Battery charger part

/2=A0 =A0 =A0 Plate

THe 'plate' is the rectifier. The two brown plasitc 'pucks' are the actual diodes. The 'diode' in your photo is the circuit breaker [12A]. Very typical simple full wave rectifier battery charger.

Neil S.

thermal self-reset circuit breaker

be selenium

Plate

So I should be able to replace the 'plate' with a solid state full wave bridge rectifier? What causes the tapering down effect as the battery charges?

You have solid state full wave rectifier. A bridge will give you twice the output voltage.

The output voltage of the charger is fixed, as the battery voltage comes us (charges) the current goes down.

Jeff

Are you sure?

Yes, the key here is the word bridge.

There are two diodes in a simple full wave rectifier. AND a center tap on the transformer. Only one half of the winding is conducted through a diode at a time. But since they alternate, it's a full wave rectifier.

A bridge rectifier has four diodes, and goes across the entire output winding. So you'll get twice the voltage.

Jeff

Not if you hook it up simply as a full wave rectifier -- leaving the other 'side' (probably marked with a minus ("-") sign) floating.

Jonesy

The wave pulse is doubled but the voltage is not I believe. The peaks remain at the same level in my experience.

Plate

No. You could replace the 'plate' with two stud mount rectifiers in a suitable heat sink. However, the voltage drop of of the silicon rectifiers is much less than the voltage drop of the original selenium rectifier. This will result in overcharging, and release of hydrogen and oxygen gas as the water in the electrolyte is disassociated.

Good grief, don't you even understand lead-acid batteries? As the battery charges the voltage across it increases (although the voltage is not linear with percent of charge). The output of the battery charger is the 'upper' portion of the sinusoidal AC waveform. Current will flow into the battery only when the instantaeous voltage out of the rectifier is greater than the charging voltage of the battery. As the battery voltage increases, the fraction of the waveform actually charging the battery decreases.

The transformer was designed taking into consideration the characteristics of the original rectifiers. In other words, as the battery became fully charged, only a small 'trickle' charge would flow.

PlainBill

Yes, he's sure. So are most of the rest of us. Two diodes in a full wave rectifier.

I am not sure what you are saying here. Jeff is quite correct. With the existing rectifier you get one positive peak of DC output every half cycle, therefore it is 'Full Wave'. There is little point in replacing the rectifiers since thay appear to be [or similar to] Motorola MR type 50A silicon diodes attached to the heat sink. anything you replace them with is not likely to be any better and could be much worse.

Neil S.

The part in the first pic ("diode") is a circuit breaker. The two ceramic buttons on the "plate" are the diode rectifiers, that'd be a half-wave bridge configuration.

H> Can somebody tell me what this part is?

But you could replace the diodes with a bridge, just ignore either the plus or minus terminal on bridge and use centre tap of transformer as per existing circuit. Those high current large bridges are probably easier to buy and mount on the heatsink than a pair of power diodes.

The unused pair of diodes in the bridge wont do any harm.

Grant.

As mentioned, the 12V 12A part is a circuit breaker, a self-resetting type which opens when heat from high current causes it to open, then closes again when it cools. The opening action will repeat if the current draw/load is excessive.

The plate assembly has 3 connections, not 2. A (black?) wire lead is connected directly to the plate. The plate is a heatsink for the 2 red diodes. The red diodes are silicon types with terminals/contacts on each side/end. Two diode terminals contact the plate, and the other terminals contact the metal holders that the other

The diodes need to be in their original specific orientations for the correct output of the charger. Mark the diodes before removal, or make sure of their orientation before removal, as the anode and cathode may not be marked. The part number or other printing around the body may be the only indication of the diode polarity. With the printing viewed right-side-up, the top contacts should be the same terminals.. A or C. Or, there may be a diode symbol printed to indicate A and C. Or use an ohmeter to determine polarity of the diodes.

-- Cheers, WB .............