Please help me with this simple circuit, which is creating lots of confusion due to different results from theoretical calculation and SPICE simulation.

Ques: In this circuit switch is closed for time t = 1sec, and then switch is opened and we have to find current through iductor and diode after time t = 2sec when switch was opened.

I solved it using V=L*(ΔI/Δt), and when simulated the circuit in LTSPICE the results were completely different. my calculation gave value of current after time t=2sec from opening the switch as I=7A through inductor whereas in LTSPICE simulation value came out aound 4A.

it's L=1 Henry only, sorry it was typo before

Circuit description: Ideal inductance (L=1 Henry), ideal diode and resistance of R=1ohm.


simulate this circuit – Schematic created using CircuitLab

THIS IS THE SIMULATION RESULT OF LTSPICE enter image description here

enter image description here

  • 1
    \$\begingroup\$ Is L1 3H or 1H? Your description contradicts the schematic \$\endgroup\$
    – Frog
    May 26, 2021 at 7:36
  • 1
    \$\begingroup\$ Beware SPICE inductors, check there's no default R included in the model, but hidden from you. \$\endgroup\$
    – Neil_UK
    May 26, 2021 at 7:54
  • 2
    \$\begingroup\$ Would you mind showing us those calculations that you say you did? It's much easier to point potential errors. \$\endgroup\$ May 26, 2021 at 11:42
  • 1
    \$\begingroup\$ The inductor in LTSpice is not ideal unless you specifically make Rser=0 (default is 1 mohm). \$\endgroup\$
    – qrk
    May 26, 2021 at 17:11
  • 1
    \$\begingroup\$ @Vikas no need to apologise, just need to be clear what we’re dealing with \$\endgroup\$
    – Frog
    May 27, 2021 at 11:00

2 Answers 2


It's a screwy problem. It only makes sense if V1 is a 3 volt step with the step occurring at t=0. If that's the case, with SW1 closing at t=0, then the current through the 3H inductor (if it is 3H) will climb at 1 ampere/sec until SW1 opens at t=1. Then the voltage across L1 will be 2 volts with the diode and V2 clamping the bottom (as it's drawn) of L1 to 1 volt. The current through it will continue climbing at the new rate of 2/3 amperes/sec. R1 doesn't enter into it, at all. When SW1 opens, it sees a 1 volt step and the current through it goes to 1 amp, but its presence has no significant effect on the rest of the circuit. L1's current at t=1,2,3 seconds is 1,1.67,2.33 amps. If L1 is 1H, its current at 1,2,3 seconds is 3,5,7 amps.

In any case, there must be differences between what's here and what was given to Spice and what OP was analyzing.

  • \$\begingroup\$ L is 1 henry only, sorry for typo in question. but wouldn't the current through inductor after 2 sec will be 7A? because i applied the same logic of not considering the effect of resistor on the circuit. then with L=1H after switch release, ΔI/Δt = 2A/s and after 2 sec current will be 4A and adding previous current i.e. 3A gives 7A?? @john \$\endgroup\$
    – Vikas
    May 26, 2021 at 15:47
  • \$\begingroup\$ @Vikas: yes. 2 seconds after the switch opens is 3 seconds after the start, what I called t=3 seconds. \$\endgroup\$
    – john
    May 26, 2021 at 16:04

Considering L= 3H.

When switch is closed the inductor is charging with 1A/s rate (V/L = 3V/3H = 1A/s). So at 1sec the inductor is charged with 1A.

When switch is released all inductor current (1A) starts to flow to R what couses 1V drop on it. Thats exact voltage what your diode is opening so at this time the diode current rise from zero. This voltage will be presented all rest of time so the R will take 1A from inductor current. The inductor current rate is 2V/3H = 0.67A/s now so after 2sec it will be incresed by 1.34A.

Adding together with current at t=1s the inductor current at 2s after releasing switch is 2.34A. The diode current is 1A less due to constant R current, so 1.34A.

  • \$\begingroup\$ sorry for typo in question, value of inductor is 1 henry. But why current won't flow to diode first, when switch will be released? while solving with L=1H ,(V/L= 3A/s) and when switch will be released 3A current will be flowing through inductor and making diode forward biased and 1 volt will appear across the resistor thus making 1A through resistor. and adding current at t=1sec with the current through inductor after 2 sec is coming 7A. whereas spice is showing 4 only. i guess it is not adding the previous current? @Michal Podmanický \$\endgroup\$
    – Vikas
    May 26, 2021 at 16:36
  • \$\begingroup\$ @Vikas The results you added seems good to me. 3A at t=1s , 5A at t=2s , 7A at t=3s. \$\endgroup\$
    – user208862
    May 26, 2021 at 17:34
  • \$\begingroup\$ According you comment, the current thru diode will flow only if anode point (without diode connected) is higher than 1V whats the case when resistor current is higher than 1V/1ohm=1A. \$\endgroup\$
    – user208862
    May 26, 2021 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.