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interestingfellow 5th August 2011 12:25 AM

OT: Noob needs help with voltage divider basics
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OK, here we go!
Total noob here, and I'm trying to teach myself electronics 101. It's overwhelming, btw.

I'm trying to make an analogue voltmeter (of sorts) from a few LEDs and resistors. I'm actually building this to monitor my e-cigarette voltage. I got it from this thread, but know you all over here are pretty sharp at these sorts of things.

Long story short, every LED has a turn on voltage, and you can exploit that by setting up a voltage divider to create steps, causing the LED's to turn off at certain voltages.

I know the formula for a simple divider is Vin=((R2)/(R1+R2))*Vout
But figuring out what value resistors I need is just kicking my ace. I understand the principle, but don't know how to figure out the specifics.

My schema is attached, and here is some data on the LED's
(I measured a cut on of 2.7v @ 31uA)
Size (mm) : 3mm
Peak Wave Length (nm) : 395 ~ 405
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <= 30
Luminous Intensity Typ Iv (mcd) : 3000(Typical) ~ 5000(Max)
Max Power Dissipation(PM): 80mW
Max Peak Forward Current(IFP): 75mA
Max Continuous Forward Current(IFM): 20mA

Thanks, ya'll!

nigelwright7557 5th August 2011 12:32 AM

Look up bargraph driver IC's.
You should be able to scale the input to suit your application.

interestingfellow 5th August 2011 12:39 AM

I know it can be done without using an IC, and the fact I can't figure it (voltage devider calculations) out is just ******* me off.

stratus46 5th August 2011 06:49 AM


Originally Posted by interestingfellow (
I know it can be done without using an IC, and the fact I can't figure it (voltage devider calculations) out is just ******* me off.

Nigel gave you excellent advice. What you're trying to do is actually tricky since the load keeps changing as LEDs turn on. Can you do it your way? Yeah. Is it a PITA? Yeah. If you insist on avoiding IC's I would recommend a 2 resistor divider for each LED to avoid the varying load. With no comparators to detect thresholds, your LEDs are ALL going to vary in brightness as the currents go up with varying voltage and there is no way to avoid that 'quirk' unless you use comparators to keep the LED drive constant. That is how the LM3914/3915/3916 chips (that you don't like) work. If you want to change the LED color you'll have to re-calculate all the resistors because different color LEDs have different operating Voltages. Or you could go to Jameco and pay the $2.75 for the LM3914, use any color LED - even mixed colors for different voltages - and move on to something more fun. Or not.


interestingfellow 5th August 2011 01:07 PM

My apologies, I wasn't trying to be a dbag.
It's not that I don't like the IC's, I'm just frustrated; electronics is something I've dabbled in here or there on a very basic level. but the last several projects I've sought out to accomplish have been beyond my immediate capabilities. No one likes not being able to do something.

As far as the changing current dimming the LEDs, cool; as the voltage creeps down from use, they will dim their way down and turn off as voltage drops. That's what I want it to do; I"m not trying to measure the voltage on some critical lab equipment, just a general idea of where my battery is at (+/- 0.1v).

I though it would be fairly easy to do this one; guess not.

Thanks ya'll.

sofaspud 5th August 2011 02:27 PM

I'm surprised. I'd of thought a microcontroller would have been suggested over a bargraph chip. Anyway, your schematic is not correct for what you want to do. If all the LEDs are identical, the 4V one will be the only one to light up. To be more precise, the LED with the lowest Vf will be the one that lights, regardless of battery voltage. Try thinking of the resistors not as a voltage divider but instead as ballast. They drop excess voltage, and limit current thru the LED. It may then become apparent that your voltmeter would be easier to implement with 5 LEDs with different forward voltages. They would be lowest to highest moving right from the battery and thru a single resistor. +/-0.1V accuracy is asking a lot from a small passive circuit, not to mention the error generated by the mismatched load.
Which reminds me... you really ought to determine the load on the battery during normal use, then design a voltage tester around that load figure. Hope this helps.

interestingfellow 5th August 2011 04:59 PM

Thank you.
To keep it short, I'll just get too it (not to be rude)
Yes, and IC would be easier to implament; follow the wiring diagram and throw it in. But now it's just the point (for me) that I can't figure this out. I do intend on doing a 101 course at the local community college.
I meant 0.2v +/-0.1v at each LED. Maybe I'm just not reading it right, here's the idea behind it as posted on that other forum:

Originally Posted by user RJG at ECF
As for a how to - it's simple but requires some work.
All LED's have a "turn on" voltage. You can measure it with most multimeters with a diode setting. Or you can rig up an LM317 with a pot and slowly creep up to the voltage until it turns on.

I did that, it's 2.7v

Then make a row of LED's and create a voltage divider with a couple resistors that makes exactly that voltage at each point you want to monitor.
I think this is where I get lost. Is he saynig just make a voltage devider for each one, or string them together? My intent on my attached schema was to string them together, maybe I just don't get it.

Using small clear red LEDs like this you can easily get away with a 1K + x dividers. The entire string lit up full bright on a brand new battery draws only 9mA.
That's .009 amps. The atomizer draws about 1.5 amps. Wired to the push switch, it has no effect on battery life whatsoever.
On this one, the first
LED starts dim at 3v, and is full bright at 3.3 volts.
The next is dim at 3.3 and is full bright at 3.6 volts.
The next is dim at 3.6 and is full bright at 3.8 volts.
the last is dim at 3.8 and is full bright at 4 volts.

When the battery is full, all four are on bright. When it's dead, only the bottom one lights up. I chose 4, since all four LED's, from dim to bright, overlap slightly into the range of the next one, so it's a full scale. You can tell whatever volts it is very accurately - not just the 4 points.
My e-cig draws 2.8amps (3.6v, 2ohm load), and I was trying to do a <3v to 4v scale (the 18650 "3000mah" battery runs about 4v, no load fresh charge. and yes, I know it isn't 3000mah, but $10 for 2 from china, who cares, it will still last longer than the 14500 700mah battery I'm using in my other ecig).

Fivetide 5th August 2011 05:53 PM

QED P300 service manual ..very long shot but
so sorry though I was starting a new thread .. doh !

interestingfellow 5th August 2011 05:56 PM

not 2.8amps, 1.8.

Upon closer inspection, I see 9 resistors, and 4 leds.... 4 seperate voltage dividers, and (probably) 1 resistor to limit the current to the whole circuit?
Off to the breadboard!

sofaspud 5th August 2011 11:20 PM

Sorry, but I'd have to see a schematic, at the very least, to accept this as a working idea. Or I'm totally missing something.

I think this is where I get lost.
Don't feel bad. There's only one point you want to monitor - the + terminal of the battery (ground referenced). And "a voltage divider with a couple resistors that makes exactly that voltage" only does so at an equally exact voltage from the battery. Obviously, that's counter to the entire point of the circuit.
"When it's dead, only the bottom one lights up."
A dead battery that lights an LED? To repeat, the first LED to turn on and the last to turn off will be the one with the lowest Vf.
I'd love to see the working prototype.

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