Diode in parallel with resistor on battery charger
Good day folks. Today I had to work on a power supply/battery charger which had some parts arranged in a manner which intrigued me. It is a proprietary board that's part of an alarm panel and it also acts as a charger for a backup SLA battery. It's a very basic supply, with no proper battery management: have a look at the terminals and you'll immediately see how it works. There's a 14v output and there's a "BAT" output, which basically just ties the battery in parallel to the supply, but not directly, instead through a diode and resistor, as outlined on the doodles I did on the back. The E-C terminals correspond the emitter and collector of an optocoupler which signals the panel when mains power is present, so don't bother with that.
When I got it, the power LED would flash intermittently as if there's a high load on the output or a short. I found D11 shorted - it is the reverse polarity protection diode which forward-biases if the battery is connected in reverse by accident, blowing the fuse (the fuse is housed inside that black box on the secondary side). I replaced it, the supply now turned on without a battery attached, but when I connected a battery, it went back to blinking again. Turns out the adjacent diode, D4, was ALSO shorted. It's paralleled with that large ceramic 10ohm resistor. I replaced it and it works, but my question is why is there a diode in parallel with the resistor and why the supply failed when it shorted ? I was expecting the diode to act as a wire between the supply's output and the battery, so I didn't expect it would shut off. I suppose bypassing the resistor (which is what the shorted diode did), placed too much of a load on the supply, even though the battery was new and full and I used to connect SLAs directly the output of power supplies before and never caused them to shut off, so perhaps I don't understand the principle of this one...it's only a handful of components - no current sensing as far as I can tell...the cluster of SMDs on the left is for feedback.
If I'm not mistaken, when mains is present and the supply is on, the 14v from the regulated output flows through the resistor and into the battery (since D4 is reverse-biased and current can't flow through it). When the supply is off and the battery takes over, D4 is now forward biased since current flows back in the opposite direction and the rail which is normally powered by the rectifier is now powered by the battery. Still, what happens when D4 shorts ? If the battery is dead or shorted, a high load is placed on the output I believe, because the dead diode shunts that resistor entirely and places the battery (the "DUT") directly on the regulated output which doesn't appear like it can supply too high of a current, so it just shuts off. With the resistor in place, if my math is correct, shorting the battery output would result in a maximum current draw of 1.5A through the resistor: 14v/10ohms = 1.4A
When I got it, the power LED would flash intermittently as if there's a high load on the output or a short. I found D11 shorted - it is the reverse polarity protection diode which forward-biases if the battery is connected in reverse by accident, blowing the fuse (the fuse is housed inside that black box on the secondary side). I replaced it, the supply now turned on without a battery attached, but when I connected a battery, it went back to blinking again. Turns out the adjacent diode, D4, was ALSO shorted. It's paralleled with that large ceramic 10ohm resistor. I replaced it and it works, but my question is why is there a diode in parallel with the resistor and why the supply failed when it shorted ? I was expecting the diode to act as a wire between the supply's output and the battery, so I didn't expect it would shut off. I suppose bypassing the resistor (which is what the shorted diode did), placed too much of a load on the supply, even though the battery was new and full and I used to connect SLAs directly the output of power supplies before and never caused them to shut off, so perhaps I don't understand the principle of this one...it's only a handful of components - no current sensing as far as I can tell...the cluster of SMDs on the left is for feedback.
If I'm not mistaken, when mains is present and the supply is on, the 14v from the regulated output flows through the resistor and into the battery (since D4 is reverse-biased and current can't flow through it). When the supply is off and the battery takes over, D4 is now forward biased since current flows back in the opposite direction and the rail which is normally powered by the rectifier is now powered by the battery. Still, what happens when D4 shorts ? If the battery is dead or shorted, a high load is placed on the output I believe, because the dead diode shunts that resistor entirely and places the battery (the "DUT") directly on the regulated output which doesn't appear like it can supply too high of a current, so it just shuts off. With the resistor in place, if my math is correct, shorting the battery output would result in a maximum current draw of 1.5A through the resistor: 14v/10ohms = 1.4A
Attached Files
The fuse was blown when the first diode shorted, so I replaced it and both diodes and everything's working, but I was just curious why shunting the resistor (due to the shorted diode in parallel with it) would cause a shutdown of the supply. It's more of a speculative discussion, since any component that is found faulty must replaced, even if you don't know what it does, but it sure helps to know what its purpose is and why its failure made the device not functional.
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