*Automatic Cut-OFF Ckt. using BJTs

Discussion in 'The Projects Forum' started by JUAN DELA CRUZ, Mar 8, 2010.

  1. JUAN DELA CRUZ

    Thread Starter Active Member

    May 27, 2008
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    Greetings,

    I am constructing a LI-ion battery charger using BJTs because some charging IC is to expensive and some are not available here in my country. That is why I am constructing a charging ckt. using BJTs.

    I need to cut the charging voltage when the battery reaches 4.1V to avoid overcharging. I need also to add an LED indicator when the battery is fully charge.
    I am using small high frequency transformer for small and compact application that suits to my need.

    Please help me to modify this circuit.
    Thank you in advance.
     
    Last edited: Mar 14, 2010
  2. SgtWookie

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    Jul 17, 2007
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    The 1N1519 is a 4.7v 40mA Zener diode. Did you mean a 1N5817 through 1N5819 Schottky instead?
     
  3. SgtWookie

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    I guess that's something you threw together from the other circuit. It's not going to work well at all the way you have it drawn.

    You only have one rectifier diode (the one closest to the transformer). You really need a bridge rectifier. However, that will drop the available output voltage by roughly 0.8v if you're using Schottky diodes. The 1k resistor limits maximum charging current to around 30mA, if it weren't for the MPSA56 sucking it all up because there is no base resistor.

    What were you going to use to drive the input side of the transformer?

    What is your desired charging current?
     
  4. SgtWookie

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    What is your desired charging current?

    Or don't you care about the charging current?
     
  5. ifixit

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    Nov 20, 2008
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    It is not clear from your circuit how much DC voltage is available at the 220u capacitor, I'm assuming 6 volts. Your circuit has too many flaws so I started from scratch. See attachment.

    With this example circuit you can adjust R4 to get the 4.1V battery limit needed. The charge current is set to 100mA by R2 since you indicate a 960mAh battery, which is getting charged at approximately one tenth its Amp hour rating in this case. This circuit can be modified to use different parts depending on what you have available.


    Questions:
    1. What is the voltage on the 220u cap?
    2. What current is the battery rated to be charged at?
    3. What rate do you want to charge it at?
    If your interested, I can give more details.

    Regards,
    ifixit
     
  6. SgtWookie

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    Ifixit,
    I'd assume that the output from the transformer is as stated; 6v p-p @ 350mA prior to rectification. Juan apparently has access to 1N5819 Schottky diodes. If a FW bridge were constructed from them, roughly 0.7v would be dropped across the bridge.

    No frequency for the xfmr output was mentioned, but I'd imagine that it would be in the realm of 60kHz to maybe 400kHz or so.
     
  7. SgtWookie

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    Ifixit,
    I did a little tweaking on your circuit to add in a simulated transformer output, increase output current, add another charge indicator LED and changed some resistor values to more standard values. I decreased the resistance values in the "rubber diode" divider to make cutoff a bit more positive.

    [​IMG]

    In order to decrease time required for the simulation to run, I arbitrarily used 10kHz for the simulated transformer output, and decreased the simulated battery cap to 10mF. Since the unloaded P-P output voltage of the transformer was not specified, I just took a wild guess at transformer impedance.

    Power dissipation in the battery being charged is of concern. I do not know what the battery package power dissipation rating is. Right now, dissipation is roughly 630mW; if the battery is unable to dissipate heat that fast, there will be a melt-down or rupture of the battery package.

    It would help a lot if Juan could give more specifics about the battery that he wants to charge.
     
  8. ifixit

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    Nov 20, 2008
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    I will investigate what the minimum input voltage my circuit needs and if the charge current can be 200mA.

    I'll be back!
     
  9. ifixit

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    Nov 20, 2008
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    Hi Sgt,
    I just missed your last post. Thanks for the review. I'll download the .asc and try it out.

    I'll be back
     
  10. SgtWookie

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    Jul 17, 2007
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    Juan,
    I AM trying to help you as much as I can. However, I do not know the manufacturer and part number of your battery, as you have not posted it.

    I made some changes to Ifixit's circuit so that it would charge the battery faster, light an LED when the battery was roughly 90% charged, and "float" at 4.1v a bit more accurately.

    However, since you have not posted your battery specifications, we really do not have any idea if it is safe to charge them as quickly as the circuit will perform.

    The charger I posted works pretty well in simulation, but I have not tested it with real components. I do not have any batteries of your missing specifications to test it with.

    I am concerned about the power dissipation in the battery during charging. It may not be safe to charge it that rapidly. It would be a big help to have access to the battery specifications.

    You might think that we are "dragging our feet" to help you. This is not the case. We wish to help you, but we wish for you to be safe, too. Your safety is more important than charging a battery. If you do not appreciate our concern for your safety, then you are on your own.
     
  11. ifixit

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    Nov 20, 2008
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    Juan,

    It looks like that transformer would work okay with my circuit. The smart charge controller for the Nokia battery is likely in the phone itself and only the simple raw power source is outside in the charger you show in the picture.

    You can use it to power my circuit, which regulates current flow to the battery and turns off at 4.1V as requested. Sgtwookies improved version charges at 270mA to start, ramps down to 155mA as the battery voltage approches 3.9V, at 4V its 46mA, and at 4.1V volt its only a trickle at 1mA. You can tweak the charge current by changing the R2 value.

    It is possible that too much power can be dissipated in Q1 if the current is too high so monitor that closely during testing. It is good that the raw voltage drops with the more current that is used.

    I believe, my circuit is safe to use once you have calibrated and tested it well. It is best if you can verify the recommended max. charge rate for the BL-5C. Right now we're using approximately 0.3C, which is below what the Nokia charger transformer is capable of. However, because it is not a smart charger there is still a small risk of overcgharging, although, not by much.

    SgtWookie & Juan,

    The power used by the discharged battery goes into a chemical conversion process, not heat. Once the battery is fully charged then the power is converted to heat, which is bad for the battery. This, of course is why it is important to remove the charge current as the charge process completes.

    It is also important not to charge at a higher current than the manufacturer recommends. What can happen is, one of the cells in the pack can become fully charged before the other one(s) and start dissipating heat, which as I mentioned, is not good.

    A smart charger circuit determines the exact moment of full charge very well... my circuiit will not work nearly as well, but it will have to do since Juan has limited resources available.

    Good Luck, be safe.
     
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  12. SgtWookie

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    Ifixit,
    Thanks for your circuit, and for sharing your expertise.

    I don't know if Juan understands the circuit, or will be able to successfully implement it. However, I think you've come up with a rather good compromise for a charger in what might be considered a primitive environment.

    Thanks for your input.
     
    Last edited: Mar 10, 2010
  13. JUAN DELA CRUZ

    Thread Starter Active Member

    May 27, 2008
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    Sir, I realIy appreciate your concerns :) . But I will assure that I will always observe "safety first" as I always did in dealing electronics.
    I have search through google and found this data here www.rockbox.org/mail/archive/rockbox-dev-archive-2009-12/0028.shtml regarding the specs. of the battery that I will use. I think it is better to charge the battery with 1/10 of the charging current (more charging time approx. 2hrs) to avoid heating of the battery.

    The charger that I need to to modify (limiting ckt. Needed) is inherently designed to charge the battery that I will use (Rated 3V7/ 950mAh). All I need is to add a limiting ckt. with indicator and charge the said battery maybe in low current to charge it effectively.
    Thank you.
     
  14. SgtWookie

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    No, a green LED will not work.
     
  15. ifixit

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    Nov 20, 2008
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    Use the circuit attached to this post. The attached circuit should not be taken as a "plug-n-play" solution for your problem. You will likely have to tweak the values smomewhat to suit your components on hand. It is a flexible design.

    Circuit description for a 100mA, 3.7V lithium battery charger...

    • The LED D2 is used as a cheap voltage reference so you need to select one with a foward voltage drop of approximately 1.8V to 2.5V. Higher is better for more precise control of battery voltage. D2 is biased through R1 from the 6 volt supply at 3.5mA.
    • When the battery voltage is low the base of Q5 is biased off so no collector current flows. This leaves D2, Q5, & R3 as a constant current source of 13mA to bias D1 and Q2 on. This current is determined by (V(D2)-V(Q5be))/R3.
    • LED D1 is also used as a 1.8V reference for the current source made up of R2, Q2, & D1. D1 sets up a reference voltage across Q2b-e junction and R2. Current flows through R2, & Q2e-c, which increases the V drop across R2 and will tend to reduce the b-e voltage and cut Q2 off thus regulating the current constantly. This current is determined by (V(D1)-V(Q2be))/R2. Change R2 to get the inital battery charge current of 100mA.
    • The charge current will slowly raise the battery level. The voltage divider of R4 & R5 will then bring the Q3 base to a point where Q3 collector current flows. This raises the emitter of Q5 up causing its current to decrease and therefore the battery charge to decrease also. Collector current in Q5 moves over to Q3 as Q3b increases. Eventually, a stable point will be reached where the charge current is minimal and battery voltage has stopped changing. Adjust the value of pot R4 to get this stable point at a battery voltage of 4.1V.
    • D1 can be used as an indicator of when the charge is complete since it will slowly go out as charge completes. SgtWookie had added another LED to the collector of Q3, which would slowly come on as charge completed. However, this biased Q3c very near its saturation point so I removed it. It could be added back in if D2 was also a 1.8V LED, but this would reduce the accuracy of the battery voltage limiter. Hopefully D2 will do as an indicator of charge complete.
    • The charge termination point is effected by ambient temperature. The coefficient is positive at ~ 8.5mV per C. for the components I used. So adjust R4 on a warm day. I guess all days are warm in the Philipines.
    • If you try to charge at higher current, keep an eye on the power on R2 and Q2.
    Setup and Calibration...

    1. To help calibrate the circuit, use a 3.9K resistor in place of the battery. This will be equivilent to a fully charged battery. The circuit should hold at 4.1V once you have adjusted the R4 pot. R4's theoritical ideal value is 960 ohms. You can use a combination of resistors and a pot on hand to set this up.
    2. Next replace the 3.9K with a 39 ohm 1W resistor and adjust R2 (current limit) for 3.9 volts across the 39 ohm resistor (100mA). This sets the circuits current limit. You can use pots, or series/ parallel resistor combinations to get this set up.
    3. Try charging a battery. Monitor the R2 voltage to see the charge current, and the battery voltage at the same time if you can. The current in R2 also includes Q3, R4, & R5 current aswell, so you'll have to subtract around 14mA.
    4. Do not put a current meter in series with the battery for this testing.
    Once you have a circuit based on components you available, post it and we can review it.

    Good Luck,
    Ifixit
     
    Last edited: Mar 11, 2010
    JUAN DELA CRUZ likes this.
  16. ifixit

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    Nov 20, 2008
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    I'm not sure if that is possible. Give me some time to investigate. In the mean time, start looking for a small 5V relay with at least one set of normally opened contacts.

    I'll be back.
     
  17. ifixit

    Distinguished Member

    Nov 20, 2008
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    Juan,
    A new circuit, version 3, is attached. Dispose of all others.

    I don't like to put anything in series with the battery because it makes getting the battery voltage set accurately more difficult, but version 3 of my schematic has a solution that I think will work quite well. No relays required.

    However, during test and calibration, measure battery voltage between bat_com and 3.7V_bat so as to avoid including the Vd drop of D4 in the measurement. The circuit feedback via R4 includes the Vd drop of D5 to compensate for D4 drop and temperature effects.

    If the 6V power source disapears, D4 will reverse bias and disconnect the battery. Why didn't I do this in the first place.

    You can connect a load directly across the battery (bat_com to 3.7V_bat) while it charges. The charger will supply the load and charge the battery at the same time. Just add the load current to the charge current. When the charger is unplugged, only the load will drain the battery.

    What is the load current? It can't be too high or we'll have to pick a bigger transistor for Q2.

    Regards,
    Ifixit
     
    Last edited: Mar 11, 2010
  18. SgtWookie

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    Jul 17, 2007
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    How in the world do you expect to power something that requires 1A current, when the transformer only supplies 350mA current at 6v? Do you expect Ifixit to build a nuclear power plant in the middle of the charger circuit? :rolleyes:

    He already gave the formula for calculating charging current, and how to adjust the final voltage. However, if the load is significant, the battery will never fully charge.
     
  19. SgtWookie

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    You didn't just edit your post, you quoted it and deleted the original, so that now MY post looks like it is out of context. Just for clarification if anyone reading this thread might be confused, Juan wanted to know how to power a 1A load from the existing circuit, which Juan originally specified with a 6vac @350mA transformer. That simply won't happen.
     
  20. JUAN DELA CRUZ

    Thread Starter Active Member

    May 27, 2008
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    0oops. . I am really sory about that sir. I just want to repost my reply with complete details. For the reason that, my previous post was lacking of full details. That is why I DELETED IT, and repost it with more details.

    --> I am planning to use a higher rating xformer in the future use. That is why I am wishing for the complete computation so that I can use it as a guide coz' I am not a professional.

    --> But, for now I will be using the xformer which I have said 350mA rated according to the charger casing. But, I think it can handle up to stable 600mA coz' before I revise it to be regulated powersupply and it can supply up to 800mA to power an array of LEDs.
     
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