Help with Voltage Boost Circuit (ambiguous datasheet)

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#1
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So I've never did a boost converter circuit before. I have this 3 volt watch batter I'm gonna use to power my infrared device. I'm gonna turn 3V battery into 3.3V steady DC. I'm laying out the circuit right now and I'm stumped.

The data sheet shows a circuit for a 5V boost. I bought the 3.3Volt boost model of the chip. On the 5V circuit, it shows a pin called LX attached to output with a Schottky diode (whatever that is). It says no need for diode in 3.3V circuit.

Do I take the diode out and break the circuit between Vout and LX pins? Or... Do I take the diode out and connect the Vout and LX pins with a trace?

This is the component I'm using, the AS1325 for 3.3V.
http://www.austriamicrosystems.c...

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I think it's a bootstrap diode. Take it out, don't connect LX to OUT.

- S

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Hahah thanks you just saved me an hour of work =) Of course I designed it backwards... Exactoknife a trace and I'll be good to go.

Thanks!!!!!!!!1

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Actually, after thinking about it a bit more, the external diode is probably not for bootstrap, but for efficiency, and to limit the voltage on LX. I've seen similar diodes on other synchronous boost regulators (e.g. the ltc3401).

Whatever the case, not connecting LX and OUT seems to be the right setup.

- S

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I'm running this thing without the guy connected, it seems to be working good. funny thing is now I have a 3.6 volt battery hooked up to it instead of a 3.0 battery and it's still doing its job of giving me 3.3. When Unloaded it outputs 3.5 V steady but when I connect stuff it does 3.3.... Any harm in this? Data sheet says it can handle it but I don't know if they are talking about the 5V version or both... First time stepping up voltage here...

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Quote:

I have a 3.6 volt battery hooked up to it instead of a 3.0 battery and it's still doing its job of giving me 3.3. When Unloaded it outputs 3.5 V steady but when I connect stuff it does 3.3.... Any harm in this?

It's a bit hard to tell. For a one-off project for myself, I would probably be OK with a 3.6V input, but the manufacturer wouldn't have tested it completely at that input, so there may be unexpected surprises.

- S

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You run the circuit in a dangerous area. The internal diode is open all the time now, so any increase on the input voltage will be seen at the output.
If you plan to use this batteries, better add a LDO at the input or output or a Schottky diode.
Some 3.3V IC's are rated only up to 3.45V.