(11-18-2023, 08:31 AM)Mister.E.M.F. Wrote: [ -> ]Hello Everyone ...
... took a look at Joel Lagaces One Wire circuit from Jan23 and he refers to I=V/A for the X factor. Anyhoo i have interpreted to mean Capacitive Reactance formula and calculate ... seem to work out to 41 milliamps. Its just a wild guess !
Picture attached below.
O.K. THANK YOU admin ... i never used these forums things before still gettin the gist of it but i do have Flikr acc. ... (i just seen video earlier today where tuber cuts to the chase a inserts a different voltage feed setup between two car batteries ... link below)
Carlos Benitez - Simplified system : https://www.youtube.com/watch?v=HappI1xXDII
Forced Oscillator running threw transformer and cap dump : https://www.youtube.com/watch?v=qnqYovhZVLY
Before selecting the dropping capacitor, it is necessary to understand the working principle and the operation of the dropping capacitor. The X rated capacitor is designed for 250, 400, 600 VAC. Higher voltage versions are also available. The Effective Impedance (Z), Rectance (X) and the mains frequency (50 – 60 Hz) are the important parameters to be considered while selecting the capacitor.
Calculating reactance and frequency
The reactance (X) of the capacitor © in the mains frequency (f) can be calculated using the formula:
X = 1 / (2 ¶ fC )
For example the reactance of a 0.22µF capacitor running in the mains frequency 50Hz will be:
X = 1 / {2 ¶ x 50 x 0.22 x( 1 / 1,000,000) } = 14475.976 Ohms 0r 14.4 Kilo ohms.
Reactance of the capacitor 0.22 uF is calculated as X = 1/2Pifem>C
Where f is the 50 Hz frequency of mains and C is the value of capacitor in Farads. That is 1 microfarad is 1/1,000,000 farads. Hence 0.22 microfarad is 0.22 x 1/1,000,000 farads. Therefore the reactance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.To get current I divide mains Volt by the reactance in kilo ohm.That is 230 / 14.4 = 15.9 mA.
Effective impedance (Z) of the capacitor is determined by taking the load resistance ® as an important parameter. Impedance can be calculated using the formula:
Z = √ R + X
Suppose the current in the circuit is I and Mains voltage is V then the equation appears like:
I = V / X
The final equation thus becomes:
I = 230 V / 14. 4 = 15.9 mA.
Therefore if a 0.22 uF capacitor rated for 230 V is used, it can deliver around 15 mA current to the circuit. But this is not sufficient for many circuits. Therefore it is recommended to use a 1 uF capacitor rated for 400 V for such circuits to give required current of 41ma in our case with my values.