I suggested a pull-down resistor. When the LTC4057 is "off", the datasheet shows 2 uA at the BAT pin. !SHDN current is shown as 5 uA. So, a simple connection between the two may be enough to enable the device . A resistor, say 10K from BAT to ground would ensure the device was shut down (2uA x 10K Ω = 0.02V). Now, when the battery is attached, the voltage at the BAT pin would be 4 V or more. The resistor itself would draw 0.4 mA, which is wasted energy, but the voltage at the pin would be enough to enable the charger. I am not sure whether that is a workable solution, as I have no experience with that chip.

Your schematic didn't show the battery, but it does show "3.7V/700mAh" which I assumed was the battery and power to the instrument was intended to be at CON4, pin1. (BTW, what is the purpose of the FDS6375? If it is a high side switch, the drain and source seem to be reversed. On P-channel mosfets, the body diode is forwarded biased D-->S. Moreover, the gate is grounded, so it should be on all the time. Is there another connection to the gate pin that is before R38?)

John

Hi John,

Yes, You are correct U20(FDS6375) connection is reversed in schematic.
But actual in PCB it is connected properly. D is contented battery pack +Ve and source is connected LTC4057 BAT pin.
For you kind information-
CON4(J22)- A battery pack is connected through wire. That is in pin 1 of CON4, battery pack +Ve and pin 2 of CON4 battery pack -Ve.
FDS6375 is connected between battery pack and BAT pin of LTC for switching purposes. I do not why it is connected.
SW5 is ON/OFF switch between battery and instrument(load).

Let me test your recommendation and get back to you with results.

Thanks !!!

 

I suggested a pull-down resistor. When the LTC4057 is "off", the datasheet shows 2 uA at the BAT pin. !SHDN current is shown as 5 uA. So, a simple connection between the two may be enough to enable the device . A resistor, say 10K from BAT to ground would ensure the device was shut down (2uA x 10K Ω = 0.02V). Now, when the battery is attached, the voltage at the BAT pin would be 4 V or more. The resistor itself would draw 0.4 mA, which is wasted energy, but the voltage at the pin would be enough to enable the charger. I am not sure whether that is a workable solution, as I have no experience with that chip.

Your schematic didn't show the battery, but it does show "3.7V/700mAh" which I assumed was the battery and power to the instrument was intended to be at CON4, pin1. (BTW, what is the purpose of the FDS6375? If it is a high side switch, the drain and source seem to be reversed. On P-channel mosfets, the body diode is forwarded biased D-->S. Moreover, the gate is grounded, so it should be on all the time. Is there another connection to the gate pin that is before R38?)

John

Hello John,

I have tested your recommendation but unfortunately it does not work.
Please find the correction made by me in existing one.

1st Scenario-:
1. 10K resistor is connected between BAT pin of LTC4057 and ground.
2. Now i have connected charger through charging jack means input of LTC.

Observation -

1)Battery in place and not connected to charger -: Device works
2)Battery in place and being charged-: Device works and battery is being charged
3)Battery removed and device attached to charger :- Device does not work

2nd Scenario - 10K connected between !SHDN and BAT pin LTC4057.
Observation-
1)Battery in place and not connected to charger -: Device works
2)Battery in place and being charged-: Device works and battery is being not charged
3)Battery removed and device attached to charger :- Device does not work

means we unable to achieve the objective.

Hope this help !!!

 

2nd Scenario - 10K connected between !SHDN and BAT pin LTC4057.
Observation-
1)Battery in place and not connected to charger -: Device works
2)Battery in place and being charged-: Device works and battery is being not charged
3)Battery removed and device attached to charger :- Device does not work

At #2, what is the voltage at the !SHDN pin relative to ground, and if that is very low (<0.5V), what is voltage relative to the BAT pin and VCC (charger pin)?

Your 1st Scenario: Is that something you did or is that how you want it to work. It sounds like something you did and is how you want the circuit to work.

Re: Mosfet
I understand that the D and S are reversed in the schematic. There is still the issue with the gate. Your schematic shows it grounded permanently. There is no way to turn it off.

John

 

At #2, what is the voltage at the !SHDN pin relative to ground, and if that is very low (<0.5V), what is voltage relative to the BAT pin and VCC (charger pin)?

Your 1st Scenario: Is that something you did or is that how you want it to work. It sounds like something you did and is how you want the circuit to work.

Re: Mosfet
I understand that the D and S are reversed in the schematic. There is still the issue with the gate. Your schematic shows it grounded permanently. There is no way to turn it off.

John

Lets leave scenario 1.

Now come to your point- Yes Mosfet's Gate G is connected to one end of resistor 100K as you can see in schematic and now one other end of 100K is permanently grounded.

1. Tested only with charger connected and battery pack removed-
All voltage measured relative to ground
VCC- 5.18V
!SHDN- in uV
VBAT- 4.17V

2. Tested with charger connected along-with battery pack (Both connected at a same time)
VCC- 4.56V
!SHDN- 4.11V
VBAT- 4.15V

Hope this helps in our further analysis .

 

@#1 Not too surprised. Charger is off, but without a substantial load on the BAT pin, there may be voltage. To confirm that, add a 500 Ω resistor from BAT to ground and retest voltage (do not leave that resistor in place).
@#2 That is as suspected. You may need a higher charger voltage to charge the battery.

In summary, those results are not surprising and may indicate correct operation. The problem with charging may be resolved with a higher input voltage. I need to review that datasheet again. Will edit later.

John

EDIT:
1) Just checked the datasheet. Problem is not input voltage (4.25 to 6.5 V allowed)
2) Maybe failure to charge battery is due to battery already being fully charged? Try a discharged battery.

 

@#1 Not too surprised. Charger is off, but without a substantial load on the BAT pin, there may be voltage. To confirm that, add a 500 Ω resistor from BAT to ground and retest voltage (do not leave that resistor in place).
@#2 That is as suspected. You may need a higher charger voltage to charge the battery.

In summary, those results are not surprising and may indicate correct operation. The problem with charging may be resolved with a higher input voltage. I need to review that datasheet again. Will edit later.

John

Thanks for your guidance !!!

You can see at no load voltage at VCC pin is 5.18V.

I want to let you know that with respect to above test results .

test result 1- My instrument does not work
test result 2- my instrument works and battery is being charged .

please make sure that 10K connected across !SHDN and VBAT.

In the test results 2 my instruments works and also battery is being charged, now if i remove the battery and charger is remain connected then even still it works. This can do for my application.

I am not sure how much it technically perfect .

Please find the revised schematic attached herewith. I still do not understand the purpose of MOSFET connected .


hope this helps.