Need Help on reading circuit

The circuit shown is a half wave capacitive drop circuit that is essentially a voltage divider such that a series capacitor drops an inputs voltage to more usable level.

Some circuit description,
Positive Half Cycle
-Ac Current flows through Rf,C1,D and parallel of C2 and RL.
-C2 been charged to one diode below the zener voltage, the current will have another parallel path in which to flow, C2 remains charged.
-C1 being charged during this time to a high voltage state

Negative Half Cycle
C1 discharged from the forward bias of zener diode

As an engineering rule of thumb, this approach can provide a load current of 10mAdc for each 1micro farad of ac capacitance.

Above are the information provided on the web, here is my question,
For positive half cycle,
1. How a zener diode is being turned-on? it mentioned there that is due to the C2 but i cant really understand it.
2. What is the another parallel path it mentioned? the loop that passed through the D,C2,and zener diode?
3. C1 is being charged to a high voltage state, the high voltage here meant te voltage that is nearly equal to input voltage?

For the circuit, isnt it the charging of C2 is only purposely for turning the zener diode on? How it discharge?

may i know what it meant by "according to the enginnering rule of thumb", it is a formula for calculating the value? what is the formula?

OK that's the problem i facing now, Pls help me... i am a newbie in this electronic things. Thankz...

Bump (sry for that)......

Noone can help me? it is lack of information or what?

ahkyo said:

may i know what it meant by "according to the enginnering rule of thumb", it is a formula for calculating the value? what is the formula?

Click to expand...

A "rule of thumb" is an approximate guide. There is no mathematical formula - it is knowledge gained from experience.

And it doesn't always work.

To keep it simple we are going to assume the lower AC input is at ground potential. It doesn't have to be, but it'll make it easier to explain.

1. How a zener diode is being turned-on? it mentioned there that is due to the C2 but i cant really understand it.

Click to expand...

At the start of the positive half of the AC cycle, C1 charges through RF, D, C2 and RL. It will take a few cycles, but once the voltage on C2 = Vzener - 0.7V then Z will take some of the current and the voltage on C2 will stabilize at Vzener - 0.7V
On the negative half of the cycle, C1 will discharge through RF & Z. Z will only have a voltage drop of 0.7 across it on the negative cycle.

3. C1 is being charged to a high voltage state, the high voltage here meant te voltage that is nearly equal to input voltage?

Click to expand...

By nearly, on the positive cycle, we mean minus Vzener and the voltage drop across RF. Negative cycle, we mean minus 0.7V (forward bias on zener) and the voltage drop across RF.

For the circuit, isnt it the charging of C2 is only purposely for turning the zener diode on? How it discharge?

Click to expand...

C2 turns the pulsating rectified current from D into a semi-steady DC. It can only discharge through RL because D blocks any current in the reverse direction.

C1 is just a way to use capacitive reactance to limit the current into Z and the load RL. You could use a larger resistor for RF instead, but you would create a lot more heat for a given output on the supply. C1 essentially returns the unused energy to the AC mains instead of wasting it as heat like a resistor would do.

may i know what it meant by "according to the enginnering rule of thumb", it is a formula for calculating the value? what is the formula?

Click to expand...

It is probably this: C = 1 / ( 2 pi f XC ) Where XC is the reactance you need to limit the current though RF to the proper level.

Thankz a lot, kchriste.....