Sensor - Calculate the Capacitance

[2013]

The capacitors are switched in series and I wanted to determine the total capacity to come up with the extra capacity over the load.
How do I do that then?

 

[gneill]

The capacitors are switched in series and I wanted to determine the total capacity to come up with the extra capacity over the load.
How do I do that then?

The capacitors are neither in series nor in parallel in the given circuit. That is why you need to apply circuit analysis methods other than simple series/parallel reductions. Now, if you apply superposition then series/parallel opportunities will appear and then you can use the capacitive divider rule.

Is there some reason that you're avoiding applying superposition? It is usually one of the first circuit analysis theorems/techniques introduced to students.

 

[2013]

To your question: No, we have not yet worked on the topic in our course.

C1 and C2 are switched in series so:
C = 1 / ((1/C1) + (1/C2))
C = 1,107 * 10^-11 As/V

Cc = V * (C/Vc) - C = 12V * (1,107 * 10^-11 As/V / 0,15V) - 1,107 * 10^-11 As/V = 8,74 * 10^-10 As/V

Is it right?
I`m so sorry but our prof never speak with us about circuit analysis, he expects that we can solve the problem that way. As a final test, so to say.

 

[gneill]

To your question: No, we have not yet worked on the topic in our course.

C1 and C2 are switched in series so:
C = 1 / ((1/C1) + (1/C2))
C = 1,107 * 10^-11 As/V

Cc = V * (C/Vc) - C = 12V * (1,107 * 10^-11 As/V / 0,15V) - 1,107 * 10^-11 As/V = 8,74 * 10^-10 As/V

Is it right?
I`m so sorry but our prof never speak with us about circuit analysis, he expects that we can solve the problem that way. As a final test, so to say.

No, it is not right. As I said, those capacitors are NOT IN SERIES. The third capacitor, your C_c, is connected to the same junction where they join.

Barring the use of superposition, you will have to fall back on KVL and KCL, and the fact that for a given capacitor Q = VC. You'll end up with several equations to solve simultaneously.

If you haven't been taught (or taught yourself) circuit analysis, then you cannot solve this problem.

 

[2013]

Node rule:
-Q1-Q2+Q3 = 0

Mesh rule:
1) Q1/C1 - Q2/C2 - Q2/C3 = -U1 - U2
2) Q2/C2 + Q2/C3 + Q3/C4 = U2+U3

Is it right? How should I go on? Could you give me a formula as a tip?

 

[gneill]

Node rule:
-Q1-Q2+Q3 = 0

Mesh rule:
1) Q1/C1 - Q2/C2 - Q2/C3 = -U1 - U2
2) Q2/C2 + Q2/C3 + Q3/C4 = U2+U3

Is it right? How should I go on? Could you give me a formula as a tip?

I cannot say with certainty, since you haven't specified what the variables are (which charges belong to which capacitors, what U1, U2, and U3 are... ). However, if I were to assume that Q1, Q2, and Q3 are the charges on the three separate capacitors of the circuit, then I cannot see how you justify placing the same charge (Q2) on two different capacitors. And where did C4 come from?

Please strive to be clear. The variables and component designations have changed several times over the course of this thread, and it is becoming frustrating. You are coming up with formulas and asking if they are correct without defining the variables, showing the basic work, or describing the logic behind them. This makes it difficult to follow and judge their value.

 

[2013]

Is it right? Can I go on with the node and mesh rule?
How can I set it up?

 

[gneill]

Is it right? Can I go on with the node and mesh rule?
How can I set it up?

You've placed some labels on the diagram:

What do they mean? The appear to indicate potentials across the capacitors, but what are the variables? To me, "C1 - V1" seems to indicate a capacitance minus a voltage, which doesn't make sense in terms of units.

If you want to go with KCL (node) and mesh (KVL) methods, I suggest that you start by assuming that there is some charge q1 on capacitor C1, q2 on capacitor C2, and q3 on capacitor C3. Then write KCL equation that ties these three together. After that write KVL equations for the individual loops, using the relationship Q = V/C for each capacitor's potential in those equations. Recognize that you want the potential on C3 to be Uc = q3/C3, where Uc is the trigger potential. Use these equations that you've written to solve for C3.

 

[2013]

C1 = 2,01*10^-11 As/V
C2 = 2,46*10^-11 As/V

Q = C * U U=12V

Q1 = 2,01*10^-11 As/V * 12V = 2,41*10^-10 As
Q2 = 2,46*10^-11 As/V * 12V = 2,95*10^-10 As

Uc=0,15V Cc=?

-Q1+Q2+Qc=0
-2,41*10^-10 As + 2,95*10^-10 As + Qc = 0
Qc = 5,4*10^-11 As

Cc = Qc/Uc = 5,4*10^-11 As / 0,15 V = 3,6*10^-10 As/V

Is it right? Is that the additional capacitance?

I have got another question:
I first viewed on the drawing, the three panels, the middle plate moves around to the left or right and thus the capacitance of the two capacitors changes. A is increased, the other decreases. I have calculated. What's with the wavy line under the middle plate and the capacitor underneath? Can you explain to me how this relates? I think that's the reason why I do not get ahead.

Thank you for all your help and your patience with me.

 

[gneill]

C1 = 2,01*10^-11 As/V
C2 = 2,46*10^-11 As/V

Q = C * U U=12V

Q1 = 2,01*10^-11 As/V * 12V = 2,41*10^-10 As
Q2 = 2,46*10^-11 As/V * 12V = 2,95*10^-10 As

Why do you assume that the potentials on the capacitors are both 12V?

You'll need to use KVL to write equations relating the potentials on all the components.

Uc=0,15V Cc=?

-Q1+Q2+Qc=0

That looks okay, being based upon KCL applied at the common node.

-2,41*10^-10 As + 2,95*10^-10 As + Qc = 0
Qc = 5,4*10^-11 As

Cc = Qc/Uc = 5,4*10^-11 As / 0,15 V = 3,6*10^-10 As/V

Is it right? Is that the additional capacitance?

No. But you're applying more thought to the problem, which is good. Write the KVL equations for the loops. Use Q1/C1, Q2/C2, Qc/Cc, for the potentials on the capacitors to begin with, then substitute for them from other known relationships (such as your KCL equation above, and Qc/Cc = Uc, for examples).

I have got another question:
I first viewed on the drawing, the three panels, the middle plate moves around to the left or right and thus the capacitance of the two capacitors changes. A is increased, the other decreases. I have calculated. What's with the wavy line under the middle plate and the capacitor underneath? Can you explain to me how this relates? I think that's the reason why I do not get ahead.

Thank you for all your help and your patience with me.

I believe that the wavy line is meant to represent a flexible connection to the moving plate. It has no electrical significance.

 

[2013]

I used before U=V for the voltage.
I really don`t have any good ideas anymore.

maybe it is:
- Q1/C1 + Q2/C2 + Qc/Cc = - V1 + V2 + Vc

Can you show me your formula for this. So I have a chance to continue, this would be very nice.

 

[gneill]

I used before U=V for the voltage.
I really don`t have any good ideas anymore.

maybe it is:
- Q1/C1 + Q2/C2 + Qc/Cc = - V1 + V2 + Vc

Can you show me your formula for this. So I have a chance to continue, this would be very nice.

The forum rules forbid directly providing solutions. But consider the following circuit diagram depicting the circuit under study:

Assume that the sensor plate has moved in such a way as to cause a brief current to flow, redistributing charges on the capacitors. Since the node labeled with the potential Uc is isolated by capacitors, the total charge associated with the node must always remain zero. So by KCL you have written:

-q1 + qc + q2 = 0

or,

q1 = qc + q2

It is also true, and you have essentially written this before at various times, that

V1 = q1/C1 ; V2 = q2/C2 ; Uc = qc/Cc

Where V1, V2, and Uc are the potentials on the capacitors. It is important that you pay attention to the polarities of the potentials on the capacitors; the plate with the +q will be positive with respect to the plate with the -q on it for this arrangement of charges.

Apply KVL to the two loops to obtain two independent equations. With these equations and the ones above you can solve for Cc.

ALTERNATIVE METHOD:
As I mentioned previously, superposition is a much easier path to solution. Suppress one 12V voltage source at a time and solve for the resulting voltage Vc at the node. Sum the results to arrive at the expression for actual potential at the node. For example, the circuit with the right-hand side 12V source suppressed becomes:

You will have to draw the circuit corresponding to having the other source suppressed, and solve again for Vc; You should be able to find the two expressions for Vc using the capacitor voltage divider method. The expression will have Cc in it as an unknown, and you already know that sum of the two Vc's will be your Uc, the trigger voltage.

 

[2013]

KVL:

1. V1+Vc=12V => Q1/C1+Qc/Cc=12V
2. V1+V2=24V => Q1/C1+Q2/C2=24V

Are this the two equations?Thank you so much for your help! :)

 

[gneill]

KVL:

1. V1+Vc=12V => Q1/C1+Qc/Cc=12V
2. V1+V2=24V => Q1/C1+Q2/C2=24V

Are this the two equations?

Sure; they are two independent KVL equations for the circuit.

Thank you so much for your help! :)

No problem; I'm happy to help.

 

[2013]

1.Q1/C1+Qc/Cc=12V
2.Q1/C1+Q2/C2=24V
3.-q1 + qc + q2 = 0 => Qc=Q1-Q2

in 1.
Q1/C1 + (Q1-Q2)/Cc=12V

How can I replace Q1 and Q2? Or do I already know the charge?

 

[gneill]

1.Q1/C1+Qc/Cc=12V
2.Q1/C1+Q2/C2=24V
3.-q1 + qc + q2 = 0 => Qc=Q1-Q2

in 1.
Q1/C1 + (Q1-Q2)/Cc=12V

How can I replace Q1 and Q2? Or do I already know the charge?

You don't know the charges, they're unknowns. But you do know that Uc = Qc/Cc, and you are given the value of Uc. That leaves you with unknowns Q1, Q2, and Cc.

You have enough equations to solve for the unknowns.

 

[2013]

1. Q1 = (12V - Vc) * C1 = (12V - 0,15V) * 2,01*10^-11As = 2,38 * 10^-10 As
V1 = Q1/C1 = 11,85V

2. V2 = 24V - V1 = 24V - 11,85V = 12,15V
Q2 = V2 * C2 = 12,15V * 2,46*10^-11As/V = 2,99*10^-10As

3. Cc = (Q1-Q2)/Vc = ( 2,38 * 10^-10 As - 2,99*10^-10As) / 0,15V = -4,047*10^-10As/V

Is this right?

 

[gneill]

1. Q1 = (12V - Vc) * C1 = (12V - 0,15V) * 2,01*10^-11As = 2,38 * 10^-10 As
V1 = Q1/C1 = 11,85V

2. V2 = 24V - V1 = 24V - 11,85V = 12,15V
Q2 = V2 * C2 = 12,15V * 2,46*10^-11As/V = 2,99*10^-10As

3. Cc = (Q1-Q2)/Vc = ( 2,38 * 10^-10 As - 2,99*10^-10As) / 0,15V = -4,047*10^-10As/V

Is this right?

Well, since capacitance can't be negative, no it's not right.

However, that may be because the plate can move in either direction; That means the voltage on Cc can swing positive or negative from its neutral position (by symmetry that potential will be zero when the plate is centered). Thus, if the Δx used to calculate the capacitances accidentally had the wrong sign then the calculated values for capacitances C1 and C2 would be swapped.

To correct this, use a negative value for Uc or interchange the values of the capacitances C1 and C2 and recalculate the above.

 

[2013]

CC= (2,99*10^-10As - 2,38 * 10^-10 As) / 0,15V = 4,07*10^-10As/V

Alternative:
Cc = (Q1-Q2)/Vc = ( 2,38 * 10^-10 As - 2,99*10^-10As) / (-0,15)V = 4,07*10^-10As/V

Is this right? Did I make the right moves? Did you figure out the same thing?

 

[gneill]

CC= (2,99*10^-10As - 2,38 * 10^-10 As) / 0,15V = 4,07*10^-10As/V

Alternative:
Cc = (Q1-Q2)/Vc = ( 2,38 * 10^-10 As - 2,99*10^-10As) / (-0,15)V = 4,07*10^-10As/V

Is this right? Did I make the right moves? Did you figure out the same thing?

No, not quite. But I think I may see a way to sort things out. Assume that your Δx should have made the plate move to the left. Then C1 should end up larger than C2 since its plates will be closer together. So, using your calculated capacitance values let's set:

C1 = 2.459 x 10^-11 F = 24.59 pF
C2 = 2.012 x 10^-11 F = 20.12 pF

Then using your first KVL equation,

Q1/C1 + Qc/Cc = 12V
V1 + Uc = 12V
V1 = 11.85V
and so
Q1 = V1*C1 = 2.915 x 10^-10 C

just as you did before.

For determining Q2, use your second KVL equation:

Q1/C1 + Q2/C2 = 24V
V1 + Q2/C2 = 24V
Q2 = (24V - V1)*C2
Q2 = [STRIKE]2.915 x 10^-10 C[/STRIKE] 2.445 x 10^-10 C [EDIT]

Now use your third equation to find Qc. You know Uc. So determine Cc.

 

[2013]

Qc = Q1 - Q2 = 2.915 x 10-10 C - 2.915 x 10-10 C = 0
Cc = Qc/Vc = 0/0,15V = 0

I don´t think that this is right. I do not think that the extra capacity is not zero, right?

 

[gneill]

Qc = Q1 - Q2 = 2.915 x 10-10 C - 2.915 x 10-10 C = 0
Cc = Qc/Vc = 0/0,15V = 0

I don´t think that this is right. I do not think that the extra capacity is not zero, right?

Oops. My mistake. I copied the wrong value for Q2 (I used my value for Q1 by mistake). Take a look at my edit of my previous post.

 

[2013]

Qc = Q1 - Q2 = 2.915 x 10^-10 C - 2.445 x 10^-10 C = 4,7*10^-11As
Cc = 4,7*10^-11As / 0,15V = 3,33*10^-10As/v

Have you got the same additional capacitance?

 

[gneill]

Qc = Q1 - Q2 = 2.915 x 10^-10 C - 2.445 x 10^-10 C = 4,7*10^-11As
Cc = 4,7*10^-11As / 0,15V = 3,33*10^-10As/v

Have you got the same additional capacitance?

I kept extra decimal places for intermediate results and got a slightly smaller value (3.13 rather than 3.33) but, yes, that looks much better.

You should get in the habit of using standard units for electrical quantities. "As" is Coulombs (C), "As/v" is Farads (F). Use metric prefixes as appropriate.

 

[2013]

Thank you very much.

Can you also explain me the second method?
How can I apply the voltage divider method?

 

[gneill]

Thank you very much.

Can you also explain me the second method?
How can I apply the voltage divider method?

Take a look at the diagram in post #42. You should be able to see the voltage divider (For the case where the 12V supply on the right is suppressed). We went over the capacitor voltage divider earlier in the thread. You should know how to combine capacitors in parallel.

 

[2013]

$$V_{C1} = V \frac{C}{C1 + C}$$

Capacitors in parallel: C2+Cc=C

I don´t think that is right, but I don´t have any other idea.
How should I go on?

 

[gneill]

You need to make sure that you understand how a capacitor voltage divider works. For capacitors, the value in the numerator of the fraction is the "other" capacitance, not the one you want the voltage across. So, for example:

Then

$Vout = Vin \frac{C_1}{C_1 + C_2}$

(NOTE that this is an example separate from the thread's problem, so the capacitor names don't necessarily pertain directly to the ones in the thread's problem. What is important is to recognize which capacitances go where in the voltage divider equation in order to find the appropriate output potential).

Notice how the expression for the potential across $C_2$ (the bottom capacitance) has the upper capacitance $C_1$ in the numerator of the fraction. This is the 'opposite' of the situation for a resistor voltage divider.

 

[2013]

Vc=V*C1/(C1+C)

C=C2+Cc

Is this right? And then?

 

[gneill]

Vc=V*C1/(C1+C)

C=C2+Cc

Is this right? And then?

It depends. What is V?

Also, expand C in your expression, don't drag along another equation. You want ONE expression for Vc that includes Cc in it.

Then do the same again, suppressing the other voltage source.