Can Kirchhoff's voltage and current laws apply to circuits with capacitors?

[psparky]

Gotta run for an hour...I'll be back...not done with this.

If the diode is on at .5 volts...and has a trickle current...it also has a resistance.

V=IR~!

 

[DragonPetter]

If the diode is on at .5 volts...and has a trickle current...it also has a resistance.

V=IR~!

That's where you're mistaken. Ohm's law is linear, the I-V relationship of the diode is not. It only has an incremental, small signal resistance and that is not defined by Ohm's law. In a diode, $V != IR$, but rather $V = nV_{T}ln\left(\frac{I}{I_{s}}\right)$and so ohm's law is not valid.

I'm not trying to argue that ohm's law is not useful or not essential to learn circuits, but its misleading to say it always holds true and is universal.

Another example: Replace the diode and resistor in the above circuit with an open. What is the resistance of an open circuit? Even if you let the resistance be infinite, ohm's law will not necessarily give you the correct voltage across the open.

Any time in a circuit when there is a condition for V, I, or R = infinity, ohm's law will fall apart (the voltage across the open will be undefined, which disagrees with what KVL tells you). There is no such thing as infinite resistance, infinite current flow, etc. in the real world anyway and most of the cases where you would approximate as open are actually nonlinear i-v relationships (e.g. dielectric in a capacitor will breakdown at a high enough DC voltage for example, even if we like to say the DC impedance is infinite).

Perhaps you could say Ohm's law is always true in LTI systems, but that is a very specific class of circuits, and so it is not universal. It would make sense why textbooks and professors stress the meaning of continuous, linear, time invariant, dynamic, and causality in circuits because they know that these conditions are necessary for our assumptions (V = IR) to be true.

 

[Jony130]

Comon!
https://ece.uwaterloo.ca/~dwharder/N...wton/diode.png

KCL, KVL and V=IR are all true! You just proved my point.

Just because the voltage source is not strong enough to turn on the circuit...that does not mean the laws are cancelled!

It is a FACT you may not be able to solve all circuits with those three tools...but those three tools are at your disposal at any times...and there laws always hold true!

AHHHHHHHHHHHHHHHHHHHHH!

But for this circuit normally it is impossible to find analytical solution.
We a force to use numerical method or further simplify the circuit.

 

[psparky]

You made the point that the diode has a non IV relationship. Ok...so what.

At any point in time that voltage or current is based on the resistance of the diode.

In the circuit you showed...let's say there is a trickle current for the diode.

This trickle current will also trickle thru the resistor.

The current thru the diode multipled thru the resistance at that time...is the voltage drop.

The trickle current thru the resistor has a voltage drop based on same rules.

The sum of the two voltage drops equals the voltage source (KVL)Current into the source equals current out.(KCL) V=IR on all things described.

You have not convinced me...nor will you ever.

 

[DragonPetter]

You made the point that the diode has a non IV relationship. Ok...so what.

At any point in time that voltage or current is based on the resistance of the diode.

In the circuit you showed...let's say there is a trickle current for the diode.

This trickle current will also trickle thru the resistor.

The current thru the diode multipled thru the resistance at that time...is the voltage drop.

The trickle current thru the resistor has a voltage drop based on same rules.

The sum of the two voltage drops equals the voltage source (KVL)Current into the source equals current out.(KCL) V=IR on all things described.

You have not convinced me...nor will you ever.

So you say a diode still obey's ohm's law? In that case, you must be able to tell me the resistance or impedance of a diode. Can you show me a datasheet that tells what the R in ohm's law is for a diode? What your point is that incremental, small signal resistance exists, and I won't argue that. I will argue that a diode does not follow Ohm's law and you cannot apply Ohm's law to a diode.

 

[DragonPetter]

Since I don't have much credibility, I will use a reference:

Read the 3rd paragraph here: http://www.phy.syr.edu/courses/PHY222.07Spring/manuals/ohm.pdf

 

[psparky]

After really thinking about it...I will have to agree that diodes do not follow ohm's law.

Why? Because if I have a 12 volt battery in series with a diode and resistor...I can change the value of the resistor and I will always have .7 volts dropped across the diode (assuming a .7 volt diode)...this clearly violates OHMS LAW..not to mention the couple sources you referenced.

Therefore, I clearly stand corrected. I'm so used to just dropping the .7 that I never stopped to think about it.

Are there any other devices that violate OHMS LAW?

 

[Antiphon]

Linearity of indivdual components is not required for KVL and KCL to work. They work just fine on Diodes, FETs, whatever.

 

[DragonPetter]

Linearity of indivdual components is not required for KVL and KCL to work. They work just fine on Diodes, FETs, whatever.

Yep, the linearity discussion was only with regards to Ohm's law.

 

[rbj]

Lumped circuit-theory is already based on approximations of Maxwell's equations (e.g. capacitor current I = C*dV/dt is derived by ignoring the effect of the time-varying magnetic field in Faraday's Law). So within lumped circuit theory, I'd say KVL and KCL are always true.

pretty much agree.

What is this epsilon you mention?

any vanishingly small real and positive number. you know: "Given an $\epsilon>0$, find a $\delta$ such that ..."

KVL and KCL ain't the perfect truth, but they are within $\epsilon$ of it.

r b-j

 

[Kholdstare]

After really thinking about it...I will have to agree that diodes do not follow ohm's law.

Why? Because if I have a 12 volt battery in series with a diode and resistor...I can change the value of the resistor and I will always have .7 volts dropped across the diode (assuming a .7 volt diode)...this clearly violates OHMS LAW..not to mention the couple sources you referenced.

Therefore, I clearly stand corrected. I'm so used to just dropping the .7 that I never stopped to think about it.

Are there any other devices that violate OHMS LAW?

I think you are forgetting that Ohm's law is the law of a material property. If You say "a diode does not violate ohm's law" it would certainly be true for the material inside that diode given they conduct only by drift. But you know about the depletion layer and carrier injection and so and so. Do they violate ohm's law? Yes they do. As they conduct by diffusion as well as drift. However having solved this issue we can also say, diode's won't violate laws of electromagnetism and Poisson's equation. After all Ohm's law is just a special case. Are there other devices that violates ohm's law? Well roughly wherever depletion region is involved it will be violated.

EDIT: I just remembered you can roughly use ohm's law in linear portion of tunnel diode although it has depletion region. A very special case.

 

[NascentOxygen]

Are there any other devices that violate OHMS LAW?

Incandescent light globes are non-linear, and useful as a non-linear resistance in the region well below where the filament glows red. Carbon rods have a negative resistance with temperature, as I recall.

 

[Femme_physics]

Since there is a capacitor I must take time into consideration. Does this mean I must use this formula?

http://img525.imageshack.us/img525/9917/icicicicicicic.jpg

 

[I like Serena]

Yes, the formula for voltage looks like that.
Myself, I would use a different representation of the same formula.

Do you have a method to find out what I, R, and A in that formula should be?

 

[Femme_physics]

That's exactly what I'm trying to figure out...no findings so far

 

[I like Serena]

Typically, you would first simplify the circuit as much as possible.
The first step is to use what vk6kro showed in post #24.

The method I know (after the simplification), is to set up a differential equation with KVL and KCL and solve that (which I did).
But I'm pretty sure, that's not what you are expected to do, since you do not know (yet?) how to set up and solve a differential equation.

So you should have a different method in your textbook... but what is it?

 

[vk6kro]

That formula seems unduly complex.

This one is more usual:

where

which is the time constant of the circuit. R is in ohms. C in Farads. (A convenient shortcut is to use Megohms and Microfarads here.)

For example,
if Vin =100 volts and the capacitor has no charge on it to start with.
C= 5 uF
R = 100 K (so RC = 0.5 seconds)
t = 0.6 second

V capacitor = 100 * (1 - 2.71828 ^ ( -0.6 / (5 * 0.1 )))
V capacitor = 100 * 0.6988 = 69.88 volts

 

[rbj]

That formula seems unduly complex.

This one is more usual:

where

which is the time constant of the circuit. R is in ohms. C in Farads. (A convenient shortcut is to use Megohms and Microfarads here.)

i think if Femme blast out her equation, it will become equivalent to your more usual one.

 

[Silverlining]

You can use KCL and KVL on a capacitor. You represent it'd impedance with a phasor or in the laplace domain.

This is a snap shot at a certain time... you set the source to have a phase of zero and then each component has a different phase

 

[Silverlining]

I agree. Kirchoff's laws and Ohm's law apply to 100% of the cases.

Adding a capacitor, inductor, diode...etc does not change that.

Accepting these laws seems to take time...but it shouldn't.

KVL, KCL and V=IR

Just accept it now.

only if Voltage = current * impedance. You need to translate into phasor form to analyze any non linear circuit.

 

[Femme_physics]

I'd like to explain how the original circuit I posted works. I'll repost its image if I may:

http://img853.imageshack.us/img853/9302/diodescabal.jpg

Once the switch is closed (and the push button remains unpressed) all the current flows towards the capacitor until it fully charges, then afterwards current flows through the resistor and LED's to the ground. When it passes the 5 mA for each diode they light up (in different times since the resistors don't have the same value). P is there to cause a shortcircuit thereby discharging the capacitor.

How did I do?

 

[vk6kro]

Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.

When you actually see the LEDs lighting up depends on the efficiency of the LED and the background lighting.
In a dark room, you may see the light at 1 or 2 milliamps.

Is that capacitor actually 500 MILLI Farads? ie half a Farad?

 

[Femme_physics]

Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.

Let me see if I have it straight. At the first moment the switch is closed, all the current flows towards the capacitor. The more time goes by, the more current starts flowing through the diodes (and resistors) until all the current flows through them. So the capacitor in fact just steals a bit from the current influx in the beginning until the diodes get it?

Is that capacitor actually 500 MILLI Farads? ie half a Farad?

Yes

 

[vk6kro]

The capacitor will charge up to almost the supply voltage because the diodes draw very little current compared to the 3.75 amps that the capacitor initially draws.

The final voltage across the capacitor will be caused by the supply voltage (15 volts) minus the voltage drop due to the very small current drawn by the diodes. This will be something less than 32 mA so the drop will only be about 128 mV.

If you were to really build this, you would need to limit the current in the switch across the capacitor because this could be very large (depending on the internal resistance of the capacitor) and may damage the switch.

A good way to do this would be to put a small resistor between the capacitor and the 4 ohm resistor, then shorting that junction of the two resistors to ground. About 1 ohm should be adequate protection for the switch.

This way, the capacitor could fully discharge in a few seconds.

 

[Femme_physics]

OK, thanks for this, but I'm trying to gauge this process based on time. What happens in the first moment it's closed, till the steady state. I was hoping my explanation above is correct?

 

[vk6kro]

In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.

 

[Femme_physics]

In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.

How can you say that there is not enough voltage at the first moment if voltage is constant at 15 V?

 

[NascentOxygen]

How can you say that there is not enough voltage at the first moment if voltage is constant at 15 V?

Hi Femme_physics! vk6kro is saying there is not enough voltage on the capacitor plates to send current through the LEDs. It is the capacitor voltage which supplies the LEDs and their current-limiting resistors.