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August 2, 2011
about 3 years ago
Reading through the comments here, the datasheet, and the schematic, I think I have found a few issues:
The datasheet recommends using a reverse-bias protection diode to prevent the battery from discharging into the outlet if plugged in backwards:
“The optional reverse-blocking protection diode,
depicted in Figure 6-1, provides protection from a
faulted or shorted input, or from a reversed-polarity
input source. Without the protection diode, a faulted or
shorted input would discharge the battery pack through
the body diode of the external pass transistor.”
This may have been the cause of some smoking and flames.
Also, while the chip itself can handle an input voltage of 4.5 to 12 volts:
“4.5 – 12 V”
The power indicator LED setup (LED red in the schematic) does not allow for usage above about 10.75 volts.
I assumed they either used an indicator LED that was 2.0V@20mA or 2.5V@25mA. Using their 330 ohm resistor. I calculated that the LEDs would be hitting their 20 or 25mA limit at 8.6 and 10.75 volts respectively, but would be safely below that when given 5 volts. I derived this from the equation that:
V(supply) = V(LED) + I(LED)*R where R = 330 ohms
When given 12 volts, the current across the LED would be:
12 = 2 + I(LED)*330 I(LED) = 30mA enough to potentially damage the LED
or for 2.5V@25mA
12 = 2.5 + I(LED)*330 I(LED) = 29mA again potentially enough to damage the circuit.
While this might not cause smoking or flames, the power going through the LED resistor would be approaching ¼ Watt which would definitely be bad for a ¼ rated resistor after a few hours.
P = I(LED)2*R
P = .0282330 = .259 watts or P = .032330 = .297 watts
The LED resistors are the likely cause of any flames.
Another issue is the R(sense) resistor. In the schematic it shows 2 .110 ohm resistors in parallel for a total of .055 ohms which according to the datasheet would put out approximately 2 amps instead of 1. Fortunately, pictures of the actual circuit show only one resistor, so it is likely that this was a simple mistake.
The datasheet calls for the drive transistor and current management chip to be placed as close as possible to the battery outlet for proper voltage regulation. I think that Sparkfun could have moved it over an inch to get better results.
Finally, everyone should take note that this is designed for Lithium Polymer batteries which take a final charge of 4.1 volts as opposed to the standard 4.2 for Li-Ion. This means that if you attached a Li-Ion, you would get about 87% of a full charge according to table 2 on this page made by battery university:
Overall, if you are looking for a complete Li-Ion solution without worrying about input voltage and bias. Make your own charger on the cheap by following the datasheet for the chip and using a reverse-bias diode, a transistor with a higher wattage rating, and LEDs and caps to match your voltage and capacity specs. The major parts are all about 50 cents each and OSHPark will make you 3 copies of a board for $5 per square inch. You also don’t have to pay for a pretty piece of injection molded plastic.
Great resource and good circuit, but the devil is in the details.
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