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Another transformerless power supply design question!

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poodlenuggets

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I'm working on a transformerless power supply - I've read all the old posts, I know the risks - disclaimer for everyone else, this circuit could electrocute you.

I was wondering, what is the smallest resistor (R1) in the circuit that will effectively limit inrush current? I'm trying to get ~35mA out of the circuit, so I used 68Ω, 1W, but I'm worried it might be on the small side.

https://www.electro-tech-online.com/custompdfs/2008/11/00954A.pdf
 

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Yes, R1 limits the inrush current.

The circuit you posted can supply a maximum of 15mA.

Use a bridge rectifier and this circuit will be able to deliver about 30mA.

Always put the resistor and capacitor in series with the live conductor for safety's sake.

If you want more current then add a resistor in parallel with the capacitor or use a larger capacitor.

The paper you posted privides all the formulae you should need to calculate the current a capacitor value.
 
Thanks - I actually just posted that circuit for reference. I'm changing the capacitor to 1.5µF, and I wanted to change the resistor to 68Ω - I'm just wondering if that's enough to limit inrush current. How do you know how much inrush current is too much?

But you raise an interesting point:

Always put the resistor and capacitor in series with the live conductor for safety's sake.

Would that make it safer? Wouldn't the resistor limit the current no matter where it is in the circuit?

I asked the Microchip tech support people if this would work if the polarity was switched, and they said yes, which I will be trying Monday with a fuse on the Line power.
 
The live is at 230V with respect to earth and the neutral is near 0V with respect to earth.

The inrush current can be calculated using Ohm's law.

Mains rated 1.5:muF capacitors are quite big bulky and expensive, you're better off using a full bridge rectifier and halve the size of the capacitor.
 
Hi there,

To calculate the inrush current you short out the capacitors and use Ohms Law
with the resistor(s). To be really safe, you should double the line voltage peak
too.
For example,
with 120vac line voltage and 170 ohm resistor, the peak is 170v so double
that and get 340v, divided by 170 ohms equals 2 amps surge. The more
typical surge is half that, or 1 amp.
If you have other significant voltage drops you have to subtract them
from the peak first.

That's not the end of it however, because now that we know the resistance
value is 'ok' we have to calculate the power. For this you have to
estimate the power in the resistor when the load is drawing its normal
current level. One way would be to calculate the power knowing the
average load current which flows through the resistor, then calculating
the power. The resistor power rating has to be sized accordingly.
Note that if the power rating isnt high enough it will burn up.
Also, if your load is half wave you have to double the load current
to figure out the power in the resistor.
 
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To be really safe, you should double the line voltage peak
too.

Why do you need to double the line voltage?

The capacitor starts off at 0V so the maximum possible inrush current is the peak voltage over the series resistor.

The inrush current isn't normally an issue anyway. With a 10Ω resistor and 230V the maximum inrush current would be 23A which wouldn't blow a 13A fuse (I doubt it'd even blow a 3A fuse) as with 470nF the RC time constant is only 4.7µs. You could probably even use a 1Ω resistor, 230A for an RC constant of 470ns won't do any damage.
 
Why do you need to double the line voltage?

The capacitor starts off at 0V so the maximum possible inrush current is the peak voltage over the series resistor.

The inrush current isn't normally an issue anyway. With a 10Ω resistor and 230V the maximum inrush current would be 23A which wouldn't blow a 13A fuse (I doubt it'd even blow a 3A fuse) as with 470nF the RC time constant is only 4.7µs. You could probably even use a 1Ω resistor, 230A for an RC constant of 470ns won't do any damage.

Hi Hero,

The voltage across the resistor for short time periods is:
vR=Vpk-vC
where
vC is the voltage across the capacitor.
Current through the resistor for short time periods is:
vR=(Vpk-vC)/R

Doubling the line voltage peak helps to figure out the worst
case surge current, which is not Vpk/R but could be 2*Vpk/R.

The capacitor starts off at 0v only when the circuit has been unplugged
and left alone for some time so that the cap discharges to 0v. If the
circuit is for example plugged in and then unplugged when the line
phase is at 270 degrees then the cap is charged to -Vpk, and if plugged
back into the wall a short time later while the line phase is at 90
degrees, the total current through the resistor is:
i=(Vpk-vC)/R=(Vpk-(-Vpk))/R
which is the same as:
i=2*Vpk/R

This means that there very well could be times when the surge
current is two times what the line voltage itself could produce
in the resistor. How often this occurs depends on how many times
the circuit is plugged in and unplugged and how unlucky the user is.

Interesting, when the unit is unplugged at say 90 degrees and plugged
back in at 90 degrees there is zero surge current because Vpk-vC=0.

The inrush current is an issue when you have components downstream that
have to take the whole surge current. Sometimes LEDs are connected
as a load and there is no smoothing capacitor, so the LED sees the full
surge current and often this is beyond its peak current rating.
Adding a smoothing cap obviously helps in that case, but how much
it helps also depends on the ratio of the value of the series cap and
the value of the smoothing cap, and so it may or may not be ok for
the LED. In order to be sure another calculation has to be done.
 
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hi Al,:)

The UL requirement for the capacitive driven mains psu is that a 470K or 1M0 must be connected across the capacitor.
Also a 47R or 100R must be connected in series with the capacitor.
 
hi Al,

The UL requirement for the capacitive driven mains psu is that a 470K or 1M0 must be connected across the capacitor.
Also a 47R or 100R must be connected in series with the capacitor.

Is there a UL standard/requirement or some other literature that covers transformerless PSUs? I know there's a section at the end of the app note, but I haven't found anything outside of that. Thanks
 
Hi there Eric,


hi Al,:)

The UL requirement for the capacitive driven mains psu is that a 470K or 1M0 must be connected across the capacitor.
Also a 47R or 100R must be connected in series with the capacitor.

So what?
 
Is there a UL standard/requirement or some other literature that covers transformerless PSUs? I know there's a section at the end of the app note, but I haven't found anything outside of that. Thanks

hi,

Look at this link and pdf.

Search
 

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Hi there Eric,




So what?

If you connect a 470k resistor in parallel with the 470nF capacitor the time constant will be 221ms and I doubt the user is going to be able to unplug and plug it back in fast enough to get double the peak mains voltage across the capacitor.

Also a 47R or 100R must be connected in series with the capacitor.
47R?

That can't be a UL requirement because it will limit the inrush current to only 7A, many appliances take larger surges than that.

My bipolar power supply uses a 80VA torroidal transformer with 4,700:mu:F capacitors on the rectifier. The primary has a DC resistance of just 27R giving a maximum surge current of just over 12A. It is protected by a 500mA slow blow fuse and a 3A plug fuse and niether of them blow when the power is connected.
 
hi Al,
So 'builders' are aware that these components are required, if their designs are to meet the UL safety codes.

Hi again,

Oh ok, good idea there.

Hero:
Well, a human may not be able to plug and unplug the cord
within 200ms or so on purpose, but they could easily do it
without even trying. The way this happens is the plug acts
as a poor switch and so creates contact bounce as it is being
plugged in. If it bounces between the two phase extremes, the
voltage could be 2x that of the single line voltage. I agree that
it wont happen every time, but it could eventually happen and
so it's a good idea to take that into consideration and design
for 2x the line voltage. It doesnt take much to do this anyway.

As far as the 47 ohm series resistor, i dont design things to make
sure they have a surge current of less than 15 amps, but instead
i design so that the device itself doesnt blow out the first
time the end user plugs it in :) This means sometimes less than
1 amp surge max, but it all depends on the exact design and
how much max current the various components can take.
For some LEDs this would be 100ma, for others 500ma, but yes
by design sometimes you can lower this other ways too, but
that's my main concern. This means i'll almost always have
a resistor of 100 ohms or more.
The LED night light circuit i had up on my website used a 100 ohm
for example.

Of course i'd like to also caution everyone about this surge limit
resistor, in that when the device it is used in is running normally
it dissipates power too. This has to be looked at and calculated
too in order to use the right size resistor.
I'll see if i can throw together a few equations to make it fast
to calculate the power dissipated. If the value is kept low
it works out pretty well anyway though, but some people get
carried away and try to stick a 1k or 2k resistor in series.
 
47R?

That can't be a UL requirement because it will limit the inrush current to only 7A, many appliances take larger surges than that.

hero,
Look at this extract from a microchip application note:
Posted earlier.
 

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The series resistor to the capacitor is there to limit surge current to the capacitor, which has a detrimental effect on its useful life.

There are many reported cases(mine experienced twice during repair, and no doubt others too) that the capacitance of the capacitor dropped to less than half it rated value after months or years in service. These designs had no or low value of series resistor.

I can only speculate that the surge current has some how vaporized the electrode of the internal foil and caused the diminishing of capacitance. Luckily, because all these X2 capacitors have good "self-heal" properties, none of them become short-circuited.

I would use a value of over 270 ohms or even 470 ohms.
 
There are many reported cases(mine experienced twice during repair, and no doubt others too) that the capacitance of the capacitor dropped to less than half it rated value after months or years in service. These designs had no or low value of series resistor.

I can only speculate that the surge current has some how vaporized the electrode of the internal foil and caused the diminishing of capacitance. Luckily, because all these X2 capacitors have good "self-heal" properties, none of them become short-circuited.

That was my main concern - how much do I need to limit inrush current to prevent damage? I'm planning to use 68Ω-100Ω series resistor. I was also intending to use an X2 capacitor, but they're huge, so I looked at another device which has a metallized 250VDC polyfilm capacitor, which has UL approval. Do I need an X2, or will any metallized polyfilm cap work?

This cap is cheaper and smaller, but it's only rated for 250VDC (U.S. device) - if it's used on a 120VAC line, 170Vpk, wouldn't they need at least a 340VDC rated cap? Or 250VAC rated?

The LED night light circuit i had up on my website used a 100 ohm
for example.
What's your website?
 
I was under the impression that the series resistor was in case the capacitor goes S/C, and act's as a kind of fuse.
I've heard it both ways - the resistor is both the poor man's fuse and limits the inrush current.
 
I've heard it both ways - the resistor is both the poor man's fuse and limits the inrush current.

In that case for the cost of less than 1p for the resistor, I would always fit it.:)

I would recommend a good quality capacitor rated at least for the local mains supply.
 
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