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# calculating inrush current for reactance type power supply

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#### ltkenbo

##### Junior Member level 1
I have designed the following transformerless power supply (based on many sources around) but I'm not sure how to calculate the proper size of the resistor to limit inrush current. Will a 1W 47 ohm resistor be sufficient? This is the zener diode I want to use:
https://www.mouser.com/Search/ProductDetail.aspx?R=1N5245B-TRvirtualkey61370000virtualkey78-1N5245B

And I want my circuit to be able to supply around 100 mA. Couldn't I use an inductor to limit inrush current instead? Any tips would be appreciated.

A 1W 47 ohm resistor should be sufficient if C1 is 3.3uF. However if you want to draw 100mA DC at the output, I think you will have to at least double the value of C1, which will increase the power dissipation of the resistor a lot.

Don't forget to put a bleed resistor across C1 to discharge the high voltage across it when the circuit is disconnected from the mains.

I suspect if you use an inductor there will be a problem with ringing of the tuned circuit it makes with C1.

The problem of inrush current could be neatly avoided by using some kid of zero-crossing switch.

My simulation shows 3.43 A inrush current surge.

Just for curiosity, here is a screenshot of your schematic.

As you can see, your safety resistor is subjected to power peaks of 2.1 W.

The zener threshold needs to be 0.6 V higher than your desired output volt level.

I found it better to reduce the load to 50 mA. If you try to make it 100 mA, it will cause a decline in output voltage.

There will be 211 mA (peak) going through several components. You may get by with a 500 mW zener, but it will get hot.

There is a way to use a full-wave diode bridge, for better efficiency.

- - - Updated - - -

You can use a 15V zener here. Furthermore the zener is not subjected to as much current here.

The load can draw 100 mA and you get 15V.

Cool thanks guys I think I'm going to use the bridge, but how do I actually calculate the inrush current?

Cool thanks guys I think I'm going to use the bridge, but how do I actually calculate the inrush current?

If you switch on when the mains is at 0V, then there is no inrush spike.

If you switch on when the mains is at peak, the inrush is:

A = Vpeak / R

So for 120VAC, the peak is 170. Your resistor is 47 ohm. (The other components have little effect here.)

The calculated answer is 3.6 A. About the same as the simulated response.

Screenshot:

According to the simulation, the surge only lasts less than half a millisecond. The capacitor quickly charges to the incoming supply level.

The resistor momentarily carries 600 W.

I have designed the following transformerless power supply (based on many sources around) but I'm not sure how to calculate the proper size of the resistor to limit inrush current. Will a 1W 47 ohm resistor be sufficient? This is the zener diode I want to use:
https://www.mouser.com/Search/ProductDetail.aspx?R=1N5245B-TRvirtualkey61370000virtualkey78-1N5245B

And I want my circuit to be able to supply around 100 mA. Couldn't I use an inductor to limit inrush current instead? Any tips would be appreciated.

View attachment 84782

What inrush current are you worried about ? Your main workhorse in your circuit is the C1 cap, but as Godfreyl noted, this needs to be increased if you want to draw ~120mA at 26volts across 220ohm. I think 10uF would be sufficient.

Note that since C1 is in "series" with the 2mF smoothing cap, and its much smaller, then THIS cap controls the current, not the larger 2mF cap.

Imho you don't need any resistor.

Screenshot of the simulation with no surge-limiting resistor (or rather the resistor was reduced to 2 ohms to keep in touch with reality):

Now the surge is 84 A. It lasts a few uSec. The resistor carries 14 kW. (This assumes 84 A is even available through house wiring.)

Although this will not heat up components, it is likely to cause immediate catastrophic breakdown somewhere.

Ok yes, because the capacitor ideally allows infinite current at startup right, so I just calculate it as if it wasn't there. But like you said it will briefly carry around 600 watts (because of the surge), so should I size the resistor based on the constant current and then bump it up a tad (so .47 watts for constant current and bump it up to 1 watt?) since the surge is very brief?

Ok yes, because the capacitor ideally allows infinite current at startup right, so I just calculate it as if it wasn't there. But like you said it will briefly carry around 600 watts (because of the surge)

To figure out what is an optimum resistance, it all depends on what amount of surge current you think could destroy components. And then if any of them fail, could your load be subjected to dangerous voltage / current?

The capacitor drops 90 percent of the mains AC in normal operation. That seems to fit the intended purpose of reactive drop.

so should I size the resistor based on the constant current and then bump it up a tad (so .47 watts for constant current and bump it up to 1 watt?)

My second schematic in post #3 shows the resistor carrying 2W peaks in normal operation. About 1 or 1.2 W average. I would use at least a 2W rating. Or 4W if you follow the customary advice to rate for 2X the anticipated conditions.

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