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Car Battery Bounce Eliminator

Discussion in 'Power Electronics' started by flippineck, Nov 18, 2013.

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  1. flippineck


    Sep 8, 2013
    Problem I have is this:

    I connect the 12V lead acid battery in my car by bolting on the terminal. I think the manual says always connect the negative terminal last.

    When I do, there are loads of sparks and sometimes even the alarm goes off. I imagine the main supply rails all over the car transition from a floating state to a constant, stable 12 Volt potential difference via a period of time where there is significant noise / voltage bounce.

    Everytime I do this, the car's computer subsequently reports errors with the onboard computer network. From talking to many experienced auto technicians familiar with the model of car, these errors are ubiquitous

    I'd like to design a simple circuit which operates something like as follows:

    The device has two jumper leads terminated in crocodile clips.
    You connect one croc clip to battery -ve.
    You then connect the other croc clip to the car's -ve battery terminal lead.
    The circuit then applies a clean bounce free connection, 'soft starting' the car from zero Volts across the battery leads, rising cleanly to the full 12 Volts.

    Once the supply voltage has been cleanly stabilised onto the car's power terminals, one can bolt up the negative terminal to the battery without sparking & bounce etc.

    Once everything's tight, you can then un-clip the croc clips & remove the device.

    I'm wondering if something could be done with a handful of capacitors, resistors, transistor or two?

    It wouldn't have to handle much current, just whatever the car's permanently connected electrics would require.. immobiliser, clock, alarm, ecu etc

    The soft-connect device would be removed before any attempt to draw starting current etc. were made.

    Any ideas how to accomplish this or something like it in the easiest way?

    Would a massive capacitor across + and - leads before connecting them to the battery work or would that be too simple
  2. Boltar


    Nov 18, 2013
    I'm by no means an expert but I believe a capacitor in parallel will block DC until its charged and then allow it to flow normally, essentially causing a gradual rise in supply? Based on that I would think that a sufficiently large and discharged capacitor connected to the +ve of the battery and to the -ve lead before you attach the -ve lead would do the job. When you attach the -ve lead the capacitor should start to charge and cause a gradual rise to the car. I would have no idea how to work out how large the capacitor would need to be though. I may even be totally wrong with what I'm saying, I am after all only a beginner.
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    I'm not sure that you've identified the problem, but if so, that's a good idea, and it's not hard to do what you want. The switching can be done easily by a large MOSFET like this one:

    You just need a drive circuit that will generate a voltage that quickly and cleanly rises from 0V to about +10V while driving a capacitive load (the MOSFET's gate capacitance is about 10 nF). This can be done with a couple of transistors and a toggle switch. I would power it from eight C-cells to ensure there's plenty of current to switch the MOSFET quickly.

    I may be over-engineering here; that MOSFET is rated for 200 amps and 500 watts. But I haven't done this before and I want to make sure it won't fail, even if that means a few more dollars.

    Shall I draw up a circuit diagram? Can you build up a circuit with about ten components on stripboard?
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    How about simply putting a resistor between the crododile clamps to limit the inrush current? Wait a few secconds, then connect battery directly, in effect short-circuiting the resistor.
    As for the value of the resistor, you'd have to experiment a bit.

    Of course, Kris' solution is more elaborated.
  5. duke37


    Jan 9, 2011
    Use a sidelight bulb as the resistor then you can see what is going on.
  6. BobK


    Jan 5, 2010
    Isn't there some kind of problem here, if the car is drawing that much current when it is allegedly turned off?

  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    I thought so too, Bob. That's why I said I wasn't sure he had identified the problem. I guess it depends on what "loads of sparks" means. Everything is relative :)
  8. flippineck


    Sep 8, 2013
    I'm just talking about the normal 'standby' stuff that most modern cars have that is powered even when the key is removed from the ignition. I'm not sure of the exact figures and details but there's a baseline draw of a few tens? hundreds? of milliamps taken by the gubbins that listens for your remote central locking, the alarm, godknows what other computerised stuff.

    When I say 'loads of sparks' - more of a light crackle pop & fizz really. But qualitatively it seems to me that if when you connect the negative terminal, there's a few pops and fizzes and the kind of sparks you'd maybe expect to see if you shorted a pp3 battery, then the supply probably switches on in a very jingle-jangle manner with tons of noise.

    The exact problem with the car (Ford Focus) is that I can't get rid of Ford Diagnostic Trouble Code D900.

    This isn't a code that you read off the car's OBD port - there's no fault being reported there (although I've fixed a bunch, mainly to do with a dodgy diesel particulate filter) - it's a code that comes up on the driver-accessible "Dash Diagnostics" which you access by fiddling with the indicator stalk.

    D900 is listed as "non specific can communications error"

    i.e. The ECU saw that the car's computer network went a bit mental, but it doesn't know why, when, how, why.

    Chatting to various people on forums, my garage, Ford, the RAC man etc, it seems to be the case that pretty much all Ford Focusses report this D900, even when they're otherwise faultless. The conclusion that's been reached is maybe, everytime you connect the battery (disconnecting it for a while and shorting the car's battery leads apparently allows the ECU to reboot to a default state), the noise caused by a ham fisted connect, might itself result in the reporting D900 everytime.

    So I figure if I make sure the connection is accomplished as cleanly as possible, with a smooth uninterrupted rise from 0V to full battery voltage that doesn't ring around, maybe I can avoid the D900 and know I don't need to hunt further for a cause.
  9. flippineck


    Sep 8, 2013
    Yes I should be able to do that.

    The device would only have to handle the same level of current that the car would draw when stood still with the ignition key out. No need to handle the hundreds of amps needed for starting, heated windows etc since, the idea would be to remove the device entirely once the main connection had been established.

    So what would the current requirement be? I'm guessing let's say, 2 or 3 amps max once established.. plus a margin for any initial inrush as capacitors etc get charged. 50A peak? That's a wild guess...

    If I get a chance tomorrow I'll put my ammeter in series with the car's battery & try and find out exactly what the 'static' current draw is when the car's quiet & still.
  10. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    May 8, 2012
    If it's throwing sparks when everything is supposed to be off then I'm in the "There's something wrong" camp too. If you use a brake light in series with the positive battery cable and it doesn't go out in a few seconds then I'm sure there's something wrong.

  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    OK, here's my suggested circuit.


    If anyone suspects that it may be over-engineered, I can put that thought to rest right now: it IS almost certainly over-engineered. I've done that because I really don't know the variables involved in this application and I just want that MOSFET to turn ON as QUICKLY as possible!

    Here's a circuit description.

    Q2 and Q3 form a latch that can be forced ON and OFF by SW1. When SW1 is in the middle position, the latch holds its current state because of the positive feedback via R2; it also holds its state while the contacts in SW1 are bouncing.

    SW1 is a three-position toggle switch with a momentary-off-momentary configuration, i.e. it has a centre position, and the UP and DOWN positions are spring-loaded momentary positions, so that when the lever is released it returns to the centre position. You could also use a standard two-position toggle switch; the important thing is that it's a NON-SHORTING switch, also called BREAK BEFORE MAKE.

    The output of the latch on Q3's collector swings almost fully from +12V to 0V depending on the state of the latch. It drives the indicator LED, and is buffered by Q4 and Q5 and fed to the gate of the switching MOSFET via a 15 ohm resistor which limits the current to about 0.8A to suit the BC337/BC327 transistors specified. If the alternatives, SS8050 and SS8550, are used, this resistor can be reduced to 10 ohms for a slightly faster turn-ON/OFF action.

    The main switching MOSFET, Q1, is specified to switch 100V and 200A (one-off, not sustained) with an instantaneous power dissipation of 500W. Its gate-source capacitance is about 10 nF and it will turn ON or OFF in around 0.2 µs, or somewhat faster if R7 is 10 ohms.

    The Q4/Q5 buffer stage can be omitted if only the transition from OFF to ON is used, as in this case. I included it because others may find this thread later and may want to use the circuit to switch circuits OFF as well as ON.

    Here's a component list with Digikey references.

    BT1: eight C-cells (carbon-zinc or alkaline) in a battery holder
    C1: 1000 µF 25V electrolytic:
    LED1: 5 mm red LED:
    Q1: IXYS ITXQ200N10T:
    Q2,4: BC337: or SS8050:
    Q3,5: BC327: or SS8550:
    R1~4: 10k resistor:
    R5,6: 1k resistor:
    R7: 15 ohm resistor: or 10 ohm resistor:
    SW1: momentary-off-momentary toggle switch: or two-position toggle switch:

    The circuit draws no significant current from the battery when it's switched OFF, so there's no need for an ON/OFF switch.

    Attached Files:

  12. flippineck


    Sep 8, 2013
    Looks good! Seems like just the sort of thing needed but am bleary eyed, just woke up with a rampaging and uncooperative 4 year old to take to nursery here, so I'll come back and have a real good mental munch on it later. Thanks :)
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