2
\$\begingroup\$

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....?enter image description here

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – jonk Nov 28 '16 at 18:50
  • \$\begingroup\$ What is your estimated load current? I can see the zener and the LED... but what about the rest? \$\endgroup\$ – jonk Nov 28 '16 at 18:56
  • \$\begingroup\$ 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. \$\endgroup\$ – brhans Nov 28 '16 at 19:07
  • 1
    \$\begingroup\$ You do know that 1N4007s will be happy with 30A for a half-cycle? \$\endgroup\$ – Neil_UK Nov 28 '16 at 19:10
  • \$\begingroup\$ @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 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 28 '16 at 19:41
1
\$\begingroup\$

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:

$$ I = \frac{340V - 2V}{100\Omega} = 3.38 amperes $$

|improve this answer|||||
\$\endgroup\$
1
\$\begingroup\$

Either of these will work up to 47uF @240Vac while you are using 2.2uF with a temp rise of 90'C/W being regulated at the Curie Temp with free air.

enter image description here

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.