Low voltage cutout Circuit Help

low voltage cutoff circuit

Hey there freaks.

I'm having some troubles trying to get my low voltage protection circuit going. The circuit is attached below.

The circuit simply drives a 10W 6v Halogen lamp off a SLA battery. When the battery voltage drops to less than a voltage the zener sets (D1) I want the Mosfet (M1) to turn turn off. I have had success getting this circuit to work on the simulator, however when I attempt to get it going in real life, since the battery voltage is dropping slowly, I hit a point where Q1 is acting as an amplifier, this in turn makes Q2 also behave like an amplifier. The result, when the low voltage protection is kicking in I get around 3V at the Mosfet's Gate, causing it to be not fully swiched on. This causes the Mosfet to heat up to unacceptable levels.

Is there a simple solution to this circuit, ie to make it switch off at a very fast rate so I see 0v at the gate of the Mosfet? Or do I have no choice but to go to a comparator based (ie LM358) solution?

low voltage cut out circuit

Your comparator (with hysteresis) will work better. You could add some transistors to the circuit shown but that would cost more.

low voltage cut out switch

Try to make the following chages:
- connect R3 between B and E of Q2 and use 10kΩ,
- change R2 to 10kΩ,
- change R1 to 10kΩ,
- add new resistor (4.7kΩ - 10kΩ) between B and E of Q1,
and try again ..
This circuit is not ideal switch and you may still need a comparator wirh a small histeresis to properly switch off Q3 .. but first try to implement the suggested changes ..
Regards,
IanP

low battery cuout circuits

You need a little bit of hysteresis. That is, the circuit should turn off the lamp at a certain voltage, but should turn it back on at a voltage slightly higher than the turn-off threshold.

The first thing that comes to mind is connect a resistor from the source of M1 to the base of Q1. That way, when the circuit turns off the lamp the voltage across the 0.1ohm resistor will drop and so Q1 will be "helped" to turn off. The input voltage will then need to increase a little over the previous threshold for the lamp to turn back on. When the lamp turns on, the voltage drop across the 0.1ohm resistor will now "help" Q1 to stay on.

A good hysteresis value would be about 0.5~1V. So the lamp would turn off at about, say 9V, but would only turn back on if the input voltage is greater than 10V.

low voltage cutout circuit

Wow!

That was quick.

I'll try these tips and see where things end up. I'll post my results here.

Thanks for the help ppl

oddbudman

low voltage battery protection circuit

Check out this link:

It is about a cutoff circuit with independently adjustable On and OFF levels (histeresis) ..
Regards,
IanP

battery low voltage cut out

Ok, well I found a circuit idea here:

https://www.siliconchip.com.au/cms/A_104378/article.html

So i went to my local electronics store, bought a LM358 and had a bash.

Here is what I came up with.

V1 is my battery input simulation (Green Trace). Vlamp is the collector voltage of the drive transistor (Blue trace). R8 represents the lamp. D2 is a 4v7 zener.

I have used a BJT to drive the LAMP as that way I avoided low gate voltages that were occuring on a Mosfet (causing it to run hot).

I had troubles adding the LM358 to LTspice, so i did the simulation using a LT1013. [asside - if anyone can help me getting the LM358 model going in LTspice it would be appreciated, or even if you can point me to a link dealing with adding new models to LTspice.]

Hysteresis is achieved using R4. And by changing the Value of R4 I am able to configure the Re-light voltage of the Lamp.

Thanks for the tips people. It was much appreciated.

If you have any further suggestions on this design please post.

oddbudman