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In the above two circuits the only difference is post rectification zener and pre rectification zener arrangement is used. According to my understanding in both the cases i=240/Xc1+R1=15 ma

and voltage at C2=Vz=5.6 V in post rectification zener

and voltage at C2=Vz-Vd2=5.6-0.6=5 V in pre rectification zener. so my questiion is what is the advantage of pre rectification zener when it provides less voltage at output. does it provide higher output current, if yes then how?

also how to find the value of C2 in this case it is given as 470 uf . if I replace by 1000uf then will I get high current even though it is known that input current is limited by 0.2uF cap?

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  • \$\begingroup\$ The zener will prevent the voltage from rising to a significantly higher level in the event that your load is less than 15 mA \$\endgroup\$
    – MikeP
    Commented Mar 21, 2016 at 20:37
  • \$\begingroup\$ @MikeP yes as it requires minimum Iz knee but I'm not able to calculate what voltage will come at output if load draws higher current than 15 ma. \$\endgroup\$
    – BRT
    Commented Mar 22, 2016 at 18:02
  • \$\begingroup\$ @MikeP My load is ldo having output voltage 3.3 volts.so what is the input voltage of ldo in case load is drawing higher current.if I have a resistor as load the c1 and load resistor will form a voltage divider and we can calculate here it is input to ldo i so don't know.how can I model ldo as resistor for that load current. Or if you know any other method. \$\endgroup\$
    – BRT
    Commented Mar 22, 2016 at 18:20
  • \$\begingroup\$ If the load is higher than 15 mA, it won't make any difference. The only time it might turn on is if the load is less than 15 mA \$\endgroup\$
    – MikeP
    Commented Mar 22, 2016 at 18:55

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The only advantage of having zener before the rectification diode is less component count..!

Output voltage is the function of choice of the zener and any voltage dropping elements e.g. the diode as in the top circuit.

In this types of circuit the continuous current limitation is always dictated by the input Xc and nothing else, but peak momentary output current is determined by the final output capacitor.

For example if you use 1000uF low ESR capacitor for the output, you may able to draw even 100 Ampere at 5Volt for a brief 1 millisecond.

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  • \$\begingroup\$ My bad... I wanted to say if the capacitor is discharged very rapidly even hundreds of amperes can be drawn from it. A 1000uF capacitor charged to 5V holds energy of 12.5 mili Joules. Assuming the capacitor is very low ESR type, if this capacitor is forced to discharge into 5 mili ohm resistance the discharge time constant (T=RC) will be 50 micro second. Peak initial current of 100A will decay almost linearly within that time constant and the capacitor will be near empty in 1 mili second. \$\endgroup\$
    – soosai steven
    Commented Mar 22, 2016 at 18:55

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