# Voltage spike at the coil when opening the switch [duplicate]

I've the following circuit:

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?

• Why do you think that this is not true? electronics.stackexchange.com/questions/288380/…
– G36
Commented May 18, 2019 at 14:40
• 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?
– JRE
Commented May 18, 2019 at 14:40
• 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. Commented May 18, 2019 at 19:48

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.