8A load transistor/FET for 555 output

Discussion in 'The Projects Forum' started by lvgforums, Jul 12, 2015.

  1. lvgforums

    Thread Starter Member

    Jul 12, 2015
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    I'm very frustrated at this point. First I'm hardly educated in electronic circuit design so that doesn't help, second I've been searching for several hours without any luck so I've decided to post.

    I have a 555 circuit I'm working on which will take a 1/2 second input 12V pulse and turn on 4 lights for 1/2 second as well. The draw from these 4 lights is 8 amps for about 1/2 of a second.

    My pin 3 output obviously cannot handle 8A load, so I've discovered I need a transistor or pMOSFET on the output of pin 3 to trigger something (acting like a relay) that will provide 12V and can handle 8A draw. Radio shack does not carry a pFET, and I want to get this going immediately so I read that a transistor will do the same thing. But what kind of transistor? There are so many.

    simplified:
    555 pin 3 send it's little 200ma 12V signal to (insert what I need here) to open a circuit that sends 12V to 4 incandescent lights that draw 8A. The 555 is only "on" for 1/2 of a second from it's own trigger and will turn off after the initial pulse (except to discharge the 1/2 output)

    An N mosfet won't work because I can't turn on lights with a ground. I need a +12V signal.

    Thank you all, hope to get to radio shack soon.
     
  2. pwdixon

    Member

    Oct 11, 2012
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    If you wanted to use a n channel mosfet you could invert the 555 output with a second smaller n channel mosfet first.
     
  3. lvgforums

    Thread Starter Member

    Jul 12, 2015
    51
    1
    Hello, thank you.

    I eventually would like to take this to PCB, so I'm trying to keep it simple for my inexperienced mind.

    If I was forced to get something from radio shack, what transistor would I use? I understand you are telling me I can go there and get to n channel MOSFETs, but what can I use that's one piece and cheap (at this stage)
     
  4. lvgforums

    Thread Starter Member

    Jul 12, 2015
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    Two*, sorry.
     
  5. pwdixon

    Member

    Oct 11, 2012
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    Radio Shack is not likely to be the cheapest supplier, having said that your 8A capable mosfet will be perhaps $1-$2 from a real supplier and a small mosfet to invert the signal would be a few cents.
     
  6. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    What provides this pulse? You could perhaps use it directly to turn on a N-FET, thus avoiding the need for a 555.
    Or do you want the lights on for 1/2 sec after the 1/2 sec pulse has terminated?
     
  7. wayneh

    Expert

    Sep 9, 2010
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    I'm with Alec – can you show us a schematic or at least a block diagram of what you want to do? There might be an easier approach.

    I also recommend getting away from Radio Shack. Try Mouser, Newark or DigiKey. With just a little effort to order more items all at once to save on shipping, you'll get a huge range of modern parts at competitive prices. Think of the gas you'll save not driving to the Shack.
     
  8. ian field

    Distinguished Member

    Oct 27, 2012
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    If you mean filament lamps; you should be aware that the filaments have a PTC characteristic, which means the resistance when cold will be much lower than you calculate from Amps & Volts when the lamp is at running temperature - this means there is a turn on surge somewhere around 12x the normal operating current at the instant of switch on.

    Bearing in mind the switch has to overcome a turn on surge, a straight bipolar transistor will probably be close to the minimum gain you can get away with, a Darlington would be a better choice.

    A MOSFET would be easier, its gate takes no current to drive but you must have at least 8V of drive to ensure the device switches fully on.

    N-channel MOSFETs are better, cheaper and easier to get than P-channel types - so if you can modify your 555 circuit to invert the mark space ratio, you can use N-channel instead of P-channel and get a better end result.
     
  9. ian field

    Distinguished Member

    Oct 27, 2012
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    The TS is working at 12V - easily within the 30V or so rating for MOSFETs on scrap motherboards - they are usually rated at least 50A, 70A is quite common and I've even seen one or two rated 80A.

    Possibly free - unlikely more than scrap metal price for a pile of old boards from a small independent computer shop.
     
  10. lvgforums

    Thread Starter Member

    Jul 12, 2015
    51
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    I've explained this to about 7 people now on various forums, but they all drop out leaving me hanging, so I'm hesitant to do it all again, so I've been shortening the example of what I'm doing with each pass. Maybe the best minds are here and can help me out so I'll write it all out.

    I'm too inexperienced to write out the circuit. I don't know what I'm doing. I wish I could talk to someone over the phone because I'm going in circles.

    Here's my best explanation.

    My car has remote unlock and lock. My first revision of this project, I just took a wire from the 12V door unlock solenoid pulse in the kickpanel. It's 12V for about 1/4 of a second, if that. So I took the wire, ran it up to the turn signal + wires, put 2 diodes on it so the 12V wouldn't jump over back to the lock when I used turn signals or the emergency flasher. It worked fine. When I unlocked the car, all hazards would flash like on a modern car. The hazards draw about 2A each (like the previous poster said, it may be much higher when the car is cold, so that's news to me), but I'm only flashing them for less than 1/2 second.

    I then thought, well on newer models, the door ECU has a 12V out to flash the hazards, and it flashes twice on unlock, and once on lock (Forget the lock for now). So I asked around, and someone said use a 555 time in monostable mode, find the right caps and resistors to add a delay and flash the lights with a second pulse.

    However I'm aware enough that I know that a 555 timer can't output 8A, so I need a transistor or MOSFET to act like a relay, and take the 12V that can handle 8A and turn on from the 555 output.

    It will work something like this.

    Door unlocks w/ remote
    12V pulse goes from unlock wire to hazard lights drawing from the same wire (first flash)
    That same 12V pulse, also turns on the 555, which adds (approximate) 1/8 delay and then flashes the lights again for 1/4 second, imitating the modern factory unlock system. The key here is that the second light pulse needs to come from the car's 12V source not the 555 since the amperage load is so high.

    So people have been telling me all kinds of variations of this, don't really have an exact answer, and unfortunately I'm an extreme novice with this kind of stuff. I went out bought a breadboard and some resistors and capacitors and have been tinkering with youtube videos which don't really explain how this works. I've read SEVERAL articles about the 555, but I', just not familiar enough with these concepts to grasp it.

    The "positive" side to this is that I'm driven to do this and want to get it done myself no matter how many tries it take. I want to solder it together on a hobby board at the end too.
     
  11. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    For clarity, let's restate the problem. Your system has two parts that can be considered independently of each other to get a handle on things, then maybe combined later to simplify implementation.

    First, you need a current amplifier stage. In post #1 you were clear about switching the +12, not GND. If that still is true, then switching the 8 A directly with a 555 will take either a P-channel power MOSFET or a PNP bipolar power Darlington transistor like a TIP145-147. Another option is for the timing stage to drive a relay with an N-channel or NPN transistor, and let the contacts handle the 12 V.

    Second, you need a timing stage. You actually need two pulse circuits. The trailing edge of the car's +12V door lock signal starts a 0.2 s delay timer, and the trailing edge of that delay starts the 0.5 s output pulse for the current amplifier stage. This can be done with two 555's or one 556 (yawn), a CD4093 (classy), or a ULN2004 (*major* geek cred!).

    ak
     
  12. lvgforums

    Thread Starter Member

    Jul 12, 2015
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    OK, I'm starting to throw money at this because I don't understand it and I'm getting frustrated. I just went to radio shack and bought the following:

    Before I list all the stuff, are you sure I need two 555's (or a 556) ? I ask because I'm already getting ("stealing") my first pulse from the unlock +12V signal to make the light flash the first time, that same pulse will be used to turn on the 555, generate the delay and flash the 2nd time. Assuming I do need two, why can't a capacitor be a source of that 0.2s delay? You can see how little I grasp these concepts with my questions, I'm sure.

    I'm normally a patient person with people and situations, but I'm completely past level 10 with how complex it is to make a light blink once. Why is it so hard to make a light blink?

    Onto the list:
    1N4001 diodes
    IRF-510 MOSFET N channel
    555 (I have two now)
    556 (just in case I do need 2 and can simplify)
    20 pack capacitor variety
    100 piece carbon 1/2W resistor pack
    TIP120 darlington transistor (because someone mentioned it)
    0.001uf ceramic cap
    +some other stuff I needed 20150712_164042.jpg
     
  13. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    This should do what you want :-
    DoubleFlash.gif
    U1 is triggered by the trailing edge of the Unlock pulse and generates a ~0.25 sec pulse. The trailing edge of that triggers U2 to generate another ~ 0.25 sec pulse. The latter is diode-ORed with the Unlock pulse to turn on a N-FET and so pull in an automotive-type relay to switch the lights.
     
  14. lvgforums

    Thread Starter Member

    Jul 12, 2015
    51
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    Thank you for your time.

    Is the first capacitor 33uf? I looked on wikihow and N denotes a 0.05% range of what is appropriate as a substitute. Is that correct? Can I ask why not just write 33uf instead of n?

    Now, that is 3 flashes? or 1 input pulse, then 2 output? Like I said, I'm already getting my first flash just by tapping into the initial pulse itself, then just need one more flash from the 555.
     
  15. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    33n denotes 33 nanofarads. That's 0.033uF. Note there are also 100n caps used.
    When the circuit is first connected there is an additional flash. The 3 flashes initially are only at power-up. Extra components on the Reset input of the 555 could be used to prevent that if desired (but I wouldn't bother).
    A possible problem with just tapping the lights off the Unlatch motor wire is that you will be asking that wire to carry an estimated ten or more times the current that it was designed to carry. So far you've been lucky but, depending how generous (or not) the car manufacturer has been, there is a slight risk of overheating the wiring harness.
    The circuit I posted was done in a bit of a hurry, late at night. Here's an improved version which has been beefed up to allow for inevitable load-dump spikes on the 12V supply.
    DoubleFlash2.gif
    TVS is a transient-suppressor diode. The generated pulses are also slightly longer in case the relay is sluggish. R3/6/8 may need changing if the timing isn't quite what you want. In principle the relay could be replaced with a P-channel MOSFET, but that would need protection from voltage spikes so might complicate the circuit further.
    BTW, you should make certain that your car insurance is not invalidated by the wiring modifications.
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    I was going to point out that the input pulse negative edge is coming from whatever the load on that 12 V circuit is when the 12 is disconnected by the door unlock timer, and its impedance will affect the fall time of the edge that triggers the first 555, when another thought hit me. If the input 12 V pulse is coming from a relay or motor, then the trailing edge that triggers the first 555 will be a negative voltage spike from the inductive kick. You might want to diode-clamp the input.

    ak
     
  17. lvgforums

    Thread Starter Member

    Jul 12, 2015
    51
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    This is how far I got last night, took about 2 hours. Again, this is all way over my head and I don't know if I'm even connecting these things correctly. One problem is that these capacitors (which are the wrong ones anyway) are +/- and I have to figure out how to stretch them across the board. Also, I'm running out of plug holes where R3 and R4 resistors are. I can't fit 3 resistors, 2 capacitors and a diode in 4 holes. It just doesn't fit. I don't understand.

    I went to RS, but bought the wrong capacitors again, I bought .22 instead of .022. This is quickly turning into a failure. Maybe I should just go back to my old single flash design and put a relay on it so the amperage draw isn't on the wire and forget this circuit. I'm just not educated enough about electronics. I said I wouldn't give up on it, but I'm just in over my head.

    About the wiring amperage draw, the "dome" fuse controls the locks and is 10A. Also, the door unlock wires in the harness are much larger than the other wires in the block.

    Good note about the insurance, but I suppose that would lock out anyone who installed a custom radio / headunit. I wonder how they deal with it.
     
    Last edited: Jul 13, 2015
  18. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    When a circuit node has more connections than there are holes in a socket strip, cut a short piece off of a resistor and bend it over to make a jumper to connect two rows. Now there are 6 free holes electrically connected. If you need more, do it again. Each additional jumper adds 2 holes.

    Buying parts before the design is fixed achieves little except creating the added frustration of wasted time and money. Many of us here have basements full of parts because we've been doing this for a decade or 5, so we can putter around until things work. You don't have that luxury. Slow down, keep working with us, get a circuit design we agree on, and *then* go shopping. Some of the threads on this forum are over 100 messages. At the risk of sounding like a pompous ass,

    Patience, grasshopper.

    ak
     
  19. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Good point, AK. Don't you just love all the things that automotive can throw at you ;). I'm guessing the unlocking motor draws an amp or two which drops fairly sharply at switch-off, so that's another 1N4007 (or similar) diode needed, reverse biased, from the 'Unlock' input to the ground rail.
    Neither of those values is in the circuit I posted. There is, however, a 220uF electrolytic cap (the only polarised capacitor and which should be 25V rated).
    Just use a jumper wire to link two rows of holes.
     
  20. lvgforums

    Thread Starter Member

    Jul 12, 2015
    51
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    I'm been sitting at my table for almost two hours now, pulling my hair out trying to connect whatever I have according to the first diagram and I'm stuck.

    Here's what doesn't make sense on the diagram - if you don't want +12V from the VCC touching pin2, why is there a diode AND a resistor on the +12V rail? The diode isn't doing anything to prevent 12V from coming in from the VCC because the resistor is just passing the VCC over.

    Here's an edited version showing how the electricity passes - step 1, goes from VCC hits the diode, step 2 gets goes over to the resistor, step 3, goes right into pin 2 through the resistor.
     
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