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I have four 12 V lead-acid batteries in series and want to measure the total voltage. I'm interested in the 48 to 58 V range. How can I convert the range to represent it with a 0 - 10 V volt meter?

The setup is fixed, the voltmeter is a bar graph with 0 - 10 V input.

With a voltage divider I only get the range 0 to 58 V not 48 to 58 V. To explain, 48 V should result in 0 V and 58 V should result in 10 V.

ocrdu
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ThomasDr
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    Do you already have a meter picked out? – vir Apr 07 '23 at 19:14
  • Dividing by either 10 or 6 sounds easy (with a very high "impedance" (resistance, here) meter even trivial). Subtract 45 and divide by 2 should improve resolution/relative precision. Decide whether you want *battery* voltage or *load* voltage. – greybeard Apr 07 '23 at 19:37
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    Please _edit your question_ so it says how much precision you need, what delay and sampling rate, and whether you will be looking at it visually or piping it to a computer. And since it keeps raising itself in my head, address the issue of why you can't just buy a USB-capable 3-1/2 digit voltmeter and wire it into your setup permanently. – TimWescott Apr 07 '23 at 19:45
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    Your input range is 10 V. Your output range is 10 V. All passive resistor networks attenuate, so it's not possible to achieve using resistor divider techniques even with unholy voltage rails. (Current source perhaps.) So, an active circuit. Assuming you deal with @TimWescott 's questions adequately, it might only then be appropriate to suggest an active circuit. But everything has tolerances and perhaps initial accuracy (that drifts and is noisy.) So, much detail should be provided about your purposes and goals. To "measure" is not sufficient. So go beyond that by a fair mile or two. – periblepsis Apr 07 '23 at 20:35
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    You need to be more forthcoming on the specifications of your meter. – RussellH Apr 07 '23 at 23:43

3 Answers3

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A rather obvious and simple solution is just to add a series zener diode, although it may not provide sufficient accuracy.

schematicenter image description here

A better and still simple solution would be something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Input and output voltages

The input is 40 to 60 VDC, but the output is 0 for 48V and 10 for 58V.

PStechPaul
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    Note that I used four zeners in series because CircuitLab does not have a 48V zener in its library. The 10k resistor emulates an analog voltmeter with 1000 Ohms/volt (1 mA movement). Zeners usually will not work well at such low currents. Best performance for, say, a 200 mW zener will be around 100 mW. – PStechPaul Apr 08 '23 at 04:35
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    You can manually edit the parameters of the Zener - voltage, resistance, and Zener current. The library is there as a starting point, but more often than not one ends up with custom parameters. The op-amp approach would be great with a micro power high voltage op-amp and a low Zener current. That R1 will be discharging the batteries nicely, although perhaps for chunky lead-acids it’s all irrelevant. – Kuba hasn't forgotten Monica Apr 08 '23 at 15:22
  • I considered doing that, but I wasn't sure about other parameters. There are micropower voltage regulators that work up to 60 or 72 volts, but the ones I found are about $6. https://www.mouser.com/datasheet/2/609/MAX6765_MAX6774-3129954.pdf – PStechPaul Apr 08 '23 at 19:14
  • Here's one that's less than $1 and works up to 100V, with about 500 uA quiescent current: https://www.diodes.com/assets/Datasheets/ZXTR2012K.pdf It's just a 150k resistor, 12.7V zener, and series pass NPN BJT. Output is 12V +/- 10% – PStechPaul Apr 08 '23 at 19:32
3

The basic approach is simple: throw a 48V voltage source between the battery and the output:

schematic

simulate this circuit – Schematic created using CircuitLab

Of course, the problem is then: "how do we get an accurate voltage source like that?".

We could use an ATL431LI shunt reference, since it can operate on a meager 80μA:

schematic

simulate this circuit

That one is very accurate and works down to about 48.01V.

One other approach uses a resistor loaded with a current source as a voltage offset generator that is then buffered:

schematic

simulate this circuit

Adjust R1 for 5.000V on the voltmeter when the battery is at mid-voltage, ie. 53V. You'd want to use an adjustable power supply for that purpose.

RV models the voltmeter's input resistance. It can be pretty much anything, without introducing much error, thanks to the mosfet output buffer.

A mosfet's threshold voltage drops with temperature - in the very rough ballpark of 6mV/K per mosfet. That's not a big deal for such battery monitoring, most likely.

I1 needs to be stable with temperature. It would be lovely to use an LM334, but that won't work, as it's a PTAT current source - proportional to absolute temperature - and the tempco at the gate of M1 would be about -0.1V/K. That'd be quite bad.

Instead, we need something much more stable. REF200 would do the job, resulting in a tempco of about ±1mV/K. That's better than the buffer stage.

M2 provides additional voltage drop to keep M1's gate voltage within I1's compliance range, i.e. at minimum 2.5V.

R3 provides short-circuit protection on the output, and dissipates 0.06W when the output is shorted.

Kuba hasn't forgotten Monica
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3

You need a 48V reference voltage source. Then the negative probe of the voltmeter is placed on the reference. The positive probe is placed on the (48V to 58V) battery node.

The meter will read 0V if the battery is 48V and 10V if the battery is 58V.

schematic

simulate this circuit – Schematic created using CircuitLab

Any low current commercial reference supply will work as long as you are confident of its value.

RussellH
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  • This is OK if the meter can operate with common mode voltage of 48V. And it will read a negative voltage below 48V. – PStechPaul Apr 07 '23 at 23:38
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    @PStechPaul: True. But the principle of measurement is the same. It needs the zero part of a spa zeo circuit. Whether it will display negaive voltages or not depends on the meter, which there is little information given. As far as the question implies my answer is complete. – RussellH Apr 07 '23 at 23:49