Low Battery Voltage Alarm

[elfvenlord]

Hi All

Im trying to build and understand the attached circuit but im a bit lost on how it works.Basically all it is is a low voltage alarm that sounds the buzzer when the battery voltage is below the set voltage alarm level,which i am guessing is set with the voltage divider from the two resistors.when the voltage is below the set alarm voltage the comparator realises that,and switches to the +V.The problem is how can the voltage divider stay constant if the input voltage from the battery is dropping,unless im not looking at it the right way.

This circuit is supposed to be for my rc plane battery wich is 5V and is supposed to trigger the alarm when the battery voltage drops past about 4.0V.

Can anyone just briefly explain how it really works,I would most gratefull.
Thanks :wink:

 

[Nigel Goodwin]

Hi All

Im trying to build and understand the attached circuit but im a bit lost on how it works.Basically all it is is a low voltage alarm that sounds the buzzer when the battery voltage is below the set voltage alarm level,which i am guessing is set with the voltage divider from the two resistors.when the voltage is below the set alarm voltage the comparator realises that,and switches to the +V.The problem is how can the voltage divider stay constant if the input voltage from the battery is dropping,unless im not looking at it the right way.

You're almost right, the LM311 is a comparator chip, it compares the voltages on it's +ve and -ve inputs. The input with the two resistors will fall as the voltage on the supply fails, you quite rightly pointed out that this is a voltage divider. However, the other input won't fall, it has a voltage reference source on it - the LM336 - it will keep the voltage on the -ve pin constant. Once the +ve input drops below the -ve the buzzer will sound.

 

[elfvenlord]

Hi Nigel

Thanks allot for your reply.I didn't know that the -ve stayed constant as the battery voltage dropped,thats handy to know

So if i wanted the alarm to trigger at say 4.0V then i would change the 3k resistor so as to achieve that voltage at the -ve.But how come you would need a voltage divider? or have I gone off track again

 

[Nigel Goodwin]

Hi Nigel

Thanks allot for your reply.I didn't know that the -ve stayed constant as the battery voltage dropped,thats handy to know

So if i wanted the alarm to trigger at say 4.0V then i would change the 3k resistor so as to achieve that voltage at the -ve.But how come you would need a voltage divider? or have I gone off track again

No, you change the resistors in the potential divider, changing the 3K would only alter the current in the LM336.

 

[elfvenlord]

Thanks allot,I think I now fully understand how it works.This forum is brilliant.

 

[Nigel Goodwin]

Thanks allot,I think I now fully understand how it works.This forum is brilliant.

I've only just noticed your location! - how come we're communicating via a BBS the other side of the world, when we live in the same county?.

 

[elfvenlord]

Nigel

Sorry but whats a BBS?
Have pitty on me Im a n00b...lol.

 

[elfvenlord]

By the way I built and tested the circuit but it hasn't worked and I found out why.As the voltage drops the voltage at the -ve drops even though the zener voltage is 2.4V.I read up on zener diodes and I know that it is supposed to only allow 2.4V unless the input voltage drops below that but it hasn't seemed to work.I did connect it up as shown in the original attachment,maybe the electronics store gave me the wrong type of diode?

does the 3k resistor have any effect on the voltage at the -ve?

If it did would i have to calculate the voltage at the -ve by using the voltage divider formula and if so how do I find the resistance of the diode or do I use another formula?

Thanks for the help!!!

 

[Nigel Goodwin]

By the way I built and tested the circuit but it hasn't worked and I found out why.As the voltage drops the voltage at the -ve drops even though the zener voltage is 2.4V.I read up on zener diodes and I know that it is supposed to only allow 2.4V unless the input voltage drops below that but it hasn't seemed to work.I did connect it up as shown in the original attachment,maybe the electronics store gave me the wrong type of diode?

Sounds like he may have, the voltage on the top of the LM336 should remain quite stable as the supply voltage varies.

does the 3k resistor have any effect on the voltage at the -ve?

Only a very slight effect, unless you change it massively!.

If it did would i have to calculate the voltage at the -ve by using the voltage divider formula and if so how do I find the resistance of the diode or do I use another formula?

You calculate it for the current you want through the voltage reference, it's not a potential divider at all.

The LM336 isn't actually a zener, it's a precision voltage reference, similar to the one used in my analogue tutorial board. Have you got the correct part?.

BTW - BBS is "Bulletin Board System", the thing we're communicating through - it runs on PHPBB (PHP Bulletin Board).

 

[elfvenlord]

Thanks Nigel

I asked the man at the electronics store for the LM336 but they didnt have any but he said i could use a zener diode instead wich had a zener voltage of 2.4V.I connected it in the same way as in the diagram.but the voltage at the -ve was more than 2.4V.Allot more.

I dunno,what do you think-They must of given me the wrong diode or a faulty one. :roll:

 

[Nigel Goodwin]

Thanks Nigel

I asked the man at the electronics store for the LM336 but they didnt have any but he said i could use a zener diode instead wich had a zener voltage of 2.4V.I connected it in the same way as in the diagram.but the voltage at the -ve was more than 2.4V.Allot more.

I dunno,what do you think-They must of given me the wrong diode or a faulty one. :roll:

Possibly so, but a zener is a much poorer device in the first place, but probably good enough for a battery indicator.

 

[elfvenlord]

Ok thanks for your help,i'll have to fing the LM336 diode and try it again.

 

[elfvenlord]

Well I looked briefly for the lm336 diode but I have found another low voltage alarm circuit wich might be better,but like the first im a bit lost on how it works.I know that when the first transistor is saturated the second one switches on the buzzer,but how does the current to the base of the first transistor increase as the voltage of the battery drops.I know it has something to do with the zener diode.Could anyone help explain how the first transistor works.Thanks Allot.

 

[John Sorensen]

The circuit works by holding the second transistor (the one on the right, call it Q2) off when the battery voltage is high enough.

Q2 is held off by the voltage on the 10k resistor below Q1, via the 4.7k resistor (which may not be needed). So, when Q1 is on, the voltage on the top of the 10k is the same as the battery voltage (near enough) so that the voltage from the emitter to base on Q2 is nearly 0 (much less than the 0.7 Vbe needed for it to conduct), so it is off and the buzzer is off.

So what causes Q1 to turn off (and turn on Q2)? Well, the voltage across the 3.3k resistor is fixed at 0.7V by being in parallel with the Vbe on Q1. So the remainder of the battery voltage (Vbat-0.7V) is across the zener and 10k resistor. As long as that voltage is great enough to cause the zener to conduct (a little over 3.3V) enough current through the 10k and zener to cause Q1 to turn on (saturate) then the buzzer is off (as described above). The 10k in series with the zener is to limit the current from the emitter through the base of Q1.

Questions?

j.

 

[John Sorensen]

Maybe you want to do some experimenting with zener diodes apart from these circuits to see how they work.

Try a simple series circuit of a resistor and a zener (the resistor "above" the zener). Experiment by changing the resistor out with different values, calculate (or measure) the current by measuring the voltage across the resistor (I=Vresistor/R), then measure the voltage across the zener and plot out I vs Vzener.

If you figure out the zener and get a relatively stable voltage in your first ciruit, the battery low alarm level is easily adjusted by the voltage divider. Changing the voltage divider doesn't change the threshhold level of the comparator: it's fixed by the zener, it just changes how the battery voltage compares to that threshold.

j.

 

[teckshen]

I have a Sealed Lead Acid rechargeable battery, 6V, 4.5AH. I have tried to make a tester which works like this: First, I discharge the battery 1 min, if the battery voltage drop to 6.2V(if the fully charge voltage is 6.45), a indicator or buzzer would turn on. if the voltage drop to 6.3V then nothing will happen. What can I do to?

 

[McGuinn]

IMHO... One point to add on this subject is to mention the fact that the addition of a 'low battery alarm' designed around a buzzer will increase the current drawn from the battery, and worsen the situation in some cases.
Rechargable batterys have a tendancy to deplete very quickly, and the addition of this may only cause a the loss of the device (if it's an RC boat or such).
Recommendations would be to utilise a piezo buzzer or flashing LED...

 

[Crofty]

Try the MAXIM MAX917. This nano power comparator has an on-chip 2.45V bandgap ref. Just add 2 resistors to set the trigger point. All in a SOT23-5 SMT package.

Steve

 

[neil_m23]

transistor basic help needed

In the ckt u described, if Q1 becomes ON, then will zener vtg come across Q2 base or not?

Q1: left hand side transistor of yours
Q2: right most

If yes then how come vtg at Q2 emitter(E2) and collector (C2) will be same ???
sorry, but i dont have strong fundamentals/basic clear abt transistor.

When transistor is conducting, ON can we replace it by short ckt(neglecting emitter-collector Vce voltage drop, usually small)

In that case, can we replace transistor by simple nodal point ( 3 lines comin together B,C,E )

regards, Neil.