[SOLVED] capacitor charge with current source + resistor

[c0x]

Hi all,


I know that I should be able to do this within 5 minutes, but I am not.
I simply do not remember my mathematics well enough anymore! :_(

I have a capacitor C.
It is charged using a constant current source (Ik) and
parallel to that source is a resistor R from supply Vcc, charging it with current Ir.
Total charge current through the capacitor is hence Ic = Ik + Ir.

I want to deduct an equation telling me what the voltage Vc across a capacitor is, at any specific time t from time = 0.
Charge is 0 at time = 0.

I can easily simulate it, but now I want to have that equation, because I want to calculate it.

If you can help, please do.



/c0x

 

[barry]

I'm very confused. Can you show a schematic?

But if what you say "Ic=Ik+Ir" is true, then your equation is simply:

Vc=1/c*Integral(Ic dt)

But Ir is going to vary as the capacitor charges, right?

We need a schematic.

 

[c0x]

Hi Barry,

I'm very confused. Can you show a schematic?

Absolutely. Here you go:

But if what you say "Ic=Ik+Ir" is true, then your equation is simply:

Vc=1/c*Integral(Ic dt)

But Ir is going to vary as the capacitor charges, right?

Correct.

I am certain it is piece of cake for someone who knows his or her mathematics.
But I just spent several hours without cracking it, which is why I now ask for help. Sad but true.



/c0x

 

[Ratch]

Hi Barry,



Absolutely. Here you go:
View attachment 144214



Correct.

I am certain it is piece of cake for someone who knows his or her mathematics.
But I just spent several hours without cracking it, which is why I now ask for help. Sad but true.



/c0x

This can be solved by either the node method, loop method or superposition method. Are you familiar with Laplace transforms?

Ratch

 

[c0x]

Hi Ratch,

This can be solved by either the node method, loop method or superposition method. Are you familiar with Laplace transforms?

I know the basic component equations in time and s domain (for the capacitor i=C*dV/dt and X=1/s/C), but transformations I haven't done for some 10 years. I might be able to catch up, though.

 

[CataM]

This can be solved by either the node method, loop method or superposition method. Are you familiar with Laplace transforms?

Superposition is the easiest method for this specific circuit. The circuit implements 2 simple first order LPF for both sources, .
Here is your solution. Just apply superposition and see page 4: **broken link removed**

 

[c0x]

Hi CataM,

Superposition is the easiest method for this specific circuit. The circuit implements 2 simple first order LPF for both sources, .
Here is your solution. Just apply superposition and see page 4: **broken link removed**

Sounds simple, but not so easy for me. Maybe I should go back to school.

Am I on the right track if I think that one way to go is:
* generating expressions for Ic in s domain for both sources separately,
* then add the expressions to get combined expression for Ic,
* reduce and
* finally do inverse Laplace to get a time domain expression?


/c0x

 

[Ratch]

Hi CataM,



Sounds simple, but not so easy for me. Maybe I should go back to school.

Am I on the right track if I think that one way to go is:
* generating expressions for Ic in s domain for both sources separately,
* then add the expressions to get combined expression for Ic,
* reduce and
* finally do inverse Laplace to get a time domain expression?


/c0x

You are correct in acknowledging that you need a good review of linear differential equations. I think the best way to do this problem is by node analysis, because you need only one equation with only one unknown, specifically the voltage across the capacitor. Call that voltage "v".

The node equation is:


Solving for v, we get:


Substituting c = 1 nF :


Find the inverse Laplace:


Plot the equation for 50 microsecconds:


Any questions?

Ratch

 

[c_mitra]

Ik=100 uA =0.1 mA; given.

Ir=5/10k =0.5 mA; calculated (valid at t=0)

Const current source has infinite impedance. Ir-Ik must flow through the capacitor. (t=0)

The capacitor charges exponentially to final potential of 4V with a time const of 10k * 10nF

How do you get 6V at t = infinity?

 

[FvM]

How do you get 6V at t = infinity?

5V + 10k*100µA according to the source polarities in the post #3 schematic

 

[Ratch]

Ik=100 uA =0.1 mA; given.

Ir=5/10k =0.5 mA; calculated (valid at t=0)

Yes, that is true.

Const current source has infinite impedance. Ir-Ik must flow through the capacitor. (t=0)

No current exists through the capacitor. If it did, the capacitor would be defective. Charge accumulates on one plate and depletes on the other plate for a net charge gain/loss of zero. The capacitor separates the charges and thereby becomes energized. That means the capacitor stores energy, not charge. Current exists up to one plate for a limited time and away from the opposite plate for a limited time, but charge does not go through the capacitor.

The capacitor charges exponentially to final potential of 4V with a time const of 10k * 10nF

Capacitors don't charge, they energize.

How do you get 6V at t = infinity?

Mother Math gave me the answer. Did you see the plot above? After the capacitor is fully energized, the CC source has nowhere to go but through the 10k resistor and through the voltage source. That is a one volt drop across the resistor plus 5 volts from the voltage source gives six volts final steady state value.

 

[c_mitra]

5V + 10k*100µA according to the source polarities in the post #3 schematic ...

Thank you; I goofed up!

 

[c0x]

I think the best way to do this problem is by node analysis
--- cut ---
Any questions?

Seeing the solution I can accept the procedures. Some I recognize, some is forgotten long time ago. Maybe some of it I never actually learned.
I acknowledge that I should practice this sort of circuit analysis more, and I will try and do so, because this is so simple and I want (and need) to be able to do it.

For now, I can move on. Thank you Ratch, and everyone else who contributed.

Yay, my LTSpice and Console Calculator finally agree.



/c0x

 

[CataM]

Hi CataM,



Sounds simple, but not so easy for me. Maybe I should go back to school.

Am I on the right track if I think that one way to go is:
* generating expressions for Ic in s domain for both sources separately,
* then add the expressions to get combined expression for Ic,
* reduce and
* finally do inverse Laplace to get a time domain expression?

Yes but it is to much time consuming.

An easier method is like this:
1) Transform the voltage source and resistor to current source in parallel with resistor. https://en.wikipedia.org/wiki/Source_transformation
2) Sum up both current sources.
3) Convert the summed current sources in parallel with the resistor to voltage source in series with resistor
4) The remaining circuit is already solved in the link of post #6.

 

[c_mitra]

No current exists through the capacitor. If it did, the capacitor would be defective. Charge accumulates on one plate and depletes on the other plate for a net charge gain/loss of zero. The capacitor separates the charges and thereby becomes energized. That means the capacitor stores energy, not charge.

Capacitance is the amount of charge required to increase the potential by 1 unit. Charge must flow. The capacitor stores charge. The energy is actually stored in the dielectric (it can be a vacuum) in the electric field.

Flow of charge is current. The current through the dielectric is called "displacement current"- see for details: https://en.wikipedia.org/wiki/Displacement_current

Alternating current flows through a capacitor just like a resistor: except for the phase.

- - - Updated - - -

Some I recognize, some is forgotten long time ago. Maybe some of it I never actually learned...

The problem is badly defined; let me explain. Please see the diagram in post #3.

The initial condition is not defined: it will be tempting to assume that you want a solution of the steady state (which is trivial).

To give a possible idea (there are several) consider a switch on the left of the top node connecting the const current source to the rest. The time starts when you close the switch and you wish the voltage and currents as a function of t.

If you consider the switch on the capacitor side (top central node; connecting the capacitor) then you may get a different result.

But if you consider the switch connected to the resistor side (right of the node) then you get another result (that can be rather interesting)...

And there are other possibilities.

But the problem can be more complex in real life.

 

[Ratch]

Capacitance is the amount of charge required to increase the potential by 1 unit.

Capacitance is a proportional measure of the amount of energy a capacitor can store at a specified voltage. A 1 farad capacitor can store a half joule of energy at one volt.

Charge must flow. The capacitor stores charge.

The net charge of a capacitor is the same at 0,100 or 1000 volts, specifically zero. That is because when a voltage is applied across a capacitor, the same amount of charge that enters at one plate leaves from the opposite plate. The charge is separated and unbalanced between the plates, but the net charge change is still zero. This separation causes a electric field to form between the plates, which is where the energy is stored. Therefore, the capacitor is "charged" with energy, not charge. So you might as well say the capacitor is "energized". Energized means imbued with energy, and is completely unambiguous and correct.

The energy is actually stored in the dielectric (it can be a vacuum) in the electric field.

The energy is stored in the electric field which permeates the dielectric. The dielectric changes the permittivity constant which affects the characteristics of the field.

Flow of charge is current.

Yes, that is the definition of current.

The current through the dielectric is called "displacement current"- see for details: https://en.wikipedia.org/wiki/Displacement_current

That is Maxwell's misnomer. Although the term he added to the field equation has the units of current, it is really not current at all. His equations are correct, but his terminology was wrong. Just as torque and work have the same units (newton-meters) but are different entities, so is his false "displacement current".

Alternating current flows through a capacitor just like a resistor: except for the phase.

If current actually flowed through a capacitor, it would be discarded as leaky. It would also dissipate heat, which a perfect capacitor never does. A capacitor stores and gives back energy to the circuit twice a cycle. In conclusion, current exists up to one plate of the capacitor and is measurable by an ammeter. Current also exists from the opposite plate and is measurable by an ammeter. At no time is current present through the capacitor. The charge accumulates on one plate and depletes on the opposite plate. That is what makes the current appear as though it is existing through the capacitor.

- - - Updated - - -

c0x,

Thanks for your thanks. The conditions I assumed was that the capacitor was initially unenergized, and both the current and voltage sources were switched on at the same time.

Ratch

 

[c_mitra]

The net charge of a capacitor is the same at 0,100 or 1000 volts, specifically zero...

A conducting sphere of radius R has a charge Q and a surface potential of Q/(4*pi*epsilon0*R) and a capacitance of 4*pi*epsilon0*R.

Capacitance of earth is about 100 uF

 

[Ratch]

A conducting sphere of radius R has a charge Q and a surface potential of Q/(4*pi*epsilon0*R) and a capacitance of 4*pi*epsilon0*R.

Capacitance of earth is about 100 uF

I am referring to a capacitor energized in an electrical conductive circuit, not by a charge transported by mechanical physical means such as a static generator.

Ratch

 

[tinhkt5491]

Can you show a schematic?