From: praetzel@maxwell.uwaterloo.ca (Eric Praetzel)
Subject: Re: Help! Trickle-charging a gel-cell?
Organization: University of Waterloo
Date: Tue, 12 Jan 1993 21:16:00 GMT

In article <1993Jan12.163316.29715@osf.org> david@postman.gr.osf.org (David George) writes:
>Lead Acid battery charger (from Horowitz and Hill:)

                             - | / | + D1
                     ----------|<  |------------
                    |          |  \|            |
		    |                           |
>                   |     LM317T       0R2      |
>                   |     +-----+     ------    |
>          Vi o-----+-----|i a o|-----|    |----+-------o  Vo
>                 + |     +-----+     ------    |
>           1000uF ===       |                 | |
>            elec   |        |                 | | R1
>                   |        |                  |
>                   |        +------------------|
>                   |                           |
>                   |                          | |
>                   |                          | | R2
>                   |                           |
>                   |                           |
>           0 o-----+---------------------------+-------o
>
>
>The LM317T tries to maintain 1.5v between the adjust and output
>terminals.  With the simple resistor divider network shown:
>
>
>        Vo = 1.5(1 + R2/R1)
>
>The adjust terminal draws a minimal 50-100 uA.
>
>They add a 0.2 ohm resistor in line, presumably as some current
>limit ?
>
>For a nicad charger you could replace R2 by the nicads, the current
>would then be  I = 1.5/R1.  Vin would govern the number of nicads
>which could be charged in series.
>
>You may wish to replace/add to R2 with a variable resistor, or add
>a 12v/6v changeover.
>
>David.

I added the diode to protect the voltage regulator.  ie if you hook a battery
up without having the power supply turned on you would blow the regulator.
This way it will survive.  It may also help if you add a small capacitor
( 0.1uF) across the input capacitor and another one between the output and
ground.  The size of the input capacitor only needs to be 10uF if the DC
source is stable.

Note: The above design does include current protection.  The voltage regulator
has its own current protection and overheating protection.  The extra 0.2 ohm
resistor can be used to adjust the current.  By making it larger you will 
reduce the maximum current.  By selecting it properly you can make this into
a current source.
  For example:  Q = R2 / (R1 + R2)
    1.5 V = [Vreg - I * Rx] * Q 
      where I = current out, Rx = 0.2 ohm series resistor (in this case)
      Vreg = voltage regulator output voltage (maximum = Vin - 3V since the
	voltage regulator needs 3 volts to work)
    Then:
      Vo = 1.5 / Q = Vreg - I * Rx

    The upshot is that if I increases enough so that Vreg approaches Vin - 3V
    the regulator is runing out of voltage and the output voltage has to drop.
    This means that the current will become fixed (sort of) and the output
    voltage will drop  ie you have a current source.

    You can forget the math and get yourself a 10 ohm variable resistor that
    can handle about 1A (if you are using 1A max).  Then you will have an
    adjustable current limit ie you can control that point at which the 
    regulator runs out of voltage.  This has major problems though. If the
    input voltage fluctates then the current will change.

    A better solution is to remove that resistor totally and use the regulator
    in its "standard" configuration.  In front of this you put a nearly
    identical circuit.
    In this ciruit (the current regulator) you replace R1 with a short circuit,
    remove R2 and then you have:
      1.5V = I * Rx  where Rx is the series resistor.  
    Because the regulator tries to keep 1.5 V across the resistor Rx it will
    keep a constant current (until it runs out of voltage).
    So all that you need is a 10 ohm variable resistor that can handle the
    current (and power).  Perfect parts for this are 5W wire wound resistors
    .  Watch out beause as Rx goes towards 0 ohms the current will increase
    to infinity (or whatever the regulator will put out [ie normally 1.5A]).
    It is usually better to add a resistor in series so that if the variable
    one goes to 0 ohms the current will be 1A (or some usefull maximum).
    For my charger I can go between 100mA (nicads) and 500mA (gel cell).
    That means that:  (using R = V / I)
      1.5 V / 500mA = 3 ohms (my minimum resistance).
      1.5 V / 100mA = 15 ohms (maximum R).
      So I use a 3 ohm fixed resistor (500 mA * 1.5 V = 0.75W)
      and a 15 ohm variable one (pick a 0.75W as above to be sure its safe).
Thats about all.  Generally the higher the power rating of those two resistors
the better the circuit will be (ie if they can't handle the power they will
go up in smoke).

For added protection you may want to put another diode in series with the
positive output lead.  This will protect the regulator from people plugging
the battery in backwards and destroying it or the power supply.  In both
cases the standard 1N4001 (aprox 1A) diode works well.
- Eric 
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