Sensor - Calculate the Capacitance

AI Thread Summary
The discussion focuses on calculating the capacitance of an airbag sensor that uses accelerometers to trigger deployment. The sensor consists of three parallel plates, with the middle plate able to move due to acceleration, affecting the capacitance. Participants are trying to derive the maximum deflection of the middle plate under sudden braking and how this relates to the additional capacitance needed to trigger the airbag at a specified voltage. Key concepts include using conservation of energy and analyzing the circuit with capacitors in series, while addressing the correct formulas for capacitance and voltage division. The conversation highlights the need for clarity in circuit analysis and proper application of physics principles to solve the problem effectively.
  • #51
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?
 
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  • #52
2013 said:
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.
 
  • #53
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?
 
  • #54
2013 said:
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.
 
  • #55
Thank you very much.

Can you also explain me the second method?
How can I apply the voltage divider method?
 
Last edited:
  • #56
2013 said:
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.
 
  • #57
gneill said:
$$V_{C1} = V \frac{C}{C1 + C}$$

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?
 
  • #58
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:
attachment.php?attachmentid=60281&stc=1&d=1373835248.gif


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.
 

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  • #59
Vc=V*C1/(C1+C)

C=C2+Cc

Is this right? And then?
 
  • #60
2013 said:
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.
 
  • #61
1. Vc = Vin * C1/(C1+Cc)

2. Vc = Vin * C2/(C2+Cc)

Are this to equations are correct? What should I do then?
 
  • #62
2013 said:
1. Vc = Vin * C1/(C1+Cc)

2. Vc = Vin * C2/(C2+Cc)

Are this to equations are correct? What should I do then?

No, they are incomplete. Both expressions should contain all three capacitances, and you must put a value to Vin for each. Look at the circuit and identify all the capacitances that correspond to the variables in the voltage divider equation.
 
  • #63
1. Vc = C1/(C1+C2+Cc) (I don´t know how I put a value Vin for each. How should I do that?)

2. Vc = C2/(C1+C2+Cc) (I don´t know how I put a value Vin for each. How should I do that?)Can you show me, how I should do that?
 
  • #64
2013 said:
1. Vc = C1/(C1+C2+Cc) (I don´t know how I put a value Vin for each. How should I do that?)

2. Vc = C2/(C1+C2+Cc) (I don´t know how I put a value Vin for each. How should I do that?)


Can you show me, how I should do that?

Can't you just look at the circuits and tell?
 
  • #65
I really don`t know. Maybe:
1. Vc = C1 * Vin/(C1+C2+Cc)

?
 
  • #66
2013 said:
I really don`t know. Maybe:
1. Vc = C1 * Vin/(C1+C2+Cc)

?

You've just rearranged the formula; you haven't replaced Vin with its numerical value.
 
  • #67
Ahh, okay sorry I don´t understand what you mean.

1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = 12V * C2/(C1+C2+Cc)

Is this right?
 
  • #68
2013 said:
Ahh, okay sorry I don´t understand what you mean.

1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = 12V * C2/(C1+C2+Cc)

Is this right?

Not quite. Did you pay careful attention to the polarities of the sources in the circuit?
 
  • #69
Can you show me what you mean?

Do you mean, that:
1. Vc = -12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)

?
 
  • #70
2013 said:
Can you show me what you mean?

Do you mean, that:
1. Vc = -12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)

?

Surely you can read a circuit diagram? Do both sources have the same polarity with respect to the node where Vc is?
 
  • #71
1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)?
 
  • #72
2013 said:
1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)


?

Yes. Now each of those expressions for Vc contribute to the actual Vc (you need to sum them; that's how superposition works). Solve for Cc.
 
  • #73
You mean:
1. + 2.

2Vc = (C1+C2) / (2C1+2C2+2Cc)

Is this right? Have I understand you right?
 
  • #74
2013 said:
You mean:
1. + 2.

2Vc = (C1+C2) / (2C1+2C2+2Cc)

Is this right? Have I understand you right?

No. First, each of the expressions contribute to the total of Vc. Think of them as Vca and Vcb, and so the total is Vc = Vca + Vcb. That's how superposition works. You sum up the contributions due to each source to yield the result of both sources acting together. (Surely you must have covered superposition in your circuit theory classes?)

Second, you have to add the fractions properly. You don't add the common denominators; Is 1/3 + 1/3 equal to 2/6?
 
  • #75
We have done superposition, but only in a short launch. We have set different priorities in the first semester.

I`m so sorry that I have no idea what I exactly should do.

Can you show me the beginning?
 
  • #76
2013 said:
We have done superposition, but only in a short launch. We have set different priorities in the first semester.

I`m so sorry that I have no idea what I exactly should do.

Can you show me the beginning?

Sum the two expressions (their right hand sides). That will yield the final expression for Vc. This is just algebra; you should be able to sum fractions, especially when they have the same denominators.
 
  • #77
12V * C1/(C1+C2+Cc) + (-12V * C2/(C1+C2+Cc)) = Vc
Vc = (C1+C2) / (C1+C2+Cc)

?
 
  • #78
2013 said:
12V * C1/(C1+C2+Cc) + (-12V * C2/(C1+C2+Cc)) = Vc
Vc = (C1+C2) / (C1+C2+Cc)

?

What happened to the +12V and -12V terms :frown:
 
  • #79
12V * C1/(C1+C2+Cc) - 12V * C2/(C1+C2+Cc)) = Vc
12V * (C1/(C1+C2+Cc) - (C2/(C1+C2+Cc)) = Vc

Vc = 12V * ((C1-C2)/(C1+C2+Cc))

?
 
  • #80
2013 said:
12V * C1/(C1+C2+Cc) - 12V * C2/(C1+C2+Cc)) = Vc
12V * (C1/(C1+C2+Cc) - (C2/(C1+C2+Cc)) = Vc

Vc = 12V * ((C1-C2)/(C1+C2+Cc))

?

Yes.
 
  • #81
Vc = 12V * ((C1-C2)/(C1+C2+Cc))
Vc*Cc=12V * ((C1-C2)/(C1+C2)
Vc*(Qc/Vc)=12V * ((C1-C2)/(C1+C2)
Qc=12V * ((C1-C2)/(C1+C2)=12V*((2,46*10^-11As/V - 2,01*10^-11As/V) / (2,46*10^-11As/V+2,01*10^11As/V)
= 1,21As

Cc=(Qc/Vc)=1,21As/0,15V=8,05As/V

Is this right? Why is it not the same capacity as in the other method?
 
  • #82
2013 said:
Vc = 12V * ((C1-C2)/(C1+C2+Cc))
Vc*Cc=12V * ((C1-C2)/(C1+C2)
Vc*(Qc/Vc)=12V * ((C1-C2)/(C1+C2)
Qc=12V * ((C1-C2)/(C1+C2)=12V*((2,46*10^-11As/V - 2,01*10^-11As/V) / (2,46*10^-11As/V+2,01*10^11As/V)
= 1,21As

Cc=(Qc/Vc)=1,21As/0,15V=8,05As/V

Is this right? Why is it not the same capacity as in the other method?
The answers should be identical. Your algebra has errors.
 
  • #83
Can you say me at which step, I have done a mistake.

?
 
  • #84
2013 said:
Can you say me at which step, I have done a mistake.

?
Your second line is incorrect. You cannot eliminate a single additive term from the denominator by multiplying the fraction by that term. Also, why are you introducing Qc at this point? You have Cc alone as a variable in the expression, and all other quantities have known values. Don't make things harder! Just do the algebra to isolate Cc.
 
  • #85
Vc=12V*(C1-C2)/(C1+C2+Cc)
Vc*(C1+C2+Cc)=12V*(C1-C2)
how can I get C1 and C2 to the other site?
 
  • #86
2013 said:
Vc=12V*(C1-C2)/(C1+C2+Cc)
Vc*(C1+C2+Cc)=12V*(C1-C2)
how can I get C1 and C2 to the other site?
Elementary operations: Divide both sides by Vc. This will leave the LHS (Left Hand Side) as a simple sum of terms. Move everything that isn't Cc to the RHS.
 
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  • #87
I have to say thank you to you, I have already the solution and it is the same as in the other method.
Thank you, Thank you for helping me :) :)
 
  • #88
2013 said:
I have to say thank you to you, I have already the solution and it is the same as in the other method.
Thank you, Thank you for helping me :) :)

*Whew*! I'm glad we got there in the end :smile:

You're welcome. Good luck in your studies.
 

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