working of under voltage protection circuit.

[nitin.maurya1]

please help me by your valuable views.

For the circuit of "Automatic LED Emergency Light with Under Voltage Cut Off Protection" in the following url,

Automatic LED Emergency Light-Modified Version - Electronic Circuits and Diagram-Electronics Projects and Design

how the chain (of T3 bc 548, R18 1.2k, Z02 5.1volt 0.5 w) is working as under voltage protection circuit for battery(as shown in circuit) of 6 volt, for the under voltage of 5.7volt for battery.

 

[Pjdd]

When the battery voltage is higher than about 5.7V, the zener diode Z02 breaks down, i.e., it passes current. The diode current goes into the base of T3. This switches T3 on. T3's collector current goes to the base of T2 via R17. This in turn switches T2 on, bringing its collector voltage almost up to its emitter and thus supplying power to the LEDs.

If the battery voltage is too low, Z02 does not break down and acts like an open circuit. There's no base current going to T3 and it's turned off. This also turns off T2 and the LEDs.

 

[nitin.maurya1]

@ pjdd: but when zener is in breakdown, then can't it shot circuit battery through R 18 and base emitter junction of transistor, so that no current goes to power LEDs

---------- Post added at 13:47 ---------- Previous post was at 13:43 ----------

If i use 5.8volt zener in place of 5.1 volt, then will the LEDs get off earlier than it is with 5.1 volt zener?

 

[Pjdd]

@ pjdd: but when zener is in breakdown, then can't it shot circuit battery through R 18 and base emitter junction of transistor, so that no current goes to power LEDs

When the zener diode is in breakdown, it still maintains a voltage of 5.1V across it. The difference between that and the battery voltage is taken up by R18 and T3's Vbe. This allows less than 1mA to flow through the series circuit of Z02, R18 and Vbe. That's only a small part of the load on the battery when it's on. In other words, it's not a short-circuit.

If i use 5.8volt zener in place of 5.1 volt, then will the LEDs get off earlier than it is with 5.1 volt zener?

Yes. But then you'll need a higher battery voltage to let the whole circuit turn on. You need a minimum of the zener breakdown voltage (5.8V), about 0.6V for T3's Vbe and at least a small amount of voltage for R18. That's a total of more than 6.4V. Only a fully charged or nearly fully charged 6V battery will be able to supply that and the whole circuit will switch off too early. By the way, 5.8V is not a standard zener voltage. The nearest one is 5.6V, but even that's still too high for normal use.

This is a simple design (but workable). A more sophisticated design will have an adjustable cut-off voltage, better charging control, etc.

 

[Raza]

The following circuit is the Battery charging circuit and has the cut off voltage arranged for the battery over charging:

The low voltage battery protection is established by T2. When the battery voltage hits 5.4 volts (the selected low limit for battery voltage) T2 will go off and the battery supply will be cut off to the LEDS putting them off. The value of TWO RESISTORS on the base of T2 are selected for this function.

Hope it helps.

 

[nitin.maurya1]

in ckt when transistor T3 is ON, then according to me current through R17 will be flowing away from base of T2 towards the negative terminal of battery through collector of T3, so potential across R17 will be such that it will be positive (bcoz of flow of current from right to left through R17) at base of T2 , so how will T2 (PNP) get turned ON?

---------- Post added at 22:14 ---------- Previous post was at 22:09 ----------

in ckt when transistor T3 is ON, then according to me current through R17 will be flowing away from base of T2 towards the negative terminal of battery through collector of T3, so potential across R17 will be such that it will be positive (bcoz of flow of current from right to left through R17) at base of T2 , so how will T2 (PNP) get turned ON??

 

[Syncopator]

..... so potential across R17 will be such that it will be positive,

The right hand side positive with respect to the left, yes.

..... so how will T2 (PNP) get turned ON?

Because its emitter is more positive than its base.

The current flow is, battery positive ---> ON/OFF switch ---> T2 emitter ---> T2 base ---> R17 ---> T3 collector ---> T3 emitter ---> battery negative.