Emitter Follower Battery Charger Circuit

Discussion in 'General Electronics Chat' started by Joster, Aug 13, 2014.

  1. Joster

    Thread Starter Member

    Jun 12, 2013
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    Hi All,


    let's say I have designed myself a 30V 3A Regulated DC power supply and want to charge a battery bank with it (bank is about 250 Amp/hr). I would likely need to put an emitter follower stage between the regulated supply and the battery bank as the bank is a very low impedance load.

    I have simulated the power supply portion but my sim fails when I decrease the load resistance to less than 1 Ohm.

    I know emitter followers are used for driving low impedance loads but how low is too low..the impedance of this battery bank will be in the low milliOms


    Any tips would be great and thanks so much,

    Joster
     
  2. crutschow

    Expert

    Mar 14, 2008
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    What exactly do you want the emitter follower to do?

    Does your power supply have an adjustable current limit? If so that should be all you need to drive the low impedance load battery load (if it's a lead-acid battery).
     
  3. Joster

    Thread Starter Member

    Jun 12, 2013
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    no the power supply has no adjustment...I just need it to be able to drive the battery impedance and not kill my power supply

    I also need to find a way to throttle back the curret once the battery is approaching full charge about 15V
     
  4. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    And a low impedance load, driven through a low impedance emitter follower, is going to look like what, to your power supply?
     
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Ask yourself, if I put an emitter follower behind a power supply, where does the current come from???
     
  6. Joster

    Thread Starter Member

    Jun 12, 2013
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    the supply can handle it but it is not designed for 0.0023 Ohms of load resistance
     
  7. Joster

    Thread Starter Member

    Jun 12, 2013
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    here is the regulator circuit...just need the emitter follower and a circuit to measure battery impedance
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    That is not a real regulator...

    It is in-fact a Darlington emitter follower..
     
  9. to3metalcan

    Member

    Jul 20, 2014
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    Shhhh...you'll hurt its poor underdesigned feelings. Zener followers still pass as regulators in lots if lower-tech devices (I see 'em as +/-15V rails in guitar amps all the time.)

    But Joster, Mike's right...and the Darlington pair already has much lower output impedance than a single transistor emitter follower would. If you plugged this into another emitter-follower, you'd be increasing the output impedance. And your statement that the supply "can handle it" is incompatible with your statement that it can't deal with the battery impedence. It sounds like your supply CAN'T actually source the current you need.
     
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    No, a 3A supply cannot possibly "handle" driving a 0.0023Ω load.

    To drive 0.0023Ω to 24V, the supply would have to be capable of delivering I=E/R = 24/0.0023 = 10435A!!!.

    Conversely, if you can only get 3A out of a current-limited supply, the voltage across a 0.0023Ω resistor would be only E=IR = 3*0.0023 = 0.0069V = 6.9mV.

    Where did you get the 0.0023Ω from?

    The correct way of describing what happens when you charge a mostly-discharged 24V lead-acid battery bank is the following:

    Initially, the charger needs to be "current-limited", mostly to protect the charger itself (from excessive heat produced in the regulator, blowing up the rectifiers, overloading the transformer, etc).

    The initial magnitude of the charging current is chosen primarily by the components in the charger, and to a lesser extent, by the maximum charging current the battery maker allows, and finally by how long you are willing to wait for the battery to become charged...

    The job of the "current-limiter" part of a battery charger is to hold the current at the selected value while the battery terminal voltage slowly climbs from 1.9V/Cell (~22V for a 12 cell battery) to about 2.28V/cell (27.4V). It is the current-limiter that prevents the charger from blowing up when it is connected to a discharged battery bank...

    As the battery accumulates charge, the battery terminal voltage slowly climbs. After reaching 2.28V/cell, the battery is about 85% charged. To get it the rest of the way to full charge, at this point in the charge cycle, the charger should automatically become "voltage-limited" instead of being current-limited...

    If the battery voltage is now held constant at 2.28V/cell (27.4V), the charging current will naturally gradually taper to a much lower value which in the steady state is determined only by the losses and leakage in the battery. The time it takes for the current to taper to the final value could be quite long.

    There is a way to shorten the time it takes for the battery to charge the last 15% of the charge cycle, but that is beyond the level of this discussion...

    A 3A charger is not up to the task of recharging a mostly discharged 240AH battery bank. The best you can ask of a 3A charger is to maintain (float charge) the bank near full charge while the battery bank is idle for long periods...
     
    Last edited: Aug 13, 2014
  11. crutschow

    Expert

    Mar 14, 2008
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    As Mike noted, what you want is a current-limit circuit to control the current when a discharged battery is initially connected to the charger. That way it makes no difference how low the battery internal resistance is or how low the battery voltage is.
     
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