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| Resistance of MOSFET vs IGBT/SCR ? |
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| Zero999:
--- Quote from: ricko_uk on March 30, 2020, 03:16:56 pm ---Thank you again Duak, You are right about "Oddly enough, reducing the series resistance causes the time constant to increase so that it takes longer to discharge the inductor."! I didn't think of that. I assume that is the reason why they make spark plugs or spark plug cables with resistor inside...? I thought it was only to reduce the electrical noise in the car's electrical system... But then if I add R1 in the circuit as shown in the new schematic attached, then the same resistor increases the time constant of the capacitor and decreases the time constant of the inductor... How does it work? I assume there is a "minimum" point of balance? If so how do I find the correct value? With calculation and experimentally (as calculations might not take in consideration all parassitic effects)? Thank you again :) --- End quote --- Yes, lower resistor values will increase the time constant. An inductor with zero resistance will have an infinite time constant and current will circulate perpetually. This is what happens in superconducting magnets in MRI scanners. Adding a resistor dissipates energy. The higher the resistor value, the quicker it will dissipate the energy. Capacitors are the opposite to inductors. A capacitor with an infinitely high resistance between its plates will remain charged perpetually. Adding a resistor will dissipate energy. The lower the value, the quicker it will dissipate the energy. An RLC circuit will have a resonant frequency and will oscillate for a short period of time, when the switch is closed. The resistor will damp the resonance, causing the energy to be dissipated more quickly. The Wikipedia arcical has the relevant equations, but you need to be familiar with calculus to solve them and my calculus is quite rusty so I cheat and use a computer. https://en.wikipedia.org/wiki/RLC_circuit |
| duak:
G'day Ricko, I've been working with cars and engines as a hobby since the 70's. I've learned a few things about automotive ignition systems over the years. By the way, since you're in the UK, you might appreciate that I'm helping a neighbor campaign his 1960 Austin Healey Sprite at the local vintage races. He was doing well until, in an unsupervised moment, tried to run it on diesel. (sigh). Indeed, resistance wire or spark plugs with integral resistors are used to suppress RFI. They do it by limiting current risetimes. Consider the secondary circuit. The ignition coil generates a voltage that charges up the capacitance of the coil and the ignition wire until the voltage is great enough to cause the spark plug gap to break down and conduct, discharging the capacitances and the ignition coil inductance. If the breakdown voltage is 20 KV, the wire and coil are 4000 ohms and the spark is 1000 ohms, the peak current could be 5 A. The current risetime would be short, on the order of a few microseconds. Worse, the ignition wire forms a loop with the engine block and the car's structure that makes a dandy transmitting antenna. Resistor wires slow down the rate of rise of current and the peak current. If the wire's resistance is 20 K, the current risetime extends to 10 microseconds and the peak current to under one amp. This reduces the amplitude of the transmitted RFI and shifts the spectrum downward in frequency. As Zero999 says, there is an interplay between the resistances, inductances and capacitances. The actual equivalent circuit is quite complicated and requires differential equations or a circuit simulator to solve accurately. Your simplified circuit does not show the ignition coil secondary, an important part. If you could tell us more about the application, what the requirements are and if you have some parts already chosen. ie., the ignition coil. One other consideration is that if the spark plug fails to spark or if the ignition lead falls off, the stored energy will have no place to go and may stress the primary components. I repaired a CDI unit once, but I'm not sure of the waveforms across the coil primary and the switch. Looking at prior art, I see that CDI systems arrange the capacitor, switch and ignition coil differently, see attached figure. I'm not sure why it was arranged this way but I believe it allowed for a simple switch that could bypass the CDI if something went wrong, and use the original points and condensor. It may be possible to rearrange the circuit closer to yours. The circuit shows an SCR for the switch. I haven't seen any other type of device used here, but it's possible. SCRs have a handy feature in that they will self commutate, ie., once triggered, they will stay conducting until their anode current drops to zero. SCRs don't generally conduct in the reverse direction so if that's important, it rules out using a MOSFET without some additional diodes to isolate the body diode. Some IGBTs, but not all can operate with high reversed C - E voltages. Gotta run, Cheers, |
| ricko_uk:
Thank you Duak and Zero999 :) Duak, the applications vary. It is for various experiments including spark discharge like a tesla coil as well as automotive ignitions. It is a bit of general circuit to have in the lab to test various ideas. You mentioned "see attached figure" but I think you might have forgotten to attach a file perhaps? Regarding the secondary, you are correct. The reason why I didn't show it was only because in one application I want to try is to see how fast the magnetic field changes just by abruptly opening the transistor. Many thanks again :) Riccardo |
| duak:
Riccardo, you are indeed correct. I've added the file to the above posting. The rate of change of the magnetic field is a function of how quickly the switching device can be shut off, the stray capacitance across the coil and the maximum voltage that the circuit can tolerate. If you have something like a 10 uH inductor and a few amps of current, and maybe 100 pF of capacitance, the voltage can easily spike up to a few hundred volts if the switch shut off time is on the order of 100 ns. I'm going from memory with these values from when I last worked with a circuit like this - don't hold me to them. |
| ahbushnell:
--- Quote from: ricko_uk on March 27, 2020, 12:58:09 pm ---Hi, with reference to the attached schematic, the capacitor is charged to high voltage and then discharged into the coil. As switching device I can use a MOSFET, an IGBT or an SCR. I need to keep the rate of change in current across the coil as high/fast as possible. If I use a mosfet it obviously has a internal resistance (albeit very low) which would decrease the rate of change of the current through the coil. Using an IGBT or a SCR has a voltage drop across it but not a resistance as such. Two questions: 1) does the IGBT/SCR have some equivalent characteristic to the mosfet internat on resistance? 2) does the IGBT/SCR cause the rate of change of the current to decrease like the on resistance of a mosfet would (compare to an ideal "mechanical-like" switch)? UPDATE: there is a type in the schematic. The cap is mF not F! :) Many thanks :) --- End quote --- you say fast. How fast microseconds, ms, nanoseconds? What is the value of the inductor. The inductor will probably be the limiting factor unless it's small. The capacitor and inductor will form a tank circuit. It will go to peak (ignoring resistance) in pi/2*sqrt(L*C). The peak current will be 300V/sqrt(L/C). If the switching resistance and wire resistance starts to approach sqrt(L/C) then that will start limiting the rate of rise. |
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