On the filter capacitance of rectifier bridge

The rectifier bridge circuit is a rectifier bridge made up of diodes. When the input voltage is in the positive half cycle of AC voltage, diodes and load resistance form a circuit. The output of the rectifier bridge circuit is also a pulse voltage with constant direction, but the pulse frequency is twice that of the half wave rectifier.

The basic characteristics of the rectifier bridge circuit are as follows:

1. The output of rectifier bridge is a DC pulse voltage.

2. The AC utilization rate of rectifier bridge circuit is 100%.

3. Capacitor output rectifier bridge circuit, the diode bears a large reverse voltage of 2 times the AC peak voltage.

4. The load current of diode in rectifier bridge circuit is only half of that of half wave rectifier.

In the rectifier bridge, the appropriate filter capacitance should be selected, so how to select the filter capacitance? Usually, the filter capacitance is calculated according to the formula of power filter output ripple coefficient, which needs to be based on the two values of large output voltage and current of bridge rectifier. C》0.289/{f×（U/I）×ACV}。 C is the filter capacitance, 0.289 is the constant deduced from the ripple coefficient of the half wave resistive load rectifier circuit, f is the pulse frequency of the rectifier circuit, u is the large output voltage of the rectifier circuit, I is the large output current of the rectifier circuit, and ACV is the ripple coefficient.

Can we do without filter capacitor after rectifier bridge? What are the consequences? After rectifying, it is pulsating DC without filter capacitor, and its value is measured by multimeter, and its peak value is still about 22V. If your load is electric tools, electric heating equipment, etc., without filter capacitor is OK, but it can not be used in electronic equipment, audio equipment, etc. If your electrical equipment belongs to the latter, you can reduce the rectified voltage through 7-9 rectifier diodes in series and then filter. Reduce the voltage of each diode by about 0.7V. The rated current of the diode is more than twice the load current.

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