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Suggestions for using high value resistor

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mrinalmani

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Hi!
I am designing a battery string stabilization circuit with 20 individual Lithium cells in series. The cells combine together give an output of nearly 80V.
The voltage of each individual cell needs to be monitored, so a resistor divider is needed for each tap-off.
To minimize leakage current, I am thinking to use resistors in the range of 500K to 1M Ohm. (Each divided voltage is then buffered with a precision amplifier with low bias current.)

Is 1Mohm high enough to take extra precautions? How high can I go without complicating things?
Many people are of the opinion of using multiple series resistors to make up a large value resistor instead of using one singe large resistor. (Example 2x 500Kohm in series rather than a single 1MOhm) Should I do the same?

Thanks
 

Hi,

I assume you are overcomplicating things.

* 80V isn´t that high voltage...
* 1M gives a current of just 80uA...this will be less than the self-discharge current of a Li cell. So don´t care too much about it.
* Low bias current: 0.8uA (which is far beyond low bias current) just causes 1% of error. Even a 741 is below 0.8uA. Any mediocre OPAMP will do.
* instead of "individually" buffer the signals you could use a MUX, and then use 1 (or 2) buffer in total.
* "precautions" about what? overheat? health? --> don´t worry. You talk about 80V and not about 80kV.
* "multiple resistors": Why? It´s a waste of time. And a doubtful benefit. I use a single resistor for 2500V AC in industrial environment without problems. You have to read the resistor´s datasheet about voltage rating.

I recommend to use identical voltage dividers. Calculated for the highest voltage. Add a capacitor in parallel to the lower end resistor to get a low pass filter..
Then a MUX. Then a cheap delta sigma ADC with line frequency suppression.
ADC resolution: for 0.4% resolution on each cell you ned just 8 bits.
For the series string of 20 cells you need additional 5 bits.
So 13 bits should be sufficient. --> A 16 bit ADC is more than sufficient.

Klaus
 
Hi!
Thanks for the quick reply.
I very much admire the idea of using multiplexers and single opamp at he output of multiplexer. However is it OK to send a weak signal to multiplexer without strengthening it with a unity follower or so? Since the current is only about 80uA and series resistor is high, will the multiplexer distort this weak signal voltage?

Also by precautions I meant PCB layout precautions such as guard ring around the high input resistance voltage divider. Is 1Mohm not so high a value to consider distortion due to several factors that need some PCB design precautions?

With the same resistor divider at each tap off, as we go lower in the string the instrumentation current increases for each cell. With the top-most cell bleeding 80uA and bottom most 80uAx20 = 1.6mA. Is this acceptable? s 1.6mA current not too high to be drained out of cells continuously?

Thanks again!

- - - Updated - - -

Just realized that the bottom most cell will not bleed as high as 1.6mA. But certainly it will be close to 1mA
 

Hi,

Is MUX OK:
every MUX has it´s datasheet. Read them.
Especially:
* introduced charge
* leakage currents

I assume the sampling rate is very slow, therefore the introduced charge may cause no significant error because it is compensated by the source impedance.
I assume the leakage currents of analog MUX will also have no signaficant influence on your measurement precision.

Source impedance:
Maybe you use 1M - 50k resistor divider. Then the source is about 50k .. not that high when buffered with an OPAMP
The 1M - 50k divider brings down 80V to about 4V. On a 5V system this leaves some headroom for high charging voltage.

Guard ring:
For yure you can do.
I call it useless, because with a usual (clean) PCB one can expect leakage currents in the range of some 10 nanoamperes.
It won´t cause much error. I´d keep the sensitive node short and increase the distance to other tracks.
(If you use a guard ring, then you need an actively driven around the center node of the voltage divider. Too much effort)

1M and PCB:
1M is the input resistance... but there is no node that is that high ohmic. (unless the input is left OPEN ... but then the measurement is invalid)
Your highest ohmic node will be around 50k (multiplied with 50nA is 2.5mV, x 20 gives about 50mV cell voltage error...worst case)

1.6mA.
Your math is not correct (you realized it already). It is about 0.84mA.
What is the expected charge to charge period time in hours? Multiply this value with 0.8mA and get the drained out charge.
What is your cell charge? Calculate the relation of both values.

Klaus
 
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