capacitors acting as short circuit

Joined Jan 24, 2017
153
if capacitors acts as short circuit when first connected, how do you avoid that to happen if it may possibly damage the cicuiutry ?

Joined Feb 20, 2016
3,564
You could add a low value resistor in series with the power supply to limit the initial current. But usually it is not needed. The power supply should be designed to handle the transient current. Most times the capacitors are not just put onto a full voltage but get charged as the supply voltage rises after power is applied.
Some mains switch mode supplies have negative temperature coefficient surge limiting resistors in line with the mains for slowing down the inrush current.
So it depends on the size of the capacitor and the application.

shortbus

Joined Sep 30, 2009
7,756
if capacitors acts as short circuit when first connected,
You may have the wrong idea on that statement. They aren't really a short circuit in the sense of a real short. They only look like one until they get charged up. Many "delays" in a circuit are done using the principle of the effect, of it looking like a short.

OBW0549

Joined Mar 2, 2015
3,441
if capacitors acts as short circuit when first connected...
They do not.

Of all the possible ways to describe or understand how a capacitor behaves, "acts as a short circuit when first connected" is absolutely the worst because it is the most misleading.

A far better way is to simply understand the basic relationship between voltage, current, capacitance and time in a capacitor: the current through a capacitor (in amperes) is equal to the capacitance (in farads) times the rate of change of voltage (in volts per second) across the capacitor.

In other words, I = C * dV/dt.

And that's it. Everything else follows logically, including what happens if you attempt to suddenly charge a discharged capacitor by connecting it to a powered circuit, what happens if you suddenly discharge a charged capacitor through a very low resistance, what happens if a small amount of current is made to flow through a capacitor, and what happens when the voltage across the capacitor is not changing at all.

I = C * dV/dt. Always.

Whether or not a circuit needs some sort of "protection" from any of these consequences, depends entirely on the details of the circuit and the capacitance and voltages involved.

Joined Jan 24, 2017
153
You may have the wrong idea on that statement. They aren't really a short circuit in the sense of a real short. They only look like one until they get charged up. Many "delays" in a circuit are done using the principle of the effect, of it looking like a short.
What kind of delays do u mean?.can u extrapolate on that?

Joined Jan 24, 2017
153
They do not.

Of all the possible ways to describe or understand how a capacitor behaves, "acts as a short circuit when first connected" is absolutely the worst because it is the most misleading.

A far better way is to simply understand the basic relationship between voltage, current, capacitance and time in a capacitor: the current through a capacitor (in amperes) is equal to the capacitance (in farads) times the rate of change of voltage (in volts per second) across the capacitor.

In other words, I = C * dV/dt.

And that's it. Everything else follows logically, including what happens if you attempt to suddenly charge a discharged capacitor by connecting it to a powered circuit, what happens if you suddenly discharge a charged capacitor through a very low resistance, what happens if a small amount of current is made to flow through a capacitor, and what happens when the voltage across the capacitor is not changing at all.

I = C * dV/dt. Always.

Whether or not a circuit needs some sort of "protection" from any of these consequences, depends entirely on the details of the circuit and the capacitance and voltages involved.
Will the balancing resistor(equaliser) do any good?

OBW0549

Joined Mar 2, 2015
3,441
Will the balancing resistor(equaliser) do any good?
What "balancing resistor"? "Do any good" against what???

I have absolutely NO idea what you are talking about.

Joined Jan 24, 2017
153
I
What "balancing resistor"? "Do any good" against what???

I have absolutely NO idea what you are talking about.
I meant equalizing resistors usually connected across capacitors in series

OBW0549

Joined Mar 2, 2015
3,441
I meant equalizing resistors usually connected across capacitors in series
Oh. If that's what you meant, you should have said so. Otherwise, I have no way of knowing what you're referring to.

Yes, resistors connected across capacitors in series will tend to equalize the DC voltages across the capacitors (provided the resistors are of equal resistance). Other than that, there's not much to say about them.

AnalogKid

Joined Aug 1, 2013
8,487
It is common to make a simple voltage divider out of two resistors in series. Something often forgotten is that you can make a simple voltage divider with two capacitors in series, even at DC. It rarely is done this way because it sucks, particularly with large value capacitors. Large electrolytics have very poor initial value tolerance, and the actual value changes significantly with temperature and with aging. One way to help stabilize the midpoint is to bypass the capacitors with resistors, and one effect of those resistors is that they will discharge the capacitors completely when power is removed.

ak

BobaMosfet

Joined Jul 1, 2009
1,117
if capacitors acts as short circuit when first connected, how do you avoid that to happen if it may possibly damage the cicuiutry ?
No. A capacitor does not EVER act as a short circuit when first connected. Anyone who tells you this is misinformed, or a poor teacher.

"ICE" = Current leads Voltage across a capacitor. What this means is that electrons on either side of the capacitor move. On the positive side, they move away from the plate on that side, towards the power supply. On the negative side, they move in, onto the plate. At some point, the capacitive reactance (negative electrons on the negative plate pushing back against incoming negative electrons), prevents more electrons from piling on the plate. Until that happens current seems to flow through the capacitor, fast at first, until it stops, while voltage (which is actually the drop in volts, or the difference between one plate and the other) seems to climb. But again, it isn't actually flowing through it. It's just looks that way because current flowing onto the negative plate equals current flowing off the positive plate.

Does not matter if AC or DC, the capacitor behaves this way, and this way only. It's just that in an AC environment, you can use the capacitive reactance field to impede the flow of current without much in the way of thermal loss, and in DC you use a resistor (but with thermal loss) to achieve the same thing.

This is why bypass caps appear to 'stop' DC but not AC, because DC fills the negative plate, and no current will flow.

DickCappels

Joined Aug 21, 2008
6,454
It is very common to limit the inrush current in power supplies by including a small temperature dependent resistor in series with the AC line. When cold the resistance is relatively high and a short time after power is applied the resistor is hot and the resistance is substantially lower.

This is a refinement of dendad's solution in post #2.

awright

Joined Jul 5, 2006
91
I question the authoritative statements disparaging use of the terminology, "short circuit" to describe the initial charging of a capacitor upon application of a voltage to a discharged capacitor.

While the term, "short circuit" is only applicable for the initial microseconds or milliseconds or seconds after application of the voltage source (depending upon the time constant of the circuit), the fact that the initial current is usually limited only by the stray inductance, source circuit resistance, and internal resistance of the capacitor qualifies the initial condition as a "short circuit" to me and conveys the importance of attention to the power input circuit.

The actual duration of what could be described as a "short circuit" depends, for all practical purposes, on the time constant of the circuit. If the device is line powered and the power input consists only of a rectifier bridge feeding a capacitor and it is switched on at the instant that the line is at peak voltage, the initial current surge can be significant and, in extreme cases, can cause damage. In most cases there is low enough capacitance and sufficient stray resistance, inductance, and safety factor in the circuit to avoid degradation or failure.

Note that, while OBW0549's equation, I = C * dV/dt is accurate, the factor dV/dt can be very large because the application of voltage AT THE SWITCH CONTACT is essentially instantaneous (that is, dt is extremely small) and the dV/dt at the capacitor is limited only by the stray circuit factors described above.

Whether it is advisable to add protective circuit elements is totally dependent upon your circuit. In almost all common device power supplies no supplementary protective circuit elements are necessary. However, if you have unusually high capacitance for some reason (e.g., trying to get very low ripple by brute force) and low safety factor in your power input circuit (e.g., selecting fuses or diodes without sufficient surge current capability), you can have degradation and failures.

There are various approaches to avoiding problems with the initial "short circuit" current at switch closure, including time delay fuses, low value series resistors between the power input and the capacitors (fixed or negative temperature coefficient), and circuit components with sufficient surge current rating. More complex solutions, (if any are required) can include time delay relays or circuits that allow relatively gradual charging up of the capacitors via a resistor then shorting out the charging resistor for continuing operation.

Joined Jan 24, 2017
153
It is common to make a simple voltage divider out of two resistors in series. Something often forgotten is that you can make a simple voltage divider with two capacitors in series, even at DC. It rarely is done this way because it sucks, particularly with large value capacitors. Large electrolytics have very poor initial value tolerance, and the actual value changes significantly with temperature and with aging. One way to help stabilize the midpoint is to bypass the capacitors with resistors, and one effect of those resistors is that they will discharge the capacitors completely when power is removed.

ak
Wow..haven't really tried that. I mean capacitors as voltage divider.not very common at all..could you please explain the drawbacks of using caps as VD further..thanks in advance..

ian field

Joined Oct 27, 2012
6,539
if capacitors acts as short circuit when first connected, how do you avoid that to happen if it may possibly damage the cicuiutry ?
There's some instructive insights on websites about restoring old radios - people are often tempted to upgrade the reservoir cap, and the increased charging current doesn't do the rectifier much good.

There's bound to be some useful discussion on the topic. On solid state amplifiers, there's reluctance to include surge limiting resistance in the rectifier/reservoir because it impairs transient response - I'm sure you'll find information on how they cope with charging current.

ErnieM

Joined Apr 24, 2011
8,045
falade47: Could you give us some insite as to what you are working on?

I see answers spinning wildly around lots of concepts giving answers that contradict each other with some disparaging the concept of a momentary short and others disparaging the disparagers.

The problem with universal answers is nothing is really universal. While it is true I = C dV/dT for a pure capacitance you simply cannot buy a pure capacitor. It takes quite a long time (decades? learning never stops!) to learn where concepts are and are not applicable.

So what are you doing buddy?

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WBahn

Joined Mar 31, 2012
25,904
Wow..haven't really tried that. I mean capacitors as voltage divider.not very common at all..could you please explain the drawbacks of using caps as VD further..thanks in advance..
They are used in many, many products, particularly those that have fairly small and constant power requirements. The big disadvantage is that, if they fail they often fail in such a way as to expose the low voltage circuitry directly to the mains voltage. This poses a significant safety hazard, particularly for inexperienced hobbyists who often get lulled into thinking that they can poke around in powered, low-voltage circuits with impunity.

BobaMosfet

Joined Jul 1, 2009
1,117
Wow..haven't really tried that. I mean capacitors as voltage divider.not very common at all..could you please explain the drawbacks of using caps as VD further..thanks in advance..
In most cases, it takes a while (if you're learning electronics), to begin to see the 'sub-assemblies' of individual components for what they are doing in a circuit. Your statement may change, when you think about filters- signal filters, for example. In almost any kind of band or notch or other kind of 'filter' that filters based on voltage, you're going to have a capacitor and a resistor, or an inductor and a resistor. Indeed, these are voltage dividers. Or a combination of all 3, even perhaps multiple times in combination with each other. All based on whatever you're trying to manipulate.

Adding to WBahn's statement about the dangers, I think clarification is in order- capacitors are very effective primarily in AC environments for controlling current, because they can do it with a field, instead of thermal dissipation. In fact, capacitors are the only way in many cases to do such things in a feasible way (cost, expense, component size, etc)-- but capacitors can fail. The take away is this- know your capacitor types and how they fail. For example, most electrolytics (ubiquitous as they are), dry out-- and when they do, they can alter the circuit as WBahn suggests in some circumstances and create a danger. For any project dealing with MAINS, there is no substitute for a fuse in the project.