Characteristics and component selection of capacitor step-down circuit

Since R is much smaller than XC1, the voltage drop VR on R is also much smaller than the voltage drop on C1, so VC1 is approximately equal to the power supply voltage V, that is, VC1=V. According to electrical principles, it can be known that the relationship between the average value of the rectified DC current Id and the average value of the AC current I is Id=V/XC1. If C1 is in uF, then Id is in milliamps. For 22V, 50 Hz AC, Id=0.62C1 can be obtained.

From this, the following two conclusions can be drawn: (1) When using a power transformer as a rectified power supply, after the parameters in the circuit are determined, the output voltage is constant, while the output current Id changes with the increase or decrease of the load; ( 2) When using a capacitor to step down the voltage as a 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, but the output DC voltage varies with the size of the load resistor RL. changes within a certain range. The smaller RL is, the lower the output voltage is, and the larger RL is, the higher the output voltage is.

The value of C1 should be selected based on the load current. For example, the load circuit requires a 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 voltage regulator tube VD5, C1 can be taken as 1.5uF. At this time, the actual current provided by the power supply is Id=93 mA.

The voltage stabilization value of the voltage regulator tube should be equal to the working voltage of the load circuit, and the selection of its stable current is also very important. Since the capacitor step-down power supply provides a constant current, which is approximately a constant current source, it is generally not afraid of the load being short-circuited. However, when the load is completely open, a full 93 mA current will pass through the R1 and VD5 loops, so the maximum stability of VD5 The current should be 100 mA. Since RL and VD5 are connected in parallel, while ensuring that RL draws 75 mA of operating current, there is still 18 mA of current passing through VD5, so its minimum stable current must not be greater than 18 mA, otherwise the voltage stabilizing effect will be lost.

The value of the current limiting resistor cannot be too large, otherwise it will increase the power loss and also increase the voltage resistance requirement of C2. If R1=100 ohms, the voltage drop on R1 is 9.3V, then the loss is 0.86 watts, and a resistor of 100 ohms and 1 watt can be used.

The filter capacitor is generally 100 microfarads to 1000 microfarads, but attention should be paid to the selection of its resistance. As mentioned before, the load voltage is 9V, the voltage drop on R1 is 9.3V, and the total voltage drop is 18.3V. Considering the remaining There is a certain margin, so the C2 withstand voltage should be above 25V.