Battery Check Function

Howdy,

As a minor feature in a gamma scintillator I'm building, I would like to provide a battery check function. The values may seem funky but they are a function of certain design points. I've muddled together a theoretical circuit to do this, but if people could look it over and let me know a couple things, I would appreciate it. I need to know if:

1) Will this work?
2) Is there an easier/better way of doing this?
3) It would be nice to do this without a negative ref...

Basically, three different key battery voltage levels provide three distinct output currents to a 50uA wiggle stick, which sports a pre-existing scale which makes sense what I'm doing.

thanks,

I would have used 3 terminal adjustable voltage references for the 2.7 and 3, unless you don't need much accuracy or your power supply voltages are precise and well-regulated.

Willbe said:

I would have used 3 terminal adjustable voltage references for the 2.7 and 3, unless you don't need much accuracy or your power supply voltages are precise and well-regulated.

Click to expand...

Right.

How I drew the references at this point was just for clarity sake. My supplies are indeed well regulated, but I'm not adverse to providing more precise refs, if needed. (After all, it's just a battery check, so I'm not TOO geeked out about adding more precision than required...)

Any way to make it simpler/smaller?

Thanks for your input!

Your offset voltages don't do what you want because the 1k resistor value is much too low and the 2M resistor and pot values are much too high.

audioguru said:

Your offset voltages don't do what you want because the 1k resistor value is much too low and the 2M resistor and pot values are much too high.

Click to expand...

Cool, that's useful. So how do I fix it?

(I was scrounging through books trying to figure out how to do offsets with unity gain, guess I didn't get what they were saying...)

AG,

So, how's this? (The referenced app note indicated a slight change to my previous circuit for non-inverting offset was in order.)

Is there a fundamentally better approach for this than what I'm trying?

thanks,

The value of the 1k feedback resistor is so low that your offset parts don't do anything.
The values of your offset parts are so high that they don't do anything.

Your first and third opamps are simple followers of the voltage at their (+) input.

You don't want and don't need a negative supply.

audioguru said:

The value of the 1k feedback resistor is so low that your offset parts don't do anything.
The values of your offset parts are so high that they don't do anything.

Your first and third opamps are simple followers of the voltage at their (+) input.

You don't want and don't need a negative supply.

Click to expand...

So how do I make this work? With the first and third opamps I was trying to get offset without gain. Is my approach flawed?

What's the best way to translate the three Vbat levels to their respective, indicated output currents? (With a single resistor between Vbat, say 155k, and the 50uA meter, I can satisfy the first of the three. But then I don't get the spread to fullscale with the highest indicated Vbat...)


thanks,

If my above approach doesn't work, what would? I'm looking for a circuit to give me the following:

INPUT OUTPUT
5.4V --> 35uA
...... --> .......
6.4V --> 50uA

thanks,

Hi Corey,

Check this out.

on1aag.

on1aag said:

Hi Corey,

Check this out.

on1aag.

Click to expand...

Thanks for your response. Unfortunately, the 50uA meter is single-ended (grounded) for its main function in the circuit, and I don't have sufficient switch contacts to "unground" it for use as you have it wired.

I'll look at your approach more thoroughly shortly, however.

thanks,

Okay, I'm back from the drawing board, here's my third try. Please let me know whether you think this will work, or if there a fundamentally better way to do this:

saturn1bguy said:

INPUT OUTPUT
5.4V --> 35uA
...... --> .......
6.4V --> 50uA

Click to expand...

So you have the 5.4 V source in series with a 154 KΩ resistor in series with the 50 µA meter, and you get 35 µA.

Now, you shunt this 154 kΩ resistor with a two terminal or three terminal Zener that fires at 5.4 V in series with "a resistor".

When the voltage is 6.4 V the meter is required to pass 50 µA.

It gets 6.4 V/154 KΩ = 42 µA from the existing circuitry so it only needs 8 µA from "the resistor".

1 V/8 µA = 1/8 MΩ.

Classical piece-wise-linear analysis.

For extra credit, what is the square root of 69?
If you've already heard this one don't post the answer.

Willbe said:

So you have the 5.4 V source in series with a 154 KΩ resistor in series with the 50 µA meter, and you get 35 µA.

Now, you shunt this 154 kΩ resistor with a two terminal or three terminal Zener that fires at 5.4 V in series with "a resistor". Classical piece-wise-linear analysis.

Click to expand...

I like all that, and I understand it, but I'm really hoping for all the values in between, too. That is, I'm looking for a circuit to translate any voltage between 5.4V and 6.4V to a current output of between 35uA and 50uA, respectively (even below and above that a bit.) Which means a circuit that carries off the math consistently; at least that's how I see it.

Thanks everyone for your help! Perhaps it's not do-able

I also had a go at it but it didn't work.

Your latest circuit has an input to the first opamp of 2.4V when the battery is 5.4V. The first opamp has a gain of 3.5 which changes a little with the setting of the pot.
The output of the first opamp is 3.6V.

The second opamp attenuates and level-shifts the output of the first opamp then multiplies it by 2. Its gain is not unity it is a little more.

If you want zero current into the meter below 5.4 V input here's how I think it can be done.

For some reason I can't upload the schematic so I will describe it to you. Photobucket used to work for me and now it won't upload either.

The first section switches on 35 µA to the meter for Vin > 5.4 V and the second section adds 15 µA per volt of Vin for 5.4 V < Vin < 6.4 V.
The - power supply terminal of the IC mentioned below is at ground and Vcc > 7 V.

Vin goes into the + input of a comparator (really an opamp running open loop) and the - comp input is connected to Vz, a 5.4 V zener powered through a resistor from Vcc.

The comp output connects to the meter through a resistor of such value that when the comp output goes to Vcc the meter is fed by 35 µA.
For Vin < 5.4 V the comp output is 0 V and the meter sees 0 µA.



The second section is an opamp set to, let's say, +3 gain, with the gain setting resistor that normally goes to ground going to Vz.

Vin goes into the + input, and the opamp output goes through a resistor and then to the meter so that the meter sums the current from both the 35 µA resistor and this one.

This resistor is of such value that with Vin = 6.4 V the opamp output voltage is at +3 V and another 15 µA is fed to the meter.
With Vin = 5.9 V the meter sees 35 + 7.5 µA.

Of course, I have to believe that with Vin below 5.4 V the output of this section will be at 0 V.
With dual supplies and a Vin of 4.4 V the opamp output would be at -3 V, but the output in this case cannot go below 0 V.

Hope it works!

I thought I'd report back before closing this thread. My circuit works perfectly. I finally breadboarded it and it works as expected. It did require a few tweaks, however, including bumping up the first stage gain a bit, Av=3.9 instead of 3, dropping the second stage gain a little, from 2 to 1.85, and to crank both pots just a bit from their calculated values.

I tested the first stage alone, setting Vbat to 5.40V, and trimmed the offset pot (1.90V instead of 2.0V) until I got the desired output. All subsequent Vbat values followed perfectly once the gain was corrected (empirically derived). The second stage then tested fine once I dropped the gain and tweaked its offset pot (1.64V instead of 1.4V).

The drawing here is the final version with the gain changes. Also, the pots in the previous versions have been replaced by fixed resistors here, with their actual measured values.

Again, thanks for your participation! This was a good exercise for me.