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What does IC555 do in this circuit?

  1. Apr 27, 2017 #1
    This is the circuit diagram of a solar charge controller. This circuit is fed by a wind mill or a solar photovoltaic panel. Either wind mill or solar PV panel work at any instant controlled by switch SW1. I don't understand what is the working of IC555 here. The main objective of this circuit is to charge a 12V 7AH battery. The whole circuitry is designed to prevent overcharging. IMG_20170427_170233_349.jpg
     
  2. jcsd
  3. Apr 27, 2017 #2
    It's a pulse generator, to increase the charging voltage. When the transistor is open, C3 gets charged; and when the transistor is closed, the C3 voltage is added to the initial voltage, so the battery is charged even when the initial voltage (the potential of the upmost wire on the diagram) is much less than 12V.
    Overcharging is prevented by the choice of the R5 value limiting the current so it be harmless even when the battery is full.
     
    Last edited: Apr 27, 2017
  4. Apr 28, 2017 #3

    jim hardy

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    The only thing in it that could possibly prevent overcharging is D3 if it is a Zener of around 15 volts. Whenever solar cell or windmill voltage exceeds battery voltage plus three diode drops, current flows into the battery via D1, D2 and D4.
    D3 can steal that current to prevent overcharging.

    Without pin numbers we have to assume your 555 is wired as a free running oscillator.
    From its datasheet at http://www.basicanalogcircuits.com/Session_4_files/LM555.pdf
    looks as if it'll oscillate at a few hundred hz.
    555charger.jpg

    Can you figure out its duty cycle from that datasheet? You really ought to learn the 555, it's a fabulously handy device.

    Your circuit uses the 555 just to switch that un-named transistor on and off a few hundred times per second.
    Alex described that,
    but
    his choice of words "Open" and "Closed' are not typically what we use in electronics.
    That transistor is just a switch.
    A switch that is "Open" will not pass current, unlike a water faucet that is open. A switch that is "Closed" will pass current, again unlike a faucet.
    To avoid that confusion we use ON and OFF or CONDUCTING and NOT CONDUCTING when speaking of electronic switches,
    we reserve OPEN and CLOSED for mechanical switches and relay contacts.
    SO ----- Alex's otherwise excellent description might lead you to the wrong conclusion.

    To your question - the 555 only drives that transistor switch.

    As Alex said, when that transistor switch conducts the 47 uf gets charged to supply voltage.

    When that transistor turns off, R5 pulls capacitor's left side up to supply voltage AND its right side rises by an equal amount.* Therefore it will exceed supply voltage, reverse biasing D1 and overcoming battery voltage pushing charge into the battery.
    * (That's the secret behind all charge pumps. )

    So the purpose of the 555 is NOT to prevent overcharging, D3 should do that. Check its part # and see if it's not about a 15 volt zener .
    The purpose of the 555 is to make the 47uf capacitor pump charge into the battery especially when solar cell or windmill voltage isn't enough to do that by itself.
    Here's an example at 8 volts, maybe a cloudy windless day:

    555charger2.jpg

    Write KVL around the current loop both ways and you'll see the circuit can make almost 16 volts from an 8 volt supply.
    Let's say that 500 times a second we charge 47 uf to 8 volts then dump it into the battery. How much current is that ?
    47 uf X 8 volts = 376 microcoulombs per cycle X 500 cycles per second = 0.188 amps
    Not Bad, eh ? It'll trickle charge on a cloudy day !
    Actually you get not quite that much current because the capacitor can't completely discharge, and KVL will tell you by how much. Ahh the power of Kirchoff !

    We call that circuit a "Bucket and Ladle", as if we were filling a bucket (the battery) with a ladle(the capacitor) from a reservoir below(the solar panel or mill) by repeated lifting of charge .
    Electronics guys use word pictures like that to remember concepts. You should work through the circuit using your circuit analysis skills to cement it in your mind.

    Good luck with your project. Develop the habit of working circuits in your head, pretend you're inside them to understand the mechanism, then apply math to put numbers on it.

    @Tom.G Can you polish this scraggly sprout of an explanation ?

    old jim
     
    Last edited: Apr 28, 2017
  5. Apr 28, 2017 #4

    jim hardy

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    Thanks Alex !
     
  6. Apr 28, 2017 #5
    Thank you so much Jim. Can you explain a bit more about preventing the battery from overcharging. And kuddos to you. That actually is a Zener diode of 15v
     
  7. Apr 28, 2017 #6
    So basically, the IC 555 is used to generate the pulse required for making the transistor ON and OFF.
     
  8. Apr 28, 2017 #7
    Are you kidding, Jim? You nailed it again.

    The only "improvement" I can suggest is in the circuit. To protect the circuit from a battery being installed backwards, install a 1/2 Amp to 1 Amp fuse in series with D4.
     
  9. Apr 28, 2017 #8

    Averagesupernova

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    The way I see it you cannot get much current through the 1K R5 resistor. ALL of the charging current needs to go through this resistor. Making it smaller will allow more current but this scheme is wasteful to start with. It is a starting point for learning, that's about it.
     
  10. Apr 28, 2017 #9
    Ooops, you are right.
    (talk about overlooking details! :frown:)
     
  11. Apr 28, 2017 #10

    jim hardy

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    What is duty cycle? How far can the 47 uf discharge in one transistor OFF period ?
    Time constant is 47uf X 1k = 47milliseconds. Looks like there's no benefit operating it above maybe 20 hz .

    That's how we learn. Get a circuit built, then get it working, then get it working great.

    Sorry for missing that detail .

    old jm
     
  12. Apr 28, 2017 #11

    jim hardy

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    One charges Lead-acid batteries by applying constant voltage. The zener diode clamps the node at about 15 volts , D4 drops about 0.6 volt, leaving ~14.4 across the battery. That's just abut right for a 12 volt lead - acid, 2.4 volts per cell
    from Battery University at http://batteryuniversity.com/learn/article/bu_214_summary_table_of_lead_based_batteries
    upload_2017-4-28_20-40-43.png



    . Check your car voltage with engine running....
     
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