Thank you for great videos. Small correction, for the free wheel diode the inductor energy doesnt go back to the supply but dissipated in the diode itself.
@@FoolishEngineer i am going to create pure sin wave inverter using sine wave oscillator, class D amp. and low pass filter, i need your help. Could you help me?
Great video! Any reason why I’d be getting a similar phenomenon when a solenoid is turned on. I see pretty significant spikes when I first turn on a solenoid and turn it off (with a button)
Im trying to use this aspect of inductors to create voltage spikes in a controlled fashion. How do I read the value of di/dt from the oscilloscope measurement
That HUGE spike Of energy is the inductor wanting to oscillate You're preventing energy from doing what it Naturally WANTES to do instead of what you forced to do. That's why everything nowadays HEATS UP.
At 5:00, the description is somewhat flawed I think. The protection diode just dissipates the flyback voltage as heat without diverting it back to the main power supply.
The voltage spike generated across the coil when the MOSFET goes OFF is the opposite polarity of the supply voltage. Therefore, the diode is conducting when this happens and dissipates this energy into itself and NOT as mentioned, into the supply.
Hello. I want to make a fake load circuit to test the battery that can adjust the current passing through, my idea is to use a mosfet. load directly from the mosfet. I hope you can help me calculate the tolerable capacity of the mosfet through its datasheet and how to dissipate heat for durable use. Thank you very much
Actually It can do it, because spike Voltages from inductor has more potential than the battery that makes a circuit acting like a charge battery with 2 sources.
Electronics seems like a game of chess mixed with magic the gathering. This has these basic moves. You pair this with that, and it has this unique move, but now it has a sudden weakness.
I'm trying to make mosfet spot welder after blowing mosfets i searched alot about this failure and found voltage spike is killing mosfets this is my mistake to not knowing about mosfet properly after watching your video i found that my wires are inductive load which creates positive spike at drain so i need to place diode between mosfet's DRAIN (before wire bcz wire is inducing positive spike) to "+" Terminal Of My Battery. ANODE leg to DRAIN CATHODE leg to "+" terminal of battery. Correct me if I'm wrong?
Nothing new under the sun, cause the diode parallel, i.e. to the relay coil, is common since semiconductors was introduced, but very, very well explained. Good teacher.
Thanks for the informative video. But there is correction in waveform, when vgs is high at that time vds won't be 0 it will be equal to the external supply as inductor will become short circuit (constant current). So full external supply will be shared to vds of mosfet
I want to replace the ignition system of my old car with a transistor (TCI) instead of a breaker point contact switch. is it necessary to also add a diode in this case on the ignition coil, to protect the transistor from damage. will it not affect the output of the spark plug flame
Thanks for the nice video. It helps to explain about inductive spiking for switch FET applications but why does this then result in a ringing effect? Why not just a large spike and then the FET settles back to the supply voltage?
The ringing is because of the LC circuit that's formed with the output cap, now recall a series RLC circuit's step response. The R is summation of the dcr of the inductor, trace resistance, the source resistance in the path.. hope that clears your doubt
Sinceres thanks for your lettures. But, why most half bridge topologies converter use just a resitor and capacitor to reduce the spikes during off time?
thanks for this video but i have a doubt that ,i have a gate driver circuit and i have also used a diode as u taught and iam still getting high spike after some time my simulation runs? why
When the MOSFET is OFF, the back EMF from the inductor will cause a current to flow through the diode and back to the supply, but in addition to that it can also flow through the inductor itself. So there will be two paths for current to flow, right?
When MOSFET is off then diode act as close switch for flyback voltage so diode short circuit to the inductor during this time period so current flow only in inductor not to power supply because ground path of power supply is not complete . During this time inductor and freewheeling diode will heat so generally one snubber circuit i.e. parallel combination of a resistor and a small value ceremic capacitor add in series with freewheeling diode.
@@pushpendraaryan5551 Yup you're right about the diode part. The circuit is actually complete, since the MOSFET has a body diode, the only thing is that the body diode will be reverse biased.
I was using inductors and high speed switching with a 27,000MHZ quartz crystal to create a swing for a Tesla coil and I was testing it with a dead 9v battery that had about 6v left in it and it went into a resonance cascade and it fried the whole ciruit also it wasn't hooked up to a coil yet .. but ya my goal was to creat a specific resonance only I got wayyyy more than I expected . If I can recreate it again it will be tricky to handle the continuing increase to find out far it can actually go without frying it again ,, not sure what to do next since it isn't ready for the coil or , maybe I need something else . I had originally thought I would create two circuits and apply them against each other with another crystal at 28,000MHZ to get the resonace started and then figure ways to adjust it provided it's close enough to correct parameters .. 🤔 I was able to run another circuit like it with an earlier SCR without it frying and it runs off of 5 v and of course the goal here is to keep it running after the 5 volts is gone if it can continue to remain regeneratively and if it can pull what it needs from the air and keep the LEDs lit up ....
The capacitor could get damaged because the potential difference between the + of the back EMF (i.e. Voltage between the Drain and the Voltage Source when the MOSFET turns OFF) and the positive terminal of the voltage source will be large. So, we might need a very high voltage capacitor.
@@electroquests Once the diode is added to the circuit, the back emf can never exceed the positive supply by more than one diode drop (not more than a volt or so, worst case). You only need a capacitor with a voltage rating a few volts more than the supply voltage. You really need a return path to ground via the supply in any scenario, so you're going to put a capacitor across the supply in any case.
