Calculate the capacitor capacity

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  • #1
oph
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Homework Statement



Consider a simple defibrillator consisting of a capacitor which discharges across the two associated with the patient's chest electrodes in a period of 150 ms to about 5% of the fully charged voltage of the capacitor. The resistance of the chest between the electrodes amounts to about 100 Ω and the energy required for defibrillation be 200 J.

Since the voltage V0 of the battery is lower than the capacitor voltage required, it must be upconverted. One potential avenue a so-called step-up converter.

The switch S opens and closes periodically, where he closed a share g of the period and a fraction 1 - g is open. The period should be very small compared to the time constant of the capacitor-resistor system. The quantity g is called duty cycle.

The high-impedance resistor R represents the resistive behavior of the capacitor. All components must be assumed to be ideal.

Task (1):Estimate the capacitance of the capacitor and what voltage it needs to be charged for the operation at least.

Task (2):Derive is an expression for the maximum adjusting capacitor voltage after some time, depending on the sizes which occur.


2. The attempt at a solution

Task (1):
I don`t know how I should estimate the capacitance of the capacitor.

I can determine the voltage with this equation:
V=√(2Eel)/C

Task(2):
Value of the output voltage is greater than the input voltage

I can be derived, the output voltage across the inductor electricity:
ΔIL = (1/L) Ue t1 = (1/L)(Ua-Ue)(T-t1)

Vout=Vin ·T/(T-t1)

Is this right?
It would be very nice if somebody could help me.
 

Answers and Replies

  • #2
lewando
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Your post is a little incoherent, but focusing on Task 1, consider using these two equations:

Eq-1: Energy stored in the capacitor, UC = 0.5CV2

and

Eq-2: Capacitor Voltage, discharging in a RC circuit, VC(t) = V0e-t/(RC)

These 2 equations will allow you to determine C and V0.


Task 2 is not clear at all. Some questions for you:
--What do you mean by "maximum adjusting capacitor voltage after some time" and "the sizes which occur"?
--Is there anything you are not posting? Like the charging circuit?
--Where is "L" coming from?
--If "all components must be assumed to be ideal" then why do you care that "the high-impedance resistor R represents the resistive behavior of the capacitor"--a property of real capacitors.
 
  • #3
NascentOxygen
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Hi oph. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Is there supposed to be a pic attached to your post?
 
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  • #4
oph
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To your questions:
(1) What do you mean by "maximum adjusting capacitor voltage after some time" and "the sizes which occur"?

I mean V(out).

(2)Where is "L" coming from?

L=inductance of the coil

Could you please help me?
 

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  • #5
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Could you please help me?
I think his post is helpful.

For task 2, what can you say about the time where the switch is closed? What happens in the coil?
In equilibrium and after a full cycle, the system will return to its initial state. Together with the answer to my previous question, what can you say about the coil when the switch is open?
What does that mean for the whole system (with capacitor and resistor)?
 
  • #6
oph
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What can you say about the time where the switch is closed? What happens in the coil?

During the the switch is closed, the voltage Vin decreases to the inductance L and the current IL increases linearly.

In equilibrium and after a full cycle, the system will return to its initial state. Together with the answer to my previous question, what can you say about the coil when the switch is open?

Turns off the switch, the current IL flows through the diode on and charges the output capacitor.

What does that mean for the whole system (with capacitor and resistor)?

Energy consideration, describe:
During the activation energy is loaded into the inductor.
This is transferred to the output capacitor during the blocking phase.
 
  • #7
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During the the switch is closed, the voltage Vin decreases to the inductance L and the current IL increases linearly.
The voltage where decreases?
Current increases linearly, right.

Turns off the switch, the current IL flows through the diode on and charges the output capacitor.
Can the current be constant during that period?

Energy consideration, describe:
During the activation energy is loaded into the inductor.
This is transferred to the output capacitor during the blocking phase.
Well, in equilibrium (at the maximal capacitor voltage), the energy stored in the capacitor does not change any more. The energy gets dissipated somewhere else (where?).
 
  • #8
rude man
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During the the switch is closed, the voltage Vin decreases to the inductance L and the current IL increases linearly.

Turns off the switch, the current IL flows through the diode on and charges the output capacitor.

Energy consideration, describe:
During the activation energy is loaded into the inductor.
This is transferred to the output capacitor during the blocking phase.

That's about it.

When there is no load R applied, VC charges to V0 as a function of g.
When the load R is applied, C discharges per V0exp(-t/RC).
So you have 2 unknowns: C and V0, and you have the discharge equation and the energy equation during discharge.

Hint: energy dissipated during discharge ~ initial energy in C since at 5% practically all the energy in C is dissipated (energy in capacitor goes as V2).
 
  • #9
oph
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When there is no load R applied, VC charges to V0 as a function of g.

equation1: ???


T
When the load R is applied, C discharges per V0exp(-t/RC).

equation2: Vc=Vo*e^(-t/R*C)

???
 
  • #10
rude man
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equation1: ???

You should derive this yourself. But today I'll be a nice guy and give you what I calculated:
Vc = V0/(1 - g), g < 1

I threw out my notes since this was posted some time ago so I may not want to re-derive it.
equation2: Vc=Vo*e^(-t/R*C)

???

Why the "???" ? It's just the equation for discharging a capacitor C with initial voltage V0 by shunting it with a resistor R.
 
  • #11
oph
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1. Vc=Vo/(1 - g)
2. Vc=Vo*e^(-t/R*C)
3. Vc=√(2Eel)/C


Could you please help me? What have I to do first?
Vc=Vc*(1-g) * e^(-t/(R*(2Eel/Vc^2)) Is this right?
 
  • #12
rude man
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You need to solve two equations simultaneously, reflecting the required discharge energy and duration.
 
  • #13
oph
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1. = 3.
Vo/(1 - g)=√(2Eel)/C
Vo^2 * C= 2Eel * (1-g)^2
Vo= √(2Eel * (1-g)^2)/C

and than into the 2.?
 
  • #14
rude man
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1. = 3.
Vo/(1 - g)=√(2Eel)/C
Vo^2 * C= 2Eel * (1-g)^2
Vo= √(2Eel * (1-g)^2)/C

and than into the 2.?

Look at post #2. Those are the two equations you need to solve simultaneously. Except correct equation 2 by replacing V0 with V. V = Vc(0) is the max. capacitor voltage after charge-up & before discharge. V0 is the battery voltage. Solve for Vc(150 ms.).

I can't make any sense out of the second part of the question.
 
  • #15
oph
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You mean I need this two equations?
1.Vc = 0.5CV^2

and

2.VC(t) = V0e-t/(RC)

Do you know a website where it is explain how I can solve the two equations simultaneously?
 
  • #16
lewando
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Use the "Go Advanced" button to access superscripts and subscripts. They will make your communications much more readable and less prone to error and misunderstandings Your text will look a little weird with all the additional style tags, so use the "Preview Post" button to see if what you did is what you want.

VC(t) = V0e-t/(RC), which is unclear, becomes VC(t) = V0e-t/(RC) which is much more clear.

Per Mr. Man's advice, use VC(t) = Ve-t/(RC), where V = VC(0).

Regarding the solution of simultaneous equations-- locate your algebra book or search for "simultaneous equations" or "system of equations" on the internet.
 
  • #17
oph
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Do you mean this:

1. Vc(t) = Vo*e^(-t/(RC)) by V=Vc(0)
2. Vc = 0.5C*V^2 -> C=2Vc/Vo^2

in 1.
Vc(t) = Vo*e^(-t/(R*(2Vc/Vo^2)))

4. Vc=Vo/(1 - g) -> Vo=Vc*(1-g)

in 3.
Vc(t) = Vo*e-t/(R*(2Vc/(Vc*(1-g)^2))

is this the right way?
 
  • #18
lewando
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There are 2 parts to this problem. The charging part and the discharging part (the small contribution of the charging circuit during discharge can be ignored for simplicity).

If you were to focus on the discharging part, then your job is to determine C (the capacitance of the capacitor used to deliver the shock) and V (the initial voltage across this fully-charged capacitor).

That can be done without any consideration of the charging circuit, so V0 (the battery voltage) and g (switch-on part of duty cycle) are not required for this job.


Do you mean this:

1. Vc(t) = Vo*e^(-t/(RC)) by V=Vc(0)

Why is "Vo" there? It should be exactly as stated in the previous post:
lewando said:
Per Mr. Man's advice, use VC(t) = Ve-t/(RC), where V = VC(0).



oph said:
2. Vc = 0.5C*V^2 -> C=2Vc/Vo^2
Not really even close. Your units should match.

I believe what you may have meant to say was:

2. Uc,fully charged = 0.5C*V2 -> C = 2Uc,fully charged/V2



oph said:
in 1.
Vc(t) = Vo*e^(-t/(R*(2Vc/Vo^2)))

4. Vc=Vo/(1 - g) -> Vo=Vc*(1-g)

in 3.
Vc(t) = Vo*e-t/(R*(2Vc/(Vc*(1-g)^2))

is this the right way?
Focus on the discharging job, get rid of V0 and g and try again.
 
  • #19
oph
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1. Vc(t) = Ve^-t/(RC), where V = VC(0)
2. Uc,fully charged = 0.5C*V2 -> C = 2Uc/V2

=> Vc(t) = Ve^-t/(R*(2Uc/V^2))

Is this right? How must I go on?
 
  • #20
lewando
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oph said:
1. Vc(t) = Ve^-t/(RC), where V = VC(0)
2. Uc,fully charged = 0.5C*V2 -> C = 2Uc/V2

=> Vc(t) = Ve^-t/(R*(2Uc/V^2))

Is this right? How must I go on?

Yes, that looks right. If your intention is to isolate V, then isolate V.
 
  • #21
oph
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how can I isolate V?
Is this useful?
how must I go on to solve the two tasks?



I want to say thank you for all your great help!!! :)
 
  • #22
lewando
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Isolate V algebraically. You should be evaluating this at t = 150ms. Can you form an equation with V0(150ms) and V using the information given in the OP?

Yes this is useful. Determining V allows you to determine C (task 1). Knowing V and C helps with the charging circuit as well (task 2).
 
  • #23
oph
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i´m not sure, how I can isolate V

Vc(t) = Ve^-t/(R*(2Uc/V^2))
ln(Vc(t))=V*-t/(R*(2Uc/V^2))
ln(Vc(t))*(2Uc/V^2)=V*(-t/R)
ln(Vc(t))*2Uc=V^3*(-t/R)
ln(Vc(t))*2Uc/(-t/R)=V^3

is this right? what have i done wrong?
 
  • #24
lewando
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i´m not sure, how I can isolate V

Vc(t) = Ve^-t/(R*(2Uc/V^2))
ln(Vc(t))=V*-t/(R*(2Uc/V^2))

ln(ab) = ln(a) + ln(b), so the right-hand side is not right... no sense going further.



Have you given any thought about the relationship between Vc(150ms) and V, considering from the origninal post:

oph said:
Consider a simple defibrillator consisting of a capacitor which discharges across the two associated with the patient's chest electrodes in a period of 150 ms to about 5% of the fully charged voltage of the capacitor.
?
 
  • #25
oph
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maybe:
ln(Vc(t))=ln(V)*(-t/(R*(2Uc/V^2)))

What do you mean with the relationship?
Where do i need it?
 
  • #26
lewando
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maybe:
ln(Vc(t))=ln(V)*(-t/(R*(2Uc/V^2)))

That is a correct statement, but will not help you simplify (sorry if you think I was encouraging you to take this path).

Before you take the ln of both sides, consider dividing both sides by V.

On the left-hand side, you will have Vc(150ms)/V.

What can you do with that?

Vc(150ms) and V can be "related" by a fairly simple equation. Hint: it will involve the "5%" piece of information. Re-read that part of the original question.
 
  • #27
oph
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Vc(150ms)/V = e^-t/(R*(2Uc/V^2))

Is this right?



Consider a simple defibrillator consisting of a capacitor which discharges across the two associated with the patient's chest electrodes in a period of 150 ms to about 5% of the fully charged voltage of the capacitor.


I don´t know how i should determine an equation.
maybe: V=t*Vc
 
  • #28
lewando
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If V = Vc(0), the maximum initial voltage, and Vc(150ms) = 5% of V then you could say that Vc(150ms) = 0.05V, or Vc(150ms)/V = 0.05, would you agree?
 
  • #29
oph
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yes, i would agree but how should i go on?

should i insert it in the equation?
0.05=e^-150ms/(100 Ω*(2Uc/V^2))

and then??


thank you for your great help!!!!
 
  • #30
lewando
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In post 23, you must have seen e() and thought about using ln() to isolate the exponent--only V was in the way, but now its not...
 
  • #31
oph
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ln (0.05) = ln(-150ms/(100 Ω*(2Uc/V^2)))
(2Uc/V^2)*ln (0.05) = ln(-150ms/100 Ω)
(2Uc/V^2) = ln(-150ms/100 Ω) / ln (0.05)
(2Uc/V^2) = (-150ms/100 Ω) / 0.05
(2Uc/V^2) = -3000ms/2000Ω

is this right?
and then?
 
  • #32
lewando
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I said in an earlier post that you should review algebra. I am now even more certain that you should. Your last post has way too many errors.

I will mark up your last post and let you try again. If you are seriously having trouble with algebra (as opposed to just not trying very hard), then it will continue to be a problem until you do something about it. Talk to your teacher, get a tutor, ask a classmate. Go to Khan...

Anyway, this is the last known good equation that I can see:

0.05=e^-150ms/(100 Ω*(2Uc/V^2))

Note: Uc (really Uc(0), the energy in the capacitor at t=0) is going to be 200J based on the "rude man approximation" offered, without objection, for the sake of simplicity, in post 8.

[1] ln (0.05) = ln(-150ms/(100 Ω*(2Uc/V^2))) <-- not right--when you take ln(e(whatever)) the ln and the e anihilate each other and you are left with whatever. Formally, ln(e(whatever)) becomes (whatever)ln(e), where ln(e) = 1

[2] (2Uc/V^2)*ln (0.05) = ln(-150ms/100 Ω) no sense in going further until you get [1] right

[3] (2Uc/V^2) = ln(-150ms/100 Ω) / ln (0.05)
[4] (2Uc/V^2) = (-150ms/100 Ω) / 0.05 please don't ever do this again (going from [3]->[4]): ln(a)/ln(b) a/b !!!
[5] (2Uc/V^2) = -3000ms/2000Ω -(150/100)/0.05 -3000/2000
 
Last edited:
  • #33
oph
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(1) ln(0.05)=-150ms/(100 Ω*(2Uc/V^2))
(2) ln(0.05) * (2Uc/V^2) = -150ms/100 Ω

are this two equations right?
 
  • #34
lewando
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Yes.
 
  • #35
oph
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(3) ln(0.05) * 2Uc = -150ms/100 Ω * V^2
(4) V^2 = ln(0.05) * 2Uc * 100Ω/(-150ms)
(5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms)

is this right?
 

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