Sensing battery voltage ....battery is at the end of wire lead.

[Flyback]

Hello,

We are designing an Emergency Light, which powers a lamp from a 6V battery when the normal power supply fails. Otherwise, the Emergency light PCB charges the battery.
The battery is at the end of a wire lead, which is half a metre long. The resistive drop in this lead during battery charging (1 Amp) and discharging (400mA) means that we are not correctly sensing the battery voltage.
Therefore, we wish to run separate “sense” wires from the battery terminals to the PCB, so that we can more accurately sense the actual battery terminal voltage.

During discharge, the battery voltage needs sensing because the battery is composed of five NiCd cells in series, and it is dangerous to discharge such a battery pack to below 5.25V as individual cells may become damaged.

During charging, the battery terminal voltage must also be known, because the battery terminal voltage must always be below 8V.

The internal impedance of each cell is 15mR. (so that’s 75mR for the battery, -unfortunately, we don’t know what the resistance of the cell to cell interconnects is)

Schematic of Battery voltage sense circuitry.
https://i49.tinypic.com/e9t8rd.jpg

(Also attached as PDF)


..Please forgive me using LT1006 opamps in the schematic, in the actual circuit , MCP601’s would be used……………..

MCP601 datasheet:
https://www.electro-tech-online.com/custompdfs/2013/02/21314g-1.pdf


Please could you state how accurate the sensed battery voltage would be with this setup?……we would prefer not to have the expense of an (opamp) instrumentation amplifier if at all possible.

 

[dougy83]

Why don't you just use the differential amp and get rid of that odd amp (U2). To make sure the input voltage is above 0V, add some common-mode offset by connecting two [equal-value] resistors from the supply to the opamp inputs.

 

[Flyback]

U2 is needed because otherwise the impedance of the lower divider resistor is compromised..........the source impedance is too high and thus the opamp buffer is needed.

Sorry, i dont understand, why is this needed?.......

add some common-mode offset by connecting two [equal-value] resistors from the supply to the opamp inputs

 

[dougy83]

U2 is needed because otherwise the impedance of the lower divider resistor is compromised..........the source impedance is too high and thus the opamp buffer is needed.

The "lower divider resistor" should be part of the differential amplifier instead of being an extra stage. A diff amp can have a gain different to unity. Your diff amp can just have R3 = R4 = 12k and R5 = R6 = 1k86.

Sorry, i dont understand, why is this needed?.......

Quite right, it's not. Both inputs will always be in the correct range with the battery connected.

 

[Flyback]

The "lower divider resistor" should be part of the differential amplifier instead of being an extra stage. A diff amp can have a gain different to unity. Your diff amp can just have R3 = R4 = 12k and R5 = R6 = 1k86.

though i think that will make for inaccuracies, as you have R3=R4=12K, and 12K is not that much bigger than 2K2 (R2). [its bigger but not enough bigger]
Also, since you have R5=R6=1K86 , the gain will be 0.155 which is too low, i'm going to start getting troubles with getting too near the VOL of the opamp.

Is there a cheap opamp Instrumentation amplifier solution here?, my source resistance is just too high, i fear, for this differential amplifier?

 

[dougy83]

though i think that will make for inaccuracies, as you have R3=R4=12K, and 12K is not that much bigger than 2K2 (R2). [its bigger but not enough bigger]
Also, since you have R5=R6=1K86 , the gain will be 0.155 which is too low, i'm going to start getting troubles with getting too near the VOL of the opamp.

Why is the gain suddenly an issue? You proposed a 12k/2k2 divider (which is a gain of 0.155) followed by a unity gain diff amp using 1k resistors. 1k is less than 1k86 and would therefore have more loading on the opamp output.

Anyway, 0.155 x 5V is 0.775V, which should be the minimum output voltage with a flat battery pack. Which VOL value are you using?

Also what do you mean by "bigger but not enough bigger"?

 

[Flyback]

Sorry, my misunderstanding, you mean like this.

https://i46.tinypic.com/351btye.jpg

I suppose this would be accurate?
The MCP601 opamp supply rail is just 3V3, and the battery is anywhere between 1.2V and 8V, so when its 8V, i hope this won't be a problem?

 

[dougy83]

Sorry, my misunderstanding, you mean like this.

Yes

I suppose this would be accurate?
The MCP601 opamp supply rail is just 3V3, and the battery is anywhere between 1.2V and 8V, so when its 8V, i hope this won't be a problem?

It's as accurate as a standard differential amplifier. The fact that the gain is 0.155 means that the input voltage offset error is not amplified, so if the resistors are accurate, then the output voltage will be accurate. An input of 1.2V (very low for a 6V battery pack mind you) should produce an output voltage of 186mV, which is quite possibly >VOL for a 1k8 load: which would be fine. If VOL is >186mV, then you could perhaps use a pull-down resistor on the output.

When the battery has 8V, the output will be 0.155 of that, which is 1.24V

 

[Mr RB]

...We are designing an Emergency Light, which powers a lamp from a 6V battery when the normal power supply fails. Otherwise, the Emergency light PCB charges the battery.
The battery is at the end of a wire lead, which is half a metre long. The resistive drop in this lead during battery charging (1 Amp) and discharging (400mA) means that we are not correctly sensing the battery voltage.
...

Just use thicker wires? 1A and 1 metre are fairly insignificant amounts and with decent wire the voltage drop would likely be less than 0.05v, and even at a worst case of 0.1v it's going to be insignificant for charging and discharging setpoints and/or could be compensated in software!

If it was 5 amps and 5 metres that would be a different matter.

 

[MikeMl]

Does anybody feel like Flyback is wearing out his welcome on this forum? After all, he is being paid to engineer this product, but I'm getting the feeling we are doing his job...

Just asking?

 

[ChrisP58]

I agree with Mr. Black that thicker wires is the way to go.

20 AWG (0.8mm dia) is 33mOhms per meter, so your voltage error would be 33mV at 1 amp.

For a target voltage of 8V, the error is 0.4% so, unless you will use high precision resistors or trimming, the error from your remote amplifier will be greater than the error from the wiring.

Also, due to the wire resistance, the battery voltage will always be lower than the charge output, so you are safe there.

 

[Mr RB]

Does anybody feel like Flyback is wearing out his welcome on this forum? After all, he is being paid to engineer this product, but I'm getting the feeling we are doing his job...
...

He's getting paid to design a commercial emergency lighting product and strangers on the internet are doing all the work for him for free? How rude.