I found this LOW voltage cut out circuit diagram with very little information. I want the voltage to cut off at 10.6 Volts/ Therefor D1 is shown as 8.2 Volts I thought it should be 9.4 Volts.

I don't understand why we have to use a Darlington transistor and what stops the relay switching on and off when you remove the load.

Actually really confused.

Any help will be appreciated

 

darlington transistors have more gain. and the load isnt connected to anything but the relay contacts, it has nothing to do with the circuit operation at all.

 

The value of R1 is Shown between 10R and 47R why is this and what is its purpose

 

I found this LOW voltage cut out circuit diagram with very little information. I want the voltage to cut off at 10.6 Volts/ Therefor D1 is shown as 8.2 Volts I thought it should be 9.4 Volts.

I don't understand why we have to use a Darlington transistor and what stops the relay switching on and off when you remove the load.

This is a primitive, simplistic circuit. There is no hysteresis, so the relay can repeatedly cycle on/off as battery voltage decreases.

I can provide you with a much better circuit that uses a TL431 (voltage reference), is adjustable, has adjustable hysterisis.

What is the load current? Do you really want/need a relay for switching the load?

 

This is a primitive, simplistic circuit. There is no hysteresis, so the relay can repeatedly cycle on/off as battery voltage decreases.

I can provide you with a much better circuit that uses a TL431 (voltage reference), is adjustable, has adjustable hysterisis.

What is the load current? Do you really want/need a relay for switching the load?

I would really like the better circuit. The load is 10 Amps and if possible I would prefer not to use a relay. THANK YOU !!!

 

How is your parts availability? Can you obtain a TL431 and a Large TO220 PFET ?


Is this a load-disconnect from a discharging battery type of circuit?

 

I can get the TL431 from Mantech in South Africa will take 10 days to arrive. However the local suppliers may just surprise me. I am pretty sure I can get a PFET no problem as well

 

Sorry The Load is a 300Watt Inverter. I am going through batteries as they go past there depth of discharge which is killing me. With 15 hour daily power cuts I rely on my Inverters

 

At 25A, the PFET will have to be on a Heatsink. How big depends on the PFET's Ron. The lower, the better.

 

View attachment 76166

At 25A, the PFET will have to be on a Heatsink. How big depends on the Ron.

Thank you so much. I will build this and let you know how I get on. Your help is greatly appreciated!

 

I found this LOW voltage cut out circuit diagram with very little information. I want the voltage to cut off at 10.6 Volts/ Therefor D1 is shown as 8.2 Volts I thought it should be 9.4 Volts.

I don't understand why we have to use a Darlington transistor and what stops the relay switching on and off when you remove the load.

Actually really confused.

Any help will be appreciated

While browsing through an archive of Popular Electronics, I spotted a low voltage alarm based on a Lambda diode (a pair of JFETs) - what little I read into the article claimed that the Lambda diode doesn't draw any current as long as the voltage is above the trip point - not so far fetched since the JFET is cut off by increasing the gate reverse bias. A search for "Lambda diode voltage monitor" may produce a few Google results.

 

I have managed to Buy everything locally The FET is and IRF9540 is that suitable

 

I have managed to Buy everything locally The FET is and IRF9540 is that suitable

That Fet is not up to switching 25A. Refer to this data sheet.
Look at Fig 1 or Fig 2. In the circuit I posted, the Vgs will be ~-10V. Look at the voltage drop Vds at 25A. I read 3V @25degC and 6V@175degC, which is way too much. It would drive a relay just fine, but not the 25A load...

 

I found this LOW voltage cut out circuit diagram with very little information. I want the voltage to cut off at 10.6 Volts/ Therefor D1 is shown as 8.2 Volts I thought it should be 9.4 Volts.

I don't understand why we have to use a Darlington transistor and what stops the relay switching on and off when you remove the load.

Actually really confused.

Any help will be appreciated

a dpdt relay should solve the repeated on off problem. this circuit is configured in active on mode -i.e. current must be going through relay to keep it "on" . low voltage switch off should also move the other pole/terminal of the dpdt relay, which should disconnect the source, permanently turning off the circuit.

but it's not illustrated in the diagram, and also the induction surge protection diode is missing across the relay coil.

& i guess transistors matching with relay and base currents should work without any issues.

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if u want to modify this circuit to use to cut off 10.6V, then i will suggest using a cheap comparator ic like lm393 with a voltage reference like lm385 @ one pin and a 100k range voltage divider to source @ the other pin to drive the transistor. that will be much convenient and precise.

 

View attachment 76166

At 25A, the PFET will have to be on a Heat sink. How big depends on the PFET's Ron. The lower, the better.

Hi Mike

I have built this circuit I could not get not get a 5K Preset so I used a 4k7 and I could not get the 27k so I used a 33K.

When I first switched on I had the supply voltage at 3 volts and I used a 12 volt lamp as my load. I gradually turned the voltage to 12 volts at which time the lamp was bright.

I then turned the voltage down to 10.6 volts which will be my cut off voltage. I turned the Preset slowly clockwise and then anti clock wise to see when the lamp would switch off which I assumed would be the position to leave the Preset at.

Unfortunately the lamp never switched off.

The voltage on the Reference of the TL431 (pin 1) does not vary it stays on 2.47 Volts whilst 1 adjust the Preset.

I checked my board and everything seemed right so not really sure what to do now.

 

Hi Mike

I have built this circuit I could not get not get a 5K Preset so I used a 4k7 and I could not get the 27k so I used a 33K.

When I first switched on I had the supply voltage at 3 volts and I used a 12 volt lamp as my load. I gradually turned the voltage to 12 volts at which time the lamp was bright.

I then turned the voltage down to 10.6 volts which will be my cut off voltage. I turned the Preset slowly clockwise and then anti clock wise to see when the lamp would switch off which I assumed would be the position to leave the Preset at.

Unfortunately the lamp never switched off.

The voltage on the Reference of the TL431 (pin 1) does not vary it stays on 2.47 Volts whilst 1 adjust the Preset.

.

If your TL431 is in a closed loop system and working correctly - it will hold its reference pin somewhere pretty close to 2.5V.

 

If your TL431 is in a closed loop system and working correctly - it will hold its reference pin somewhere pretty close to 2.5V.

Not in this circuit. The TL431 is used as voltage comparator; not a regulator.

 

Hi Mike

I have built this circuit I could not get not get a 5K Preset so I used a 4k7 and I could not get the 27k so I used a 33K.

By Preset, you mean U1, the 5KΩ trim pot? If so, a change from 5K to 4.7K wouldn't matter.
There is no 27K in the circuit; only R6 which was supposed to be 26.1K, a standard 1% value. The ratio of R8/R6 is critical. If R8 is really 100K +- 1%, and R6 = 33K+- 1%, then parallel a new resistor Rn across R6. Rn should be 133K +- 10%.

When I first switched on I had the supply voltage at 3 volts and I used a 12 volt lamp as my load. I gradually turned the voltage to 12 volts at which time the lamp was bright.

I then turned the voltage down to 10.6 volts which will be my cut off voltage. I turned the Preset slowly clockwise and then anti clock wise to see when the lamp would switch off which I assumed would be the position to leave the Preset at.

Unfortunately the lamp never switched off.

The voltage on the Reference of the TL431 (pin 1) does not vary it stays on 2.47 Volts whilst 1 adjust the Preset.

I checked my board and everything seemed right so not really sure what to do now.

With R8=100K, R6=33K//133K, with the trimpot=4.7K, I reran the sim, plotting V(ref) for three positions of the trimpot wiper.
Note the variation of V(ref) as V(IN) is reduced, and then the little step as it switches. If you are not seeing V(ref) do that in the actual circuit, you either have something miswired, or you have a bad trimpot.

 

By Preset, you mean U1, the 5KΩ trim pot? If so, a change from 5K to 4.7K wouldn't matter.
There is no 27K in the circuit; only R6 which was supposed to be 26.1K, a standard 1% value. The ratio of R8/R6 is critical. If R8 is really 100K +- 1%, and R6 = 33K+- 1%, then parallel a new resistor Rn across R6. Rn should be 133K +- 10%.


With R8=100K, R6=33K//133K, with the trimpot=4.7K, I reran the sim, plotting V(ref) for three positions of the trimpot wiper.
Note the variation of V(ref) as V(IN) is reduced, and then the little step as it switches. If you are not seeing V(ref) do that in the actual circuit, you either have something miswired, or you have a bad trimpot.

View attachment 81816

Thank you very much Mike.

I am going to get the exact components that you have specified at 1% tolerance. It will take me about 10 days to get them to Zimbabwe. Once here I will re build and let you know the results.

Best wishes

Rodney

 

Dont wait. Go through your collection of resistors. If you can find one that measures close to 100K, use it for R8. Take your 33K and parallel it with another higher value resistor to get something closer to 26K for the parallel combination. Use that for R6.

Another idea: leave R6 and R8 as is, and replace the trimpot with a 10K. Now you should have enough resistance range to hit your trip point.