That is what R9 (the hysteresis setting) is supposed to take care of. As designed, if this circuit is connected between a discharging battery bank and a load, the load will be disconnected (PFET turns off) when the voltage decreases below Vlow. As the battery voltage increases, the voltage at V(in) will have to go up by some ΔV to Vhi (controlled by R9) before the load reconnects to the battery (PFET turns on) automatically. This is the way the circuit is now, but it sounds like I guessed too low for ΔV

If the load consumes many amps, the voltage drop from the internal resistance of the battery can be larger than ΔV, in which case the circuit will "chatter", meaning it will cycle on and off repeatedly. This happens because as the load is turned off, the battery voltage comes up because it got relieved of the large load. That higher unloaded battery voltage is higher than Vhi, causing the circuit to turn back on...

Seems like we have several choices:

1. Increase ΔV. You would measure the actual voltage drop at the battery terminals when the inverter load is turned on/off, and then we will modify the circuit to increase the ΔV (by playing with R9) so that Vhi comes only after the battery has been charged. i.e. Vhi must be higher than the voltage jump as the PFET turns off...

2. Modify the circuit such that it requires an extra input to start/arm it. That way, it would turn off the inverter load automatically the first time the voltage dips below Vlow, then just sits there until you push a "start" button. You would wait to push the start button after the battery bank has been sufficiently recharged so it is ok to run the inverter again.

3. Make the cycle totally automatic. Modify the circuit so that it has two-independently adjustable trip points: Vlow similar to now. Vhi has its own trim pot. You would adjust Vhi to turn on the inverter only when the battery has been recharged to a voltage like 14.2V

Please advise how you would prefer to proceed.

Friendly dig: You see now why I always breadboard a circuit before making it "permanent". Here is an example where the circuit design met the original specs, but you hadn't considered all of the requirements, so coming up with a final circuit is an iterative process. Much easier to move a wire or to plug-in a new resistor on a white board.

 

Mike

All the above options sound good. However I have learnt a very good lesson this week. Like you said way easier to change something on the bread board so you can change it as the project develops. So, I am going to test this circuit in the actual environment its going to eventually work and see exactly what changes may be needed. At least when I ask for help to modify this I will have the correct detail.

Rod

 

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

Yeah the TL431 is excellent. Like 5-cents each on eBay if you buy say 100 pcs.

 

Here is yet another sim of the circuit. Here I am using time as the independent variable, and I am using it to sweep V1 from 13V down to 10V and then back up to 13V. Because I modeled V1 (analogous to the battery in your system) with 100mΩ of series resistance, you can see V(in) jump as the load turns on/off. This drop in V(in) is what causing the chatter in your actual system.

Note that I am guessing what the drop is. It is a function of condition of your batteries, wiring resistance, and how big the load (R5 in the sim) is. I suspect that the drop in your system is larger than what I am using here for illustration.

I repeat the sim with various values for R9, from 2.2megΩ down to 680KΩ (the lower the value of R9, the more hysteresis). With the original 2.2meg, and with the pot centered (green trace), the downward cutout voltage is 11.23V (cursor1) and the upward cutin voltage is 11.71V, for a ΔV of 489mV, which is not enough...

Note that with R9=680KΩ, with the pot centered (dark blue trace), cutout = 10.47V, cutin=12.0V, for a Δ = 1.53V. Note that changing R9 effected the actual cutout voltage, so if you change R9, you will have to reset the pot to get your desired cutout voltage.

You still must decide if you want to make the cutin automatic, or manual...

 

Should be connected to the drain, see my schematic.

View attachment 82045

In normal usage of an NFET or PFET, the little arrow, which represents the body-diode inside the FET, should never conduct.

Hi Mike

What is the best way of Calibrating this Circuit to switch off at the 10.6 volts. I have been adjusting the trim pot until the load goes off but it does not seem to be very accurate. Would it be better To connect a multimeter to the output and adjust the trim pot until the voltage cuts out.

Best wishes

Rodney

 

The only way of adjusting it is to connect a variable power supply in place of the battery, and then slowly reduce the supply voltage until the load turns off. Measure the supply voltage with an accurate DMM as you slowly reduce the voltage, and note what the voltage was just before the trip. Adjust the trimpot. Reset the supply to ~14V, and then slowly reduce the supply voltage again to see where the trip point moved.

 

Rodney asked:
"If the load is switched off through the FET is there still a current draw? I ask this because the load switches out at the cut off voltage, however the battery continues to drain. "

If you built the circuit I posted in #44, the current drawn from the battery after the circuit switches the load off is only 77uA, which should not be noticeable. How big is your battery?

 

Rodney asked:
"If the load is switched off through the FET is there still a current draw? I ask this because the load switches out at the cut off voltage, however the battery continues to drain. "

If you built the circuit I posted in #44, the current drawn from the battery after the circuit switches the load off is only 77uA, which should not be noticeable. How big is your battery?

The Battery in this application is a 7Ah 12 Volt battery. I don't have this problem with the first circuit you helped me with for my inverter. I will replace the battery and try it with a new one and see what happens.

 

The Battery in this application is a 7Ah 12 Volt battery. I don't have this problem with the first circuit you helped me with for my inverter. I will replace the battery and try it with a new one and see what happens.

With all of the trouble you had getting the PFET connected right, are you sure it is really working?

Use the variable DC power supply in place of the battery for testing. If the supply has an ammeter, when the supply voltage is initially set to ~15V, the FET will be on, and the supply current will be whatever the load draws (several Amps). AS you slowly reduce the supply voltage (simulating the battery discharging), the circuit turns off the PET, and the supply current should go to ~0A (77uA to be exact).

If the supply doesn't have a built-in ammeter, then put a multimeter in Amp mode in the positive lead of the supply.

 

With all of the trouble you had getting the PFET connected right, are you sure it is really working?

Use the variable DC power supply in place of the battery for testing. If the supply has an ammeter, when the supply voltage is initially set to ~15V, the FET will be on, and the supply current will be whatever the load draws (several Amps). AS you slowly reduce the supply voltage (simulating the battery discharging), the circuit turns off the PET, and the supply current should go to ~0A (77uA to be exact).

If the supply doesn't have a built-in ammeter, then put a multimeter in Amp mode in the positive lead of the supply.

I will do it and get back to you