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Troubleshooting switching regulator

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Rusttree

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I put together a simple battery-powered LED circuit using a 5V booster switching regulator. It's powered by two AA rechargeable batteries. At most only two LEDs plus the uC and passive circuit components will be energized at one time, so it's relatively low power. Maybe 60mA max.

I picked up a cheap 5V switching regulator here:
http://search.digikey.com/us/en/products/NJU7261U50-TE1/NJU7261U50-TE1CT-ND/673933

Under near-zero load (uC on, but no LEDs), the regulator holds the promised 5V. When I activate two LEDs, however, it drops to about 4V. Note, only two LEDs are ever on at the same time. That's well within the current rating of the regulator.

Any ideas how to troubleshoot this? The circuit is attached.

Thanks!
-Dan
 
Hm, immediately after posting this, I realized my inductor is off by an order of magnitude. The application circuit in the datasheet uses a 100uH and I picked out a 10uH. Oops. Probably just a typo on my part.

Can anyone help me with the theory behind the inductor on a switching regulator to determine if that's definitely the problem?

EDIT: Found a couple inductor selection guides online and definitely confirmed the 10uH is way too small. Looks like I'll need something closer to 200uH.
 
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In a nutshell: a boost converter stores energy when the transistor is on in the inductor in the form of:

1/2 x L x I squared

Obviously if the inductor is too small, you can't store enough energy to feed the load.


NOTE: make sure the inductor is rated for the peak current it has to handle (which is much higher than the load current). Running an inductor above it's rated current means the inductance drops off rapidly.
 
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Thanks both. Very helpful information.

Simon, question about a part in your tutorial. When the MOSFET turns off, "the right hand side of the inductor has to increase in voltage with respect to the left hand side. The left hand side is connected to the input voltage (so cannot change), thus the right hand side voltage increases above the input voltage and continues to do so until something conducts."

From the circuit on your webpage, why doesn't current back flow into the input source as soon as any part of the inductor exceeds 3.3V? Obviously, it doesn't (or boosters wouldn't work), but I don't see why it doesn't.
 
I see your problem... here goes... Once you have set up a current in the inductor, it effectively becomes a current source. When the FET is switched off, that current has to keep flowing and in the same direction. it is just the nature of inductor and yes it kinda defies Ohms law. The way it keeps the current flowing is that it effectively turns itself into a battery. The right hand end goes positive (and pushes current out of that node) and the left hand end goes negative (to sink current into that node). That is the best way of explaining it. I hope this helps
 
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