Current Through an Inductor at time t

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Homework Statement


LTspice schematic: pulse source V1 PULSE(-1 1 1n) drives series R1 1k into inductor L1 15µ, with a switch S1 controlled by V2 1V, and directives .tran 100n, .model sname sw, .ic I(L1)=0.


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

Simulator plot of current titled I(L) over 29.999998 ns to 30.000002 ns, showing a nearly flat horizontal trace at about 854.94 µA on a y-axis from 854.64 µA to 855.24 µA.


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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  • LTspice schematic: pulse source V1 PULSE(-1 1 1n) drives series R1 1k into inductor L1 15µ, with a switch S1 controlled by V2 1V, and directives .tran 100n, .model sname sw, .ic I(L1)=0.
    upload_2018-11-3_21-9-58.png
    19.6 KB · Views: 984
  • Simulator plot of current titled I(L) over 29.999998 ns to 30.000002 ns, showing a nearly flat horizontal trace at about 854.94 µA on a y-axis from 854.64 µA to 855.24 µA.
    upload_2018-11-3_21-20-40.png
    3.5 KB · Views: 553
Last edited:
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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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...
 
I think we'll need to see some more details of your calculations. Here's what I get using Mathcad:

Mathcad worksheet showing tau := 15 microhenry over 1 kilohm, tau = 15 ns, delta t := 1 ns labelled start time of pulse, I(t) := 1 V over 1 kilohm times [1 minus exp of negative (t minus delta t) over tau], and I(30 ns) = 855.335 microamps.
 
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  • Mathcad worksheet showing tau := 15 microhenry over 1 kilohm, tau = 15 ns, delta t := 1 ns labelled start time of pulse, I(t) := 1 V over 1 kilohm times [1 minus exp of negative (t minus delta t) over tau], and I(30 ns) = 855.335 microamps.
    upload_2018-11-4_14-52-50.png
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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.