simple PSU

[johnny78]

hi Guys

what is the name of this circuit & how to calculate the resistor & capacitor values for 24v 100mA output ?

 

[KlausST]

Hi,

The voltage drop is due to a capacitor (C1). Thus let´s call it "capacitive power supply"

as a very raw "over the thumb" estimation:
Calculate C1 to pass a current of 100mA at (220V - 24V).
Then add some headroom.

*****
C2 needs to be charged by C1. There will be only short pulses for charging... up to V_Zener.
The rest of the time it will be discharged by the load current ... and voltage will drop according standard capacitor formula.

***
I recommend to use a simulation tool to see more details about timing and ripple voltage.

Klaus

 

[FvM]

Series resistor required to limit inrush current and effect of harmonic grid voltage and transients.

 

[johnny78]

Hi,

The voltage drop is due to a capacitor (C1). Thus let´s call it "capacitive power supply"

as a very raw "over the thumb" estimation:
Calculate C1 to pass a current of 100mA at (220V - 24V).
Then add some headroom.

how to calculate C1 ?

*****
C2 needs to be charged by C1. There will be only short pulses for charging... up to V_Zener.
The rest of the time it will be discharged by the load current ... and voltage will drop according standard capacitor formula.

***
I recommend to use a simulation tool to see more details about timing and ripple voltage.

Klaus

--- Updated May 15, 2024 ---

Series resistor required to limit inrush current and effect of harmonic grid voltage and transients.

how to calculate this Resistor value?

 

[KlausST]

Hi,

how to calculate C1 ?

What's the exact problem?

As a long term member ....

Klaus

 

[johnny78]

Hi,

What's the exact problem?

As a long term member ....

Klaus

im not tht good with formulas or remembering how to calculate values
& as i said i dont know the name of the circuit which you called capacitive power supply
so i couldnt search for a way to calculate the values
this is the exact problem
i want some guide to calculate the values of this circuit

thanks

 

[KlausST]

Hi,

Honestly .... I will not do this simple math for you.
And if you are not good in remembering (as many other people, including me) .... you should find ways to overcome this. Other people write down, use books, use the internet (like Wikipedia). So not remembering is no excuse.

Ever heard about "Ohm's law"?
If not, you need to look it up.

The result of Ohm's law is "resistance" when dealing with resistors only, but it is called "impedance" when "non resistors" (capacitors, inductors, mixture) are involved.

So I expect you to calculate the (necessary for your application) capacitor impedandace according post#2.
The short for impedance is Z ... and especially capacitor impedance is X_c

****
Also all over the internet you will find the capacitor impedance formula.

Re-organize it and calculate C.

That's it.

*******

as i said i dont know the name of the circuit which you called capacitive power supply
so i couldnt search for a way to calculate the values

What?
You are not able to do an internet search for "capacitive power supply". Really?
Then it's high time to learn!

When I type in "capacitor power su" ... it even automatically recommends "capavitor power supply calculation". You even dont need to type every letter in ... just press the link.
--> numerous hits will show up, from simple hobbyists documents to detailed semiconductor manufacturer application notes.

Your "I could not search" ... means, you not even tried ... I don't support this l.......

i want some guide to calculate the values of this circuit

I've guided you, there is more than enough guidance in the internet, even videos.
But if you need a guide to hike up a mountain you can't expect him to carry you the whole way. You need to do the steps on your own.

Klaus

 

[BradtheRad]

As mentioned there is a formula to calculate capacitive reactance (capacitive impedance):

Xc= 1/ (2 π F C)

Thus the .68 uF value presents 4700 ohms impedance at the moment it conducts. It conducts at the 330V peaks. It lets through 70mA (330/4700). As the circuit runs this is averaged to a level about right to light an LED (20 mA).

It's a basic method which covers the main factors. You can do similar calculations to obtain an average of 100 mA which you desire in your project.