Doubt in circuit

Discussion in 'General Electronics Chat' started by dennis.roshan, Aug 3, 2011.

  1. dennis.roshan

    Thread Starter New Member

    Aug 3, 2011
    bq 2002 IC and party analyzed circuit

    BQ2002 IC decides wheather charging current flows from the DC source to the battery or not.

    pin 4 is GND and pin 6 is Vcc. for the ic to work pin 6 is given to 6V and pin 4 to GND

    Input pins
    Vbat(pin 3) is monitored if this voltgae exceeds 2V then bq2002 stops the charging current. (i.e. max voltage in battery per cell. )
    Vtemp(pin 5) : the battery temp is measured via thermistor. this has a negative temp cofficient, so if temperature increases voltage decreases. if this voltage drops below 1.5v(excessive heating of battery has occurred) then bq2002 stops the charging current.
    INH (pin7): this is an input pin, if this goes high then charging is stopped. if this is low then charging is continued. (bassically works like an intterupt).

    if the charging is stopped by meeting any of the parameters then powers has to be restarted for the charging to continue.

    Output pins
    if BQ2002 decides to allow the charging current then the output pin8 CC goes to high impedance state(5.9V).
    if BQ2002 decides to not allow the charging current then the output pin8 CC goes low (0V)
    Ignore LED pin
    TM decides the charging rate (C) in our case TM is connected to Vcc/2 to achieve C/4 charging current.
    In the C/4 charge rate, if the charge time exceeds 320 min or 5hrs and 20 min bq2002 terminates the charging current.
    in our case as seen in the schematic TM (pin1) is attached in between 24Kohm resistance between 6v and GND. so if the charging time exceeds 5hr and 20min the charging current is stopped.

    U14B is a comparator circuit. battery temp voltage is connected to pin 5(+ve terminal) if U14B, 2V is given to -ve terminal. this generates a low on pin 7 of U14B which is connected to INH pin. If a non rechargable is connected here then INH will go high, else it will be low.

    D11,R73, R74, R75 is used to generate 5.5V which is given to pin 3 of U14A.

    we know that CC (Pin 8) - if high impedance(5.9V) to allow charging current - if low stop charging current.

    U14A is a comparator. it compares cc(Pin8) with 5.5V. if CC is high impedance then the output on pin 1 of U14A goes low which turns on the pnp power transistor which allows the charging current flow from the dc source to the battery. if CC is low then the output on pin 1 of U14A goes high which makes the pnp power transistor off which stops the charging current flow from the dc source to the battery.
    U14A also works as an opamp as a current source. follow the link and compare with U14A circuit
    current = (+6V - Vref)/1ohm = .5A or 500 mA
    Vref = 5.5V which is labeled as 0.5V(pin 3 of U14A) which means Vref is 0.5V below the +6V

    this is the rought idea, the link below is the data sheet for BQ2002 it tells u what I summarized above

    I have attacthed pdf shematic & in the jpg file I have marked the part of the circuit to be analysed

    I have connected a ammeter(multimeter in current mode) to measure to current in the circuit
    Ammeter is connected between +6V & Vbat
    according to the design 500mA current shd flow thru and the ammeter shd read 500mA
    but this is not happening
    Can anyone please explain this?

  2. SgtWookie


    Jul 17, 2007
    Your 5.5v reference design is poor, as it's dependent upon the stability of the 6v source. The reference voltage will drift with the 6v supply.

    One of the big problems is R64 being 15k. Q3 might have a minimum hFE of 100, but the most current you'll get through the base is (6v-0.62)/15k - 358.7uA, which means ~35.8mA. Granted, it might be higher than that, but it won't likely be above 600 @ 25°C - and in an ideal world, that would be about 200mA.

    You have a bunch of voltage dividers for reference voltages, and they're all tied to 6v. Where that 6v is coming from is not clear; just that it's not in the schematic. Meanwhile, there are a number of other things being powered by the same 6v, and they are really noisy items like switching regulators.

    You don't have enough bypass capacitors in your circuit.

    You say that the current should be 500mA, but it is "not happening" - do you not get ANY current?

    Without looking at the circuit in too much detail because I've already found several problems, I think you need to take a hard look at it yourself, and give us some more data about where you have taken readings.

    Document the voltages on the schematic itself, and please use .PNG format, as it is compact and not "lossy" as .jpg files are.