Simple Circuit using LED to alert low voltage - But I need lower threshold

Thread Starter

Lumenosity

Joined Mar 1, 2017
614
Hello
I'm wanting to use this circuit in a small, 2 battery, AA powered vacuum.

I want it to light the LED when the combined battery power of both batteries drops below (approx) 2.5v

The problem is, this circuit is set up for a minimum of 6volts. It seems that all I need to do is use a higher value resistor for R3 but I'm not sure and what resistance.

Can you help?
Thanks

Data Sheet for TI TL431
http://www.ti.com/lit/ds/symlink/tl431a.pdf

 
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Audioguru

Joined Dec 20, 2007
11,248
First you must fix the problem where the inductance of the motor boosts the voltage. Why do you have two threads about this vacuum cleaner?? A fix might be a filter capacitor parallel with the battery. Try 470uF.

A 3V battery that has dropped to 2.5V is still almost new. It is about 2V when it is almost dead. But the TL431 does not work below 2.5V.
 

DC_Kid

Joined Feb 25, 2008
1,072
how about just a 2v Vf diode to turn on/off led when V drops below 2v (or whatever v you need)?? std alkaline cells are about "dead" when they reach 1.2v. i guess i would also question if the batts have enough mAh left to power the LED as i assume this LED indicator is in parallel to the vacuum motor, hence, the motor is still a load to the batts when batt voltage drops to almost their dead value. this is in fact a power wasting ckt, and if power consumption is a key factor then waste in uA units is important. mA in waste is not good for small (small Ah) batt ckt's, etc.

 
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ebeowulf17

Joined Aug 12, 2014
3,307
I feel like a dummy here. Can someone explain to me how this circuit works? It's not making sense to me.

I must be misreading this circuit, but it looks to me like it's configured to work as a voltage regulator for the LED. So, if the 431 has a reference voltage of 2.5V, it will shunt any voltage above 2.5V to ground, maintaining 2.5V across the LED and keeping the LED lit at all times.

Or, realistically, the LED will probably have a Vf that's lower than 2.5V, so it will act as a shunt for anything above its Vf, and therefore be lit as long as the battery voltage is above its Vf, and the Zener effectively does nothing.

What am I missing?
 

KL7AJ

Joined Nov 4, 2008
2,229
Hello
I'm wanting to use this circuit in a small, 2 battery, AA powered vacuum.

I want it to light the LED when the combined battery power of both batteries drops below (approx) 2.5v

The problem is, this circuit is set up for a minimum of 6volts. It seems that all I need to do is use a higher value resistor for R3 but I'm not sure and what resistance.

Can you help?
Thanks

Data Sheet for TI TL431
http://www.ti.com/lit/ds/symlink/tl431a.pdf

Indeed, that can be a bit tricky with low voltages. Perhaps you could replace the Zener with a precision voltage reference:
http://www.ti.com/lit/an/slyt183/slyt183.pdf
 

crutschow

Joined Mar 14, 2008
34,285
but it looks to me like it's configured to work as a voltage regulator for the LED.
No, it's configured as an open-loop comparator as discussed below from the data sheet.
To operate as a regulator, R2 would need to be connected to the TL431's cathode.
upload_2017-11-3_16-18-40.png
upload_2017-11-3_16-20-24.png
So, if the 431 has a reference voltage of 2.5V, it will shunt any voltage above 2.5V to ground, maintaining 2.5V across the LED and keeping the LED lit at all times.
Yes, it shunts the current to ground for any voltage above 2.5V at pin 1, but it's fully on (not regulating), so that keeps the cathode voltage at a couple volts or so, keeping the LED off.
 

ebeowulf17

Joined Aug 12, 2014
3,307
No, it's configured as an open-loop comparator as discussed below from the data sheet.
To operate as a regulator, R2 would need to be connected to the TL431's cathode.
View attachment 138611
View attachment 138612
Yes, it shunts the current to ground for any voltage above 2.5V at pin 1, but it's fully on (not regulating), so that keeps the cathode voltage at a couple volts or so, keeping the LED off.
I see it now. Thanks so much!
 

ebeowulf17

Joined Aug 12, 2014
3,307
No, it's configured as an open-loop comparator as discussed below from the data sheet.
To operate as a regulator, R2 would need to be connected to the TL431's cathode.
View attachment 138611
View attachment 138612
Yes, it shunts the current to ground for any voltage above 2.5V at pin 1, but it's fully on (not regulating), so that keeps the cathode voltage at a couple volts or so, keeping the LED off.
Ok, next question. What about DC_Kid's suggestion above, putting a high Vf diode in parallel with the LED? Wouldn't that circuit result in whichever device (LED or other diode) has the lower Vf taking essentially all the current, but with no switching action? In other words, either:
  • LED has lower Vf and LED is lit until battery voltage drops below LED Vf.
  • Diode has lower Vf and draws essentially all of the current, keeping LED off all the time.
Or am I missing something there too?
 

crutschow

Joined Mar 14, 2008
34,285
Ok, next question. What about DC_Kid's suggestion above, putting a high Vf diode in parallel with the LED? Wouldn't that circuit result in whichever device (LED or other diode) has the lower Vf taking essentially all the current, but with no switching action? In other words, either:
  • LED has lower Vf and LED is lit until battery voltage drops below LED Vf.
  • Diode has lower Vf and draws essentially all of the current, keeping LED off all the time.
Or am I missing something there too?
No, that sounds correct to me.
 

DC_Kid

Joined Feb 25, 2008
1,072
Ok, next question. What about DC_Kid's suggestion above, putting a high Vf diode in parallel with the LED? Wouldn't that circuit result in whichever device (LED or other diode) has the lower Vf taking essentially all the current, but with no switching action? In other words, either:
simple. if you use a LED that has lower Vf than the diode, the LED will be off when diode is conducting. it is however a power wasting ckt.

edit: sorry, this wont work because diode Vf will always be above LED Vf.
let me think about another way. i think use of npn bjt will work, use diode/resistor on the gate, when it stops the bjt stops and you have led as common collector
 
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Thread Starter

Lumenosity

Joined Mar 1, 2017
614
Why do you have two threads about this vacuum cleaner?
I believe I thought that the topic in the other thread (creating a Cut Off Circuit) was different from this one about using an LED to warn of low voltage.

But I agree all of it could have been discussed in one thread. I probably just forgot about the other thread and excitedly created this one when I found the circuit with the LED.

Thanks for all the replies.
VERY helpful thread.
 
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Thread Starter

Lumenosity

Joined Mar 1, 2017
614
Use TLV431 or ZR431L. It is reference 1.22-1.25V.
On second thought......are we considering LOADED or UNLOADED voltage because unless I'm mistaken (quite possible), the unloaded voltage is irrelevant. It's the LOADED voltage that's of concern here.

There are two AA batteries in series. So the voltage of each battery is added to the circuits total voltage.
Fully charged voltage available in the circuit is 1.4 x 2 = 2.8v
Minimum voltage per NiMh battery = ~1.0v x 2 =2.0v (unloaded)

But doesn't there have to be consideration for LOAD voltage?
Under load, the voltage may dip to 2v (or lower) total but still be enough to power the circuit. When the load is removed, the voltage will rebound.
It's the loaded voltage that I need to monitor. Not the unloaded voltage.....right?
 

ebeowulf17

Joined Aug 12, 2014
3,307
On second thought......are we considering LOADED or UNLOADED voltage because unless I'm mistaken (quite possible), the unloaded voltage is irrelevant. It's the LOADED voltage that's of concern here.

There are two AA batteries in series. So the voltage of each battery is added to the circuits total voltage.
Fully charged voltage available in the circuit is 1.4 x 2 = 2.8v
Minimum voltage per NiMh battery = ~1.0v x 2 =2.0v (unloaded)

But doesn't there have to be consideration for LOAD voltage?
Under load, the voltage may dip to 2v (or lower) total but still be enough to power the circuit. When the load is removed, the voltage will rebound.
It's the loaded voltage that I need to monitor. Not the unloaded voltage.....right?
I guess that depends on when you want the indicator light to be meaningful - when the rest of the circuit is off, or when it's running.
 

Audioguru

Joined Dec 20, 2007
11,248
Many low voltage indicator circuits oscillate on and off when the battery voltage is low because the voltage is low when loaded and rebounds up when unloaded.
 
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