Float charger project

Discussion in 'General Electronics Chat' started by tracecom, Dec 11, 2010.

  1. tracecom

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  2. Kermit2

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    [​IMG]

    ??

    I think a more useful addition than the 5 volt reg. would be a comparator and a simple temp sensor to shutdown the charger on an overheat detection.

    OR a timer circuit to stop charging automatically
     
  3. timrobbins

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    tracecom, what do you think is wrong with it? viceversa, what do you think is right with it?
     
  4. Wendy

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  5. tracecom

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    @Bill,

    Thanks. I had looked at this, but had a few questions.

    Is the final design in post 12? What are the wattages required for the resistors? And did anyone ever build it or is it a simulation only project? I would like to build something that has been proven in actual use.
     
  6. kubeek

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    If I were to buy a pretty expensive battery, I wouldn´t charge it with a circuit that merely relies on your ability to see the LED to prevent the battery from overcharging. Use some more intelligent circuit that actually stops the charging when its done.
     
  7. Wendy

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    As far as I know it has never been built, but I trust it a lot more than the design I'm seeing on the 1st few posts. You need a constant current that is limited to a set voltage, which I don't see in your original design. All I see is a voltage regulator with a back current prevention diode, and the lack of any current regulation is a serious design flaw, and the sense when charged is not automatic, which means if you forget to check on it goodbye battery. It is also depending on the overload protection on the LM317, which is not a good idea. I do not always have to build circuits to see their weaknesses, that is a very flawed design.

    This design is a constant current source until the current is cut off by the load, at which time it transitions to a constant voltage source. Max current is limited to 100ma. I added the switch so you could calibrate the battery voltage to the regulator voltage.

    [​IMG]

    What Wookie and others added was a simple temperature compensation scheme. I'm none too sure about it, but I trust the guys recommending it.

    Post 11 has a proven design, Wookie's has an enhanced version.

    Unless otherwise stated, assume ¼W for all resistors.
     
    Last edited: Dec 12, 2010
  8. SgtWookie

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    Schematic from post 12:

    [​IMG]

    That was as far as I took it.
    The charge current is determined by R4.
    R4 = 0.62/ChargeCurrent, where 0A <= ChargeCurrent <= 1.5A

    R4's wattage should be >= 0.62*ChargeCurrent*2

    Note that power dissipation in the regulator can be quite high, depending upon the input-output voltage differential across the regulator, and the charge current. You'll need to calculate it as (Vin-Vout)*(ChargeCurrent+10mA), and ensure that the regulator's heat sinking can dissipate that much heat.

    Ideally, Q2 should be thermally coupled to the battery positive terminal.

    One item that has not been included in the schematic is a diode to prevent discharging the battery if the input power is turned off while the battery is still connected.

    To set the float voltage, connect a capacitor (10uF-1000uF) in parallel with a ~3k to 20k resistor across the BAT+ and BAT- outputs; nominal float voltage should be 13.6v to 13.7v when at 25°C/77°F. See the table in post #38 in the Tips and Tricks thread, General Electronics Discussion forum for other temperatures.

    R4 needs to be calculated as described above. For higher currents, Nichrome wire wound on a ceramic cylinder works quite well if you want fairly precise control over the charge current. You could also use power resistors in a series/parallel configuration, or simply select one that's close to the charge current you need. 1/4 Watt would be fine for the remaining resistors .

    I've built a very similar circuit to charge a cell phone battery at a low current, since I didn't have a method to monitor the battery temp at the time. It worked just fine.

    It should work just fine, if the cautions about power dissipation are followed.
     
  9. Wendy

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    I was thinking of asking you to look at this thread, because I have a question.

    Where does the temperature compensation come from?

    R4 and Q1 fold the current back, this I can see. But how does Q2 sense temperature and adjust for the battery? I'm thinking thermal runaway (or something similar) but you have the appropriate feedback on the base resistors R5 and R3.

    You mentioned the current limiting on this design wasn't linear.

    [​IMG]

    I don't see why it would need to be linear on current, the idea is to trickle charge a battery in use. I accept temperature compensation is something different, but as a base charger I vaguely remember you recommending the configuration for another float charger on another thread.
     
  10. SgtWookie

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    The base-emitter junction of Q2 will have a lower Vf as the temperature increases, which will cause Vce to decrease, which will cause the regulator's output voltage to decrease. This negative temperature coefficient is just what's needed for lead-acid batteries.

    Yep, it isn't. Basically, you want roughly C/5-C/10 (C=AH capacity) current flow into the battery until it's nearly at the float charge level. This design:

    ...starts decreasing the current right away, even if the battery is nearly dead.

    Take a look at the attached schematic/simulation. The green trace is the current going into the battery, and the blue trace the battery voltage. I had to adjust R13 downwards in order to get a reasonable amount of current into the battery, and that caused the "knee" to go away completely. If you only wanted a few mA (I'm talking 10mA-20mA current) into the battery, then you could use a larger R13 - but that would not be sufficient for a gel cel or other SLA battery.

    Note also that I increased the input voltage. In order for the battery float voltage to be 13.7v, you'll need at least 13.7v+1.7v=15.4v input. If temperature compensation is desired, you'll need more like 17v in.
     
    Wendy likes this.
  11. SgtWookie

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    I updated the schematic a bit; added a 1N5820 Schottky diode between the charge circuit and the battery (you could use a 1N5817-1N5819 for the low current you'll have; the 1N5820 is a 3A 20v diode) and notes to make it easier for people to figure out what's going on; extended the plot to show the initial charge and differences over temperature, with a legend in the schematic itself so you can see which plot line represents voltage over temp.

    C1 has been decreased to 10nF in order to avoid overshoot.

    Note that the slight non-linearity in charge rate is due to the Vf of the Schottky diode over the decreasing forward current as the battery charges.
     
    Last edited: Dec 12, 2010
  12. Wendy

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    How do you tweak R2 to the correct value?
     
  13. SgtWookie

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    That's in the 2nd note down, top right.
    Connect a 100uF-1000uF cap and 2k-10k resistor in parallel across the B+/B-, and measure across them with a voltmeter while adjusting R2.
     
  14. spinnaker

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    Sorry if I my hijacking this thread. But I have another thread with very similar questions and it probably is better for everyone if these questions are all in one thread.

    What percentage resistor for R4? The smallest wattage Digikey has at 1 % is 2 watts.

    I realize I have the option to use multiple resistors in series. But if 5% works.

    I would like 1 amp, meaning a .62 ohm resistor. How do you calculate wattage for this resistor? For 1 amp is it .62 x 1 = .62 watts.
     
  15. SgtWookie

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    The current formula is not exact, as the Vbe of Q1 will vary somewhat over temperature and from transistor to transistor.

    Have a look at the attached simulation; the battery charge current is plotted over three temperatures: 0°C, 25°C, and 50°C, the variance in the Vbe of Q1 over temp accounts for the differences. I selected a 2.07 Ohm resistor for R4 to get a 300mA charge current.

    Note on the red plot (high temp, 50°C) that the power dissipation in the regulator causes thermal shut-down until it cools enough to resume charging. There are no provisions in the LM317 model to connect a heat sink.

    The LM317 SPICE model itself (not the real LM317) is limited to around 800mA output current, which is why I didn't model it for 1A output.

    An 0.62 Ohm 1 Watt resistor would be OK; 2 Watts would be better.

    You can make your own power resistors from Nichrome wire and a suitable form; preferably ceramic or other heat-resistant media. I did that for a charger I whipped up for my father-in-law's cell phone on his last visit; he'd forgotten his charger and needed his phone available on his arrival back home.
     
    Last edited: Dec 13, 2010
  16. spinnaker

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    Your amazing! :)


    What are some good common items to use for a core?


    How do you calculate the wattage? I know P=I*E but what voltage and current do I use? Do I use.62 volts * the rated current?
     
    Last edited: Dec 13, 2010
  17. SgtWookie

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    I just started in electronics a few weeks before you did. :cool:

    I picked up some ceramic tubes that have tinned ends from Skycraft Parts & Surplus here in Orlando. They were just perfect for what I was wanting to do.

    You could use a power resistor (the rectangular ceramic/cement ones) of a much higher value than you wanted, and wrap some Nichrome around it, using the square corners as a form. I suppose you could even use a piece of chalk, but that would be brittle and somewhat messy.

    If you use a 0.62 Ohm resistor for R4, Q1 will keep the voltage across R4 from rising above ~0.62V nominally; I(R4) is 1A; P=EI, so P=0.62x1 = 620mW. You double that for reliability's sake, so 1.24 Watts. 1 Watt is ~81% of 1.24 Watts, or ~19% less - so you'd really want to go to 1.5W or 2W.

    You don't have to fiddle around with making your own resistor.
    Digikey stocks these:
    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=P0.62W-2BK-ND
    That's just what you need. $0.35+ USPS 1st Class shipping, probably less than a buck total.
     
  18. spinnaker

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    I sure envy you guys that have resources like this. :)

    Just curious, how do you get an exact resistance for such low ohms when you are winding your own?


    I was thinking more like .5 amps. So maybe 5 .25 W resistors that would give me 1.24 (you mentioned that idea in the other thread) and 5 .25 would give me 2.25 watts. I could even bump those up to 1/2 watt I guess.
     
  19. SgtWookie

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    If you were thinking more like 0.5 amps, then why did you say 1 amp in post #14? :rolleyes:

    Since you already know how to get the output that you want (it's been explained several times already), I see no point in continuing this discussion.
     
  20. spinnaker

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    Because in #15 you said the LM317 was only capable of 800mA. So I figure I would dial back my expectations.


    But yes I think I have it handled now. Thanks.
     
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