# Inrush current calculation in capacitive circuite…?

I am using the following circuit to make a transformer-less AC to DC converter. It is working fine in my case. Now I need to protect my rectifying diodes from damage due to inrush current in this circuit.Here the resistor R4 is used to limit the inrush current.So to select a suitable value for R4 I need to calculate the maximum inrush current current in this circuit. So how can I calculate the inrush current in this circuit........? What about the surge power in R4? What will be the suitable value for R4....? • Inrush current will be roughly $2\pi\cdot f\cdot V_{AC p-p}\cdot C_2$. But normal peak currents will be roughly $1.5\cdot V_{ripple} / t_{charge}\cdot C_2$ and you want those to pass without significant drop. You have to make a judgment here if you will just use a resistor, $R_4$. But $R_1$ and $C_1$ are a different story and I'd like to hear your purpose there. $C_1$ certainly complicates the inrush current calcs, for example. – jonk Nov 28 '16 at 18:50
• What is your estimated load current? I can see the zener and the LED... but what about the rest? – jonk Nov 28 '16 at 18:56
• Worst-case inrush current for a transformerless supply will be when it is connected to the source at maximum peak voltage with all of its capacitors discharged. So in this case, treat the capacitors as short-circuits (so ignore R1 and everything to the right of C2) and calculate R4 with simple R=Vpk/Imax where Vpk is your peak AC line voltage and Imax is the maximum current you want your rectifiers to have to withstand. – brhans Nov 28 '16 at 19:07
• You do know that 1N4007s will be happy with 30A for a half-cycle? – Neil_UK Nov 28 '16 at 19:10
• @iqbalpalemad This offline cap charge supply must only be used for constant load apps. not relays to bulbs controlled by Uno. then use NTC with R4 – Tony Stewart Sunnyskyguy EE75 Nov 28 '16 at 19:41

If you assume that AC is 240V,RMS and can be turned ON instantaneously at any angle, that C2 is completely discharged before AC power-up, and that the the voltage drop across each of the diodes in the bridge is 1 volt, then if the mains are enerizgized at at either $90 ^{\circ}$ degrees or $270^{\circ}$ the instantaneous initial current out of the mains, on power-up will be:
$$I = \frac{340V - 2V}{100\Omega} = 3.38 amperes$$ 