Battery differential Voltage measurement help needed

sghioto

Joined Dec 31, 2017
1,436
Probably homework but OK.
Assuming input impedance of ADC is infinite and LR voltage is correct.
Voltage at input of ADC is:

Case 1....( 8.4 X 20K)/(20K+22K) -.7

Case 2.....(4.2 X 20K)/(20K+22K) + 4.2 -.7

Case 3.....(4.2 X 20K)/(20K+22K) -.7
Steve G
 

Thread Starter

mishra87

Joined Jan 17, 2016
851
Probably homework but OK.
Assuming input impedance of ADC is infinite and LR voltage is correct.
Voltage at input of ADC is:

Case 1....( 8.4 X 20K)/(20K+22K) -.7

Case 2.....(4.2 X 20K)/(20K+22K) + 4.2 -.7

Case 3.....(4.2 X 20K)/(20K+22K) -.7
Steve G
Many Thanks !
It is not homework.
 

Thread Starter

mishra87

Joined Jan 17, 2016
851
Hi Friend,

I wanted to measure battery voltage of upper cell with 4V opamp supply.
here is the file i tried with simulation but i could not see the vout as expected.

Since opamp supply is 4V i have lower down battery voltage at input Side.


Could anyone explain me how it can be done ?
Thanks in advance !
1597948206580.png1597948232112.png
 

Attachments

sghioto

Joined Dec 31, 2017
1,436
The LT1112 is not rail to rail, max output with 4 volt supply is around 3.75v depending on the load.
What output are you expecting to read with the circuit as designed?
SG
 

Thread Starter

mishra87

Joined Jan 17, 2016
851
The LT1112 is not rail to rail, max output with 4 volt supply is around 3.75v depending on the load.
What output are you expecting to read with the circuit as designed?
SG
Thanks you so much for interest and reply !

I just wanted to read the voltage of upper cell i.e. lets sat 4.2V of fully charged Li Ion battery !
I just took random opamp from LT spice.
My supply voltage is fixed i.e. 4V.
Now i wanted scale down the battery voltage at input side let say if vbat = 4.2V, opamp output is 3.6V or something like. Since i know the scaling factor i can add in calculation.

Hope now you have perception in your mind !
Regards,
 

sghioto

Joined Dec 31, 2017
1,436
Use a difference amp. The reading is the difference between the supply (V1 + V2) and V2 which = V1. You don't need a separate 5 volt supply.
SGEEE difference amp LT1112.png
 
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Thread Starter

mishra87

Joined Jan 17, 2016
851
Use a difference amp. The reading is the difference between the supply (V1 + V2) and V2 which = V1. You don't need a separate 5 volt supply.
SGView attachment 215209
Thanks for your time !

Still you did not get my points what I was trying to say !

I wanted to operate opamp with microcontroller supply voltage which is 4V. I do not want give opamp supply of 8.4V I.e battery voltage total !
 

Danko

Joined Nov 22, 2017
988
Now i wanted scale down the battery voltage at input side let say if vbat = 4.2V, opamp output is 3.6V or something like. Since i know the scaling factor i can add in calculation.
I wanted to operate opamp with microcontroller supply voltage which is 4V. I do not want give opamp supply of 8.4V I.e battery voltage total !
Do you want something like this:
1597989110817.png
ADDED:
R9/R8=R7/R10=SCALE
(V1+V2)/V3 should be < than (R7+R10)/R7
 
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sghioto

Joined Dec 31, 2017
1,436
Thanks for your time !

Still you did not get my points what I was trying to say !

I wanted to operate opamp with microcontroller supply voltage which is 4V. I do not want give opamp supply of 8.4V I.e battery voltage total !
Is this just an exercise like homework or a real circuit you are planning on using? Why measure battery voltage then have to add a "scaling" factor to get the value when you can get the reading directly as I suggested.

SG
 

ci139

Joined Jul 11, 2016
1,696
ADDED:
R9/R8=R7/R10=SCALE
(V1+V2)/V3 should be < than (R7+R10)/R7
Differential Amplifier Configuration
Figure 7 shows a differential amplifier configured for offset adjustment. ...
// the Vos in above figure (at the link) is the merging point of the Rf , the R1 and the R2
// the Vos can be set by multiple means
// the formula defines your Vo
// if your Rf = Ri then the voltage gain would be 1
// if your Rf = N·Ri then the voltage gain would be N
// the amplified input voltage N · Vin is simply shifted by the offset voltage Vos
// in the other words the output voltage Vo is referenced to Vos
// the (output voltage shift) Vos = SGND (Signal Ground) which is also the Vss (the Op Amp's negative rail voltage) ... in your case ... is ±0 (Zero) V
// e.g.-- your output signal has no offset respective to the potentials at the SGND and at the Vss
 

Thread Starter

mishra87

Joined Jan 17, 2016
851
Thanks All.
This is real circuit implementation. i am trying to think.
yes there is limitation in design to think in this way only.

MCU stand for microcontroller in schematic.
Will Q4 and Q3 will turn ON.
Note: MCU is shiftrd 4.4V above battery ground.
What signal Q4 will see wrt to MCU ground and battery ground ?
1598086329280.png
 
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ci139

Joined Jul 11, 2016
1,696
200µA (\(\rightarrow\ I_C≈40mA\)) is likely enough to fully turn on the BJT (unless it has to transmit a highr power)
you are driving the Q5(NPN) with 3.3mA ... you may do so ... but depending on how fast you need it to be switched - you might save some power there

if your output is the battery voltage reading ... then you might want to use constant current sinks instead of the Q3 Q5 ... otherwise you have the Vce drop (that may be varying) affecting your output :: about https://wiki.analog.com/university/courses/electronics/text/chapter-11 see 11.6 Basic MOSFET current mirror Figure ( Fig. 11.7 Simple MOS current mirror ) -- you may use the BJT current mirror but it is a bit more tricky to tune and it's also less stable = more dependent on the supply voltage
 
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