Parallel RC transient response

[acw260]

Homework Statement

An accelerometer is connected via a length of coaxial cable to an amplifier. The arrangement is modeled by:

(a) a Norton generator in parallel with a capacitor (CP) representing the piezoelectric crystal within the accelerometer;

(b) a lumped capacitor (CC) representing the coaxial cable; and

(c) a load (RL) representing the amplifier.

See attached for circuit diagram.

The current generated by the accelerometer (iP) is proportional to the rate of displacement of the piezoelectric crystal. Hence iP = K dx/dt. In laplace form iP(s) = KsΔx(s).

(1) Derive an expression for the laplace transfer function T(s) = ΔvL(s) / ΔiP(s).(2) Express ΔvL as a function of time (i.e. the transient response of the voltage ΔvL ) when iP is subject to a step change.(3) Using the values given in TABLE A, estimate the time taken for the voltage vL to reach its steady state value if the current iP is subject to a step change of 2 nA.

CP=1400 pF
CC=250 pF
RL=5 MΩ
TABLE A

Homework Equations

The Attempt at a Solution

Q(1).

I believe (based on internet research) that after redrawing the circuit with an equivalent single parallel capacitor, C, that the answer is

ΔvL(s) / ΔiP(s) = R / 1+sRC

However, I’m struggling with the derivation. In time-domain vL = iPR(1-e^-t/CR). But when I inverse laplace transform R / 1+sRC I get 1/C(1-e^-t/CR). I don’t understand how I can replace R with 1/C in the time-domain. Can anyone help?Q(2).

vL(t) = iPR(1-e^-t/CR).Q(3).

C = CP+CC = 1400*10^-12 + 250*10^-12 = 1.65*10^-9

Time for vL to reach steady state = 5CR = 5*5*10^6*1.65*10^-9 = 41.25*10^-3 s

However, if this is correct the magnitude of the step current is irrelevant to the steady state time for vL. Since the problem specifies a step current of 2nA I’m wary of dismissing this variable as irrelevant. Am I missing something?

Any help with the above would be greatly appreciated.

 

[Delta2]

I am abit confused with your symbolism . Δ is for the laplace transform or for the laplacian operator? And where is TableA

 

[acw260]

I've copied the delta symbol exactly as it appears in the text of the question I've been set. I've taken it to simply mean 'change in' (voltage or current in this scenario).

The three values for CC, CP and RL listed immediately above the subtitle 'TABLE A' are the values from TABLE A as they appear once the formatting of this website has been applied (ie the grid lines from the table disappear).

 

[Delta2]

Sorry i am not sure what is going on here, maybe ( i say maybe cause i am not sure at all) the mistake you do is that in equation vL(t) = iPR(1-e^-t/CR you treat P as constant while P appears to be a step function of time P(t).

 

[acw260]

Apologies, the formatting here has confused things by changing subscripts to capitals.

'vL' is voltage across the load (the L should be subscript)
'iP' is the circuit current from piezoelectric crystal / norton generator (the P should be subscript).

Neither 'L' nor 'P' are variables in the equations, they are merely subscripts to identify different voltages and currents respectively.

So, iPR means circuit current x load resistance, ΔvL(s) means change in voltage across the load (s-domain) and ΔiP(s) means change in circuit current (s-domain)

Does that make more sense?

 

[Delta2]

yes thank you that makes things a lot more clear but still i admit i do not have enough experience working with laplace transform (I usually work in the time domain for circuits and differential equations) so i am afraid i can't be much of a help here.

 

[gneill]

First find the transfer function for $\frac{v_L}{i_P}$. Begin by replacing the components with their Laplace domain equivalents (you can combine the capacitors into a single component first). Then consider how one differentiates in the Laplace domain.

 

[donpacino]

Q(1).

I believe (based on internet research) that after redrawing the circuit with an equivalent single parallel capacitor, C, that the answer is

ΔvL(s) / ΔiP(s) = R / 1+sRC

However, I’m struggling with the derivation. In time-domain vL = iPR(1-e^-t/CR). But when I inverse laplace transform R / 1+sRC I get 1/C(1-e^-t/CR). I don’t understand how I can replace R with 1/C in the time-domain. Can anyone help?

So you have a resistor and a capacitor in parallel with a current source. The derivation can be done by writing a KCL at the positive node. solve that with algebra and you will get the correct answer.

Q(2).

vL(t) = iPR(1-e^-t/CR).

incorrect, but close. Look it up in a laplace table such as this. you also forgot u(t)
http://faculty.kfupm.edu.sa/ee/muqaibel/Courses/EE207 Signals and Systems/notes/Ch5 Laplace Transform and Applications/Table 5_3 Single_sided Laplace Transforms.jpg.

In the future you can check your work by looking at the initial conditions. In your equation, the initial condition is at 1 with a iPR of 1. That is not correct for this circuit.

 

[donpacino]

However, if this is correct the magnitude of the step current is irrelevant to the steady state time for vL. Since the problem specifies a step current of 2nA I’m wary of dismissing this variable as irrelevant. Am I missing something?

that statement is correct. the step magntiude of iL will not effect the steady state time

 

[acw260]

Ok, derivation of the answer to Q(1) as follows:

Vin = Ip * Req

where
Vin is the input voltage;
Ip is the source current (from the piezoelectric crystal);
Req is the equivalent resistance of the parallel capacitor and load

In the s-domain:

Vin(s) = Ip(s) * Req(s)

therefore

Vin(s) / Ip(s) = Req(s) = (Rc*Rl)/(Rc+Rl) = [(1/sC)*Rl] / [(1/sC)+Rl] = Rl/sC / 1/sC +Rl = R / 1+sCRl

where
Rc is the resistance of the capacitor = 1/sC in the s-domain
Rl is the resistance of the load.

Thanks for your help with this one.

 

[acw260]

For the answer to Q(2) I've now found two independent texts giving the solution as V=IR(1-e^(-t/CR) where the initially open circuit with a discharged capacitor is subject to a step change in circuit current.
(I've removed the confusing subscripts, but V is still the voltage across the load and I the circuit current)

The derivation is as follows:

Applying KCL to the positive node:
Ip = Ic + Ir
where Ip is the circuit current, Ic is the current through the capacitor and Ir is current through the load

therefore

Ip(t) = V(t)/R + C*(dV(t)/dt)
Laplace transform of the above:
Ip(s) / s = V(s)/R + sCV(s)
therefore
V(s) = Ip(s) / s(sC+1/R) = Ip(s)*R* [ (1/s) - 1/(s+1/CR) ]
Inverse laplace transform of last form of the equation above:
V(t) = Ip(t)*R*(1-e^(-t/CR))

Does my answer to Q(2) now look acceptable in light of the above explanation?

 

[donpacino]

ahhh my apologies. You are correct. I forgot to add 1/s for the step function. I guess that's what happens why you try to help people solve problems without thinking!

 

[acw260]

Thanks for taking the time to look at this for me guys. This set of questions had been frying my brain for over a week. Your nudges in the right direction are very much appreciated.

 

[Electest]

I came across this thread this morning and it has confirmed my thinking for both Q1 and Q2, however I'm not 100% about Q3

I fully understand that for steady state, the time constant (t)=R*C times a multiple of 4-5 in seconds, but this is always shown for a capacitor and resistor in series, not parallel as this example is shown. I'm also dubious about the step current figure shown, surely this has a purpose.

Any pointers appreciated

Thanks

 

[gneill]

I came across this thread this morning and it has confirmed my thinking for both Q1 and Q2, however I'm not 100% about Q3

I fully understand that for steady state, the time constant (t)=R*C times a multiple of 4-5 in seconds, but this is always shown for a capacitor and resistor in series, not parallel as this example is shown. I'm also dubious about the step current figure shown, surely this has a purpose.

Any pointers appreciated

Thanks

The time constant applies to both the series and parallel arrangements. To see this, consider that for a parallel arrangement of components you are free to change the order of the component on the page so long as the same circuit topology is maintained. So if you were to exchange the positions of the R and C the circuit and its behavior is unchanged. Now you might convert the current source and the resistor to its Thevenin equivalent...

...which should look familiar.

I'm not sure what you mean about "the step current figure shown". Can you elaborate your concern?

 

[Electest]

Hi Gneill

Thanks for the reply. I questioned this as every textbook I read always stated specifically R and C in series. I was initially using my textbooks by John Bird to understand this concept.

The question states in relation to the diagram at the beginning of this thread

Using the values given in TABLE A (shown below), estimate the time taken for the voltage vL to reach its steady state value if the current iP is subject to a step change of 2 nA.

CP=1400 pF
CC=250 pF
RL=5 MΩ

Now I imagine that the 5 nA were provided for a reason?

Thanks

 

[gneill]

Using the values given in TABLE A (shown below), estimate the time taken for the voltage vL to reach its steady state value if the current iP is subject to a step change of 2 nA.

CP=1400 pF
CC=250 pF
RL=5 MΩ

Now I imagine that the 5 nA were provided for a reason?

Besides giving the reader a practical feel for the kind of current magnitude produced by a piezoelectric transducer, I suspect that it's just a distraction (superfluous information). The circuit time constant shouldn't depend upon the magnitude of its input.

 

[Electest]

Thanks Gneill, I appreciate your input.

 

[Jerremy_S]

Hi,

Would you be able to help please.

The question is 1) Draw the Laplace form of the input portion of the circuit.

The diagram shows a Norton current source (ip) in parallel witch a capacitor (Cp) ( attachment in post #1).

Is it the same as a Thevenin equivalent circuit with Cp in series with voltage source? Is the capacitor taking place of a resistor?

If so, then how do I draw the Laplace form of the input?

Please push me in the right direction.

Many thanks

 

[gneill]

Hi,

Would you be able to help please.

The question is 1) Draw the Laplace form of the input portion of the circuit.

The diagram shows a Norton current source (ip) in parallel witch a capacitor (Cp) ( attachment in post #1).

Is it the same as a Thevenin equivalent circuit with Cp in series with voltage source? Is the capacitor taking place of a resistor?

If so, then how do I draw the Laplace form of the input?

Please push me in the right direction.

Many thanks

Hi Jerremy.

This is an old thread and you apparently have a new question. Please start a new thread of your own and pose your question there, being sure to fill out the homework formatting template.

 

[MattSiemens]

For the answer to Q(2) I've now found two independent texts giving the solution as V=IR(1-e^(-t/CR) where the initially open circuit with a discharged capacitor is subject to a step change in circuit current.
(I've removed the confusing subscripts, but V is still the voltage across the load and I the circuit current)

The derivation is as follows:

Applying KCL to the positive node:
Ip = Ic + Ir
where Ip is the circuit current, Ic is the current through the capacitor and Ir is current through the load

therefore

Ip(t) = V(t)/R + C*(dV(t)/dt)
Laplace transform of the above:
Ip(s) / s = V(s)/R + sCV(s)
therefore
V(s) = Ip(s) / s(sC+1/R) = Ip(s)*R* [ (1/s) - 1/(s+1/CR) ]
Inverse laplace transform of last form of the equation above:
V(t) = Ip(t)*R*(1-e^(-t/CR))

Does my answer to Q(2) now look acceptable in light of the above explanation?

How have you accounted for the 2 parallel capacitors in this equation?

 

[MattSiemens]

How have you accounted for the 2 parallel capacitors in this equation?

Think I've answered my own question is this just the formula for 2 capacitors in parallel; C1 + C2?

 

[gneill]

Think I've answered my own question is this just the formula for 2 capacitors in parallel; C1 + C2?

Yes.

 

[MattSiemens]

For the answer to Q(2) I've now found two independent texts giving the solution as V=IR(1-e^(-t/CR) where the initially open circuit with a discharged capacitor is subject to a step change in circuit current.
(I've removed the confusing subscripts, but V is still the voltage across the load and I the circuit current)

The derivation is as follows:

Applying KCL to the positive node:
Ip = Ic + Ir
where Ip is the circuit current, Ic is the current through the capacitor and Ir is current through the load

therefore

Ip(t) = V(t)/R + C*(dV(t)/dt)
Laplace transform of the above:
Ip(s) / s = V(s)/R + sCV(s)
therefore
V(s) = Ip(s) / s(sC+1/R) = Ip(s)*R* [ (1/s) - 1/(s+1/CR) ]
Inverse laplace transform of last form of the equation above:
V(t) = Ip(t)*R*(1-e^(-t/CR))

Does my answer to Q(2) now look acceptable in light of the above explanation?

I'm struggling to see how u have calculated 'Ip(s)*R* [ (1/s) - 1/(s+1/CR) ]' from the step prior to this?

I have tried all sorts multiplying out the denominator, rearranging the formula..

 

[gneill]

Hint: Partial Fractions.

 

[MattSiemens]

Hint: Partial Fractions.

ok thanks,

Am I on the right track with ip(s)/s(sc + /r) = A/s + Bs+C/(sc + 1/r)

 

[gneill]

It'll be more straightforward to just work with the necessary portion that you want to use partial fraction on.

You start with: V(s) = Ip(s) / s(sC+1/R). Multiply though the top and bottom of the RHS by R to get:

$Vp(s) = Ip(s) R \frac{1}{s(1 + s R C)}$

Forget the Ip(s)R bit for now and work with the fraction part:

$\frac{1}{s(1 + s R C)} = ?$

 

[MattSiemens]

It'll be more straightforward to just work with the necessary portion that you want to use partial fraction on.

You start with: V(s) = Ip(s) / s(sC+1/R). Multiply though the top and bottom of the RHS by R to get:

$Vp(s) = Ip(s) R \frac{1}{s(1 + s R C)}$

Forget the Ip(s)R bit for now and work with the fraction part:

$\frac{1}{s(1 + s R C)} = ?$

1/s - 1/s+1/rc

 

[gneill]

1/s - 1/s+1/rc

I think you need more parentheses to make the order of operations clear, else it's most definitely incorrect

 

[MattSiemens]

I think you need more parentheses to make the order of operations clear, else it's most definitely incorrect

Opps..
1/s - 1/(s+(1/rc))

Thanks a lot! This was so frustrating as I new the answer was of a first order differential just couldn't get to it!

 

[gneill]

Much better.

 

[Triopas]

Hello,

I am struggling to see how this was achieved...

Ip(s) / s = V(s)/R + sCV(s)
therefore
V(s) = Ip(s) / s(sC+1/R)

Could someone point me in the right direction please?

Thanks