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Battery charging circuit - can anyone explain?

  1. Jan 11, 2013 #1
    http://electronicdesign.com/site-files/electronicdesign.com/files/archive/electronicdesign.com/files/29/1823/figure_01.gif [Broken]

    So let me just start by saying that, I want to understand how to analyze circuits and I would like to find a good source for that. If some one could point to online tutorials, or a website it would be great.

    In the above circuit (link) for example, I get the part that we step-down the AC mains, we rectify it and get a 15 V pulsating DC.

    Now, I assume capacitor C is to keep it at a constant 15V, right? Though I don't understand how a simple capacitive filter is enough for that. When does it charge, and when does it discharge in this circuit?

    Next, could someone explain the role of the diode D1, and the transistor Q1. I mean like what exactly happens with them in the positive and negative half-cycles of AC mains?

    The diode will be likeshort if forward biased, and like open if reverse biased. But if we have a rectified DC, so when will the diode be forward biased?

    Then after I am done understanding that, we could go to more complicated circuits like this one:


    If someone could elaborate/discuss these circuit (starting with the first) so that I can try and understand it, I would really appreciate. Thanks.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 12, 2013 #2


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    As to the capacitor, this little applet might help.

    Last edited by a moderator: May 6, 2017
  4. Jan 12, 2013 #3


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    The transistor turns the relay coil on and off. The diode is there to protect the rest of the circuit from the voltage spike generated by the relay coil when the transistor turns off. I would guess that you didn't recognize the relay coil. That is one of the first steps in being able to understand schematics. Know what each part is.
  5. Jan 13, 2013 #4
    Hi Jay
    First the 220v ac voltage is stepped down and rectified by bridge B1, then filtered to be almost dc by using the capacitor C1.
    Of course there will be ripple in the voltage but this ripple voltage can be calculated and ignored if found very small.
    The idea of this circuit is that the resistors (vr1,vr2, r1, r2 ,r3) are forming voltage divider which apply a voltage to between the base and the emitter of the transistor Q1.
    When the voltage of the battery is greater than some value (assume 15) the transistor is switched on then the relay is energized so switching the relay switch from closed position to open position. Thus stop charging the battery.
    You now can see that when the switch is connected to NO (normally open of the relay) the rheostat vr1 is short circuited by vbat thus making the turn on voltage of the charging state different from that of the turn off of the charging.

    But i think there is a problem in the circuit which is when the relay switch is at NC then the voltage applied to the resistors is 15v which means it will switch on the relay and disconnecting the charger.
    So i think it should take into consideration the charging current also, so after the voltage battery is 15v and the current is less than some value in mA then disconnect the charger.
  6. Jan 14, 2013 #5


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    There is a clever hysteresis function in this circuit.

    When the battery charges up to 12.39 volts, the transistor conducts and does not stop conducting until the battery voltage drops below 11.24 volts. This is assuming the rheostats are at their maximum values.

    This stops the relay rapidly switching on and off.

    However, there should be some current limiting for the battery and there should be a diode to stop the battery powering the relay if the charging voltage is removed.
  7. Jan 15, 2013 #6
    Thanks guys few more questions though :uhh:,

    I think I understood the capacitor's work, its got to do with it charging till the rectified sine reaches peak and then discharging (relatively slowly) when the rectified sine starts to drop. So that makes it DC almost.


    Could you explain how the diode protects the remaining circuit? Which components does it protect. I understand the point of connecting a diode is that we want it to act as a short in some instances, and an open in other.

    Could you explain in which instances the diode forward biases, and in which instances it reverse biases in this particular circuit?


    Can you explain why do we need two variable resistors in the voltage divider arrangement? I don't understand that. With one itself we can vary the voltage right?


    So, if I understand correctly - The relay is in the NC position first, and the circuit is completed. So the battery charges, and past a point the voltage divider arrangement sends current to the base of the transistor. This npn then acts as a short and relay is connected between positive and ground, right? And so, does the relay energize and switch to NO position which breaks the circuit? Is this right?
  8. Jan 15, 2013 #7
    As a basic BAT Charger - it does not need very "clean" DC to work with. The rectifier output will be DC with ripples and the CAP filters them, in the peaks of the ripple the rectifier supplys energy to the capacitor AND the charger, in the valleys, then the capacitor supplies the energy.
    D1 is a small free-wheeling diode to dissipate the inductive kick from when the transistor turns off - and de-energizes the relay coil. This diode is only forward biased for an instant, each time the transistor turns off ( you can not instantaneously change the current in an inductor - the relay coil is an inductor).
    The dual adjustable resistors - are a little trick to create a Hysteresis zone, when fully charged ( for a car could be 13.8V) the Transistor will turn ON (stop charging the Battery) at a higher voltage then when it turns Off ( de-energizing the relay and starting the charging again). Without this the circuit would "chatter" turning on-off-on-off too quickly as the battery comes to full charge - remember that the battery voltage is higher when charging than when not charging - even when fully charged.
  9. Jan 15, 2013 #8
    Is the relay energized when the transistor is OFF or ON?

    What I understood (if I did correctly!) is that the transistor turns ON after a point of voltage build-up from the voltage divider arrangement. This means charging has happened, and the circuit breaks. Is the relay energized or de-energized when the transistor is ON?

    Can you explain the how/why the diode gets forward-biased when the transistor is OFF and what exactly its protecting? If we didn't have the diode what would happen (so that I can understand its purpose).
  10. Jan 15, 2013 #9


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    Transistor Operation depends on the amount of current entering it's base. With no current, it's said to be in "cut-off" (i.e. there's no collector current or relay current in your case).


    With a large enough base current, it's said to be in "saturation" (i.e. there's a high collector current; causing the relay to operate or close the switch in your case).


    See: Transistor Switches

    Probably easiest explained this way:

    From: http://www.kpsec.freeuk.com/components/relay.htm

    The transistor and relay in your circuit are acting like the switch and inductor respectively in these images.

    When your transistor is in saturation,


    When you transistor is in cut-off,



    Note that when the switch is opened, current is induced (hence the name inductor which is what the relay's coil is) lighting the lamp. By having a diode in the place of the lamp, this induced current is shunted. Hence protecting the rest of the circuit from damage.
  11. Jan 15, 2013 #10
    Okay that was clear. Now when we say the relay is energized is it at NO or NC? (relay energized = when the transistor is in cut-off and the current going through the diode to the capacitor's positive place).

    The charging is happening when the relay is connected to NC, right? So this should be when the transistor base current is low and transistor in cut-off right? So the relay is energized, and current flows from diode to the positive place of capacitor? That doesn't make sense to me.

    I understood :

    Situation 1: Transistor cut-off --> Relay energized --> diode carrying current
    Situation 2: Transistor saturation --> relay just conducts --> diode not carrying any current.

    Which of these situations is when charging, and which is when discharging?
    Last edited: Jan 15, 2013
  12. Jan 16, 2013 #11


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    The relay has 3 contacts ( as looking at the diagram) a NO = Normally Open on the left, a common one in the middle and a NC = Normally Closed one on the right

    NO and NC always refer to the contacts in the relay's unenergised state
    when the relay is energised the NO contact closes with the common and the NC contact opens from the common contact

  13. Jan 16, 2013 #12
    The first rheostat is used to set the upper voltage to which you want to charge the battery (voltage at which the charging stops), while the other (var2) is used to set the low voltage at which charging is started.

    You can see that var1 is disconnected when charging stops using the relay.
  14. Jan 19, 2013 #13
    Okay, so when the relay is energized it (yes I meant the common) is at NO, right?

    So again, in which of these 2 situations is charging happening and in which situation discharging?

    Situation 1: Transistor cut-off --> Relay energized --> common closed with NO --> diode carrying current

    Situation 2: Transistor saturation --> relay just conducts --> common closed with NC --> diode not carrying any current.
  15. Jan 19, 2013 #14


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    Charging happens through the NC contacts while the transistor is not conducting.

    When the battery is sufficiently charged, the transistor turns on and this causes the relay to energise and the relay contacts cut off power from the rectifier to the battery.
  16. Jan 20, 2013 #15
    "NC contacts while the transistor is not conducting."

    I thought when the common is closed with the NC the transistor IS conducting? Are my two "Situations" correctly described?
    Last edited: Jan 20, 2013
  17. Jan 20, 2013 #16


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    A correctly drawn schematic will show relay contacts in the position that they are if the relay were pulled from the circuit and laying on a desk. In other words, unenergized.
  18. Jan 20, 2013 #17


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    Neither of your situations is correct.

    For this circuit to work, there has to be some current limiting at the right of the relay. Otherwise, the battery could draw a very large current and destroy the diodes.

    The battery voltage will rise as it is charged and the voltage at the base of the transistor will also rise.

    When this voltage reaches 0.7 volts, the transistor will turn on, activating the relay, moving the centre contact from right to left.

    This stops the battery charging.

    A fault with the circuit is that the battery supplies current for the relay and the transistor. So, if the charging power was removed, the battery would eventually be fully discharged.
  19. Jan 21, 2013 #18
    I would say the relay has 3 TERMINALS - not 3 contacts, can be confusing. The NC and NO indication refer to the relay in the de-energized state, in this case this means the the transistor not conducting or "off". The confusion here is that the transistor turns ON - to stop the charging circuit.
    The Diode only conducts for an instant - as the relay is being de-energized - the coil of the relay is an inductor and you can not instantly stop (or start) the current in an inductor.
  20. Jan 22, 2013 #19
    Hey guys, thanks for all your patience. I am really dumb at times. I find all the posts contradictory, or I haven't understood at all.

    The diagram shows the transistor in OFF (cut-off) stage, when the relay's common is closed with NC and open with NO. This is when charging is happening, right?

    After a point, the voltage divider, saturates the transistor, this is when the relay common closes with NO and opens with NC. At this point the charging has stopped. And the current goes from the battery, through the resistors and all, through the base of the transistor and then through diode, to the positive plate of the capacitor, right?

    What I didn't understand is from dlgoff post #9 in this thread, it seems like the current flows through the diode when the transistor is in CUT-OFF.

    Shouldn't the battery be charging and no current go through the diode, when the transistor is in CUT-OFF?

    If someone can explain the corresponding states of the transistor, relay position (common closed with NC or NO) and the current direction, I will probably understand. If I don't past that, forget it. I'm probably too dull to get it. Or I will give the thread another read again thoroughly.

    I am probably mistaking the use of "energized" when speaking of the relay. Energized = common closed with NO and open with NC, right? And De-energized = common closed with NC and open with NO, right? So in the diagram, relay is de-energized, right? And transistor is cut-off right?
  21. Jan 22, 2013 #20
    http://electronicdesign.com/site-fil.../figure_01.gif [Broken]

    Here is the image again
    Last edited by a moderator: May 6, 2017
  22. Jan 22, 2013 #21


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    Did you read the Inductor Link? What happens is, when the transistor goes from saturation to shut-off, the stored energy in the inductor must go somewhere. So you put in that diode to shunt that energy. This diode has nothing to do with the charging of the battery.

  23. Jan 22, 2013 #22
    " What happens is, when the transistor goes from saturation to shut-off, the stored energy in the inductor must go somewhere. So you put in that diode to shunt that energy. This diode has nothing to do with the charging of the battery."

    The link mentions how the voltage across the inductor becomes very high by quick switching and that makes sense according to V = L(di/dt). So it makes sense that its a protective element. But what happens to it ones it flows through the diode and into the capacitor plate?

    Another thing I wanted to clarify: doesn't the transistor go to saturation only when there is sufficient base-voltage due to the voltage-divider arrangement? Isn't this when the battery has reached a particular upper voltage level?

    Also, isn't the transistor in cut-off because the base voltage is not sufficient while it is still charging? That was my basis to think that charging happens when the transistor is cut-off (and so current flowing through the diode as the link shows).
    Last edited: Jan 22, 2013
  24. Jan 22, 2013 #23


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    When the transistor turns off, the inductor develops a very large positive voltage at the lower end where it joins the transistor's collector. This can be hundreds of volts more positive than the other end of the inductor.

    The diode is forward biased by the voltage and the diode conducts a current from the bottom end of the coil into the top end of the coil. The current just goes into the inductor.

    When the battery is fully charged, a current flows through the resistor chain to the base of the transistor from the battery. This keeps current flowing into the relay and stops further charging of the battery by the charging circuit.
  25. Jan 25, 2013 #24
    Thanks vk6kro. Can someone explain why the direction of the current in the image dlgoff posted when the transistor is closed seems to be from negative of the battery to the positive? Doesn't current go from positive to negative (opposite direction of electrons).
  26. Jan 25, 2013 #25
    Hello Jay - the arrows are just representing electrons - and IMO is misleading because from a circuit and mathematical perspective this is against convention. ( However it is technically correct, esp to a physicist, if you consider that the electrons are negatively charged (-). But confusing for this conversation nonetheless.)
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