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I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

In my book they stated the following sentence:

After that the switch is closed for a long time, I let go of the switch. The voltage at the node \$Y_1\$ will rise in a few miliseconds to about -90 volts.

Question: why is that true? Because the supply voltage is way way less than 90 volts. How can the voltage at the coil exceed the supply voltage soo much?

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  • \$\begingroup\$ Why do you think that this is not true? electronics.stackexchange.com/questions/288380/… \$\endgroup\$
    – G36
    May 18, 2019 at 14:40
  • \$\begingroup\$ Compare to inductor based boost converters. Also: What does an inductor do? Compare to capacitors which try to maintain a constant voltage across their terminals - what do inductors try to hold steady? \$\endgroup\$
    – JRE
    May 18, 2019 at 14:40
  • \$\begingroup\$ This is a property of inductors. When you disconnect them from their current source, the magnetic field in the coil collapses, releasing a surge of current in the opposite direction. It's pretty common when powering a diode with a transistor to connect a properly sized diode across an inductor, wired backwards from the normal direction of current flow. Then, when you disconnect the power, the diode drains off that reverse surge of current. That protects the control transistor from being damaged by the reverse voltage. \$\endgroup\$
    – Duncan C
    May 18, 2019 at 19:48

2 Answers 2

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Notice that the Current in L is 1A and as the switch opens the current is now shared with the capacitor so as they both swing own past 0A the voltage reached a maximum in the opposite polarity at a resonant frequency of \$f=\dfrac{1}{2\pi \sqrt{LC}}\$ = 223 Hz or a 4.5 ms cycle period. enter image description here

Notice the stored Energy in the inductor is also being dissipated in R1 @ 10W and the 1st peak VAR of the inductor = 50.573W and capacitor = 41.021W = 91 VAR

The ratio of total reactive power to resistor power ratio VAR/P=Q = 9.1 which is close to the voltage gain for Q>>1 thus Vi=10V is amplified x9 to almost -91V as the voltage must change polarity in the inductor for the current to change directions from +1A to some negative peak less than -1A due to losses dissipated in both R's. (-0.75A for 1st peak)

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Because the inductor stores the energy from the voltage source and it builds up (The voltage rises ) and the inductor releases it all at once

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