Hello All
My apologies if this is not the right place for my request but I am new to your forum.
I have a unit for powering our village 31V outside Christmas lights from a car battery. It uses purchased units, consisting of a timer, Low Voltage Protection Board (link) and a boost board (link)
The system works well mostly and I have the low voltage protection set to 11.9V off and 12.8V on.
My problem is that when the battery voltage drops below 12.5V, as the timer turns the lights on the voltage must drop below the off level momentarily which activates the low voltage cutoff.
I feel that a capacitor on the input side to the cut off unit would prevent this. Is this likely to work please?
The circuit draws 10ma at rest and 500ma with the lights on. If you think a capacitor will help, could you advise me on the value please.
I will post a schematic when I can get to my garage to draw it.
I have been dabbling in electronics for practical uses for many years but only with purchased units and circuit boards supplied by magazines and the internet and am not sufficiently knowledgeable to design my own systems. My friend who I used to rely on for advice now lives abroad so I am a bit lost.
Thank you very much.
Richard

 

Hi richard,
Welcome to AAC.
This link does not work for me boost board (link).

What type are the lights. eg: LED or Tungsten?
E

Update:
Clip from Cut Off Module.
Have you tried adjusting the Cut Off point to it’s lowest cut off value?

Characteristics:

UNDERVOLTAGE RECOVERY: When the battery voltage increases to a set value, the module will automatically turn on load again
UNDER-VOLTAGE SWITCH MODULE: This is a under-voltage switch module used to turn off the power of load in the event of the voltage dropping below a set value
BATTERY PROTECTION: This protection board will help you to protect battery from over-discharging and prolong battery life
TURN-ON / OFF VOLTAGE ADJUSTABLE: Both turn-off and turn-on voltage value can be adjusted by simply adjusting potentiometer
LED INDICATOR: Designed with a LED indicator to show if load is cut off or turned on

 

I would increase the reaction time of protection by adding a RC circuit on the input of sensing IC (op-amp or whatever they used).
Maybe the RC filter on input is already on the board, so just change the C.

 

Hi richard,
Welcome to AAC.
This link does not work for me boost board (link).

What type are the lights. eg: LED or Tungsten?
E

Update:
Clip from Cut Off Module.
Have you tried adjusting the Cut Off point to it’s lowest cut off value?

Characteristics:

UNDERVOLTAGE RECOVERY: When the battery voltage increases to a set value, the module will automatically turn on load again
UNDER-VOLTAGE SWITCH MODULE: This is a under-voltage switch module used to turn off the power of load in the event of the voltage dropping below a set value
BATTERY PROTECTION: This protection board will help you to protect battery from over-discharging and prolong battery life
TURN-ON / OFF VOLTAGE ADJUSTABLE: Both turn-off and turn-on voltage value can be adjusted by simply adjusting potentiometer
LED INDICATOR: Designed with a LED indicator to show if load is cut off or turned on

Thank you very much for your help.
My apologies, hopefully this link works Aideepen Boost Module, DC-DC Step-up Converter Board 400W Boost Converter Power Inverter Module 8-50V to 10-60V https://amzn.eu/d/alUX8MK
I have reduced the cut off to 11.8V but don't want to reduce it any lower as that is already a bit low to protect the lead acit battery I am using. I have just remembered I have a diode on the supply to the low voltage board to protect from incorrect polarity and wonder if that might be having an effect.
They are a string of led lights about 1100 LEDs drawing about 600ma.

 

I have just remembered I have a diode on the supply to the low voltage board to protect

Hi richard,
Link now OK.
That diode will drop approx 1V from the 12V battery to the Cut Off module, try removing it, let's know the result.

Your numbers suggest 31V at 0.6A,=19 Watts which when converted to 12V is approx 1.5A , assuming 100% conversion.
So the series diode drop could be higher, ~1.5V at switch On.
E

 

Hi richard,
Link now OK.
That diode will drop approx 1V from the 12V battery to the Cut Off module, try removing it, let's know the result.

Your numbers suggest 31V at 0.6A,=19 Watts which when converted to 12V is approx 1.5A , assuming 100% conversion.
So the series diode drop could be higher, ~1.5V at switch On.
E

Thank you,
Sorry, I have misled you. The current of 0.6A is at the 12V supply rather than the 31V output.
I will try removing the diode to see if that alters things. It is only necessary if I think others may be setting up the battery.
Thank you.

 

I have a diode on the supply to the low voltage board to protect from incorrect polarity

You can use a power P-MOSFET to protect against a reverse polarity connection with very little voltage drop (equal to the current times the MOSFET's on-resistance).
Connect the MOSFET's source terminal to the power, the gate to ground, and the drain to the load.
With that connection it acts as a near ideal diode.

 

Thank you all very much.
I have removed the diode from the supply which made no difference. Reducing the cut off voltage to 11.8 improved it slightly but didn't fix the problem. I then altered the circuit so that the timer relay contact is supplied directly from the battery with only the coil of the relay being supplied by the battery protection circuit. This made no difference but possibly because I just split the incoming leads from the battery. I will try separate leads next.
Whilst I know what an RC circuit is, Michal, I have no idea of what values to use and assume I use a parallel circuit on the input to the cut off circuit?
I can see from your suggestion, Crutschow, how the MOSFET would work but don't know if any other components are required?
I assume that the LEDs are pulling the voltage down instantaneously to below the cutoff voltage as it is not detectable on any of my meters.
I fully understand if you feel I'm on the wrong forum due to my lack of knowledge and will understand fully if you can't help me further and am grateful for your advice so far.
Thank you
Richard

 

this is the current wiring of the units. I hope it is clear.

 

I know what an RC circuit is, Michal, I have no idea of what values to use and assume I use a parallel circuit on the input to the cut off circuit?

Try 1kΩ to 10kΩ in series with the input, and a 100μF (electrolytic) capacitor across the input terminals (capacitor plus to the + input).

Crutschow, how the MOSFET would work but don't know if any other components are required?

No, it just requires the P-MOSFET, wired as I stated.

 

Thank you.
I'll have a go. Definitely got the resistor, might have the capacitor, will order the MOSFET.
Many thanks.

 

Hi richard,
As you discovered, removing the diode has no effect on the problem, neither will fitting a MOSFET.
E

 

Try 1kΩ to 10kΩ in series with the input, and a 100μF (electrolytic) capacitor across the input terminals (capacitor plus to the + input).

Are you speaking about the input to the Battery cutoff module?
If so how is that going to work since the input is the supply voltage. The relay would never energize.

 

The system works well mostly and I have the low voltage protection set to 11.9V off and 12.8V on.

The module has a connector to add a switch to "force start the load".
I would try that first and set the ON return voltage lower, maybe at 12.2 volts.
What are the specs on the Timer Relay Coil? May have another option involving said relay.

 

The module has a connector to add a switch to "force start the load".
I would try that first and set the ON return voltage lower, maybe at 12.2 volts.
What are the specs on the Timer Relay Coil? May have another option involving said relay.

Thank you for that. I hadn't realised that was what the socket is for. Unfortunately if I reduce the on voltage the system can end up in a loop of going off, then the voltage builds up when load removed switches on, cuts out and just keeps flashing.
Thank you though for the suggestion. The switch may well help me.
Thank you
Richard

 

Unfortunately if I reduce the on voltage the system can end up in a loop of going off, then the voltage builds up when load removed switches on, cuts out and just keeps flashing.

Exactly which one review on that module experienced.
My thought was to reduce the ON return lower while using the switch. Probably doesn't make any difference as I suspect the switch eliminates the return setting.
Another option is to delay the dropout of the Timer Relay Coil. Can you post the specs on this relay?

 

Exactly which one review on that module experienced.
My thought was to reduce the ON return lower while using the switch. Probably doesn't make any difference as I suspect the switch eliminates the return setting.
Another option is to delay the dropout of the Timer Relay Coil. Can you post the specs on this relay?

Thank you
I don't have the specs for the relay, only what is on Amazon I'm afraid. This is the unit: Nikou Mini CN101A Time Switch Small LCD Digital Microcomputer Control Power Timer Switch High Precision Relay Input DC 12V Anti Interference https://amzn.eu/d/gXyUAFb
Thank you for helping.
Richard

 

Thank you
I don't have the specs for the relay, only what is on Amazon I'm afraid. This is the unit: Nikou Mini CN101A Time Switch Small LCD Digital Microcomputer Control Power Timer Switch High Precision Relay Input DC 12V Anti Interference https://amzn.eu/d/gXyUAFb
Thank you for helping.
Richard

OK, I see that's the input supply voltage for the timer not a relay coil.
Any luck with the switch?

 

hi richard,
What is the battery type and rating? eg: AmpHr rating and condition.
A switch On surge of a 12V 0.3A load + plus module charging should not pull a battery low.

I did read the Amazon reviews for the cut-off module, some were critical of the performance.

E

 

Thank you Eric,
Even with the 90ah leisure battery the problem occurs when the voltage drops below 12.5V. I have the cut off set for 11.8V which is lower than I would like but it means that after the power goes off, the battery doesn't recover high enough (12.6V) to reset the switch which would otherwise cause the system to flash on and off.
I think I will have to do my own cut off circuit for next year and make do with it for this year. I just hope running my batteries down to 11.9V doesn't damage them.
I will have another look at the reviews.
Thank you
Richard