Howdy,

If the circuit has been running with the switch 'off' for a long time and then suddenly the switch is closed, there is still energy in the inductor (1/2)*L*i^2 where i is the initial current level, so this basically turns into a circuit with an inductor in parallel with a capacitor and in parallel with a resistor, with initial energy in the inductor only. So it's a parallel RLC circuit with initial conditions iL=i(0) and vC=0.

One way to solve is you can place an initial current generator in parallel to the inductor and then call the inductor current zero. The initial current generator has the value of the initial current just before the switch is turned on.
So it turns into a regular parallel RLC circuit driven with a current source at t=0.

Now to get exactly 0.12 volts DC across the capacitor constantly is impossible because eventually all the energy will be eaten up by the resistance. To force this condition you would need some additional external energy input such as another voltage source or instead of closing the switch completely use a switch with some intrinsic resistance instead. That will keep some energy pouring into the circuit, and then solving the DC solution will tell you what resistance is required in the switch to keep the cap voltage at 0.12v or whatever you need.

If this was to be 0.12v peak AC then you would have to get rid of all the resistance and solve for the steady state voltage across the cap but that would mean switching out ALL of the resistance so as to form a natural oscillator which would mostly only be practical in theory or maybe in a good laboratory.

 

By not replying the switch.
And by not changing any values.



All you need to do is move the ground to the negative terminal of the battery: switch is opened at t=0. That gets you zero volts on Vc at t=0, Vc goes positive at t>0 and this forced response settles at 0.12V based on the R2R3 voltage divider. You need to change values though.

 

All you need to do is move the ground to the negative terminal of the battery.

And then what will happen?

Thanks.

 

By not replying the switch.
And by not changing any values.



All you need to do is move the ground to the negative terminal of the battery: switch is opened at t=0. That gets you zero volts on Vc at t=0, Vc goes positive at t>0 and this forced response settles at 0.12V based on the R2R3 voltage divider. You need to change values though.

Hi,

Oh you mean we cant change or add any values? Then it is clearly impossible if it is to be a DC voltage because the inductor is a short for DC so it will always short out the cap regardless what we do, and i assume we are not allowed an infinite current either.

Once the circuit reaches steady (DC) state the cap is shorted so we need at least some series resistance in the inductor, but also we need a constant energy source which would happen if the switch also had some ESR. With with inductor ESR and switch ESR this would work, but without being able to change anything it will never work. That is, it will not work to get any constant DC voltage across the cap.

 

Even if you can change all the values and rearrange the circuit, IMO you can not achieve either of those plots. There are not enough design degrees of freedom. However, with one more resistor, and the other ideal components listed, you can get those plot shapes (if that is the actual objective) - with a settled voltage of 0.12V . Whether both are realizable with real world parts is another matter. Is the time axis of the other plot supposed to be in minutes!?

 

Even if you can change all the values and rearrange the circuit, IMO you can not achieve either of those plots. There are not enough design degrees of freedom. However, with one more resistor, and the other ideal components listed, you can get those plot shapes (if that is the actual objective) - with a settled voltage of 0.12V . Whether both are realizable with real world parts is another matter. Is the time axis of the other plot supposed to be in minutes!?

Yes in minutes, please support your decision by graphs.
Thanks

 

Yes in minutes, please support your decision by graphs.
Thanks

Are three resistors allowed or not? You still have not defined all the project goals and all the constraints in a clear, numerical and engineering fashion. If you ca nott define the problem you have no hope of solving it.

 

Are three resistors allowed or not? You still have not defined all the project goals and all the constraints in a clear, numerical and engineering fashion. If you ca nott define the problem you have no hope of solving it.

The purpose of my project is to make the same desired output by using any thing.

 

This circuit can produce a critically damped response that closely matches the plot with the ~7 minutes settling time and 0.49V peak, but requires a humongous capacitor and inductor. You can build it in a simulator but not in the lab. I would be thinking in terms of opamp differentiators/integrators.


The ground is at the bottom.

There will be a tiny hitch at switch on. If that is a concern a snubber capacitor across RL will smooth out the small voltage jump.

Once you have the topology then you need to find the equation which describes the curve which is of this form for a critically damped 2nd order ircuit

A1e^st + A2te^st + Vf = 0

The circuit is simple enough you can use text book RLC series equations to find the component values to achieve the values in the above equation.

 

This circuit can produce a critically damped response that closely matches the plot with the 7 minutes settling time and 0.49V peak, but requires a humongous capacitor and inductor.

View attachment 115212

Thanks.
And the values of components?

 

Thanks.
And the values of components?

We are here to help, not do your school work for you.

 

We are here to help, not do your school work for you.

Thanks a lot.