Resistance of MOSFET vs IGBT/SCR ? - Page 1
Forum section: electronics / projects

author: ricko_uk
author_rank: Super Contributor
timestamp: March 27, 2020, 12:58:09 pm
content: 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

author: schmitt trigger
author_rank: Super Contributor
timestamp: March 27, 2020, 01:00:20 pm
content: One thing is certain, if you employ an SCR, you won't be able to turn it off with the circuit as shown.

author: exe
author_rank: Unknown
timestamp: March 27, 2020, 01:04:25 pm
content: Quote from: ricko_uk on March 27, 2020, 12:58:09 pmWell, what I do if I need to compare mosfet and bjt: I calculate dropout at the current of interest. Dropout is a figure of merit. If you want to get "resistance" for bjt, then it's Vdropout/Icollector (or Iemmiter). Of course for bjt it varies a lot with current, but I usually calculate it for the worst scenario.EDIT: or just compare them by dissipated power, that's even better.

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 03:15:23 pm
content: To summarise the above.A MOSFET behaves like a resistor when turned on, with the current increasing linearly with current, whilst an SCR and IBGT roughly drop a fixed voltage, irrespective of the current.High voltage MOSFETs have a relatively high on resistance (if I remember rightly resistance goes up to the square of voltage rating, everything else being equal) so the higher voltage rated parts drop quite a voltage, at any significant current.For high voltages an IGBF/SCR give low on losses, than MOSFETs. The crossover point depending on whether MOSFETs or SCRs/IGBTs have the lower on loss is around 600V.

author: ricko_uk
author_rank: Super Contributor
timestamp: March 27, 2020, 03:35:53 pm
content: Thank you all!So am I correct in saying that the IGBT does not slow down the ramping up of the current through the inductor (while the MOSFET clearly does)?Thank you

author: duak
author_rank: Super Contributor
timestamp: March 27, 2020, 05:35:30 pm
content: Ricko,Both devices will have some sort of voltage drop. IGBTs have a higher initital voltage drop but a lower Rs especially at KV level. A starting point is the Output Characteristics graph (Voltage drop as a function of current and gate voltage) in the device data sheet.We don't know what the value of the inductor is nor do we know what the maximum current and standoff voltage will be so it's difficult to make a blanket statement. If the inductor is quite small with a microsecond time constant, a MOSFETs speed could be an advantage.Also, what happens later on? Does the device shut off and the voltage across it spike? Or does it stay on and reach some sort of steady state or does the inductor current fall and ultimately reverse leading to a damped oscillation? Inquiring minds want to know.

author: ricko_uk
author_rank: Super Contributor
timestamp: March 27, 2020, 07:06:43 pm
content: Hi Duak,the capacitor is isolated during discharging (I have a mosfet switch before teh diode).Then the IGBT/MOSFET is turned on just for enough time for the cap to dump all its charges into/through the inductor. Because the capacitor is isolated during discharging then no additional current flows into the inductor apart from the charges from the capacitor.As soon as that is done the mosfet/IGBT is turned off again.The inductor size varies from few nH to several hundred uH. The reason being that this is a experimental jig for various applications. It is not for any specific application.Any additional comments/suggestiong?Thank you all

author: Zero999
author_rank: Super Contributor
timestamp: March 27, 2020, 07:09:59 pm
content: Quote from: ricko_uk on March 27, 2020, 07:06:43 pmIs that 1F capacitor really rated to 2kV or is it just a typo/hypothetical simulation? That's E = ½CV² = 0.5*2000² = 2MJ which is enormous. Please post some pictures of it!

author: duak
author_rank: Super Contributor
timestamp: March 28, 2020, 01:55:37 am
content: Ricko, Thanks for the info.Well, if you're talking about discharging the cap into the inductor and then shutting off the FET/IGBT, where does the energy in the inductor go? If it doesn't get transferred to a projectile, converted to heat or magnetizes something, it will reverse direction and try to charge the capacitor with the opposite polarity.What sort of peak currents were you hoping to get? Do you have an idea as to what value capacitor you want to use? At 300 V and a few ohms series resistance, you're looking at peak currents on the order of 100 A, depending on the capacitance and inductance. Circuit loop area will be important. With low values of series inductance, including the capacitor's, suitably sized MOSFETs will switch in on the order of 100 ns. There's saying in racing: "speed costs money, how fast can you afford to go?" Here, it's perhaps less money and more effort.

author: David Hess
author_rank: Super Contributor
timestamp: March 28, 2020, 03:40:51 am
content: Most of what I would have said has been covered so I have a couple comments:1. The inductance will limit di/dt anyway until saturation. Wiring resistance should be accounted for also.2. If you want something that will switch really fast, consider a bipolar transistor in avalanche mode. Look it up.3. Consider replacing the single capacitor with a distributed transmission line consisting of several smaller capacitors and inductors. This will provide a lower peak power but longer duration pulse.

author: ricko_uk
author_rank: Super Contributor
timestamp: March 28, 2020, 03:45:52 pm
content: Thank you all!I will address each one below.Duak,Sometimes the inductor is a transformer with a large number of secondary turns like an ignition, to create a spark in a spark plug.1) Am I correct in thinking that the energy stored in the inductor is dissipated into the spark of the spark plug?2) if not, an antiparallel diode across the capacitor as shown in the new schematic attached here, would allow the energy to dissipate across the inductor's resistive part. Is that correct?3) what are the pros and cons of placing an anti-parallel diode across the capacitor vs placing it across the inductor (like in classic motors/solenoids switching)?Ref the circuit loop being important, what do you mean? I assume the smaller the loop the lower the parassitic inductance? Is that correct?Ref the peak currents, it varies on the application and the load characteristics. Peak currents anywhere from couple of amps to 40-50A.Zero999,that is just the capacitor rating. The peak voltage is 300V.David,thank you for the additional commentsThank you all

author: schmitt trigger
author_rank: Super Contributor
timestamp: March 28, 2020, 04:05:49 pm
content: What Zero was asking, and certainly I also would like to see, is the size of a 1 Farad, 2000 voltratedcap.1 Farad is supercap territory, at very low voltage.And 2 Kv is glass cap territory, at very low volumetric capacity.How do you obtain both ratings simultaneously?

author: ricko_uk
author_rank: Super Contributor
timestamp: March 28, 2020, 04:33:11 pm
content: sorry everybody, in the schematic it is a 1mF not 1F capacitor.

author: duak
author_rank: Super Contributor
timestamp: March 28, 2020, 10:17:59 pm
content: G'day Ricko,Are you working on a Ignition system then?1) Am I correct in thinking that the energy stored in the inductor is dissipated into the spark of the spark plug?Sure, but 300 V is not enough to cause a spark plug to work unless the air gap is exceedingly small. Supposing the inductor is a transformer, like an ignition coil, then a good bit of the energy is dissipated in various resistances and other loss mechanisms. It also gets dissipated in the resistances of the various other components in the circuit. Note that sparks and arcs have quite low voltage drops because of their relatively low resistances.2) if not, an antiparallel diode across the capacitor as shown in the new schematic attached here, would allow the energy to dissipate across the inductor's resistive part. Is that correct?Yes, but two things. Q1 will have to stay conducting to allow the current to continue to flow unti it reaches zero. If you shut the switch off while current is flowing, L1 will generate a voltage that will try to maintain that current. In this case positive relative to mains-gnd and could be enough to damage Q1. Secondly, inductive circuits have a time constant of L/R where L is the inductance and R is the series resistance. Oddly enough, reducing the series resistance causes the time constant to increase so that it takes longer to discharge the inductor. A diode has a low resistance, so while it prevents C1 from charging in reverse, it extends the discharge time, something that may be good thing.3) what are the pros and cons of placing an anti-parallel diode across the capacitor vs placing it across the inductor (like in classic motors/solenoids switching)?Putting the diode across the inductor creates a circuit that doesn't require Q1 to continue to conduct and also protects it from being subjected to a higher or even damaging voltage.Ref the circuit loop being important, what do you mean? I assume the smaller the loop the lower the parassitic inductance? Is that correct?Yes, the inductance of a circuit is a function of the size of the area enclosed by the circuit's loop. ie., A larger area leads to more inductance.

author: ricko_uk
author_rank: Super Contributor
timestamp: March 30, 2020, 03:16:56 pm
content: 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

author: Zero999
author_rank: Super Contributor
timestamp: March 30, 2020, 06:45:56 pm
content: Quote from: ricko_uk on March 30, 2020, 03:16:56 pmYes, 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

author: duak
author_rank: Super Contributor
timestamp: March 30, 2020, 08:56:25 pm
content: 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,

author: ricko_uk
author_rank: Super Contributor
timestamp: April 13, 2020, 08:42:24 pm
content: Thank you Duak and Zero999Duak,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 againRiccardo

author: duak
author_rank: Super Contributor
timestamp: April 14, 2020, 04:10:16 am
content: 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.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: April 14, 2020, 05:49:59 pm
content: Quote from: ricko_uk on March 27, 2020, 12:58:09 pmyou 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.