# LED Meter Qs

Discussion in 'The Projects Forum' started by realfstkid, Sep 19, 2013.

1. ### realfstkid Thread Starter New Member

Sep 19, 2013
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0
Hi all,

I'm brand new here, but I figured maybe I could learn a bit from you guys.

I've been trying to figure out how the voltage monitor system from my 1987 Volkwagen Westfalia Camper Van works. Here is the circuit board:

And this is the schematic that I've drawn out for it:

So, first things first. I assume that the blue cylindrical object between the red LED and the potentiometer is a diode (labeled D3 in schematic/picture), would I be correct? I show a .7V drop with my Fluke 113 multimeter, but there is no single beep as I get with the other 2 diodes on the board. I show 0L when I test it reverse biased.

OK, The point of the circuit is to take the voltage from the camper battery (12V nominal), and illuminate either the Green, Yellow, or Red LED's corresponding to the charge state (voltage).

So here's how I *think* this is supposed to work. The voltage drop across diode "D2" is a constant 0.578 volts (measured w/ Fluke 113). So theoretically the non-inverting voltage of the top op amp should be 0.578 volts higher than the non-inverting voltage at the bottom op amp.

Here's where I get confused: Because of the reverse biased diode, "D3", there should be no current running across this part of the circuit, and therefore the voltage at top op amp should be the same as the voltage after D1. But this would insure that the top op amp is always outputting V, and thus the Green LED would never illuminate. What am I missing?

Jumping to the inverting (-) inputs, based on the measured potentiometer values, I calculated a voltage drop of 17% of V, or Vp = 0.83*V.

For the system as a whole, this is what I gather: The difference between the inverting (-) inputs of the op amps is scaled linearly with any change in V, whereas the difference between the non-inverting (+) inputs stays constant. It's this difference that causes the circuit to change characteristic when the absolute voltage V increases/decreases.

Any help on this would be greatly appreciated regarding what D3 actually is, and what voltages are to be expected at the non-inverting (+) inputs to the op amps.

Thanks

2. ### realfstkid Thread Starter New Member

Sep 19, 2013
5
0
Could an administrator please change the title of this post to "Camper Van Voltage Meter Questions". I couldn't submit the post with any title longer than what it's titled now.

Thanks

3. ### #12 Expert

Nov 30, 2010
18,093
9,683
This would make sense if D3 was a zener diode. That would explain the stripe going the wrong way, but a common diode wouldn't support enough voltage to be in the right range, so it must be a zener. Measure the voltage or unsolder it and read the part number.

4. ### wayneh Expert

Sep 9, 2010
15,214
5,560
FWIW, that circuit looks VERY much like what you find in a cheap battery tester available in any hardware store. The LM324 is widely used for those.

If the circuit is functional, you can get the zener voltage by just probing it.

5. ### Alec_t AAC Fanatic!

Sep 17, 2013
9,002
2,124
LTspice confirms this will work as a voltage monitor if D3 is a 10V zener or thereabouts.

6. ### realfstkid Thread Starter New Member

Sep 19, 2013
5
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Thank you so much. That was exactly the information I was looking for. I plugged the system in, and the voltage across the Zener Diode (D3) is 9.01V. So from what I've been reading, the voltage at the low non-inverting input will always be 9.01, and at the high will always be 9.65, assuming there is a minutiae of current running through this circuit, as it's required to ramp up the voltage drop of the diodes (currently there's ~3mA according to the voltage drop across the 1.2kOhm resistor - which should be plenty).

This is very cool stuff. Wish I would have asked that question a LONG time ago, before spending hours trying to figure out if I had calculated something wrong.

Now, according to my measurements of voltage drop across the pot, the voltage at the inverting (-) inputs of the op amps is 84.4% of the input voltage (already lowered by .7V). So as the pot is currently set, the battery voltage that triggers a change from Green to Yellow is (9.65/.844)+0.7 = 12.14V, which, for a typical flooded battery, would correlate to about 50% charge if the voltage measured was open circuit and the battery had not recently been under load. Correspondingly, for the change from Yellow to Red LEDs, the trigger battery voltage would be (9.01/.844)+0.7 = 11.38V, which is pretty much dead assuming the same as above.

Now, assuming the battery is under some kind of load, the battery voltage will be lower for the same state of charge, so this system becomes more relevant in that case. So while it's novel how this little gauge works, it's mostly a shot in the dark without some sort of knowledge of the current rate of discharge of the battery. But it's better than nothing. If I wanted perfection, I could always measure the specific gravity of the acid... oh wait, it's a sealed AGM battery I'm using...(;

7. ### realfstkid Thread Starter New Member

Sep 19, 2013
5
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Thanks for that. I tested the system this morning, and with the current setting of the POT, the 9V drop on the zener works out just about right for the intended (circa 1987) application...

8. ### #12 Expert

Nov 30, 2010
18,093
9,683
We don't need no stinkin' Spice model.
A window comparator is second semester electronics.

Just get certain about what you want your trip point voltages to be and adjust the knob. Perhaps batteryuniversity.com would help.

9. ### realfstkid Thread Starter New Member

Sep 19, 2013
5
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Well the whole plan is actually to rip this part of the board out and replace it with a digital voltmeter display and ammeter display.... To do that I have to cut the circuit board in half to keep the other side which is another window comparator with 3 lights for the water level, but it didn't need to use the constant voltage drop of diodes, so I wasn't confused. But now I know, and knowledge is power, or at least comfort...

10. ### #12 Expert

Nov 30, 2010
18,093
9,683
Now you know where the dotted line is. Start cutting.