Voltage divider after opamp buffer for battery voltage measurement help needed ?

Thread Starter

mishra87

Joined Jan 17, 2016
759
Hi Guys,

I want to use below circuit for battery voltage measurement for two series 4.2V Li Ion battery as per below.
Could anybody body help me out here ?
My understanding was :

Without opamp : Divider leakage current = 89uA
With opamp : pA since opamp input impedance is high

Now my question is if i use opamp as a buffer will reduce the battery leakage current ?
In below circuit , will divider use current from supply or it wil still take from battery ?
1596277583503.png

Regards,
 

Thread Starter

mishra87

Joined Jan 17, 2016
759
OK,
What is the purpose of the two transistors in the 'yellow' box.?
E
Thanks,

To switch off the voltage divider circuit if measurement not needed to reduce battery leakage !
If you see path will complete via the transistor.

But i still want to understand what if i use voltage divider at opamp output.
Will this reduce voltage divider leakage ?
Will voltage divider use current from opamp supply ?


Regards,
 

ericgibbs

Joined Jan 29, 2010
10,158
hi,
If you used a OPA buffer it would reduce the 'sampling current' drawn by the resistive divider.

Consider what effect on the measured ADC voltage, having the transistor connected to the bottom of the divider, to 0V.?
ie: What will be the Vce sat of the transistor.?

Also what will the input voltage be to the ADC when the bottom transistor is Off, ie: open circuit.

E
 

Thread Starter

mishra87

Joined Jan 17, 2016
759
hi,
If you used a OPA buffer it would reduce the 'sampling current' drawn by the resistive divider.

Consider what effect on the measured ADC voltage, having the transistor connected to the bottom of the divider, to 0V.?
ie: What will be the Vce sat of the transistor.?

Also what will the input voltage be to the ADC when the bottom transistor is Off, ie: open circuit.

E
Thanks !

If you used a OPA buffer it would reduce the 'sampling current' drawn by the resistive divider.
How ? I still do not convinced and i wanted to know the reason.

What will be the Vce sat of the transistor.?
I assume 200-300mV.

Also what will the input voltage be to the ADC when the bottom transistor is Off, ie: open circuit.
ADC uses microcontroller power supply which will be above measured voltage when transistor is off.
 

Thread Starter

mishra87

Joined Jan 17, 2016
759
Do you have any other solution in your mind to reduce the battery leakage current.
or any other cell balancing circuit to reduce the because in my application i use 2 Li Ion battery.
 

ericgibbs

Joined Jan 29, 2010
10,158
hi,
Look at this LTSpice sim of the current, part section of your circuit.
It is very important to note that a OPA will have an Offset current and Voltage on its input terminals.
E

Added a Battery voltage Sweep plot.
Note: as the MCP6001 has +V supply of 5.5V Limit, I have used an alternate OPA.
 

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Thread Starter

mishra87

Joined Jan 17, 2016
759
Thanks for effort and time !

Your Simulation shows IR2 = 2uA which is quite low.
and IR3 =90uA as i calculated 89uA without opamp buffer so no point to use opamp if battery discharge current through divider with and without opamp is same.

Still i am confused this 90uA(IR3) from simulation is drawing from battery or from opamp supply.
Could you clarify this point ?
 
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Its coming from op-amp supply, but unless you have some other as yet unspecified power supply in your design, its fairly obvious its eventually coming from the battery. Why are you so worried about 89uA? What's your overall power budget?
 

Thread Starter

mishra87

Joined Jan 17, 2016
759
Its coming from op-amp supply, but unless you have some other as yet unspecified power supply in your design, its fairly obvious its eventually coming from the battery. Why are you so worried about 89uA? What's your overall power budget?
2000mAH, 8.4V , 2 series cell.

The objective was to measure the 2 cell voltage (total VBAT: 8.4V) using divider and 2nd cell voltage (VBAT2 : 4.2V) using divider.
So as per my understanding divider for 2nd cell measurement will disbalance the 2nd cell cell with 89uA and this will impact the battery life.

My Idea was either reduce the divider current some how ?
or Use cell balancing technique ? I do not find cell balancing + OVP(4.2V) chip for two cell . So i thought to reduce the current as much as possible for 2nd cell so that cell 1 and cell 2 will charge with almost same current .
 

Thread Starter

mishra87

Joined Jan 17, 2016
759
Its coming from op-amp supply, but unless you have some other as yet unspecified power supply in your design, its fairly obvious its eventually coming from the battery. Why are you so worried about 89uA? What's your overall power budget?
Vcc = 5V, VBAT = 4.2V

Your above two statement contradict itself : Once you say it is coming opamp supply and then say it is coming from battery ?
 
You're overthinking the problem; a little logical thought and some maths would show you that 89uA isn't going to significantly unbalance the pack. @2Ah, 89uA is going to take over 1.1 MILLION hours to remove 1% of the charge. The differential self-discharge rates will unbalance the pack in a few weeks...
 
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Your above two statement contradict itself : Once you say it is coming opamp supply and then say it is coming from battery ?
No, read carefully what I wrote. This is battery powered equipment, you haven't indicated any other power source. Therefore even though the current in R3 isn't coming from the battery through the measurement input at R2, the op-amp supply must be the battery. If thats not the case then run your opamp at 12v and avoid the common-mode issue.

This is what happens when you only give 1/2 the information about the problem and the reason for your question...
 

ericgibbs

Joined Jan 29, 2010
10,158
hi mishra,
A project's power supply, supplies all the current/power required for the circuits operation.
The OPA or Amplifier only controls the current flow between the power supply and the circuit load.

So for your OPA driven resistive divider, the OPA controls the current flowing from the OPA's voltage supply thru the resistive divider.

As pointed out a uAmp sampling current is a very low current drain on that battery.

You need to pay attention to the effect the transistors Vce , at the low end on the divider and also the 8.4v being applied to the ADC via the 47k when the transistor is Off.

E
 
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