 100 to 200, and the filter capacitor is generally 100 uF to 1000 uF, and the capacitive reactance is very Small, can be ignored. If R is used to represent the equivalent resistance of all components except C1, the AC equivalent circuit of Figure 2 can be drawn. At the same time, the condition of XC1>R is satisfied, so the voltage vector diagram can be drawn. Since R is much smaller than XC1, the voltage drop VR on R is much smaller than the voltage drop across C1, so VC1 is approximately equal to the supply voltage V, ie VC1 = V. According to the electrician principle, the relationship between the average DC current Id after rectification and the average value I of the alternating current is Id=V/XC1. If C1 is in uF, then Id is in milliamp units, and for 22V, 50 hertz alternating current, Id = 0.62C1.
From this, the following two conclusions can be drawn: (1) When the Power Transformer is used as the rectified power supply, when the parameters in the circuit are determined, the output voltage is constant, and the output current Id changes with the increase or decrease of the load; 2) When using capacitor buck as the rectifier circuit, since Id=0.62C1, it can be seen that Id is proportional to C1, that is, after C1 is determined, the output current Id is constant, and the output DC voltage is different from the load resistance RL. Change within a certain range. The smaller the RL, the lower the output voltage, and the larger the RL, the higher the output voltage.
The value of C1 should be selected according to the load current. For example, the load circuit needs 9V working voltage, and the average load current is 75 mA. Since Id=0.62C1, C1=1.2uF can be calculated. Considering the loss of the Zener diode VD5, C1 can take 1.5uF, and the current actually supplied by the power supply is Id=93 mA.
The regulation value of the Zener diode should be equal to the operating voltage of the load circuit, and the selection of the stable current is also very important. Since the capacitor buck power supply provides a constant current, which is approximately a constant current source, it is generally not afraid of load short circuit, but when the load is completely open, the R1 and VD5 circuits will pass all 93 mA current, so the maximum stability of VD5 The current should be 100 mA. Since RL is connected in parallel with VD5, while ensuring RL takes 75 mA of operating current, 18 mA of current is passed through VD5, so the minimum stable current should not exceed 18 mA, otherwise the voltage regulation will be lost.
The value of the current limiting resistor should not be too large, otherwise it will increase the power loss and increase the withstand voltage requirement of C2. If it is R1=100 ohms and the voltage drop across R1 is 9.3V, the loss is 0.86 watts, which can take 100 ohms and 1 watt of resistance.
The filter capacitor generally takes 100 microfarads to 1000 microfarads, but pay attention to its choice of resistance. As mentioned before, the load voltage is 9V, the voltage drop across R1 is 9.3V, and the total buck is 18.3V, considering the stay. There is a certain margin, so it is better to take C2 withstand voltage above 25V.