# Raising 1.5v DC to 5+v DC using LM358?

Discussion in 'General Electronics Discussion' started by sciguy77, Oct 21, 2012.

1. ### sciguy77

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0
Oct 21, 2012
Hi,

Anyone know if I can amplify 1.5v DC to 5+v DC (pretty much any value over 5 works) using the LM358 IC (https://www.sparkfun.com/products/9456)?

The datasheet (http://www.sparkfun.com/datasheets/Components/General/LM358.pdf) specifies the power supply range as 3V to 32V however the input voltage range is -0.3V to 32V (top of datasheet page 2). I've done some googling but I'm still not sure about the difference between these two.

Any tips or links on wiring this guy up would be awesome,

Thanks!

2. ### BobK

7,682
1,689
Jan 5, 2010
Depends on what you mean by "amplify".

It you are thiking you can start with a 1.5V battery and turn it into a 5V power supply, the answer is no, you would need a DC-DC converter to do that.

If, on the other hand, you have a power supply of somewhat over 5V and you want to turn a 1.5V signal into a 5V signal, yes it can do that.

the input voltage is different from the power supply range because well, they are two different things. The power supply range give the range of voltage you can use to power the chip. The input range tells you what range of voltages you may put on the inputs and have the chip function properly.

Bob

3. ### sciguy77

4
0
Oct 21, 2012
Bob,

Thanks for the speedy reply. Yeah, I figured as much about the LM358. Can you recommend a good, cheap DC/DC converter IC? I've been doing some research and it looks as though the MAX756 might do the trick, though I'd love to hear what the experts have to say.

Thanks again,

Sciguy

4. ### BobK

7,682
1,689
Jan 5, 2010
The MAX756 looks like it would work for up to 200ma.

Bob

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
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Jan 21, 2010
Check the specs carefully (I haven't looked at them at all). Some chips have a headline "low" voltage that they may operate from, but cannot start from.

Also remember that from a low voltage input you'll need an input current that is higher than the output current by a factor of the ratio of the output to input voltage.

So if you want 150mA @5V output, and your input voltage is 1.5V, your input current will be (5/1.5)*150 mA = 500 mA Note that the input current will actually be higher (expect around 600mA) because the DC-DC converters are not 100% efficient.

6. ### BobK

7,682
1,689
Jan 5, 2010
Yep, a look at the datasheet shows minimum startup voltage as typ: 1.1 max 1.8. So it is not guaranteed to start on 1.5V, though it typically will.

Bob