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Topic: How to get 5v down to 3.3v? (Read 12216 times) previous topic - next topic

Oddwired

Hi

I'm making a stand-alone arduino circuit powered by a 9v battery with a normal 7805 volt regulator bringing the power down to 5v.

Part of my circuit  uses an adafruit colour sensor, this is one of the sensors that Must be powered from the 3.3v source rather than the 5v.

Were I just using a normal  arduino in the circuit I would just power the colour sensor from the 3.3v pin but as I've got a stand alone circuit using an atmega 328 chip I don't think the chip actually regulates the supply down to 3.3v, it appears that that is the arduinos doing.

What is the recommended way to bring the voltage down to what I need (3.3v)?  I was thinking I might just use a simple voltage divider circuit but I have heard people saying this is not a good idea due to the circuit being dependant on current flow so are there any other recommended simple ways of doing this.

Thanks in advance

nickgammon

Get a small 3.3v voltage regulator chip.

Quote

I'm making a stand-alone arduino circuit powered by a 9v battery with a normal 7805 volt regulator bringing the power down to 5v.


You are probably better off starting with a lower voltage if possible. If not, look for a buck-converter module on eBay. I got some recently for $1.75.

Turning 9V into 5V is basically throwing away half the power as heat, and if the 9V battery is one of those small smoke-alarm batteries you don't have that much to start with.

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JChristensen

You don't say how much current is required. An MCP1700 will go up to 200mA or so and can take the 5V as input. If more current is needed, look at something like an LD1117.

As Nick says the 9V (IEC 6LR61) battery isn't the best solution unless it doesn't get used much. I had a project that I ran off a 9V battery and a 7805, but it only got used 5 or 10 minutes a day, and not every day; the 9V worked fine for that.

Running at 8MHz with 2xAA or 3xAA can be a good option too.

Oddwired

Yeah its not going to be a very frequent use project tbh so the 9v solution should be fine, I might consider using 4 aa's  at 6v but I'll see where the 7805's dropout voltage kicks in first.

The chip idea sounds interesting and I'll Have a shop round maplins to see if they have any useful iieas.

MarkT

If you have a "low-dropout" (LDO) regulator it will be able to directly drop
5V to 3.3V (some can't, you have to check the datasheet)  and it can be in
a small TO92 package since the power dissipated is likely to be small (current
times voltage drop).

You also then ensure that the sensor voltage cannot be higher than the 328's
supply at start up which is re-assuring.   If you run a 5 and a 3.3V regulator
both from 9V they are racing each other to charge the decoupling capacitors
when switched on.  Running the 3.3V from the 5V one seems a better arrangement
to me (and with luck you only need to worry about heat dissipation in the 5V one.
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cjdelphi

http://www.jaycar.us/productView.asp?ID=ZR1398&w=3v+zener&form=KEYWORD

You could use a Zener and a Pass Transistor if you wanted to do it really cheaply...

bigred1212


SirNickity

Simmuh down... Sometimes people like discrete components, especially when they're building their own.  :smiley-mr-green:

bigred1212


dc42


Yeah its not going to be a very frequent use project tbh so the 9v solution should be fine, I might consider using 4 aa's  at 6v but I'll see where the 7805's dropout voltage kicks in first.


If you want to run from 6V, then don't use a 7805, use a low-dropout 5V regulator instead.

Do you really need 5V? If not, why not run the whole device from 3.3V instead?
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nickgammon


Does nobody like $1.50 logic level coverters?

https://www.sparkfun.com/products/8745


Yes, but that board needs 3.3V input to function. It converts logic, but it still needs the 3.3V level to do the conversion. So that's a different issue.
Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

zoomkat

One could use two typical diodes in series on the output of the 7805 regulator chip to drop the voltage to 3.6v.
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polymorph

Quote
Yeah its not going to be a very frequent use project tbh so the 9v solution should be fine, I might consider using 4 aa's  at 6v but I'll see where the 7805's dropout voltage kicks in first.


Or you can look it up...
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

Page 3 indicates an LM7805 requires at least 2V (7V input) to stay within specs.

Quote
One could use two typical diodes in series on the output of the 7805 regulator chip to drop the voltage to 3.6v.


I've always thought that was a bad idea. At very low current draw, the diodes don't drop as much voltage. Having a power supply voltage that changes under load just seems like such an unnecessary thing, destined to cause odd problems at a later point.


OK, here I go, flogging the MC34063 again... There are newer switchmode regulator chips, but this one is ultracheap, comes in DIP and SMD packages, and has several design tools out there. You will need to start with more than 6V, but you aren't wasting so much power.

I favor this particular MC34063 calculator. Watch out for negative values in the answer, that signifies that what you are asking it to do, cannot be done.
http://dics.voicecontrol.ro/tutorials/mc34063/
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guix

I use this: (search on ebay LM2596 step down)


Good for 2A, 3A if you add heatsink.

nickgammon

That looks like the one I got for $1.75, mentioned above.
Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

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