Current Through an Inductor at time t

In summary, the conversation discusses a problem with calculating current and voltage through an inductor. The equations for voltage and current are provided, along with the values for initial and final current and voltage. One hypothesis is proposed, and the use of LTspice is suggested to adjust the simulation. After applying the new simulation command, the results are still different from the calculated values. It is later discovered that the mistake was made in the calculation of Tau, which was initially thought to be 10ns but is actually 15ns.
  • #1
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Homework Statement


upload_2018-11-3_21-9-58.png


Homework Equations


[/B]
V = 1/L di/dt

I = Io x (1 - e^(-t/L/R)

Tau = L/R

The Attempt at a Solution


[/B]
Ii = 0mA ; If = 1mA

Vi = 1V ; Vf = 0V

Tau = 15uH / 1k Ohms = 10ns

If my above is correct, then voltage and current equations through inductor should be as follows:

V = 15uH di/dt

I = 1mA (1-e^(-t/10ns)


The problem is when I test current I don't get the graphical answer.
For instance, I tested at 30ns:

I = 1mA (1-e^(-30ns/10ns)
I= .001A (1-e^-3)
i= .00095A or 950uA

...but on the graph the answer is just under 0.855uA

upload_2018-11-3_21-20-40.png


One hypothesis I had is that since the voltage doesn't pulse till 1ns, somehow t time or Tau is affected. But I have tried playing with the equation by adjusting t time and/or Tau by 1ns and still don't seem to be able to come to the right answer.
 

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  • #2
Lapse said:
One hypothesis I had is that since the voltage doesn't pulse till 1ns, somehow t time or Tau is affected. But I have tried playing with the equation by adjusting t time and/or Tau by 1ns and still don't seem to be able to come to the right answer.
That's certainly a good hypotheses. Clearly time needs to be shifted in some manner (not Tau, it is not affected by changing clocks). You should be able to subtract a nanosecond from time to get your equation to line up with the simulation, but there's another way, too. You can adjust LTspice to start saving data only after 1 ns. Go to the Simulate menu and choose Edit Simulation Command. Note the entry "Time to Start Saving Data". Try it out.
 
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  • #3
gneill said:
That's certainly a good hypotheses. Clearly time needs to be shifted in some manner (not Tau, it is not affected by changing clocks). You should be able to subtract a nanosecond from time to get your equation to line up with the simulation, but there's another way, too. You can adjust LTspice to start saving data only after 1 ns. Go to the Simulate menu and choose Edit Simulation Command. Note the entry "Time to Start Saving Data". Try it out.

Thanks for that tip on LTspice. After applying that new simulation command:
I = 860mV - 865mV @ t = 30ns.

As you may notice, this answer is still different from what I am getting.

When I do the calculation at t = 30ns - 1ns = 29ns, my answer comes out to 945mV.

So I am still going wrong somewhere...
 
  • #4
I think we'll need to see some more details of your calculations. Here's what I get using Mathcad:

upload_2018-11-4_14-52-50.png
 

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  • #5
gneill said:
I think we'll need to see some more details of your calculations. Here's what I get using Mathcad:

Oh... my... goodness. I said Tau = 15uH / 1k = 10ns ? It is 15ns!

Another lesson in simple mistakes can lead to greater consequences.

Thanks for your help gneill.
 
  • #6
Happy to help!
 
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FAQ: Current Through an Inductor at time t

What is an inductor?

An inductor is a passive electronic component that stores energy in the form of a magnetic field. It consists of a coil of wire wrapped around a core material, which can be air, iron, or ferrite.

How does current flow through an inductor?

When a voltage is applied across an inductor, a magnetic field is created around the coil. This field resists changes in current and causes the current to increase gradually over time.

What is the significance of current through an inductor at a specific time t?

The current through an inductor at a specific time t is important because it helps to determine the energy stored in the inductor. This energy can be used to power other electronic components or devices.

How is the current through an inductor affected by the inductance and resistance of the circuit?

The current through an inductor is directly proportional to the inductance and inversely proportional to the resistance in the circuit. A higher inductance will result in a slower increase in current, while a higher resistance will result in a smaller current.

Can the current through an inductor ever reach a steady state?

No, the current through an inductor will continue to increase as long as a constant voltage is applied. However, the rate of increase will decrease over time as the inductor reaches its maximum energy storage capacity.

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