Why doesn't my simple 4V lead acid battery charger using an LM358 op-amp work?

Question

I obtained the circuit copied below from: Instructables - Simple 4V Lead Acid Battery Charger

I changed the Zener from 4.2 V to 4.7 V.

The 7806 supplies 6 V to the circuit shown in the figure at J1-6V (TBLOCK-M2.)

A 4 V 1 Ah battery is connected at J2 (TBLOCK-M2.)

The red LED should glow when the battery is charging and the green LED should glow when the battery is full. In my circuit, the green LED is always glowing.

Across the 4.7 V 1 W Zener diode the voltage stays around 3.8 V only. The lead acid battery voltage before placing it for charging has 4.5 V and as per the opamp comparator working principle. The green LED is always glowing.

Why does the 4.7 V Zener have 3.8 V across it? I tried several resistors to increase the current through the Zener so that it will reach 4.7 V. Even though the current through the Zener is increasing the voltage across the Zener reaches a maximum of 4 V.

My questions are:

1. What is the maximum full voltage of 4 V, 1 Ah lead acid battery. Is the full voltage of battery is 4.8 V?

2. With 6 V given to one side of R4 resistor, why doesn't the Zener diode reach breakdown voltage? The datasheet says the maximum Zener current is 0.193 A, so I tried to push more than 10% of maximum Zener current using a 39 Ω resistor. The Zener voltage stayed at 4.1 V.

1) With 6 V power source to R4 and D4 (4.7 V 1 W Zener). How to achieve 4.7 V breakdown voltage? Can you point me to any website which shows practical answers for reaching the 4.7 V breakdown? I need resistor value and resistor wattage calculation also.

2) What is the full battery voltage of 4 V 1 Ah sealed lead acid battery?

Answer 1

The zener diode in this design will work as advertised, but only if the 6V source stays at 6V. I suspect that the supply is being overloaded, and its voltage is falling below the zener diode's own rated voltage. This is very likely, since there's nothing to limit battery current, as I'll explain in a moment. You'll have to measure the supply voltage too, to see if this is the case, but that's all moot anyway.

Perhaps the PCB layout is different in some way from the schematic, and behaves differently, but the design shown in the schematic doesn't make any sense:

1. The LM358 op-amp U1:A is being used as a comparator, and its only function is to switch the LEDs on/off. There's no feedback of any kind (positive or negative), and so it can't perform any regulation. It doesn't control battery current or voltage at all. Consequently, the LEDs don't indicate anything other than battery voltage. They certainly don't tell you anything about whether the battery is charging or not. Even if this circuit was otherwise good, an LM358 is not the right op-amp for this task, at least not without some modifications.

2. Transistor Q1 is configured as an emitter follower, whose base is permanently held at +6V by R1. Therefore it always applies \$6V-V_{BE}=6V-0.7V=5.3V\$ to the battery, regardless of its charge state. There is nothing in this design to limit current to 400mA, as the author claims, and the battery will always be charging with whatever current the 6V supply can provide.

3. The author states that you should use a 500mA transformer, which is nonsense. I have no idea what that even means in this context. The circuit will always draw whatever current is available from your 6V source, which unless you implement your own current limit will be more than the claimed 400mA. A transformer's current rating tells you how much you can draw without damaging it, but is in no way a physical limitation. You could draw 1A from a transformer rated for 500mA, and it will eventually overheat and die.

I suppose by "500mA transformer", the author might mean "current limited voltage source", like some cheap 6V wall-wart. If that's true, then the real "current regulation" is inside that supply, is very unreliable, and has nothing to do with this circuit. You have no guarantee of the exact voltage from such a supply, and this design depends entirely on that.

The circuit does nothing more than subtract 0.7V from the supply, and use LEDs to show whether battery voltage exceeds some value or not. The 5.3V applied to the battery certainly exceeds 4.1V or 4.7V.

Under no circumstances should you use this design, it's a fire waiting to happen, and doesn't do any of the things it claims to do. This is not an "instructable", it's a "destructable".