# Limiting current with capacitors

Discussion in 'Electronic Design' started by DillonCo, Feb 4, 2004.

1. ### DillonCoGuest

I need to limit current on "wall power" (115V, 60Hz) to 15A. A resistor
won't work because of the current and resulting voltage drop. An inductor
won't either simply because I'd need .02H (hard to get at 15A) and I have no
way of measuring inductance. I therefore decieded to use a capacitor. The
problem is that I need about 350uF in a non-polar cap at a reasonibly high
voltage (something I don't have).
I tried (as a test) connecting a 4.7uF 250V polarized cap in series with a
diode (to prevent reverseing), but this resulted in the cap exploding after
a couple seconds. There was no load (a short). I also connected a diode in
parallel with a similar cap so reverse current would flow through the diode.
I
put two of these in series swith the diodes opposing (so current wouldn't
just drain through them), and connected it again with no load. This time
the diodes (both!) exploded after a short time. I even tried two .05uF caps
in parallel, but one of these exploded when the power was turned on
(probably just caught the wrong part of the sine wave).
What am I doing wrong, and how can I successfully limit current?

Thanks.

2. ### Guy MaconGuest

If your load is drawing too much current at a given voltage,
you cannot reduce the current without reducing the voltage
Unless you are willing to reduce the voltage or change the load,
you can't.

Tell us what you are plugging in to the wall power and we may
be able to suggest ways of making it draw less current.

3. ### DJGuest

Direct-off-the-line power supplies using series capacitors are possible. They
are *practical* for currents up to a few mA, not Amps. Maxim have some chips and
excellent data and application notes.

4. ### Robert BaerGuest

Capacitors are definitely *not* the "solution", most especially at the
large 15 amp displacement current.
There are damn few capacitors that are rated at that high of a
displacement current, and the prices go way up as the displacement
current requirements go up.

The use of an inductor is a good choice, and 20mH at 15A is not too
difficult to achive with a gapped E-I core and fair-size wire to keep
the I*R drop down.
Measuring can be done (with easily available parts) in various ways:
make a bridge with 1% resistors and a 2% capacitor (the inductor is the
4th item) and tweak the resistors for the null.
There are a number of bridges that one could use, including the Owens
bridge, which is frequency independent.
Just be advised the bridge exitation frequency must be below the
inductor resonant frequency for a meaningful result.
And chokes *are* available with ratings that (may be) inthe ballpark

5. ### IDGuest

Could you let me know the model of any such chips that Maxim makes? I
looked at their website and couldn't find anything on line AC->DC
conversion.

Thanks,
-ID

6. ### SteveGuest

You're messing with lethal voltages and currents when you apparently
have little knowledge on the subject.

Did I hear someone say "Darwin awards contender"?

Tell us WHY you wish to do this and someone might be able to give you

7. ### DillonCoGuest

What am I doing wrong?
Hardly. I may have little experience as far as induction is concerned, but
I'm hardly pluging my tongue into the socket. Whenever I test a circuit
that involves mains power, I stay on the other side of the room and wear
safety goggles when I turn it on. It wasn't quite necessary in this case,
but was probably the closest I've come to actually inflicting damage with a
circuit. I plan on getting my Darwin award far more spectacuarly, thank you
very much.
Why is to make a power supply that can drive loads with little resistance.
Limit the current so nothing explodes, melts or otherwise damaged. The
current expectation is to power a flyback transformer which has nowhere near
the inductance or resistance necessary to be powered directly. It's had to
say what else it might be useful for at this point, but they would be along
the same lines (low resistance heating element, for instance).

However, that is largely just to satisfy the curiousity of those wo may
wonder, because it seems a large inductor is simply the best choice.

8. ### John WoodgateGuest

I read in sci.electronics.design that DillonCo <>
You appear to say that you are putting all these circuits directly
across the mains supply. Of course stuff will explode.

9. ### Tim DicusGuest

Hi John,

I missed the original post, so I am going to respond here.

To DillonCo:

I think what John means is that a device that has any type of impedance with a voltage applied will cause heat.

In your case, the 4.7uf cap across the 115vac 60Hz supply has a capacitive impedance of about 500 ohms. With the voltage drop of
115v, that would put the current at about 1/4 amp. That would be about 26va (watts kinda). That is gonna torch just about any device
you put in there to take its place. It does not matter if the impedance is inductive, capacitive, or resistive.

The 350uf cap will go into orbit (parts of it anyway) if you put it across the mains! The correct use is to connect the diode and
cap in series across the power supply and take the power from across (parallel to) the cap, not in series with it.

The only exception that may work is an incandescent light bulb. Maybe a 25 watt bulb in series should limit the current to about 1/4
amp. Be really careful with this. If you become part of the circuit, the bulb will not allow enough current through to blow the
power panel breaker, but enough to blow yours. May I suggest using a GFI circuit (like in your bathroom) on the mains power supply
when playing with this. Please wear eye protection, like safety glasses, and be really careful...

Tim

10. ### N. ThorntonGuest

So you dont know what youre doing. Using caps on mains is a recipe for
electrocution, mainly after the power is switched off. Seriously,
someone gave you good advice, either learn about mains and safety or
dont play with it.

These are not great choices. I used a heater as a series load for
suspect compressors, and bulbs for lower power items. A large inductor
will step your mains up to voltages that can jump through insulation
and turn you into instant toast.

What youre trying to do wont work, not with any passable amount of
safety. Learn the dangers and how to work round them or quit while

Regards, NT

PS Darwin awardees never planned to get Darwinated.

11. ### Gilbert MougetGuest

dated Wed, 04 Feb 2004 00:42:48 GMT,
DillonCo, <> says...

YOU ARE A TERRORIST.
YOU ARE A TERRORIST.

YOU ARE A TERRORIST.

12. ### Gilbert MougetGuest

dated Thu, 5 Feb 2004 11:43:48 +0000,
HE IS A TERRORIST.

13. ### Jan PanteltjeGuest

Its OK on neutral most of the time.
LOL, now you design THAT bad;-)?

Better keep doing it that way!

14. ### SteveGuest

Fair enough, glad you take it semi-seriously. It's just that when I
confidence in your design you wouldn't feel the need to stand on the
other side of the room with goggles on!!

When you say "power supply" do you mean AC at mains voltage (but
current limited)? That is gonna be hard to achieve.

Why?! You don't back up any of these statements with fact! A large
inductor across a mains supply is even MORE dangerous than
non-mains-rated caps. And it's much worse than blowing aluminium cans
across the room!

Understand that an inductor resists a CHANGE in current flow. This
means that the inductor tries to drive the back EMF to infinity as the
magnetic field collapses. Across mains it does this every cycle (60
times a second!)

Translation; it will self destruct trying to do its job unless
properly "contained", and in the process it could create incredibly
high voltages that may arc.

look at some pictures of them working, note the humungous arcs leaping
huge distances from the top of the device... read about Teslas ideas
for creating weapons from his designs... read about the bird that flew
across the beam of one of Teslas inventions and was instantly
theoretical link between a Tesla experiment and the Tunguska
large inductors are BLOODY DANGEROUS!

At very least, you should consider using a mains isolation transformer
and an earth leakage detector. Potentially, neither of these will
save you from being a Darwin awardee... but it will be a more
entertaining read when you get extra credit for ignoring warnings and
killing yourself DESPITE the recommended safety measures.

niftydog

15. ### Jim ThompsonGuest

On 5 Feb 2004 15:37:47 -0800, (Steve) wrote:

[snip]
[snip]

What do you think folks... "Dummy of the Month" or "Dummy of the Year"
Award ?

...Jim Thompson

16. ### DillonCoGuest

[snip heat calculations]
I've always been curious, what's the difference between VA and W?
I'd say so! I'm supprised it lasted that long!
I was under the impression that non-resistive impediance created no heat...
Or is that just when a load is connected?
Actually, I tried this, but as I said, my diodes exploded. Perhaps I'm
misunderstanding.

[snip incandesant bulb recommendation]
GFI's a good idea, even though I ensure the power is off before I do

17. ### Tim DicusGuest

VA is the AC version of watts.
VA = watts * 1.414
watts = VA * .707
All impedance creates heat when voltage is applied. If it creates a voltage drop while carrying current, it creates heat. It is
usually so little voltage and current you don't see or feel the heat, but it is there none the less.
Maybe I wasn't clear. The cap will go to "transistor heaven" if you connect it across the mains with no diode.

The diodes...were they like 1N4001? 1N4002? 1N4003? If so, I will bet you exceeded the PIV rating. If the cap positive terminal
(diode cathode) charges to 150v on the positive half of the AC cycle, then the diode anode goes to -150v on the negative half of the
cycle, the voltage differential is 300 volts. Say "Goodbye, diode! Hello, fire and smoke!" Use 1N4004 instead (400PIV).
It is better to be safe than sorry. I have been shocked badly by circuits I "knew" were switched off (and some were!). Watch those
caps. They have teeth! Even when the power supply is removed, the larger values can store a lethal charge.

I have almost been blinded by exploding caps that I was "certain" were polarized correctly. Doesn't happen very often anymore, but
it only takes once. That stuff that sprays out is hot and sticky, but I guess you know that by now...

Tim

18. ### DillonCoGuest

Ah, Ok.
voltage drop while carrying current, it creates heat. It is
it is there none the less.

Two questions:
Will the inductor (wire) or the core get hot?
If I use an inductor to limit the current on a transformer with little
resistance or inductance, will the inductor still get hot?
connect it across the mains with no diode.
[snip diode statement]

Well, I used a MR814 which has a reverse breakdown of 400V. If this isn't
what you mean by PIV, I don't know (don't have the data sheet handy, and
it's not on the 'net). They have a max current rateing of 1A, BTW.
I can relate. Voltage multiplier: 10nF, 1kV. Not much, but ... Suprise!
They hold charge considerbly longer than expected.
Actually no. I was on the other side of the room (I heard electrolytics can
explode viciously) when it steamed. I just pulled the power and left the
room . Didn't know it was sticky though. Hot + Sticky = .

19. ### John PopelishGuest

(snip)

VA tells you nothing about the phase relationship between the volts
and the amps. It is just the product of RMS volts times RMS amps.

Watts are the average power flow. If the volts and amps are in phase,
then the instantaneous product of volts times amps (instantaneous
power flow) is always positive (negative volts times negative amps or
positive volts times positive amps, always power in the positive
direction from source to load). But if the voltage and current are 90
degrees phase shifted, there are times when the voltage and current
have opposite signs, so those times (two quarters of each cycle) have
power going the other way, back to the source and an average power
flow of zero. Inductive or capacitive loads do not use up energy, but
just borrow and return it twice a cycle. Energy is just sloshing back
and forth. But devices like power lines and transformers still have
to handle that current. So for those devices, VA is a better
representation of what they have to deal with than watts.

20. ### John WoodgateGuest

I read in sci.electronics.design that Jim Thompson
Century.