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Home » News » industry information » A comprehensive analysis of the principle of resistive and capacitive step-down circuit(II)

A comprehensive analysis of the principle of resistive and capacitive step-down circuit(II)

Views:0     Author:Site Editor     Publish Time: 2018-04-03      Origin:Site

3. Basic design elements of resistive and capacitive depressurization.

 

In the circuit design, the maximum working current of the load should be determined first, and the capacitance capacity should be calculated by the current value, and the appropriate capacitance should be selected.

 

Here is the difference from linear transformer power supply: resistive and capacitive power supply is selected by load current capacitance; linear transformer power supply is selected by load voltage and power transformer.

 

Resistive and capacitive voltage reduction current calculation.

 

 

Resistive and capacitive step-down circuit can be equivalent to the reduced capacitance C1 and the load resistance R1, the resistance and capacitance in series partial voltage.

 

The capacitive reactance of capacitance C1 is Zc=-j/wC=-j/2 π fC.

 

The impedance of resistance R1 is Zr=R.

 

The total equivalent impedance is Z=Zc+Zr=-j/2 π fC+R.

 

So I=U/Z=U/(Zc+Zr)=U/(-j/2 π f C + R).

 

 

Because the resistive-capacity step-down power supply is only suitable for small-current circuits, the capacitance range of the selected capacitor is generally from 0.33 UF to 2.5 UF, so Zc is -1592 j to -9651 j. .

 

As a result:

 

I=U/Z=U/Zc=U/(-j/2 π fc).

 

=220*2π*f*C*j.

 

=220*2π*50*C*j.

 

J69000C.

 

I = I ~ 90 °and I _ 1 = I ~ (69 000 C). When the rectification mode uses half-wave rectification, I1=0.5|I|=34500C.

 

Design example.

 

Known conditions: load working current 15mA, working voltage 5V. How to calculate the capacity of the reduced Voltage Capacitor?

 

Using half-wave rectification, according to the formula I1=0.5|I|=34500C, C=0.43uF.


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