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This is an update to a previous post about a circuit for a 6W LED. I've decided to add a boost converter in order to be able to power the LED. I used TI's SwitcherPro Design tool and based it off TPS61087. I can't seem to get it working since the circuit isn't completed but I'm worried that the 2A in won't be enough for the circuit. Here is the circuit design so far: Circuit and the data sheet for the circuit.

Edit: Now I'm having trouble implementing the boost converter into the circuit...where should the booster be put into the original circuit, right after the battery or right before the LED?

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Link to previous post? – Brian Carlton Jul 26 '12 at 20:19
A broken link to a file sharing site to something that isn't a pdf won't get you help. – Brian Carlton Jul 26 '12 at 20:34
650kHz may give you headaches. (not literally) – abdullah kahraman Oct 4 '12 at 14:27

1 Answer

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If you only have a maximum of 2A input at 3.6V, and you need a continuous 12V @ 0.5A output then you may have problems.

If we assume around 82% efficiency (from your report) then:

0.82 * (3.6V * 2A) = 5.9W

So there will not be enough power available to provide 12V @ 0.5A (6W) at the output. In your report the calculated RMS inductor current is given as 2.63A.
Also the 3.6V is given as Vin max, so things will get worse (input current will have to rise) for the given Vin min of 3V.

To get a good idea of performance, as well as doing the calculations thoroughly, you should probably simulate this circuit in SPICE (TINA or whatever it is TI provide) or build a test jig.

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okay thanks! yea I'll prob build a test jig. Do you think having the Vin max be 4.8V would help curb the problem? Btw I'm using 3 AA Ni-Cad's (1.2V each) as my power source and they won't be fully charged at all times but I'll still need a continuous 12V and .5A for the LED I'm using. – Om23 Jul 27 '12 at 13:58
If you raise the input voltage but keep the current the same then you will have more power available, so it will help, yes. 4.8V * 2A = 9.6W. At 80% efficiency you now have ~7.8W available, so it is far more likely to work. – Oli Glaser Jul 27 '12 at 14:07
The LED needs 6W so looking back the 3.6V works. My bad I overlooked that. But I might have to get a more powerful bulb since this one might not be bright enough. But once again thanks for the help! The only problem is getting the circuit to work with the booster. I was using CircuitLab to design and test the circuit but when I connect the boost to the circuit (I attach Vin right after the battery and Vout right to the LED) but the voltage coming in the boost is 3.32V but the voltage out is only 3.27V... which means its actually decreasing the voltage and acting like a buck converter. – Om23 Jul 27 '12 at 14:17

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