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6v Battery Low Voltage Monitor

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maxeemo

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Hello and sorry for bad English.

I need a simple circuit that monitors voltage of a 6v battery. The circuit should light a LED when voltage falls below a specific (adjustable) value. I would be grateful if you upload the circuit or link to any existing ones.
 
You could build a comparator circuit using a comparator IC. Or use a TL431. do a google search for both.
 
Here is a simple circuit:

**broken link removed**
A very simple battery monitor can be made with a dual-colour LED and a few surrounding components. The LED produces orange when the red and green LEDs are illuminated.
The following circuit turns on the red LED below 10.5v
The orange LED illuminates between 10.5v and 11.6v.
The green LED illuminates above 11.6v
 
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Battery Monitor

**broken link removed**

This circuit flashes when the voltage drops to 4v. The voltage "set-point" can be adjusted by changing the 150k on the base of the first transistor.
 
I'm sorry "diy didi". I'm not skilled in electronics. I've been graduated in Computer Science, and don't know much about electronics. All I want is a circuit diagram that signals the "low voltage status" with a LED. And if possible, I want it to cutoff the load on low voltage.

This is all about adding a simple battery monitor circuit to a powerful torch (flashlight). My torch has a 6V 4.5AH SLA battery, and power of its lamp is 55 watts!!! I want to protect the battery from "over discharge". Battery voltage must not fall below 5.8 volts.
 
Battery Monitor

**broken link removed**

This circuit flashes when the voltage drops to 4v. The voltage "set-point" can be adjusted by changing the 150k on the base of the first transistor.

Thank you colin55. Cutoff voltage is 5.8V. How do I get this value? And how do I add the "Cutoff" ability?
 
Battery Monitor

**broken link removed**

This circuit flashes when the voltage drops to 4v. The voltage "set-point" can be adjusted by changing the 150k on the base of the first transistor.


I'm interested in this because I have a project using a 7.2V 2000mA NiMH pack and I'd like to know when the voltage reaches around 6V or so. I breadboarded this and hooked it up to a variable power supply, and it begins lighting the LED (dimly) around 2.1V, begins flashing the LED around 6V and the LED stays on and flashing all the way to 9V where I stopped increasing the voltage. Turning the power supply on at 9V, the LED begins flashing, reducing the voltage it keeps flashing until it dims out at 2.3V. I am using MPSA56 and A06 transistors since I don't have the BCs here in the States.

I've tried different circuits using zeners, a 741, and a LM10, but all react similarly. Does anyone have any suggestions? Thanks!
 
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It's all in the voltage divider 1M:150k. If you want to make the turn-on sharper, put a few diodes in place of the 1M and use a lower series resistor.
 
Hi,


Here's a simple circuit to monitor under voltage.
R3 sets the LED 'on' current.
 
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Hi,


Here's a simple circuit to monitor under voltage.
R3 sets the LED 'on' current.

That looks promising enough to spend $6 shipping on a $1 semi. ;)

R3 value should be calculated for the LED, correct? The project (a Byonics TT4 APRS tracker with DK display/keyboard adapter) will have a 12VDC coaxial jack to be powered from an automobile cigarette plug. When the 12V plug is disconnected, the switch in the coaxial jack, which switches the negative (sleeve) connection, allows the 7.2V NiMH pack to power the TT4. FWIW, the size of my enclosure limits me to only 6 AA size batteries or else I'd run a 12V pack.

The LED is a Lite-On rectangular, 2V 20mA, so at 12V I get 333.333 ohms, or a 348 1% 1/4W. Will that work for the LM431 also, or do I need to adjust the value for it?

What are your thoughts on blinking the LED in the attempt to reduce current draw on an already dying battery? I don't find any rectangular LEDs with that option built in, so that would take at least an R/C timer I assume.

Thanks!
 
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Hi,


Here's a simple circuit to monitor under voltage.
R3 sets the LED 'on' current.
Hi Mr Al,
I created a circuit just like yours about a year ago that didn't have the 1N4148 in series with the LED. I wondered why the diode was necessary. I looked at my notes and realized that the voltage on the cathode of the LM431 is real clear to 2.5 volts until the trigger voltage is reached, as a result the red,green and yellow LED's with a Vf near 2V will always be on without the diode. I found by using blue LED's the diode is not necessary as the Vf of the blue is over 3V.
Thanks for helping me remember the reason for the diode.
Ned
 
Hi Roger,

You brought up a good point about the dieing battery. The OP wanted a circuit to turn the LED 'on' when the voltage goes under a certain limit point, but the more usual under voltage monitor turns the LED 'off' when the voltage goes too low. This logic might be a bit better because if the battery goes really really low then the LED wont turn on anymore anyway. To get the LED to turn 'off' with low voltage the LED is placed in series with the LM431 (cathode) instead of in parallel with the LM431. The resistor is still sized for the LED.

At 12v and an LED with forward voltage of 2v nominal and nominal current of 20ma, a resistor of 500 ohms would be necessary, but i would drive it at less than 20ma so maybe 750 ohms or something like that as the life of the LED goes up sharply with lower than full rated current.

Yes a good idea with the blinking LED. An LED running at 10ma always draws 10ma from the supply (battery), but an LED at 10ma that blinks with a 50 percent duty cycle only draws 5ma average from the battery, and at 10 percent duty cycle it only draws 1ma average. I used this idea with a refrigerator monitor to reduce battery current draw and get the batteries to last for 2 years, so it make a BIG difference. They make blinking LEDs already but im not sure if you can get that low of a duty cycle from one of those.
 
Hi Mr Al,
I created a circuit just like yours about a year ago that didn't have the 1N4148 in series with the LED. I wondered why the diode was necessary. I looked at my notes and realized that the voltage on the cathode of the LM431 is real clear to 2.5 volts until the trigger voltage is reached, as a result the red,green and yellow LED's with a Vf near 2V will always be on without the diode. I found by using blue LED's the diode is not necessary as the Vf of the blue is over 3V.
Thanks for helping me remember the reason for the diode.
Ned

Hi there Ned,

Oh yeah that's one pesky little downfall about the LM431. It's not made to regulate voltages below 2.5 volts so the cathode may not go lower than 2v minimum for any connection. Yes, the diode is there for the typical red LED used with low voltage monitors, but as i was saying in another post most of the time the under voltage circuits are built to turn the LED 'off' when there is an under voltage so that if the battery runs down real low we dont have to worry about not being able to turn the LED on. With the LED coming 'on' for low voltage, there's always the chance that the battery will run way way down and thus the LED wont light up so it doesnt indicate what we really wanted it to indicate. If the LED turns 'off' with low voltage then it doesnt matter if the battery runs down to zero as it will still show the true battery condition.

I was going to mention that blue or white or some greens would work without the diode but didnt get around to it :)
Some LEDs "first light" current might be very low too, so i wanted to avoid that as well hence the diode.
I also forgot the last closing parenthesis in the formula for the low voltage set point so i revised the schematic and so also added a note about the LED color/forward drop.
Also note that the upper range limit is quite high because there is no lower resistor in series with the pot.
 
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hi Al.
There is the TLV431 1.2Vref.

E.
 
hi Al.
There is the TLV431 1.2Vref.

E.

Hi Eric,


Yes good idea, and that would eliminate that diode finally :)

I see they also now have the "H" version which goes up to 18v or so. Not too bad.
 
At 12v and an LED with forward voltage of 2v nominal and nominal current of 20ma, a resistor of 500 ohms would be necessary, but i would drive it at less than 20ma so maybe 750 ohms or something like that as the life of the LED goes up sharply with lower than full rated current.

Yes a good idea with the blinking LED. An LED running at 10ma always draws 10ma from the supply (battery), but an LED at 10ma that blinks with a 50 percent duty cycle only draws 5ma average from the battery, and at 10 percent duty cycle it only draws 1ma average. I used this idea with a refrigerator monitor to reduce battery current draw and get the batteries to last for 2 years, so it make a BIG difference. They make blinking LEDs already but im not sure if you can get that low of a duty cycle from one of those.

Hmmm. Since my math calculations on the LED resistor were so far off, care to add a flasher circuit to your original schematic? :)
 
Hi Roger,

Hey not a bad idea. That would accomplish both under voltage detection plus battery saving.
I'll take another look at this and get back here...

It looks like what might work as a minimum part count design would be as follows:
1. Connect a new resistor R4 in series with a capacitor C1.
2. Connect that network between the cathode and 'ref' terminal of the LM431.
3. Connect another cap C2 that is one tenth the value of C1 from the 'ref' terminal of the LM431 to ground.

For example with R4=10k and C1=10uf, C2 would be equal to 1uf.
The operation is such that as the monitor voltage falls, it reaches a point where the circuit starts to oscillate. The pulse width starts out very narrow so the LED would be very dim, but as the voltage falls the LED gets brighter and brighter until it reaches some limit like maybe 50 percent duty cycle.

This is still to be considered experimental.
 
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Hi Roger,

Hey not a bad idea. That would accomplish both under voltage detection plus battery saving.
I'll take another look at this and get back here...

You don't know how much I appreciate it. I'll hold off ordering the LM341 until I hear from you since I want to get new capacitors too, most of the ones in my parts bins are NOS from when I was repairing pro audio/lighting and arcade video games back in the 90s. I was pretty good at building and fixing, but not on designing circuits. :)

Eliminating the LM10 DIP-8 I was having trouble with in favor of a LM431 T0-92 gives me a bit more space on the PCB, but I'll still need to see if I should go axial or radial on the capacitors. Unless it's necessary, I'd prefer not to have components under the HD44780 display, which is spaced around .4" above the PCB. Most of the room there is taken up by header connectors and wiring routed to the top of the case for the GPS and transceiver jacks.

The PCB is cut to fit a Hammond 1592ETCDBK case. If the forum allows links, here's a pdf of what I'm dealing with:

https://www.electro-tech-online.com/custompdfs/2012/05/TT4-DK_Parts_Layout.pdf

The space above and beside the power switch is about all there is available.
 
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Hi Roger,

So you want to try the capacitors and resistors addition to get the LED to flash then?
What kind of flash rate are you thinking of here?
 
Hi Roger,

So you want to try the capacitors and resistors addition to get the LED to flash then?
What kind of flash rate are you thinking of here?

I would think a short duration blink every 3 to 5 seconds would be appropriate. Just something to grab your attention while you're looking at the information on the LCD. If I understand correctly, the rate and intensity may change as the voltage drops, which would be fine. It might actually be handy, allowing the user to make "last gasp" calculations since the discharge curve of the NiMH battery pack indicates it would drop from 7.2V to "dead" fairly quickly.

Many thanks!
 
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