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BQ24070: Input and Output Voltage for BQ24070

Ashok Kumar18
Prodigy 30 points
Part Number: BQ24070
Other Parts Discussed in Thread: BQ25896, BQ25606, BQ24195L

Hi,

We are designing a battery charger for our application device and we are planing to use BQ24070 part

Let me put forward you the requirement that I have

  • Input adapter voltage is 5V to 12V.
  • I want to charge the 3.7V Li-Poly battery as well as Power the system. The OUT pin is connected to the wide range regulator that requires 1.25A at he input side(system Load)

But there are some concerns/doubts regarding the design aspects of this IC

  • Its specified that output voltage is regulated 4.4V Vout. My doubt is whether this IC produce regulated output(4.4) for all the adapter input voltage(in my case its 12V adapter) and what will be the output current that is supported by this pin (OUT pin) at when Regulated voltage is generated Please suggest.
  • Will this IC provide the same 4.4v regulated output voltage when there is no external adapter, i.e; what will be the voltage at OUT pin when there is only 3.7V battery?Is it the battery voltage will be appearing at the OUT pin or else even the battery voltage will be regulated to 4.4V? Please clarify
  • what is the purpose of the DPPM voltage? how to set it for my application? Please suggest
  • Initial case there will be only battery connected to the system and its powered by the battery itself, now suddenly if I connect the external 12V adapter will there be any switch over delay?
  • I am not using USB input in my application So what should I connect to The ISET2 pin?
  • What will happen if the CE pin is kept floating?

Request you to address this concerns ASAP

Regards,
Ashok

  • 0
    TI__Mastermind 21590 points
    Hello Ashok,
    1) Yes, the device will regulate the input voltage to 4.4V for all input voltage above 4.4V + dropout voltage. So the output would be regulated to 4.4V for your adapter. I'll suggest adding a Zener diode to protect the IC from damage. The recommended maximum input voltage is 16V. Also keep in mind that this is a linear device so the power dissipation will increase as the voltage increases.
    2) This is a linear device. If there is no adapter, the output will be the battery voltage minus IR drop.
    3) The DPPM voltage is the point where the charger starts to reduce charging current to ensure the OUT voltage doesn't go above the DPPM voltage level You can read more about it here: www.ti.com/.../slua400a.pdf
    4) The boot-up time is 150ms.
    5) ISET2 is used to control full charge or half charge mode in your application. Please refer to section 8.4.3.4 of the datasheet
    6) CE is active high. If you leave CE floating, you won't be able to control when the charge and not charge the battery. you don't want to leave CE floating.
  • 0 in reply to Raheem
    Prodigy 30 points

    Hi,
    Thanks for your early reply.
    What is this boot up time?

    Let me clarify you what is my real concern.
    Consider that 12V adaptor is connected to the input side and the battery is being charged and the system is powered by the regulated 4.4V output from the OUT pin of the battery charger
    Now if I suddenly remove the external 12V adapter the output voltage will suddenly drop down to battery voltage and power the system?Or will there be ny delay in the transition fro one power to another?
    Because the System is connected to the OUT pin must not RESET during the transition of power sources ?

    Also please specify what is the current that is available at OUT pin when 4.4 is generated? Because in the datasheet its specified in Absolute value coloumn that OUT pin can support maximum of 4A current, so I assume that it can easily provide 2A at the OUT pin when 4.4V is there or even the battery voltage is present.
    If 4.2V is my battery voltage then at the OUT pin I can expect 4.2V-IRdrop voltage at OUT pin, is my understanding correct?

    Request you to address this issue ASAP possible.

    Regards,
    Ashok

  • 0 in reply to Raheem
    Expert 1310 points

    Hi,

    I am Karthik ( Ashok's Collegue)

    We have 3 things to be clarified about BQ24070 :

    Query 1 :
    We would like to understand about Input and Output current in detail.

    Below are the Input specification of the AC adapter source :
    Vin = 12V
    Iin=2A

    With this Input specification what is the Current value (in A) at OUTPUT LOAD and Battery Charging current ?

    We understand from the Datasheet that Battery charging current will be set by ISET1. For example we set Battery charging current as 0.5A.
    In that case whether the Load current will be 1.5A? Whether this understanding is correct?
    We just want to confirm that the input current get shared between the load current and the battery charge current without any internal Loss.

    Query 2 :
    For below specification what will be the power dissipation of the device?
    Vin = 12V
    Iin = 2A
    Vout (Load) = 4.4V
    I out (Load) = 1.5A
    Battery charging current = 0.5A

    In this case what will be the power dissipation and actual Junction temperature.
    The intention behind asking this query is to confirm whether the IC will get heated up while testing @ above mentioned parameters.

    Query 3 :

    The AC Adapter Input pin takes DC voltage upto 16V (with Mode = 1) and USB VBUS voltage of 5V (Mode = 0).
    Now say for example Mode = 1. And we connect 5V,500mA USB VBUS to Input pin. How BQ24070 will recognize whether it was USB power or DC power? Because with Mode =1 we can give 5V 500mA DC adapter voltage too which it will recognize as Adapter voltage. If I just change input source to 5V,500mA VBUS of USB in that case also the IC must recognize it as Adapter voltage right. Can you clarify.

    We are in development stage of HMI products for which we are searching a suitable low cost Linear battery charger with regulated output and we came across BQ24070. Before proceeding to further stages of the design we need to get clarification for these queries. Please address them with high priority.

    Regards,

    Karthik

  • 0 in reply to Karthik B
    TI__Mastermind 21590 points
    Karthik,
    Q1) The device will give priority to the system load on OUT. Since the input current limit is 2A, you'll only get 0.5A charging current if the system load is below 1.5A. The device will start to reduce the charge current when the load current is more than 1.5A in this scenario. When the charge current is reduced will depend on the DPPM voltage setting. Ther would be some loss due to the high input voltage but that is minimal
    Q2) Since this is a linear device, the power dissipation using equation 10 in the datasheet will be (12-4.4) * (1.5+0.5) + ((4.4V - VBAT) * 0.5) so this will depend on the battery voltage. As you can see, a lot of the power dissipation will come from the high input voltage especially since this will be using the maximum input current. The device has thermal regulation so the charging current will be reduced when the device gets too hot.
    Q3) The device does not detect if the input is AC or USB. The mode pin is only used for current limiting not whether the input is AC or USB. Changing the input source will not affect the function of the device.
  • 0 in reply to Raheem
    Expert 1310 points

    Hi,

    Thank you for the answers.

    The answer for Q1 is clear now, we dont have any doubts on it.

    Q2) In our case VBAT is 3.7V in typical. Which means power dissipation is nearly 15W. With this value whether the device is expected to get Hot and will charge current reduce? The Ambient Temperature is 30 deg C. I had used formula 9 to know Tj value with P=15W, Ta=30 degC and Theta (j-a) = 40.1 from datasheet table 7.4  this gives Tj = 600+ deg C Which the device can not afford!!!

    Does this mean whether our requirement of :
    Vin = 12V, VBAT = 3.7V, VOUT = 4.4V, ILOAD = 1.5A, IBAT = 0.5A can not be achieved by using BQ24070 as it dissipates huge power resulting in heating of the IC?

    Q3) As per pin description of Mode in Pin Functions of datasheet and as per sections 8.3.1.1 and 8.3.1.2 Mode is used for Input source selection.

    Request you to address these concerns ASAP.

    Regards,
    Karthik

  • 0 in reply to Karthik B
    TI__Mastermind 21590 points
    Karthik
    Q2) The device has thermal regulation so the charging current will be reduced when the junction temperature reaches between 115C and 135C.
    Q3) The Mode pin is used for Mode selection not input selection and the difference between the 2 modes is the input current limit.
  • 0 in reply to Raheem
    Expert 1230 points

    Thank You, I am the lead of this team (Karthik and Ashok).

    We wanted to confirm whether the calculation that we understood and mentioned in previous thread (I will copy paste them again) is correct ? This is regarding Q2.

     In our case VBAT is 3.7V in typical. Which means power dissipation is nearly 15W. With this value whether the device is expected to get Hot and will charge current reduce? The Ambient Temperature is 30 deg C. I had used formula 9 to know Tj value with P=15W, Ta=30 degC and Theta (j-a) = 40.1 from datasheet table 7.4  this gives Tj = 600+ deg C Which the device can not afford!!!


    Regards,

    Vijetha

  • 0 in reply to vijetha hn
    Expert 1230 points

    Hi,

    We read an Application note from TI : 

    As per this AN, whatever we calculated is wrong. We should not have considered Equation 9 of Datasheet to calculate Tj when Pd = 18W, TA =35deg C, R(ThetaJA) = 40degC/Watt (Table 7.4). If we follow Eq 9 to calculate Tj then it would be 600 deg C!! and AN says the parameter RthetaJA should not be cosnidered for this calculation.

    We are still stuck with Q2. We need clarification for below point :

    Pin = 24W (12V, 2A)

    Pout_Load = 6.6W (4.4V,1.5A)

    Battery Charge current = 0.5A for 3.7V Battery

    With these above parameters, power dissipation must be around 16Watt.

    So we need to confirm whether this power dissipation value is huge for the device to handle? Assuming the PCB is kept in an enclosure @ 35deg C of Ambient temperature with no airflow will 16W dissipation result in heating of the device and charge current reduction? 

    Or whether 16W of dissipation for BQ24070 need additional cooling methods so that charge current wont reduce.

    I hope I made my concern clear here. I need TI's quick support to confirm on this.

    Please respond with more details.

    Regards,

    Vijetha

  • 0 in reply to vijetha hn
    TI__Mastermind 21590 points
    Vijetha,
    That App note deals with comparing TI's R(ThetaJA) to other competitors. 16W will be too much for the device to dissipate. I tried to run your requirement on an EVM that i have and the device kept going in and out of thermal shutdown so additional cooling methods will be required.
  • 0 in reply to Raheem
    Expert 1230 points

    Hi,

    Thanks for simulation and observation.

    I just want to understand at what worst case condition the IC will function normal without going In and Out of Thermal shutdown?

    For example if you consider Equation 10 of datasheet Vin, Vout, Vbat, Iout, Ibat are the parameters to be considered for calculation of Power dissipation.

    The values considered for calculation of Power dissipation are 

    Vin = 12V

    Vout = 4.4 V (Fixed)

    Vbat = 3.5V (fixed)

    Iout = 1.5A

    Ibat = 0.5A

    Now Vout and Vbat are fixed values that could not be changed for Pd calculation. By varying which parameter we can achieve normal operation without Thermal shutdown/up. Cn you check that in your simulation?

    For ex : you changed Vin = 11v, run the simulation and you dont see any thermal shutdown. Or you change Iout = 1A and battery charge current = 0.5A with other values same as mentioned above and run thermal simulation. So in similar way can you run the simulation and give a best case situation and values for all above parameters for normal operation without Thermal shutdown fluctuation.

    I am asking this because we do not have any provision in giving cooling mechanisms for the board in our product. Kindly request you to help us with above exercise and conclude.

  • 0 in reply to vijetha hn
    TI__Mastermind 21590 points
    Hello Vijetha,
    The biggest impact on thermal performance is the input voltage. Reducing the input voltage from 12V to 5V will reduce the power dissipated by about 14W with your current load profile.
  • 0 in reply to Raheem
    Expert 1230 points

    Hi,

    We had selected this IC BQ24070 for our application initially. Now looking at this thread we dont think it will fit our design.

    The datasheet says the device supports upto 16V of Input and 2A of Input current with 4.4V fixed output.

    Now looking at the thread, if I select this IC for an application which takes 12V of Input and 1.5A of Load current + 0.5A of charge current, you say that there will be considerable amount of Power dissipation and heating of the device may switch ON/OFF the output load. So cooling is required in this case.

    This concludes that  BQ24070 supports Input upto 12V only if you have proper cooling mechanism for the IC. For applications which does not have cooling mechanism the IC BQ24070 supports only upto 5V. I hope this understanding is correct. If it is correct then I need to discard this IC from my design. If not correct please help us to understand "without cooling for normal operation of BQ24070 at Vin = 12V what should be done. 

    Regards,

    Vijetha

  • 0 in reply to vijetha hn
    TI__Mastermind 21590 points
    Vijetha,
    The BQ24070 is a linear device so the difference between input and output voltage plays a big part in thermal calculations as the device regulates voltage using an LDO. The BQ24070 can operate at 12V input voltage if the thermal calculations don't exceed the performance of the IC stated in the datasheet. This means 12V inputs can be used but only at lower input current requirements. I'll suggest you use a switching charge for your design as it will be difficult to properly dissipate the heat that will be generated on with your load and input voltage requirements. Please take a look at the BQ25606, BQ25896, BQ24195L.
  • 0 in reply to Raheem
    Expert 1230 points

    Thank You for the suggestions, that was really helpful!.

    We are checking the datasheets of the ICs as per your suggestions.

    You mentioned 12V inputs can be used but only at lower input current requirements. SO here I want to understand what is the actual maximum value of Input current I can have for 12V Input without any major Power dissipation and heating of the device. We understand that the Input current is the sum of battery charging current and Load current. I will fix the Battery charging current as 0.5A. I have option to vary my load current. So finally if you say the maximum Input current value @ 12V then I can set my load as per that by which I can still use BQ24070 in my design without heating.. I kindly request you to check my specifications in your simulator and find out max Iin current or Iload current at which the device will function normal without heating. I repeat my spec : Vin = 12V, VBAT = 3.7V, VOUT = 4.4V, Ibat = 0.5A. Iin = Iload + Ibat where In max value of Iload is to be identified.

    Regards,

    Vijetha

  • 0 in reply to vijetha hn
    TI__Mastermind 21590 points
    Vijetha,
    During my test with your specs, i see thermal regulation even with no load. The charge current is being reduced with 12V input. The switching charger will be a good choice for your requirements.