Transformer voltage too high - Page 1
Forum section: electronics / projects

author: NMNeil
author_rank: Contributor
timestamp: June 19, 2017, 01:10:51 pm
content: I'm playing around with a CDI ignition setup for my old tractor and decided to use part of a design by John Clarke.http://www.not2fast.com/ignition/cap-dis-ignit-cdi.pdfLooking at the top part of the circuit on the third page it uses an IR2155 to drive 2 mosfets which alternately switch into a transformer with a twin 13 turn primary and a single 360 turn secondary. This would give a 1 to 27 ratio meaning a 12 volt input would give a 320 volt output, allowing for losses this should give about 300 volts after the bridge rectifier, just as the article says.I wound my very first transformer exactly as described, built the DC to DC part of the circuit and started it up. 900 plus DC volts at the output!!My question is would my using IRF540 mosfets instead of the MPT3055E cause this? I used the IRF540 because the specs look better with a lower Rds and I have a big bag of them from a previous project.

author: Paul Price
author_rank: Super Contributor
timestamp: June 19, 2017, 01:43:06 pm
content: It is definitely not the MOSFET choices causing this OV problem.Something to do with how you've wired the secondary or primary or both.Look at the secondary winding again, you might have twice too many turns.Was the voltage measured without any load at all? Missing a bleeder resistor?Have you set the input voltage to >12V?Post your coil winding turns numbers and connections and also your own secondary transformer connection schematic and a diagramof your test setup.Seems almost for sure if your really read 900V it should have shorted out the CD storage capacitor in a split second.

author: tatus1969
author_rank: Super Contributor
timestamp: June 19, 2017, 02:29:19 pm
content: That is a flyback converter, it works like an isolated boost converter and can theoretically produce infinite output voltage. It needs the feedback circuit as shown in the PDF to regulate.

author: NMNeil
author_rank: Contributor
timestamp: June 19, 2017, 03:04:32 pm
content: Quote from: Paul Price on June 19, 2017, 01:43:06 pmTrust me, I counted every single one of the laborious 360 turns, no chance of piling on 600 plus, also the 360 turns only just fitted on the bobbin.The other answers is , yes I measured it with no load, the power supply was spot on 12 volt (I may drop it to 10 and see what the output is then), and I'm not sure where to put a bleed resistor in the circuit.I just can't work out how the voltage would be treble the calculation.

author: NMNeil
author_rank: Contributor
timestamp: June 19, 2017, 03:06:00 pm
content: Quote from: tatus1969 on June 19, 2017, 02:29:19 pmThanks, I'll order up some 75 volt zeners and see how it works.

author: Paul Price
author_rank: Super Contributor
timestamp: June 19, 2017, 03:50:10 pm
content: This is not a flyback converter, it is a push-pull inverter.converter.The voltage output is fixed by the turns ratio of sec/pri turns, but transformer waveform switching spikes can generate higher voltages if the secondary is unloaded.

author: Benta
author_rank: Super Contributor
timestamp: June 19, 2017, 04:14:56 pm
content: I suspect it's because it's running without load.The built-in dead time between the two channels in the IR2155 open-circuits the primary for a couple of microseconds, which is enough for the transformer to kick back and generate high voltage spikes. They have little energy, but it's enough to charge the output cap.Put a simulated load on it and see what happens.

author: Paul Price
author_rank: Super Contributor
timestamp: June 19, 2017, 04:20:46 pm
content: The circuit diagram clearly shows series 75-V Zeners used for voltage feedback to regulate the HV and function(with the help of a transistor)to turn off the inverter IC by lowering its supply voltage to trigger undervoltage shutdown(built into the IC) when the HV output exceeds 300V.This inverter chip was designed to be used with flyback converters or half-bridge converters. In half-bridge or flyback converters, undervoltage detection is necessary to allow the IC's power supply input voltage to reach the optimal value in startup before flyback switching starts. Otherwise, the IC (and the BJT or MOSFET switches) wouldn't get the normal voltage it requires. When running the IC is powered from a LV feedback winding on the flyback transformer.Undervoltage lockup is made to function when the flyback converter is first turned and allows the IC to startup only with enough (voltage) energy stored in the IC's filter cap that is charged up during turn on by a high-valued resistor from the rectified mains voltage.Once the IC's voltage rises above the undervoltage lockout level, the IC is then able to deliver the first few pulses to the BJT or MOSFET flyback switch, and then a LV flyback transformer winding delivers pulses to be rectified to power the IC..but his is another story.This IC is being used in a circuit that is not a flyback converter, but adapted to work in this inverter circuit..

author: Paul Price
author_rank: Super Contributor
timestamp: June 19, 2017, 04:38:31 pm
content: If you have connected the series 75-volt zener circuit correctly, the voltage would be regulated even with the wrong turns ratio.This means you haven't connected the feedback zener circuit:1) You have a connection fault2) Have substituted a HV rectifier diode for one or more 75-V zener dioides.

author: Benta
author_rank: Super Contributor
timestamp: June 19, 2017, 04:44:44 pm
content: Judging by the OP's post #4, there are no Zeners in the circuit at this point...

author: oldway
author_rank: Super Contributor
timestamp: June 19, 2017, 05:14:34 pm
content: QuoteThat's a dangerous game...Are sure you have enough experience and knowledge to "play" this game....Reading your post, it seems that you don't.....300V , a 1µF capacitor and an inverter of some 5W or more, this may be a little more dangerous than "a nasty shock"...Anyway, never let an inverter circuit be powered on without his protection circuit....Good practice would be to feed separately the center tap of the transformer with a low voltage (from 0 to 5V) to test safely the circuit.I note that you have not taken this precaution.

author: tatus1969
author_rank: Super Contributor
timestamp: June 19, 2017, 10:25:07 pm
content: Quote from: Paul Price on June 19, 2017, 03:50:10 pmRight, I am taking that back. Strange circuit though, don't want to see output ripple of it

author: NMNeil
author_rank: Contributor
timestamp: June 19, 2017, 11:14:30 pm
content: Quote from: oldway on June 19, 2017, 05:14:34 pmI shall heed your sage advice.

author: NMNeil
author_rank: Contributor
timestamp: June 19, 2017, 11:20:05 pm
content: Quote from: Paul Price on June 19, 2017, 04:38:31 pmI don't have any 75 volt zeners yet, so the project has been put away until they arrive.

author: Paul Price
author_rank: Super Contributor
timestamp: June 20, 2017, 03:05:42 pm
content: You can substitute the 75V series zener circuit by substituting a voltage divider for regulation feedback.1) Parallel or change the 56K resistor to ground at the base of the transistor Q3 with a resistor of 560 ohms.2) Instead of the 4 zeners, connect a 220K 1/2 W resistor in their place(or else use (2) 100K 1/4W in series).This will give you voltage regulation with some voltage change over temperature.

author: strawberry
author_rank: Super Contributor
timestamp: June 20, 2017, 05:16:11 pm
content: Poor designed PUSH-PULL converter.Missing RC snubber on MOSFETs and improper risetime can cause high voltage spike ringing

author: Buriedcode
author_rank: Super Contributor
timestamp: June 20, 2017, 06:45:05 pm
content: The transformer core material, and even size, is critical in this design. The self oscillating push pull driver with those values looks like about 21kHz. Whether or not that core saturates with the given voltage and frequency depends on its material.What transformer core are you using?

author: NMNeil
author_rank: Contributor
timestamp: June 20, 2017, 07:44:16 pm
content: Quote from: Buriedcode on June 20, 2017, 06:45:05 pmI rewound a ferrite core transformer from a 400 watt PC power supply. The bobbin had much the same dimensions as the ETD29 specified in the article.

author: strawberry
author_rank: Super Contributor
timestamp: June 20, 2017, 08:18:48 pm
content: ETD29 3C85 Al=2350nH/t^2Primari winding inductivity 395uH@13t

author: NMNeil
author_rank: Contributor
timestamp: June 20, 2017, 08:32:21 pm
content: Quote from: Paul Price on June 20, 2017, 03:05:42 pmThanks for the input. The IR2155 has been replaced by the IR2153 and I just got the ones I ordered today. The datasheet shows that a mosfet can ground out pin 3 (CT) to shut the chip down completely.So using your suggestion I will try a voltage divider adjusted to about 19 volts and use an 18 volt zener to feed the gate of a 2N7000 connected across the CT pin and ground.I have to make my setup a little safer first because when I took it apart last night I discovered that when the capacitor discharged it was similar to getting tazed.And before you jump to conclusions, no I'm not a bad guy, I was a cop and getting tazed was part of our training.

author: Paul Price
author_rank: Super Contributor
timestamp: June 20, 2017, 10:56:54 pm
content: The IRF datasheet sez grounding the Ct pin will disable both outputs, not disable the chip entirely. The shutdown is very fast <600nS and this means that double-pulsing of the same MOSFET that was switched on before shutdown would be possible(causing excessive current in the same MOSFET due to current rise to > core saturation), but this undesirable side-effect depends on how fast the chip is enabled again after shutdown by your feedback idea or if there is dithering or oscillation around the Ct output shutdown threshold voltage.Note: this new way of shutdown is different from undervoltage shutdown which has a 1-V hysteresis which slows down the action in this control game. The undervoltage shutdown is a hiccup-mode shutdown which is very much slower to turn on again once the HV output has dropped considerably below the 300V regulation voltage set point.If the core you are using has a magnetic gap, this would explain the extra high voltage reading, which can also be caused by poor magnetic coupling(insulation thickness or the bifiliar wound pri wires not wound close and tight against the core body).The IRF2153 is set to oscillate with the Rt and Ct values shown at approx 20KHz. I can only guess from experience, but I would guess that the core you salvaged from a working SMPS would have been designed to be operating at least at a switching freq >60 Khz.

author: amyk
author_rank: Super Contributor
timestamp: June 21, 2017, 02:04:01 am
content: Too high? I thought ignition systems would just use as high a voltage as needed to fire the plugs... I also seem to remember at least one car manual stating to never run the coils open-circuit.

author: Buriedcode
author_rank: Super Contributor
timestamp: June 21, 2017, 02:27:39 pm
content: Quote from: amyk on June 21, 2017, 02:04:01 amI believe the OP is referring to the 300v cap charging circuit. CDI's discharge a high voltage cap (well, medium-high voltage 200v-1kV) into the primary of an ignition coil.It is true you should never leave the secondary of an ignition coil open circuit as it can cause internal breakdown (just like a flyback, the output is a current source that will create almost any voltage necessary for current to flow). Given they are pretty well insulated, I have seen >1" arcs from these, but at some point it'll breakdown between the windings if the spark gap on the secondary is too large.To the OP. You could look for other 'cap charging' circuits to replace the self-oscilliating push--pull part. ~300V is generally what photoflash units and strobes charge to, and these tend to use flyback converters, as its generally a more efficient way of charging capacitors, doesn't require a specific turns ratio (1:6 - 1:20 will do), and alllows one to use off-the-shelf boost converter chips.The downside, again, being the transformer. It requires you to know its Al value, so you can control the primary inductance and calculate peak current for a given on-time. For a simpler approach, there are of course self-oscillating flybacks, and as long as they have enough through-put to charge your CDI caps between firing, they will do the job just fine.

author: NMNeil
author_rank: Contributor
timestamp: June 21, 2017, 03:16:52 pm
content: Quote from: Buriedcode on June 21, 2017, 02:27:39 pmThanks, I had a search for high voltage capacitor charging circuit and got lots of information.The very reason for using the circuit design from the article was that, being new to electronics it should have been the equivalent of painting by numbers or joining the dots, just follow the instructions and you can't go wrong sort of thing.But another lesson well learned because this is obviously not the case.Again thanks to all.

author: Buriedcode
author_rank: Super Contributor
timestamp: June 21, 2017, 04:08:16 pm
content: I wasn't trying to put you off by over complicating matters, or bash that CDI design - I've never designed, built or tested a CDI myselfIts quite a complicated project for someone new to electronics, and I can understand it can be a but off-putting when things dont' go to plan. Although its hardly black magic, when it comes to transformers, things can get complicated. There's the obvious turns ratio, inductance, but also saturation, core losses etc.. When I first started out in electronics I wanted to build xenon strobes and I gave up purely because the transformers I wound never worked right. Took a few years before I had working DIY xenon flashes.Winding your own should be straight forward, but rarely is. I guess if you can get the exact core size and material mentioned in the article, then it *should* work out with the given number of turns for primary and secondary. But failing that, you may have to measure the inductance of the primary you've already wound and report here. Measuring inductance if you dont' have an LCR meter is possible, but I fear it will send you off on a "tangent project". If you have access to a function generator and a scope:http://www.dos4ever.com/inductor/inductor.htmlif not, I'll see if I can find a 'quick and dirty' breadboard way of getting a rough idea of inductance.Another option would be to buy one. Chances are you won't find one with the exact spec, which is why I suggested using an off-the-shelf flyback circuit, but again, its just moving the problem elsewhere.So whilst I may be missing something (your transformer coudl be fine and I've missed some other problem) I think getting to know more about this transformer would be a way forward.I just checked that pdf again. It mentions ETD29 bobbin, with a 3C85 core.According to a philips datasheet, that core has an Al value of 2350nH. Al = L/N^2, where L is inductance, and N is turns. It says the primaries are 13 turns each.L = Al * N^2 = 2350 * (13^2) nHL = 397150nH = 397.15uH = ~400uH.That is the sort of inductance your transformer should have on one of its primary windings.