Transient protection? - Page 1
Forum section: electronics / beginners

author: somlioy
author_rank: Regular Contributor
timestamp: May 26, 2013, 12:44:00 am
content: I'm building a device that's connecting to a multimeter to read dwell angle and rpm from a vintage car distributor. The thing is when the coil is quickly turning on and off I'm getting transient voltages from about ~-120v to ~300v. The signal is a 9vdc square wave type. But how do I protect the other devices from these transients?Would something like this work?http://www.electronics-tutorials.ws/diode/diode_4.html

author: Rerouter
author_rank: Super Contributor
timestamp: May 26, 2013, 12:55:47 am
content: swap those to shottkeys and 99% of the time your ok, the main issue though is if your circuit is quite low power it can charge up your supply rails to those voltages,adding a series resistor to your signal line can generally help with this, seeing how cmos is an option 10K or more would suit you pretty well, limiting the peak current to less than 30mA,alternatively you can buy a ESD protection diode to clamp your input to ground in the same way, though i would still recommend a series resistance, they may claim very large peak wattage's but its not wise to clamp 300V to 9V when your inductor is an ignition coil (quite large)

author: Dave
author_rank: Super Contributor
timestamp: May 26, 2013, 01:55:35 am
content: In addition to the series resistor, add some capacitance to the input (picofarads). Experiment with the values and see what works best for you.Bear in mind, that when you will be probing the input with an oscilloscope, you will also be introducing some capacitance with the probe.

author: SeanB
author_rank: Super Contributor
timestamp: May 26, 2013, 06:32:25 am
content: Simpler would be to use a 2N2222 as input transistor with a diode connected across the base emitter in reverse bias, and use a 10k 0.5W resistor for the base ( and a 10k pull down on the base in case it is left floating) with a 1k collector resistor to get the signal inside. Simple, cheap and not going to fail easily with car voltages. Will work for any voltage above 4V to 400V, pulses no problem. It does invert though, so you might need an extra inverter or arrange the logic accordingly.

author: plesa
author_rank: Frequent Contributor
timestamp: May 26, 2013, 11:31:45 am
content: Use rather transils like P6K, rated for higher current peaks and available for different voltage drop.

author: Paul Price
author_rank: Super Contributor
timestamp: May 26, 2013, 07:20:03 pm
content: Just add 10k resistor in series with your input, and a single 12V zener 1/4W to ground, cathode-up, anode to ground at the right side of the resistor, at the input to the CMOS device. It will clamp transients in polarities without upsetting ignition..The 300V spike you are seeing is the ignition coil flyback voltage and clamping this voltage with anything directly connected to the points and the coil will kill the spark and dwell or no dwell, your car won't run without a spark to the plugs..

author: somlioy
author_rank: Regular Contributor
timestamp: May 26, 2013, 08:55:22 pm
content: Currently I'm feeding the input to a 555 trg input and a ttl-inverter atm. I currently don't have a CMOS NOT-gate at hand.If I can't clamp the flyback voltage, how do I protect the circuitry then?

author: Paul Price
author_rank: Super Contributor
timestamp: May 27, 2013, 12:29:51 am
content: I already see the input protection circuit with r8, D1 and d2. This should work if wasn't for the spark coil pulse being damped by the low value of ?R8. R8 will burn up at High RPM.Either the top or bottom circuits I've offered will work fine( Det begge virke fint.)You can use two protection diodes 1N4184 or the single zener 12V at the LM393 pin 2 input or at pins 1,2 of the CD4093 for input protection, Schottky diodes are not really going to work any better than any silicon sw. diode and are not needed here.These circuits replace R8, D2 and D3, everything else is the same.

author: somlioy
author_rank: Regular Contributor
timestamp: May 27, 2013, 04:29:17 pm
content: Aight, but are you sure about the pull down? Because the points are pulling the coil down when closed, which means when the points are open the signal will be floating high through the coil?

author: AndyC_772
author_rank: Super Contributor
timestamp: May 27, 2013, 04:38:35 pm
content: If your sensor has a high impedance input, then a series resistor and a TVS diode like an SMAJ series should do the trick. You want a device which can absorb significant short term energy spikes without being damaged, and I'd worry about the long term effect of using a small Schottky diode to the supply rail when the instantaneous current peaks can be so large. Ignition noise is high frequency, so whatever you use needs to be fast acting and robust.A simple R-C filter alone will take out the worst of the energy in the ignition spikes. Just make sure that the GND net within your circuit is low inductance and carefully routed.

author: Paul Price
author_rank: Super Contributor
timestamp: May 27, 2013, 04:45:24 pm
content: The pulldown. When the points are closed the input to your meter is at 0-V, but when they open the input will jump to 300V or more because of the spark coil inductance flyback pulse.You don't need schottky diodes, ordinary 1N4184 sw. diodes will work fine placed after a current-limiting isolating input resistor of 1K. You would prevent the tach from upsetting the spark and benefit with a quite a bit higher resistor (100k) than the 1K you show in your circuit. You don't need any capacitor after the resistor to protect your circuit after the resistor either.Have you considered the input buffers I attached in my last post?

author: drZoidberg
author_rank: Contributor
timestamp: May 30, 2013, 09:27:07 pm
content: OK, so I don't know anything about car distributors.But usually, when coils are involved there is a simple solution to the problem. Simply put a diode in parallel with the coil, so that the current can discharge through the diode (use a big diode :p).The problem with coils is that the currentkeeps flowingwhen it is disconnected. This current can flow through the diode, and not generate any high voltage (since the diode has a low impedance in the forward direction). I added a quick diagram.This may or may not be possible to do (depending on how that thingamajig works).Still you should also use some input protection on the board. The one proposed in the first post should work fine. If I was you, i would add a large capacitor from Vcc to GND. This is so that any spikes have a low impedance path to return through. You want to put all of this protection stuff close to the connector on the PCB btw, so that any current spikes flow back to where they came from without going through all of the other circuitry. So put diodes + big cap (or whichever solution you go for) close to the connector.

author: drZoidberg
author_rank: Contributor
timestamp: May 30, 2013, 09:37:41 pm
content: BTWIf you use the reverse protection diode (the one in the drawing), try to put it as close to the coil as possible. Maybe you can put it on the test leads or on the coil itself? The goal is to make the diode-coil loop as small as possible. This loop is a good antenna for transmitting noise. If you are counting pulses, you may end up counting a few extra :pThere are other ways to minimize the loop as well. Try not to split the test leads. Then maybe you can have the diode on the PCB =)

author: c4757p
author_rank: Super Contributor
timestamp: May 30, 2013, 10:22:10 pm
content: Yeah... somehow I don't think attempting to dump the whole transient is a good idea.

author: C
author_rank: Super Contributor
timestamp: May 30, 2013, 10:22:51 pm
content: Look at these two links for some idea.http://www.schematicsforfree.com/archive/file/Automotive/Service%20&%20Maintenance/Measuring%20Ignition%20Dwell%20Angle.pdfhttp://www.qsl.net/kd7rem/dwellmtr.htmC

author: drZoidberg
author_rank: Contributor
timestamp: May 30, 2013, 11:11:14 pm
content: Quote from: c4757p on May 30, 2013, 10:22:10 pmHow else to get rid of it? It has to go somewhere, preferably not through the input :p

author: c4757p
author_rank: Super Contributor
timestamp: May 30, 2013, 11:31:47 pm
content: With a series resistor to limit the clamping current, so the protection is only done at the input itself, as mentioned above. If you have a 10kV spike in 100kOhms, only ~100mA has to flow to clamp it. You don't have to clean the spike out of the whole system, you just have to keep it out of the input.

author: drZoidberg
author_rank: Contributor
timestamp: May 30, 2013, 11:55:49 pm
content: Those are some great links there C. Now I understand how the coil circuit is built up (assuming it's the same as figure b in the pdf).I assume you will be measuring between Y and GROUND (see attached figure). Connecting a diode with the anode at Y and the cathode at X (maybe even after the ballast resistor) would allow the coils energy to be dissipated in the diode. The diode must be large enough to handle the power dissipated in it, but since the forward voltage of the diode in only ~0.6V the power may not be as high as you think. It will still have to be pretty large though. Luckily, big diodes are not very expensive. If you have a diode in your junk box. just connect it to see if it survives. If not, try a bigger one (A good rule of thumb in any situation =) )The capacitor labeled CONDENSER is there to dump some of the transients to ground. If it wasn't there you would see the switch arc (and probably char) and the voltage spikes would be higher. It sort of "softens the disconnect".Quote from: c4757p on May 30, 2013, 11:31:47 pmYup. That's not a bad idea. Then you would only need to clamp the input terminal (after the series resistor) to Vcc and GND via schottky diodes, as in the original post. Maybe you're not as stupid as your condescending comment made you look :pAnyhow, dumping the transients through a diode is a tried and tested technique used in thousands of products.

author: c4757p
author_rank: Super Contributor
timestamp: May 31, 2013, 01:30:36 am
content: Condescending??Uh... sorry? How would you prefer I answered?

author: Rufus
author_rank: Super Contributor
timestamp: May 31, 2013, 01:38:18 am
content: Quote from: drZoidberg on May 30, 2013, 11:11:14 pmIt is supposed to go into the spark plug which is the whole point of the ignition system.The ignition coil is a transformer. If you want kV pulses on the secondary to create a spark you need 100's of volts pulses on the primary.

author: C
author_rank: Super Contributor
timestamp: May 31, 2013, 02:12:44 am
content: That condenser is just there to protect the points from arcing.As it takes a bit of energy to run that coil, and note here the secondary is not connected to ground but to the primary winding, I think they are saving power by kicking back some energy back in to the battery. Also think about the speed that the field needs to build again. Back In my hot rod days, xome of my 8 cylinder engines turned 8,000 RPMs.Not an engineer, but I would think the following about the diode you suggest.Diode is a possible failure point that was not there before.Diode on coil is in a very hot location an in my experience very hot electronics do not last long.I would also guess that a standard diode would fry before it turns on, remember that coil's whole purpose is to create a spark where a spark does not want to happen that is why it takes 20,000 volts or more to ark across a small gap of 25 thousandth of an inch. That coil is built to create a big inductive spike with a very fast transition time to fire spark plug.With a super fast Shockley diode being needed where the design not pushing a fast transition times like power supplies and motor drives, What diode would you use where the coil is built for very fast transition as it purpose in life is combined with very high voltage and very high heat environment?C

author: Rufus
author_rank: Super Contributor
timestamp: May 31, 2013, 02:18:27 am
content: Quote from: C on May 31, 2013, 02:12:44 amThe engine won't run without a condenser. The points will only take damage while you have enough battery left to crank. The engine won't run with a diode there either.

author: C
author_rank: Super Contributor
timestamp: May 31, 2013, 02:34:47 am
content: RufusI agree with your better statement, My experience is that you better use the correct condenser or soon points are needing to be replaced.C

author: SeanB
author_rank: Super Contributor
timestamp: May 31, 2013, 04:48:40 am
content: The coil primary is a series resonant LC circuit. Adding a diode will stop it oscillating, and you will get no spark. The one end of the secondary is connected to the one terminal to save an extra connection, 12V is not going to make a difference on the 8-12kV of the secondary output. You operate it by ramping up the current when the contacts close and the spark occurs as the contact opens. The contact can be a mechanical switch or a transistor or IGBT switch in an engine module, makes no difference, you still need the capacitance there, though in the modules you will find a film or high voltage ceramic capacitor with a 400V clamp on it, and a series diode to allow the current switch to go below the ground rail without issues.

author: IanB
author_rank: Super Contributor
timestamp: May 31, 2013, 05:00:14 am
content: Quote from: drZoidberg on May 30, 2013, 11:11:14 pmAs others have been saying, you don'twantto get rid of the energy from the coil. The whole purpose of the coil is to store up energy and deliver it to the spark plugs. If you short out the coil with a diode you will stop the ignition system from working and the engine will stop.

author: drZoidberg
author_rank: Contributor
timestamp: May 31, 2013, 05:01:05 am
content: Oh man...Now I see how the ignition coil works!Screw the diode.That would mess the whole circuit up. Thanks, Rufus, for pointing out how the circuit works.I was thinking of the transformer as an ordinary transformer, with a separate primary and secondary side. That doesn't work at all, of course. The current flowing in the primary windinghasto keep flowing. My solution would have stopped the engine altogetherHah!Well well..Also, C makes some good points about not putting diodes in very hot environments. To answer your question, I don't know. The diode would probably have burned up in the heat from the engine, unless it was placed sufficiently far away from the heat. I mean, there are some other electronics going in there also, and they need to be placed somewhere not so hot also. But as I said, your car would go nowhere fast.I now see that IanB has pointed out the egg on my face =)

author: drZoidberg
author_rank: Contributor
timestamp: May 31, 2013, 05:23:17 am
content: Quote from: c4757p on May 31, 2013, 01:30:36 amI would prefer you pointed out why you think I was wrong, like some of these other fine members have done, instead of rolling your eyes and saying:Quote from: c4757p on May 30, 2013, 10:22:10 pmSorry if I wasn't super nice back at you.

author: c4757p
author_rank: Super Contributor
timestamp: May 31, 2013, 05:26:25 am
content: I wasn't sure if you realized that's what you were implying, so I thought I would point it out. I'd have explained more completely, but I was typing from my phone - figured I'd at least point you in the right direction until I got to a proper computer or someone else took over.There was no eye rolling. I was trying to be as helpful as the situation allowed, and possibly misunderstood your line of thinking.

author: drZoidberg
author_rank: Contributor
timestamp: May 31, 2013, 05:27:25 am
content: OK. I understand.No worries then =)

author: somlioy
author_rank: Regular Contributor
timestamp: May 31, 2013, 10:26:45 pm
content: Just to clarify, this isnt something I'm permanently hooking up to the car. I just hook it up once in a while to check the dwell angle, which needs to be adjusted quite often with these kinds of ignitionsystems.Plus and minus to battery, coil signal to the low side of the coil (between point and coil).

author: Paul Price
author_rank: Super Contributor
timestamp: June 01, 2013, 04:48:51 pm
content: The video is almost right, but fails to explain the capacitor across the points. It doesn't just "absorb back emf", it allows the circuit to oscillate with the inductance of the transformer.Without the capacitor the voltage from the coil, when the points opened would try instantly, in a few billionths of second, to rise to 1000 or more volts, this would cause the points, still only slightly opened to arc over.With the capacitor, the coil's inductive flyback current must charge the capacitor to a high voltage at a much slower rate(over some fraction of a millisecond), so by the time the voltage is high enough to arc, the points have already moved too far apart for the voltage to jump that new point gap distance. The result, instead of creating an instantaneous arc that will just arc across the point contacts and kill all the spark coil energy, you get a lower voltage but a high voltage, a developing 400V oscillation that creates 22000 or more volts in the secondary of the spark coil.Connecting a diode across the ignition coil instantly absorbs the energy stored in the spark coil inductance, kills the spark energy but will protect your points. The side-affect is no spark to the plugs.A good dwell/tach circuit should just should sense voltage at the points-sparkcoil connection, not interfere with its operation, that's why I have suggested and offered circuits very early in this discussion to make an input buffer circuit.

author: somlioy
author_rank: Regular Contributor
timestamp: June 12, 2013, 07:01:12 pm
content: Havent had any time to work any on this but I checked out your circuits. I'm probably going for the CD4093 one, since I already got those somewhere. Is it necessary to protect the input of the CD4093 with clamping diodes or are the 100k and 1Meg doing the work?

author: Paul Price
author_rank: Super Contributor
timestamp: June 12, 2013, 11:11:07 pm
content: The inputs of the 4093 already have clamp diodes and the 100k resistor limits the current into the 4093 protection diodes to do no harm to this circuit. Just to be sure you can place a zener diode (cathode to the 4093 pins 1-2 input, anode to ground) for additional protection. The primary voltage can reach in excess of 400V to present itself at the input to your tach/dwell meter., but the 100k resistor protects the input of the 4093 and while it isolates the dwell/tach meter from upsetting the spark.A zener diode at pins 1-2 of the 4093 with the 100k resistor in the circuit will offer additional protection. A low power 8.2V zener will do.You could also just put two 1N4184 diodes at the CD4093 pins 1,2. with the top diode connecting its cathode to the +9V, its anode connecting to the cathode of the second 1N4184 and pins 1,2 of the 4093. The anode of the bottom diode connects to ground. You already show something like this in your original drawing, but again, don't bother with schottky diodes.