Hi there.

I need to elevate the steady state in this circuit

the next figure shows the output of previous circuit:

How can I modify the previous circuit to be like this output:


Thanks all.

 

hi 1991,
Have you checked the DC resistance of inductor L2.?

E

EDIT:
Checking your circuit, could explain what you mean by 'elevate' and modify.
Also indicate the node you are measuring
I may have misunderstood.?

 

Where? Which node in the circuit is the "output"?

ak

 

Where? Which node in the circuit is the "output"?

ak

Hi AnaloKid
At capacitor C2

 

hi 1991,
Have you checked the DC resistance of inductor L2.?

E

EDIT:
Checking your circuit, could explain what you mean by 'elevate' and modify.
Also indicate the node you are measuring
I may have misunderstood.?

Hi ericgibbs.
I sorry for misunderstanding.
The output at node C2 here


And my question is: how can I make the output of this circuit the same as this figure:

in other words how can I make discharge of capacitor at constant value like 0.12v

Thank you.

 

in other words how can I to make discharge of capacitor constant at 0.12v

If L2 inductor has some internal DC resistance, say about 0.2 Ohms that would give around 0.12v

Try it.
E

 

If L2 inductor has some internal DC resistance, say about 0.2 Ohms that would give around 0.12v

Try it.
E

I will keep you posted.
Thanks.

 

If L2 inductor has some internal DC resistance, say about 0.2 Ohms that would give around 0.12v

Don't think so. With the switch closed and all transients died off, there is 5 A through R3 and no current through the R2 or L2.

ak

 

I agree with @AnalogKid .. There will not be and current through R2 L2 C2 when switch is closed because It will short out the circuit and current will pass through the lowest resistance path only which means through the switch loop and not to the R2 L2 C2 loop

 

I agree with @AnalogKid .. There will not be and current through R2 L2 C2 when switch is closed because It will short out the circuit and current will pass through the lowest resistance path only which means through the switch loop and not to the R2 L2 C2 loop

Hi Mr.Mozee.
How can I solve this?

 

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@mahboubeh1991 Not really, but let me share something with you.
Assuming the switch is opened, I think that the capacitor will never charge because of the inductor in parallel with it, in DC currents the inductor acts as a wire in the end so you are basically shorting out the capacitor from the very beginning, however there will be an instantaneous reversed polarity voltage across the inductor as soon as the DC current reaches it to this phenomenon is called Reactance where the inductor will fight current change through it (which initially was Zero amps) by dropping voltage in the opposite direction but this fight will not last long as the inductor will be fully saturated and will act as a wire with a very small internal resistance and this internal resistance will result in a voltage drop across the inductor, thus, the capacitor (Ohm's law V=IR). Assuming that the internal resistance of the inductor is zero(Ideal) and the cap. is in parallel with it, it will be basically shorted out.

This is as per my knowledge! .. wait for the experts to approve or disagree

 

Without knowing the resistance of the inductor, the initial conditions in post #1 are in question.

Also, 0.125 V across L2 steady state means a significant change of the overall circuit to provide this DC path. If L2 is a large power inductor with a DC resistance of, say, 0.01 ohm, that extra 12.5 A has to come from somewhere.

ak

 

hi 1991,
Sorry misread the Switch state.!
Ignore my previous post.

E

 

hi 1991,
Sorry misread the Switch state.!
Ignore my previous post.

E

So, how can we solve this?

 

hi 1991,
This is a simulation in LTSpice of the circuit you have posted.
It shows the state of Vout before and after the U2 switch is Closed.

As you can see it does not give a response as you have posted or required.

Is this Homework or a College assignment.

E

 

hi 1991,
This is a simulation in LTSpice of the circuit you have posted.
It shows the state of Vout before and after the U2 switch is Closed.

As you can see it does not give a response as you have posted or required.

Is this Homework or a College assignment.

E

Hi.
Thank you for your interest.
This is my final graduate project.
I was used Orcad 16.6 pspice.

 

in other words how can I to make discharge of capacitor constant at 0.12v

Thank you.

Replace the switch with a BJT transistor biased into saturation at t=0 gets you close, but the circuit is all wrong to get the Vc curve you say you want.

 

Replace the switch with a BJT transistor biased into saturation at t=0 gets you close.

Hi @DGElder
What do you mean saturation? 0.12v
And if I replace the switch by the transistor can I get the desired output?

 

Play around with that topology and see what you can get. Given the only quantitative spec you provided: 0.12= Vs*R2/(R1+R2)

You have not provided a full description of the project/problem, the final specification for all the circuit performance parameters of interest nor what degree of freedom you have in your design. So we just throw up random ideas with little to no direction.

 

View attachment 114793


Play around that topology and see what you can get.

Without replacing the switch.
And without changing any values.