[Help]How to calculate the DC current...

How to calculate the DC current a transformerless power supply can source?
Thanks

Nicholas said:

How to calculate the DC current a transformerless power supply can source?
Thanks

Click to expand...

It has more than one configuration,
which one do you use? Draw its schematic, and put it here

Regards,
ahmad

ahmad_abdulghany said:

It has more than one configuration,
which one do you use? Draw its schematic, and put it here

Regards,
ahmad

Click to expand...

Thans you ahmad, please take below dwg into your concern, thanks.

The main component in this circuit that will determine the output current is the C1=1µF capacitor ..
As you kno a capacitor represents certaun impedance for AC currents, and its value can calculated as follows:

Zc = 1/(ωC) = 1/ (2Πf*C), f=50Hz, C is the serial capacitor ..

For example, if you need 100mA of 220V the required resistance (impedance) is 2200Ω .. In this case, if you use 1.45µF capacitor, it will allow 100mA to flow through the circuit ..
Also, you need to take into account the output voltage, in this case regulated by 33V zener diode, and that the voltage is recified and smoothed by a smoothing capacitor ..
33/1.4 ≈ 24Vac is required to be recified and reulated at 33V level ..

So, 220V - 24Vac = 196Vac
To drop from 220Vac to 24Vac and allow 100mA current you will need Zc=1960Ω impedance and that can be done with 1.625µF capacitor (at 50Hz) ..
C = 1/ (2Π*50*Zc) ..

R1 of 100Ω has no special meaning and is used mainly as a "fuse" .. but you can add it to Zc if you wish ..

Regards,
IanP

Above calculation is wrong. Circuit shows that rectification is half wave. So for average current of 100mA you must multiply 100mA*pi to get peak value of current. Then multiply it with 0.707 to get rms value. Now you can calculate the capacitor:
C=1/(220/(0.100*pi*0.707)*314)=3.2uF
You may choose 3.3uF/275Vac capacitor.
Zener diode must be rated for 4W if you want to run rectifier with no load.

Added after 6 minutes:

Or for above circuit:
Icrms=220*2*pi*50*1e-6=69mA
Idc=Icrms*1.41/pi=31mA
In both cases increase capacitor to 1mF to get less ripple.

My calculations were only estimations .. if I was wrong by 10% or so .. then so be it ..

Regards,
IanP

So is the RMS or the average current exactly the DC current this power supply can output?

It is the average current equal to DC.

but the output filtering capacitor should increase the average value of the output voltage up to close to the peak value, doesn't it?