6v Battery Low Voltage Monitor

[MrAl]

Hi again,


Who wrote the keyboard interface driver for the project? Maybe they could explain why it is not working or if you wrote it you might be able to trouble shoot.

 

[maxeemo]

Hi there.

The circuit that I'm looking for must protect the 4.5Ah 6V SLA battery from over-discharge. This is my design:

View attachment 64641​

I've used LM339 because its output is precise. When battery voltage falls down, circuit must light the "Low Voltage Alarm" LED, and cut the main load off. This will cause the battery voltage to rise again. So the circuit will cycle between on and off status. To prevent this, I've used a D flip-flop as a latch to keep the alarm LED on, and keep the main load off.

I built the circuit on bread board and encountered a problem; The circuit always light the alarm LED. Then I tested the LM339's output. It was OK. So I guess the problem is that the clock input of flip-flop always being triggered at the circuit's startup. Is it right? And if it is, how do I solve this?

 

[MrAl]

Hi there Maxeemo,

Why do you need a latch there? That brings up more problems. I suppose you want to make the light stay on even after the voltage recovers after the load is removed. That's a good idea, but you might get away with some positive feedback on the LM339 instead. Once the comparator tripped, it would take a lot higher voltage to reset it if the positive feedback resistance is the right value. This is typically known as 'hysteresis'.

Yes i believe the FF will be clocked when power is first applied. If you want to keep that circuit you could install a reset switch to reset the FF after power is first applied, or you could rig up a POR (power on reset) circuit with a capacitor that would automatically reset the FF when the power is first applied.

So what would you like to try first?

 

[maxeemo]

I suppose you want to make the light stay on even after the voltage recovers after the load is removed.

Thank you again MrAl. Yes. Exactly. That's why I need a latch (or any sub-circuit that acts like a latch).

So what would you like to try first?

I don't want an extra switch. Also I want most compact circuit and I don't insist on flip-flops. And if a flip-flop brings up problems, I'll leave it. Now I'm thinking on SCR (or SCR sub-circuit). Of course the same problem with flip-flop may happen with SCR. I'll test it.

 

[maxeemo]

Hi.

SCR pair behaved like flip-flop. So I used LM339 with hysteresis, and result was quite satisfying. I've attached the schematic (Pin numbers of LM339 are incorrect in the schematic). The only issue is that I'm getting a flash on white high-power LED at circuit startup (when battery is on low-voltage state), but that's OK.

And last question dear MrAl;
How should I calculate the hysteresis resistor (RH in the schematic)? In the attached schematic, I've assumed the VL*=5.5V and VH**=6.2V. Are my calculations correct (RC and RH)?

* VL: Cut-off voltage. Lower limit for hysteresis range.
** VH: Upper limit for hysteresis range.

Thank you.

 

[MrAl]

Hi again,


The hysteresis resistor RH is calculated based on how much voltage it adds or takes away from the node at the two resistor junction R1 RC it is connected to. When the comparator output is high, it adds to the voltage, and when it is low, it takes away. This is how it does it.

Thus, to calculate the junction voltage when it is high you put the RH plus the pullup resistor R3 value in parallel with the upper resistor R1, and when the comparator is low you put the RH resistance in parallel with the lower resistor RC. In each case you get a different voltage either VHigh or VLow, and the hysteresis voltage is then VHigh-VLow.
When you get the required voltage difference you then have found the right value for RH.
So we calculate (the symbols "a][b" means to calculate a in parallel with b):
Rs=RH+R3
RpH=R1][Rs
RpL=RC][RH
VHigh=Vcc*RC/(RpH+RC)
VLow=Vcc*RpL/(RpL+R1)
VHysteresis=VHigh-VLow
Do you understand how to do this?
So the value of RH comes out to be:

RH=(sqrt((Vh*R3+Vcc*R1)^2*RC^2+Vh*R1*R3*(2*Vh*R3*RC-2*Vcc*R1*RC+Vh*R1*R3))-Vh*R3*RC-2*Vh*R1*RC+Vcc*R1*RC-Vh*R1*R3)/(2*Vh*RC+2*Vh*R1)

where Vh=VHysteresis

or for an approximate value when RH is at least 10 times greater than R3:
RH=((Vcc-Vh)*R1*RC)/(Vh*RC+Vh*R1)

 

[maxeemo]

So we calculate (the symbols "a][b" means to calculate a in parallel with b):
Rs=RH+R3
RpH=R1][Rs
RpL=RC][RH
VHigh=Vcc*RC/(RpH+RC)
VLow=Vcc*RpL/(RpL+R1)
VHysteresis=VHigh-VLow
Do you understand how to do this?
So the value of RH comes out to be:
...

Yes, thank you. Equations are completely clear except one thing. What is Vcc? As I understand, VHigh and VLow are voltages of node which is connected to pin #7. And Vcc is the 'cut' voltage, not standby voltage of battery. Is it so?

An irrelevant question:
This circuit is all about protecting a 6V 4.5Ah SLA battery from 'deep discharge'. This battery powers a portable device (a camping flashlight) and it is not connected to an automatic charger all the time. Is it ever necessary to be worried about deep discharge of SLA batteries? I'm asking this because I feel doubtful. I googled the internet and did not find any circuit diagram that does this. I think that I'm the only person who is worried about this! :-??

 

[MrAl]

Hi,


First, Vcc is the positive voltage of the battery. That's the voltage that connects to R3 and R1. If the battery is 10 volts, then make Vcc=10.

Second, yes it is important not to discharge a lead acid battery too far. The usual cutoff point is 10.5 volts but even 11.0 volts is not unreasonable. You can test your battery after a full charge by discharging it and watching the voltage and calculating the ampere hour of the discharge. If you have a 4.5 Ahr battery and after 4 hours at 1 amp discharge you see 11.0 volts, then 11.0 volts is a good point to stop the discharging of the battery (that's for a new battery or slightly used).

The inverters made to run off of 12v seem to vary as to the cutoff point. Some are made to cut out at 10.5v and others around 11.3 volts. Thus the test might be a good idea. You can perform the test while you are running your load normally too if you want to. See how low it drops after you drain around 4 ampere hours out of it.

They make special batteries called "Marine Batteries" that are made to withstand more of a deep discharge.

One thing to remember though, if you see that you've drained 3 ampere hours and it dropped to 11.5 volts, and draining down to 11.0 volts only takes 1 more minute (just for example here) then it doesnt do much good to set it too far below 11.5 volts because you wont get much more energy out of it anyway. This is something to keep in mind if you do the test above.