Best way to halve 12V battery voltage for 6V device, while still being able to measure the battery level?

Question

I have a device that is powered by 4 AA (4x1.5V=6V) batteries, but I want to power it with a 12V battery. The device measures its input voltage to determine the battery level. I only need to know if the battery is low.

The question is: what is the best way halve the battery voltage so I can use it to power the device, without losing the possibility for the device to measure the battery level (e.g., when using an LDO, the output voltage must follow the input voltage)? In my case, "best" means:

1. Using THT & easy to source components (I am just a hobbyist and would like to avoid soldering SMD parts).
2. Minimal impact on battery life (low Iq).

The maximum input voltage is 9V (a higher voltage will damage the device). The device uses 50mA when active and 10uA when in sleep mode (99.9% of the time). I am able to reprogram the device (i.e., change the input voltage level software setpoint at which the device thinks the battery is low). I am only able to modify the software, not the hardware (the device is sealed in epoxy, but luckily still has its programming pins exposed).

Options I considered:

1. Use a voltage divider. However, this will waste a lot power.
2. Use an LDO with an adjustable output voltage. However, usually, the LDO Vadj pin is connected to the LDO output and a resistor to set the voltage (forming a feedback loop), not to an external reference voltage. But maybe there is a way to fool the LDO so it takes the input voltage into consideration?
3. Use a voltage tracking LDO, like TPS7B4256. But I haven't found a similar device in a THT package.

This question seems not very exotic; I can imagine that many people have solved this already but I haven't found any answers on the forum, Google/, etc. I hope some of you can point me in the right direction!

Answer 1

Sorry, mobile so no schematic. I’ll draw one later.

Edit: still mobile but stole a schematic from All about circuits (link)

Feel free to adjust up the 10 kΩ to one or two orders of magnitude to save on battery drain.

How about an ST TSX631 opamp in buffer configuration with a 1:2 voltage divider on the input? Supply Iq is low enough for most battery applications and can supply 50 mA on the output. There are others too. Search on Digikey for something which can take your entire input voltage range, can supply the current, low Iq and in a capsule you like.

It would track your input voltage with division ratio.

https://www.st.com/content/ccc/resource/technical/document/datasheet/b5/7c/ad/c8/37/4f/4b/78/DM00078052.pdf/files/DM00078052.pdf/jcr:content/translations/en.DM00078052.pdf

Answer 2

Using only easy to find parts could mean discretes. Thus here is a simple discrete voltage regulator.

R5 R6 form a 1:2 voltage divider ; D1 should be a 8.2V Zener diode to limit output voltage. The rest is a discrete opamp with a differential stage and MOSFET as an output. Any N-MOSFET from the junk parts bin should work provided it operates within its safe operating area.

Idle current use is 1mA. It can be reduced to 200µA by setting R1 to 56k instead, this will make it a bit slower which is not a problem as the output cap is quite large.

R4 is not a component you should add, it simply models the equivalent serial resistance of the capacitor which is required for stability. Thus C1 should be a general purpose cap which should give the correct amount of ESR. 100µF should work fine.

It is definitely not a low dropout regulator due to the NMOS, but you've got at least 6V to waste so it's not a problem.

There is no short circuit protection on the output.

Here's the 100µA version (90µA for the regulator and 10µA for the load):

...and the 20µA version (10µA for the regulator) which is a bit pushing it, as there is no point going below the battery's self discharge. It requires large values for R5 R6 so I switched to MOSFETs for the differential pair to avoid base current.

Transistors are BC547/557C to keep good hFe at low Ic, and MOSFETs are jellybean like 2N7000.

As expected, at lower idle currents, the error amp loses transconductance and gets slower, so performance is worse.

Answer 3

Emitter follower using NPN transistors

^{simulate this circuit – Schematic created using CircuitLab}

The output voltage will be:

$$ \mathrm{ V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2} - 2 \times V_{be} - \frac{I_{out}}{\beta^2} \times \frac{R_1 \times R_2}{R_1 + R_2}} $$

$$ \mathrm{ \approx V_{in} \times \frac{R_2}{R_1 + R_2} - 1.2\,V} $$

Main advantage is the simplicity and low quiescent current, which is 4µA with the values shown. Downside is that the output voltage will vary by about 0.5 V depending on output load and transistor temperature.

Answer 4

Oh no, this ain't easy.

The best way would be an SMPS that has an algorithm for controlling voltage output that's based on input voltage. Proportional output (let's say halve output always) won't work, it has to be a more complex linear formula.

The reason being is that 1.5 volt batteries are actually 1.6 volts full, 1.0 volts empty (well some devices may draw them down to 0.9 volts but at 1.0 volts you get a very large share of the capacity). So what you want:

• Lead-acid battery that's 13.5-14.7 volts when charged, 13 volts when full, 11.5 volts when empty.
• Output range from 6.4 volts when full to 4.0 volts when empty

So if you multiply the voltage by 0.49, then you get 6.37 volts when full (close enough), but 5.6 volts when empty (which falsely shows too much capacity). Ain't gonna work.

So the formula you want is \$V_{out} = V_{in} \cdot 1.6 - 14.4\ \mathrm{V}\$

This formula gives 6.4 volts from 13 volts and 4 volts from 11.5 volts.

However, if you must support operation when the lead-acid battery is being charged, then you must cap the output to 6.4 volts, because from 14.7 volts you would get 9.12 volts which is way too much and would probably destroy your battery-operated device.

I guess the easiest way would be to create a buck DC-DC converter with an AVR microcontroller that you can test on an Arduino and create the algorithm there. Then you can implement the formula and maybe capping the output voltage to 6.4 volts. You need some experience in creating buck converters and control algorithms for them.

I guess given the very low power draw (50 mA), you might also consider a transistor in emitter follower configuration for dropping the voltage, with the linear equation being implemented using an operational amplifier analog math circuit. However, this wouldn't cap the output voltage to 6.4 volts, you need some way to figure out how to implement the capping. Maybe some simple circuit could do that as well.

Answer 5

A trick that can be used is to use a bipolar PNP transistor as a switch in the positive side of a potential divider from the source .

With careful choice of resistor values - transistor off, leakage current only. Transistor on, drain base current + divider chain current. Voltage drop is saturation voltage of transistor..

Turn on the switch a short while before measurement, time to settle, sample, turn off switch. If I were using an STM32Fxxx MCU, this could all be programmed in hardware on a counter timer,even including the ADC trigger delay.

Then the supply voltage regulator output does not have to be exactly half the input source voltage , just sufficiently high to allow the output of the measurement potential divider to be within the ADC input range . So then a fixed 5V or 6V switchmode would do.