low voltage cutoff circuit

Thread Starter

Georacer

Joined Nov 25, 2009
5,181
Hi

I think this circuit is used to cutoff the voltage source to the output if it's the voltage is 6V or lower. Do I have it correct? I googled "6v low voltage cutoff" and found this somewhat similar looking circuit.

The red LED is for power indication in the circuit and the white LED is for low voltage indication.

Is that component on the top left a relay?

What is that IC on the top right? Someone was saying that the circuit uses operational amplifier. But someone said it could be a LM393 dual comparator. What do you think?

In what kind of applications this circuit can be used? Could you please list some examples of circuits or electronic products which can be damaged by low voltage if they don't use such voltage cutoff circuit.

Thank you for the help.

Regards
PG

This is a thread started by me on behalf of PG1995 who is unable to post a new thread here due to AAC technical difficulties.
 
Last edited:

jimkeith

Joined Oct 26, 2011
540
Yes, the item top left is a 10A relay with a 6V coil.
The IC is likely an LM358 op amp or LM393 comparator.
The item left of the IC is likely a 3.9V zener used as a voltage reference.
I cannot guess why there are two pots.
The main application for the low voltage cut-out is to protect lead-acid batteries from over discharge that tends to kill the battery.
Regarding loads, many circuits will not operate properly at low voltage, but are unlikely to be damaged.
 

Adjuster

Joined Dec 26, 2010
2,148
Yes, the item top left is a 10A relay with a 6V coil.
The IC is likely an LM358 op amp or LM393 comparator.
The item left of the IC is likely a 3.9V zener used as a voltage reference.
I cannot guess why there are two pots.
The main application for the low voltage cut-out is to protect lead-acid batteries from over discharge that tends to kill the battery.
Regarding loads, many circuits will not operate properly at low voltage, but are unlikely to be damaged.
This thing is built with 5% resistors, and the voltage reference device may well be a zener diode of perhaps 5% tolerance. Depending on the circuit used, the result could be an uncertainty in threshold voltages of the order of 15% worst-case, or ±0.9V. That dues not seem close enough to me, so trim adjustments would seem to be in order.

My guess would be that the trim pots adjust the cut-off threshold and the level at which the connection is restored. There should be positive feedback to introduce hysteresis between the two levels, as otherwise oscillation is likely: the battery voltage can rise on disconnection of the load, possibly leading to immediate reconnection...and so on. This would rapidly wear the relay and may damage the load, so should be avoided.

The pots might adjust the levels separately, or one pot may set sensitivity, and the other may set the hysteresis between them.

Of course, this may not necessarily be so - but it seems fairly likely.
 

PG1995

Joined Apr 15, 2011
816
Many, many thanks, Jony.

I was waiting for your reply! Can I get a Multisim or CW circuit for this circuit, please? If you are busy and don't have time, it's okay. I really appreciate your help.

With best wishes
PG
 

PG1995

Joined Apr 15, 2011
816
Oh, sorry. I didn't notice it.

But the LED D3 in CW doesn't light up at all even when I lower the voltage to, say, 4V. Is there something wrong? Please let me know.

Thanks a lot.

Best regards
PG
 

Jony130

Joined Feb 17, 2009
5,127
Oh, sorry. I didn't notice it.

But the LED D3 in CW doesn't light up at all even when I lower the voltage to, say, 4V. Is there something wrong? Please let me know.

Thanks a lot.

Best regards
PG
D3 is CW = clockwise? On which the diagram ?
 

Jony130

Joined Feb 17, 2009
5,127
For B1 = 9V set VR1 = 80% and VR2 = 10K.
And when D3 is "ON" the relay is also activate, and now change battery voltage from 9 to 8V ---> 7V ---> 6V. And you will see that the D3 is off and relay is not active.
And now load is disconnect from the battery (load is not included in the circuit diagram).
 

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