L vs. C

Thread Starter

Wendy

Joined Mar 24, 2008
23,065
So I am starting a discussion as to the aspects of a coil (L) vs a capacitor (C).





A capacitor will hold a charge for as long as the capacitor quality will allow and no resistance draining it. It resists changes in voltage by either charging or discharging until an equilibrium is reached. A coil will resist current change by increasing the magnetic charge or decreasing similar to a capacitor and voltage. Both The capacitor and the coil will resist change change within their domain. Both of these effects are used to reduce ripple in a power supply.





This is where it gets tricky. When you disconnect the capacitor from its voltage source it will maintain the same voltage across the leads. When you disconnect a coil from its current source it tries to maintain the same current with back EMF. If there is no resistance across the leads being present it was pulled it will create as much EMF voltage as it can to maintain this current which creates the inductive spike we are familiar with and must protect against occasionally. In the following schematic when the switch is opened The coil will discharge the current for 5 LR time constants similar to the five time constants a capacitor takes to charge / discharge a capacitor.


#1.png

so when you push S1, the resistor will get a surge 100 ma, the coil L1 will be mostly will be completely discharged in five LR units. 1TC=0.0001 seconds (100µs).

So far so good?
 
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MaxHeadRoom

Joined Jul 18, 2013
26,264
A capacitor will hold a charge for as long as the capacitor quality will allow and no resistance draining it. It resists changes in voltage by either charging or discharging until an equilibrium is reached. A coil will resist current change by increasing the magnetic charge or decreasing similar to a capacitor and voltage. Both The capacitor and the coil will resist change change within their domain. Both of these effects are used to reduce ripple in a power supply.
Also remember that the current lags in an inductive circuit, and leads in a capacitive one.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,065
At this point I am not worried about ELI the ICEman.

I'm talking about how to predict the amount of voltage you get from the EMF surge from the coil when it is disconnected from its current source.
 

Papabravo

Joined Feb 24, 2006
19,570
At this point I am not worried about ELI the ICEman.

I'm talking about how to predict the amount of voltage you get from the EMF surge from the coil when it is disconnected from its current source.
it is given by:

\( L\cfrac{di}{dt} \)

This computation happens all the time in SMPS circuits, and it depends on the nature of the disconnection and possible alternative conduction paths. For example, a diode in parallel with the inductor or the diode in a buck converter.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,065
I am all too familiar with the curves involved, I'm interested in voltages at this point. Just as a capacitor dumping its charge through a short can create some various substantial currents a coil suddenly disconnect from its current source generates large voltages similar to the current surge from capacitors. It strikes me as having some interesting ramifications that can be used for other circuits.

This discussion has been kicking around in my head for serving the last several days, and now that I have nothing but time he gives me a chance to draw and talk (and I do love talking).
 

Papabravo

Joined Feb 24, 2006
19,570
I am all too familiar with the curves involved, I'm interested in voltages at this point. Just as a capacitor dumping its charge through a short can create some various substantial currents a coil suddenly disconnect from its current source generates large voltages similar to the current surge from capacitors. It strikes me as having some interesting ramifications that can be used for other circuits.
I gave you an EXPRESSION for voltage! I don't know what curves you are talking about, and I know of no other prediction mechanism. We can simulate it, build it, or see if predictions are reasonable.

In your example circuit nothing much happens because the 10Ω resistor is continuously shorting the inductor
 
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MisterBill2

Joined Jan 23, 2018
13,712
V= L dI/Dt, that is, voltage equals inductance multiplied by the rate of change of current. THAT part is simple, but since the inductance is affected by the level of magnetic saturation, which is controlled by the current, it gets a bit complicated.So really there is a a second derivative product aside from the rate of current change. The actual inductance is changing and at the same time the current is changing. So it gets rather tricky
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,065
Which brings us to our next circuit:

#2.png
If the inducter would you try to create one amp across this resistor that would be 10 mega volts, which is a ridiculous number and therefore suspect not sure where the breakdown is.
 
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Papabravo

Joined Feb 24, 2006
19,570
Here is the somewhat boring simulation of the original circuit
1657395699802.png
As you can see the current divides between the inductor and the resistor, and the switch has no noticeable effect.
 

crutschow

Joined Mar 14, 2008
31,112
If there is no significant (or a very large) shunt resistance, the voltage of the inductive spike is likely limited by the parasitic capacitance in the circuit (both internal to the inductor and from the external connections, assuming arcing or insulation breakdown doesn't occur first).
The peak voltage occurs when the all the inductive energy has been transferred to the capacitance, with the time determined by the LC resonant frequency.
This peak is then determined by 1/2 LI² = 1/2 CV² or V = √(LI² / C).
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,065
Here is the somewhat boring simulation of the original circuit
View attachment 271149
As you can see the current divides between the inductor and the resistor, and the switch has no noticeable effect.
thing of it is, there is no AC feeding this circuit current that is divided between the the inductor which is near 0 ohms, and the 10 ohm resistor so most of the current will go through the inductor regardless how you simulate it, think DC.
 

Papabravo

Joined Feb 24, 2006
19,570
I have no idea what you are talking about. I'm just opening and closing the switch. There is no "AC here. You might want to rethink the use of a current source for your example. Maybe come up with a circuit that actually shows the effect you are looking for.
EDIT: This might be what you are looking for. The slope of the inductor current accurately predicts the voltage spike on turnoff.
1657402443536.png
 
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xox

Joined Sep 8, 2017
794
it is given by:

\( L\cfrac{di}{dt} \)

This computation happens all the time in SMPS circuits, and it depends on the nature of the disconnection and possible alternative conduction paths. For example, a diode in parallel with the inductor or the diode in a buck converter.
Doesn't the actual peak voltage also depend to one degree or another on the overall construction of the inductor? I mean a carefully wound coil with a specialized geometry would likely produce a higher voltage spike than a cheaper component, wouldn't it?
 

Papabravo

Joined Feb 24, 2006
19,570
Doesn't the actual peak voltage also depend to one degree or another on the overall construction of the inductor? I mean a carefully wound coil with a specialized geometry would likely produce a higher voltage spike than a cheaper component, wouldn't it?
No. I don't see any monetary units in the expression for voltage.
 

crutschow

Joined Mar 14, 2008
31,112
I mean a carefully wound coil with a specialized geometry would likely produce a higher voltage spike than a cheaper component, wouldn't it?
Only if it has a significantly different value of parasitic parallel capacitance (see post #11).
 

xox

Joined Sep 8, 2017
794
Only if it has a significantly different value of parasitic parallel capacitance (see post #11).
I see. So as far as DC circuits go it wouldn't have much of an effect anyway.

No. I don't see any monetary units in the expression for voltage.
Oh well isn't there supposed to be a term there to account for the distance from Shenzhen and the component's point of production? Asking for a friend.
 

Papabravo

Joined Feb 24, 2006
19,570
I see. So as far as DC circuits go it wouldn't have much of an effect anyway.



Oh well isn't there supposed to be a term there to account for the distance from Shenzhen and the component's point of production? Asking for a friend.
No units of distance from Shenzhen either. Where did you learn your dimensional analysis. Oh....wait maybe you skipped that module.
It is true that a Volt has units of joules/coulomb, so maybe if the charges started from there it would have some meaning.
 
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