# Low Battery Detector Calc?

I found this low battery detector, but with a 5.1 zener and 100kΩ pot. I learned, generally, how it works and I love it.

My detector uses a 9.1v zener and 50kΩ pot. I got it to work, but I’m hoping to learn the math to calculate for different zeners and to fine tune mine.

The original detector circuit uses a 10kΩ resistor connected to the zener, but I’ve also seen 330Ω. I’m curious about the math for this, too.

thanks!

The value of the resistor that is in series with the zener diode depends on the current required through the diode (see the data sheet) and how much current you are willing to have supplied by the 12 volt supply.

I understand the theory, but not the math, except the part about how much current I am willing to have supplied. The circuit requires a minimal current, yes?. That's the current I want.

The minimal current allowed depends upon the diode and I am not going to get into that.

The resistor R depends upon the current I and the voltage V by Ohm' s law:

R = V / I

where V is the lowest voltage out of your 12 volt supply (maybe 10 volts?) MINUS the 9.1 volts of the zener diode (so maybe V = 0.9 volts ?).

Now of course the maximum current is when the 12 volt supply is at its maximum voltage (maybe 13.2 volts ?). That current is given by

I = V / R

where V is the maximum supply voltage MINUS the voltage of the zener diode (so maybe V = 13.2 - 9.1 = 4.1 ? ) and R is the value of the resistor that you previously selected.

The power rating of the resistor should be at least I x I x R watts, where I is the maximum current and R is the resistance of the selected resistor.

Of course, these calculations make some simplifying assumptions, like the current into the inputs of the opamp is negligible ( it probably is ), and the current of the opamp and the LED may have to be added ( because I do not know what LED is used and a series resistor is not shown and I don't care to know now ) and the power to the opamp is not shown.

You lost me. Analog circuit math has always been my foil.

I was hoping for one equation with all the variables I can plug and play. Then I can use all the datasheets for their values and deconstruct the equation to understand what's happening.

The led in the circuit currently switches at 9v. I have no clue how to figure out why.

This is fine, but when I do this, I always use precision zeners, like the BZT585B series. I await the debate

You, also, probably what to put a resistor in series with that LED.

Yes, you should have a current limiting resistor in series with the LED. Without a current limiting resistor you can potentially burn-up the op-amp and/or the LED. (500-1000 Ohms if you don't know how to calculate it.)

The led in the circuit currently switches at 9v. I have no clue how to figure out why.

Maybe something is wired wrong. Maybe that's not a 5.1V Zener. Maybe the pot is wired wrong. With the 5V Zener it should be adjustable from 5.1V down to about zero.

This is a comparator circuit. When the (+) input is higher than (or more positive than) the (-) input, the op-amp output goes positive. When the (+) input is lower than the (-) input the output goes negative (or to ground if there is no negative power supply).

The Zener provides a reference voltage for the comparator.

Your LED is wired so it's off when 12V comes-out of the op-amp. (The LED will be off whenever the (+) input is higher than the (-) input.) The LED will be on when the (+) input is lower than the (-) input.

An open-loop (no feedback) op-amp has (theoretically) infinite gain (infinite amplification) and it amplifies the difference between the (+) and (-) inputs.

When the (+) input is higher than (or more positive than) the (-) input the voltage gets amplified/multiplied and the output tries to go + infinity. But, it can only go as high as the power supply so if you've got a 12"V power supply it goes to about 12V. (In the real world it may only go to 10 or 11V.)

When the (+) input is lower than the (-) input the negative voltage gets amplified and the output tries to go - infinity. If there is no negative power supply, it can't go negative at all and it simply goes as close to ground as it can.

The pot allows you to lower the voltage into the (+) input. In the original circuit with the 9V Zener, adjusting the pot to the mid point would give you 6V from a 12V battery. That's less than 9V and the LED should turn on. (Of course with a 5V zener the LED would still be off.)

This is fine, but when I do this, I always use precision zeners, like the BZT585B series. I await the debate

There's a pot in the circuit so a precision zener doesn't make any difference.

I left out the led resistor just for the drawing.

I have the BZX79...

I'm running a 12v LED strip from a deep cycle marine battery. A 9v threshold seems fine for me.

How can I adjust for a different threshold?

Also, with regard to the pot, how do higher value resistors compared to lower resistors affect the circuit here if the ratio stays the same?

You write that you want to learn the math. I simplify the math and hand it to you. The math requires only subtraction, multiplication, and division, the things that all calculators provide. No powers, no roots, no logarithms, no exponentiation. You want something else. Sigh. Never mind.

For any reasonable situation, only the ratio matters. You may choose the total resistance based on what you have handy. However...

Go too high in total resistance and the input current of the opamp starts to matter because the input current is no longer small compared to the current through the potentiometer.

Go too low in total resistance and the current from the 12 volt supply starts to get more than you want. That is, more current is used from the battery and the battery gets used up sooner. Also, with more current through the potentiometer, the power rating of the potentiometer starts to matter more.

"How can I adjust for a different threshold? " Uh, adjust the potentiometer? Use a different zener diode?
But: The zener diode voltage should always be less than the lowest battery voltage, AND the threshold voltage can never be larger than the zener diode voltage (with this circuit, of course).

No need to be rude.

As I explained, it's not the math. It's the equations to derive i, v, and r that I was hoping to learn.

KoogieBuffalo:
I'm running a 12v LED strip from a deep cycle marine battery. A 9v threshold seems fine for me.

But not for the battery.
Battery voltage is a bad indicator of remaining charge.

If you let the battery drop below ~12.2volt with a light load, than you could already have drained it too much.
Leo..

Thank you. That's helpful info that just saved me from a future issue.

What level voltage do you recommend I test for, which Zener should I use, and how will that change the circuit generously provided by ReverseEMF?

A simple voltmeter/indicator is useless while the battery is charging/discharging.
That simple circuit can only give you a rough indication the next morning (after a long rest).

Only a well designed/programmed coulomb counter can keep track of battery charge.

If this battery is only used to power a LED strip, then you can probably calculate how long the battery will last.
And use some sort of timer to warn you.
Leo…

I would be measuring it only in the morning, after the strip has turned off.

Maybe wise to buy a "3-30 volt meter" from ebay, and use your brain to estimate remaining charge.
Leo..