Engineering Laplace help on a DC transient RC circuit ? PLEASE

[willow16v]

I HAVE A RC CIRCUIT WHERE R1=100Kohm and R2=1Mohm and C=2.1micro farads with a Vin =10V,


In order to calculate the voltage through the capacitor I deduce the following formula,

Vc(s)=V/S {(((R2×1/Cs)/(R2+1/Cs)))/(((R2×1/Cs)/(R2+1/Cs)+R1) )}

I then need to produce a laplace transform of exponential growth thus, I believe to be

Vc(s)=a/s(s+a)


However I have spent a week on this now and cannot get a soloution. I would be grateful of any help or any point to know where I am going wrong. I have pages and pages of transposition but they all lead to every answer but the one I require. Many thanks



Please see attachement

Thanks Again

 

[overt26]

It's not clear what the actual circuit comprises, but reading between the lines I think the Laplace form of the voltage Vc should be as given in the attachment.

 

[Defennder]

Please upload your question in a PDF file. Word documents can contain viruses.

 

[willow16v]

thankyou overt 26, did you have a look at the word document ?

Many thanks

Chris

 

[CEL]

First multiply the numerator and the denominator by R2 + 1/Cs, to eliminate this term. You will have still terms of the form 1/Cs in both the new numerator and denominator. Multiply them by Cs, to get an expression of the form you want.

 

[overt26]

thankyou overt 26, did you have a look at the word document ?

Many thanks

Chris

Yes I did. CEL has made the right suggestion for obtaining the solution.

You should end up with something similar to that which I posted in the pdf.

 

[willow16v]

thanks to you both, i will have another look at it and take into account your suggestions and hopefully get yet another different answer but the right one.

Thanks again

 

[willow16v]

thanks i have just put the values of capacitor and resistors into the formula with the relevant time and it is about .6 of a volt away from the recorded values. Would there be a specific reason for this due to circuit tolerances etc.

I can't tell you how gratefull i am to you both

 

[CEL]

thanks i have just put the values of capacitor and resistors into the formula with the relevant time and it is about .6 of a volt away from the recorded values. Would there be a specific reason for this due to circuit tolerances etc.

I can't tell you how gratefull i am to you both

The final value should be .9V. An error of .6V is excessive.

 

[willow16v]

the voltage / time curve was between 0 and 2 seconds, where at aorund 0.6 seconds the exponential growthy curve reached its max at 8.3 V. However like i say the actual math answer compared to the measured values are different ? could it be an error with the formula or with the tolerances of the equipment.

Many thanks

Chris

 

[CEL]

the voltage / time curve was between 0 and 2 seconds, where at aorund 0.6 seconds the exponential growthy curve reached its max at 8.3 V. However like i say the actual math answer compared to the measured values are different ? could it be an error with the formula or with the tolerances of the equipment.

Many thanks

Chris

My mistake. I inverted the values of the resistances. The final value should be around 9V. Since your components have probably a tolerance of 10%, an error of 0.6V is possible.

 

[willow16v]

inverted the resistances in the formula ? do you mean there is a mistake with the formula or you made a mistake by inverting them ?.

Many thanks

 

[CEL]

inverted the resistances in the formula ? do you mean there is a mistake with the formula or you made a mistake by inverting them ?.

Many thanks

I made a mistake by inverting them. The formula is correct. As I said, if all components had 0% error, the final voltage should be 10V x 1M/1.1M ~ 9.1V.
Since there is a tolerance of 10% in the values of the resistors, the final voltage can be from
10 x .9M/(.9M + 110k) ~8.2V to 10 x 1.1M/(1.1M+90k) ~ 9.2V.
So, the 8.3V you found is a possible, but unlikely value.

 

[willow16v]

thanks very much,
i have calculated the values and they are accurate with the graphical representation. However i have another question ?

if i take the 1Mohm resistor out of the circuit what effects would it have on the exponential growth curve.

The reason i ask is that without the resistor but using the same 0 to 2s time scale the results acheived with the Laplace function of V(CR/s(s+CR)) with the time domain of
1-e(-CRt) he result is a straight line ?

what are the reasons for this or have i miss calculated

thanks again

 

[CEL]

Without the 1M resistor, the voltage source will be in series with the 100k resistor and the capacitor.
A voltage source in series with a resistor is equivalent to a current source in prallel with the same resistor, so you have approximately the capacitor being charged by a current source. The result is a ramp.

 

[willow16v]

does the circuit not act as a voltage divider network ?
Why is the resulting graphical form a straight line?

Thanks again

 

[CEL]

does the circuit not act as a voltage divider network ?
Why is the resulting graphical form a straight line?

Thanks again

Without the 1M resistor, there is no voltage divider. Of course, the current source approximation is valid for small values of time. As time goes by (pun intended), you will have an exponential and the final value of the voltage in the capacitor will be the source voltage. For small values of t the exponential approaches a straight line.