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Series Parallel diode mystery

Charcbait

New Member
Who knows where these old 120VAC-to-6.8VDC wall warts came from, but I suspect they powered mobile radio chargers. I took one apart to find out why the LED didn't light up, and I uncovered a mystery.
1642802053155.png
Measured: pri 120VAC; sec 20.8V=2x10.4V; rectified to 8.9VDC; output 8.0V; LED tested good

The full-wave rectifier is standard, but what on earth are those other diodes doing? Of course the red LED won't light up--with its 2V drop it will always lose out to the series diodes at 1.4V. Nobody includes a component that will never function, and what power supply has no regulation or filtering? Clearly there had to have been another device down the path to complete this circuit design, containing caps, coils, resistors, etc., and of course a path to ground/earth through resistance. Note also that the 8V unloaded output is well beyond the 6.8V spec marked on the case.

What additional component configuration may have been located beyond this circuit's output? Regulation and filtering no doubt, but I can't imagine what could have induced that LED to light up. Any ideas?

Chuck
 

AnalogKid

Well-Known Member
Most Helpful Member
Small cheap red LEDs clock in with a Vf in the 1.6 V to 1.9 V range. And rectifier diodes, even small ones, have a Vf above 0,7 V at any appreciable current, and this Vf increases with increasing current. Thus, the LED comes on when the current is high enough. shunting most of the current around the LED.

It is very common for unregulated wall warts to have a wide voltage range between unloaded and full load. This is due to a combination of low cost transformer steel and amplitude of the peak-to-peak ripple voltage. What is the load current on the device label?

ak
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Thus, the LED comes on when the current is high enough. shunting most of the current around the LED.
Agreed; it's a "Charging" indicator, on when the load current is high enough and off when the device was fully charged, or near so.
 

Diver300

Well-Known Member
Most Helpful Member
With no filter capacitor in the power supply, the output current will have very large ripple. In fact, there will be spikes of current near the voltage peaks, and zero current the rest of the time.

With a large peak current, the peak voltage drop on the diodes will be larger than it would if the current were steady, so that will also help get the voltage high enough for the LED to light.
 

Charcbait

New Member
Sheesh! Why didn't I think to look at the case before and chase down the model number? 3M GVP-112, Class 2 Battery Charger, Input 120VAC 60Hz 25W, Output 6.8VDC 750mA

One web hit says "...discontinuation of the 3M GVP-Series Belt-Mounted Powered Air Purifying Respirators (PAPRs)...", so now I know it provided power to charge NiMH batteries in a now-obsolete series of self-contained breathing units. However, knowing that wouldn't have solved my diode mystery.

So thank you AnalogKid for identifying the voltage drop variability. If I understand correctly you're saying that once there's actually some current flowing in the circuit then the voltages could be right for the LED to light...and thus when the battery attained full charge then the current flow might cut back and the LED would go off, indicating that the charging was complete.

Thanks again.
Chuck
 

Charcbait

New Member
Right Diver300. When writing my original post the lightbulbs were slow to go off in my head. Looking at the schematic I had drawn from a visual inspection of the internals I noticed there was no filtration...then I saw there was no regulation...then no current-limiting resistor to protect the LED...and only then did I notice there wasn't even a path to ground. I guess it takes a lot of practice to see the obvious right off the bat.
 

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