Calculate the capacitor capacity

[oph]

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?

 

[lewando]

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: U_c (really U_c(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

 

[oph]

(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?

 

[lewando]

Yes.

 

[oph]

(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?

 

[lewando]

Use parentheses to clarify the order of operations and you should be okay. What do you get for a final answer?

 

[oph]

V = √798.86 = 28.26

is this right?
and then?

 

[lewando]

Does not look right. Show the steps you took to arrive at that answer.

 

[oph]

(5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms)

V = √ln(0.05) * 2 * 200J *(100Ω/(-150ms))
= √-2,99 * 400J * (-2/3)
= √ 798,86
= 28,26

what is wrong?

 

[lewando]

You are not accounting for the m factor in -150ms.

 

[oph]

(5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms)

-150ms = -0.150s

V = √ln(0.05) * 2 * 200J *(100Ω/(-0.150s))
= √-2,99 * 400J * (-666.67)
= √ 798861.94
= 893.79

 

[lewando]

Always include units with your result.

 

[oph]

V = 895.79 Volt

Is this right?

Task 1 was:
Estimate the capacitance of the capacitor which must have and what voltage it needs to be charged for the operation at least.

Now I have got the voltage but what is with the capacitance?

 

[lewando]

Looks right. Recall that you were solving a system of equations and had decided to solve for V first. See post 19. Now that you have V, proceed with solving for C.

 

[oph]

Vc(t) = Ve^-t/(RC)
Vc(150ms)/V = e^-t/(RC)
ln (0.05) = -t/(RC)
ln (0.05) * (RC) = -t

C = -t/(ln(0.05) * R)
= -0.150/ln(0.05) * 100 Ω = 5.007 * 10^-4 F

is this right?

 

[Delphi51]

I've been following along . . . interesting question! I agree fully with your calculation for C. Note that you did not need to solve two equations simultaneously to get C. An easy way to write it up would be to begin with your post #45 calc, then use the value of C found there in E = ½CV², to find V. I get the same 895 V answer you found the long way.

 

[oph]

Thank you for all your Great help!

Have anyone an Idea how i can solve task 2?

 

[Delphi51]

Most welcome, oph!
JUST an idea - task 2 is a bit beyond my present skill level. I dimly remember that the inductor will build up a magnetic field when the switch is closed, then when the switch opens very suddenly there will be a high voltage across the inductor as the magnetic field collapses. I can't remember the formula for the energy stored in the inductor. It seems to me the voltage on the inductor will be very high compared to the voltage on the capacitor so perhaps nearly all of the inductor energy will be transferred to the capacitor if the timing is right. I don't know if you have to consider the period for the LC circuit or not. I would start by calculating the energy of the inductor and how much voltage is added to the capacitor if it is all transferred. Maybe sketch a graph of voltage vs time for the capacitor showing the jumps as the switch is opened and the energy pumped in, and the exponential decay due to the RC circuit on the right side.

 

[oph]

i really don`t know how should solve this task but, maybe with this equation:

law of induction:
ΔIL = (1/L) * Vi * t1 = (1/L) * (Vi-Vo) * (T-t1)

Vi= input voltage
Vo= output voltage

Vi=Vo ·T/(T-t1)

can somebody say me if this is the right way?

 

[lewando]

I do don't understand what you are being asked to do for Task 2. I had asked for clarification in Post 2 and your response in Post 4, "V(out)" made it even less clear.

I you could, please restate, using different words in a different order, and with as much detail and clarity as you can, exactly what it is you are being asked to provide as a result for Task 2.

 

[oph]

task 2: Derive an expression for the maximum capacitor voltage in dependence on the occurring sizes.

the task has to do with an up-converter and I should determine an equation for the maximum capacitor voltage

this are the new details:

Since the voltage Vo of the battery is lower than the capacitor voltage required, it must be upconverted. One possibility is a so-called step-up converter, as outlined in the following figure.

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 drawn, high-impedance resistor R represents the resistive behavior of the capacitor. All components must be assumed to be ideal, in particular, that the diode is fully locked in the reverse direction and causes no voltage drop in the forward direction.

I hope you understand the task now. If not please ask me



(I have already upload a circuit plan at the beginning)

 

[Delphi51]

With periodic shots of energy/charge from the inductor and losses to the resistance, there will come a voltage level when during one period the loss will equal the gain. If you can find this voltage you will have your answer!

 

[oph]

but where should I start?

 

[lewando]

I'm still not sure what, exactly, you are trying to determine.

"Derive is an expression for the maximum adjusting capacitor voltage..."

What do you mean by adjusting capacitor or adjusting capacitor voltage

If you think this somehow means V_c(t) when the capacitor is charging, then please confirm.

"...after some time,..."

I am pretty sure this means "a time when the charger is at steady state", or "at the maximum charging voltage" (there will be some ripple voltage--take the peak, take the average--your call). t=∞ works, but so will much shorter values for t. So Task 2 could be find V_c(∞), then, except you say:

...depending on the sizes which occur".

The sizes of what? There are a few things that can change "sizes", V_o (the battery voltage), L, R<sub>c, g (the duty cycle), f (the frequency of your switch control voltage)

So maybe you are looking for a general expression: Vc(t=∞, Vo, L, Rc, g, f). If so, I think that will be a very difficult thing to do.

Your charging circuit is called a "boost"-type switching regulator. I can operate in two modes: "continuous" (when there is always some current going through the inductor) and discontinuous (when the current in the inductor is allowed to go to zero, periodically). Many posts ago, an expression for Vc(∞) versus Vo and duty cycle, g, was given. This applies to "continuous mode" operation.

You should familiarize yourself with this knowledge. The best link I can find that will help the details is:
http://en.wikipedia.org/wiki/Boost_converter

Study it, let me know what you want to do.</sub>

 

[oph]

Yes, you are right.
I should determine the maximum of Vc and i`m looking for a general expression.

I have really no idea where I should start, can you please give me a tip?

 

[lewando]

Let me refine my comment: A general equation based on all the variable parameters is outside my ability to envision. Depending on what the mode of operation is (continuous/discontinuous conduction), different equations apply.

Here are some tips:
(1) Constrain the problem to simplify the general equation (consider operation in continuous conduction mode only, for example. Assume a 50% duty cycle constraint). Discuss this approach with your instructor.

(2) Have you read the wiki-link in post 54? Do you have any questions about the article? What other research have you done? If possible, share your findings and also any questions you have about them.

 

[oph]

On an other site I have found some information but I don´t know if they are useful to solve the problem.

For the continuous and steady operation the law of induction applies
ΔI_L = \frac{1}{L}V_e ⋅ t₁ = \frac{1}{L} (V_a - V_e) ⋅ (T-t₁)

V_a = V_e \frac{T}{T-t₁}

The output voltage depends only on the duty cycle and the input voltage, it is independent of the load.

Do you have an idea how i can go on?

Maybe it is helpful to know what my next task is:
Determine how large g must be chosen for a capacitance of 100uF with an resistance of 100MΩ and to charge a 12.0 V battery to a voltage of 500 V when the inductance L of the coil is 5,0 mH.

Maybe you have dot an idea