CC to PICList for critical comment Background: Electronic driver for spark ignition coil wanted, Existing 30v smd FET with no heatsink or catch diode dies rapidly. Not surprising but ... > Thanks Russell and everyone else for all the information! :) It has been > good reading! > > > Have you placed a reverse diode across the coil and/or the mosfet? > No I didn't have these.. could you explain how those two diodes function to > protect the mosfet? When the coil is de-energized and the drain side of it > peaks to 300V or so how do the diodes protect the drain of the mosfet from > seeing this voltage? I could see how a zener across the drain/source would > clamp the drain voltage to 30V or so, but this would eat up lots of the > energy that would be better to have going into the coil secondary. Firstly, not wanting to be rude but some of the advice on this has been somewhat incorrect technically. Secondly - some of my advice was imperfect too :-( A sad conclusion is that your FET is not suitable for the job (see below) For reasons explained below, you are gouing to need a MUCH higher voltage device. At least 100v (which would be marginal) and preferably many hundreds of volts !!! To battle: Draw a picture and follow the following on it and it should be moderately clear. My earlier comments (and othe rpeople's comments) about using a diode to protect the FET are correct BUT sucha diode will also prevent proper spark actoion unless certain other design features are added. See below. I'll try and explain what happens in laymans terms to make it clear what the diode does, where it should be placed and what can and can't be in series with it. As I noted in a previous post, a "catch" or protectio diode is connected from FET Drain to the postive supply rail that the coil is driven from. The diode does NOT conduct when the FET is on. ie cathode (bar end of axial diodes) goes to +ve supply and anode goes to Drain. When the FET is on current flows in the coil and builds amagnetic field. The field stores energy. When the FET is turned off the field 'collapses" and the changing field induces a voltage in the coils - both the secondary or outpur t winding AND the original input or primary winding. The physics of the situation are such that the current in the primary continues i the same direction but the voltage reverses polarity instantaneously. The collpasimg field priduces the result that the primary current or an equivalent current in another winding *MUST* continue to flow at the same value just before and just after FET turn off. If there is no means provided for the current to flow in the voltage wil increase until such a path is provided. It can rise a very very very long way in a rather short time. A 12v input can have hundreds of voltas appear on it. As the voltage at the FET drain = coil primary "bottom" voltage was BELOW the supply voltage when the FET was on, the reversed polarity means that it will rise to ABOVE the supply voltage when the FET is turned off. In the absence of a formal current path the current will flow into stray capacitance until the energy is stored between the inductance of the coil and the capacitor. Normally an energy discharge path is provided - eg a spark plug gap. For a coil with an 1:N primary to secondary turns ratio the secondary voltage will be N times the primary voltage. The coil depends on the primary voltage rising to a certain non zero voltage so that the secondary can rise to a suitably higher level. If a diode is connected as mentioned above it WILL protect the FET from the inductive turnoff spike. However, as the voltage across the diode when it is conducting (while the FET drain is above the v+ supply) is less than 1 volt the output voltage will only be N times this where N is the turns ratio. A typical coil has a turns ratio of about 1:100. So for 1 volt across the coil (due to the protection diode conducting) the out put will be about 100 x 1 = 100 volts. No spark !!!! Spark voltage is about 10,000v spark voltage anm really rather more. At 100:1 this means you need 100V (= 10,000/100) on the primary. http://www.brisk.cz/en/pro1i3.php Oh Dear !!!!!!!!!!!!!!!!!!!! The present FET is afair 30v rated. It is absolutely unsuited to this task (despite what I said recently :-( ) The good news is that you can use a much lower current rated FET ****TEST**** Take coil Take ammeter (multimeter on anmps rrange OK) Take operating voltage for coil. Connect supply via ammeter to coil. Turn in supply (briefly) Measure dc current in coil. The FET wants to be rated to at least this value. A FET so rated will survive a soimewhat higher current BUT will have higher than desirable voltage drop. A rating a few times more thgan this current will be a good idea. Note that this current is substantially higher than the current you will need when operating. This is because the DC resisatnce limits the above current whereas the current builds up as an approximately triangular waveform and may be turned off at a value lower than the mzx possible value. This is accomplished in practice by applying voltage for only a portion of each cycle. Many systems seems to use 1 to 2 amps average current but you should need rarther less. Protection. As noted above, direct connection of a reverse diode is USELESS in this application as it will clamp the coil voltage too low. IF the spark is operating as it should it will provide a maximum secondary voltage and the primary voltage will be a factor of N smaller. At 15000 volts spark the primamry will be about 150 volts. A 300v FET would be a good idea. The primary protection needs to be something that ALLOWS say 150v but stops voltag exceeding say 300v. A FET with a suitable 'avalanche" rating can disspipate such energy short term IF ADEQUATELY HEATSUNK. Onbe easy solution is a 300 volt zener (or several lower voltage ones in series) which will only conduct when drain voltage rises well above normal. Another solution (that someone else suggested) is an auxillary spark gap that will only arc when the plug is not connected. Having a capacitor in parallel with the coil will allow the energy to transfer to the capacitor (and back in an oscillaytory manner) and limit maximum voltage. This is the purpose of the points condenserm(capacitor) in a system with mechanical points. For 300 V max, 0.5 amp say, 10 mH coil 1/2 C V^2 = 1/2 L I^2 or C = i/v x sqrt (L) C =~ 0.15 uF I have no idea of how large actual points capacitors are in pointed ignitions bit around 0.1 uF at AT LEAST 300 volts or more seems appropriate here. The 3rd reference below uses 0.01 uF at 2000 volts! Note that they also use active prortection (clamp transistor to dissipate energy when voltage rsies too high). I think I'll stop about here. There will be more questions. No point in going on more than this if this is not where you want to go :-(. Your turn .... :-) Some references Basics http://www.kronjaeger.com/hv/hv/src/ign/ Typical sports coil http://www.westcoastfiero.com/electrical/Coil.html Electronic driver http://www.hills2.u-net.com/electron/ignit.htm Russell McMahon > > When the mosfet blew, it also took out the other three mosfets > unfortunately.. I put 12V on the gates of the other three mosfets (through > a 100ohm resistor) and the gate voltage drops to ~1V or so. > > > If you are driving it hard enough (10 volts on gate) it has an on > resistance > > of 0.022 ohms - say 0.05 ohms when hot. > > While they call it a logic FET (5v gate drive OK) it has a much worse on > > resistance when driven with 5v on the gate. . > > I'm using a Vgs of approximately 12Volts. > > > Posting the actual circuit may be useful - FETs have some clever ways of > > destroying themselves. What are you driving the gate from (AVR directly?). > > Is there a series drive resistor? How large? Have you got a zener or > reverse > > schottky diode from gate to ground. The miller capacitance can cause the > > gate to be driven very high without a gate zener. > > Ok I hope this is good enough.. - my real schematic isn't great either :) > > avr -- optoisolator "base" > 12v -- 100ohm resistor -- optoisolator(common) > optoisolator(emitter) -- gate > gate -- 2k resistor -- gnd > 12v -- coil -- drain > source -- gnd > > (4 mosfets are in parallel like the above) > > cheers, > Jamie Morken > > -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body