Hi guys!
Quite a long time since I last came here! I have a question.
I'm currently working on a 3DS battery extender, that is, a battery pack that goes on your pocket p.ex, and connects to the charging port of the console; effectively working as a portable charger.
This is what I'm going to do:
1) For portability's sake, battery will be limited to a 2200mAh, three cell Li-Po pack (radiocontrolled racing grade to ensure proper quality). That is about 25Wh energy storage. The 3DS pack has 5Wh, so the boost is pretty impressive. Why 11.1V? Because this way I can...
2)...embed a buck converter to step down the voltage to 4.6V with a pretty high efficiency (I'm expecting around 87-90%).
Now to the main point.
I've narrowed down to the LM2672T-ADJ with embedded MOSFET buck microcontroller. Just because it simplifies the circuitry quite a lot, and size can be kept to a minimum.
I've read through the National Semi datasheet, but I can't seem to find a sample circuit that shows how to utilize the adjustable output feature.
They talk about using a divider between Vref pin and sensing pin, but it kinda confuses me. How accurate is it, if the Vref voltage values change with the output loading characteristics? I think you get the idea.
Another thing is, I can't understand how the inductor diagrams work.
I mean, the graphs are pretty explicative, but hey, 900mA sustained load at about 11V input voltage, says 35uH. Then you go to the other figure which shows a more realistic, non-logarythmic graph, and then I need 100uH? What the...?
It's pretty obvious I don't understand the graph at all, because obviously the datasheet isn't wrong
And the third and last question is... How can I force the circuit to output 900mA, without using a resistive load and thus decreasing the efficiency? Or the 3DS will only drain the current it needs (as the balancing circuit is inside the machine)? It bothers me because the circuit has overload protection, and if the power consumption isn't in the 900mA the wall charger is rated at, I fear it might not work, or even worse; burn if overloaded (the battery/built-in charging module).
Thanks a lot!
Quite a long time since I last came here! I have a question.
I'm currently working on a 3DS battery extender, that is, a battery pack that goes on your pocket p.ex, and connects to the charging port of the console; effectively working as a portable charger.
This is what I'm going to do:
1) For portability's sake, battery will be limited to a 2200mAh, three cell Li-Po pack (radiocontrolled racing grade to ensure proper quality). That is about 25Wh energy storage. The 3DS pack has 5Wh, so the boost is pretty impressive. Why 11.1V? Because this way I can...
2)...embed a buck converter to step down the voltage to 4.6V with a pretty high efficiency (I'm expecting around 87-90%).
Now to the main point.
I've narrowed down to the LM2672T-ADJ with embedded MOSFET buck microcontroller. Just because it simplifies the circuitry quite a lot, and size can be kept to a minimum.
I've read through the National Semi datasheet, but I can't seem to find a sample circuit that shows how to utilize the adjustable output feature.
They talk about using a divider between Vref pin and sensing pin, but it kinda confuses me. How accurate is it, if the Vref voltage values change with the output loading characteristics? I think you get the idea.
Another thing is, I can't understand how the inductor diagrams work.
I mean, the graphs are pretty explicative, but hey, 900mA sustained load at about 11V input voltage, says 35uH. Then you go to the other figure which shows a more realistic, non-logarythmic graph, and then I need 100uH? What the...?
It's pretty obvious I don't understand the graph at all, because obviously the datasheet isn't wrong
And the third and last question is... How can I force the circuit to output 900mA, without using a resistive load and thus decreasing the efficiency? Or the 3DS will only drain the current it needs (as the balancing circuit is inside the machine)? It bothers me because the circuit has overload protection, and if the power consumption isn't in the 900mA the wall charger is rated at, I fear it might not work, or even worse; burn if overloaded (the battery/built-in charging module).
Thanks a lot!
