5V to 3.3V/1.8V Help Needed

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Jeff Greene

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Hello,

I need to revise a 5V design to operate with some lower voltage components. I am looking for recommendations for some good regulators to drop my 5V supply to 3.3V and 1.8V. There's literally millions of them out there and I don't want to make a bad choice.

I have 1A of power on my 5V supply to play with and I need 500mA of 3.3V and 200mA of 1.8V from the new supply.

My design criteria are simple:

Regulators must be as inexpensive as possible.
Regulators must use cheap and small (hopefully ceramic) capacitors.
Regulators must be readily available.
Regulators must be hand solderable (No D-PADS on bottom)
If possible, regulators must be multiple source components.

I think this sort of thing must come up a lot, so hopefully you folks can give me some guidance.

Thanks, and best regards!

- Jeff
 
You can use the "good old" LM317... It will do your job nicely and they are widely available, you need 2 resistors to set the output voltage (between 1.25V and 37V). You will of course need 2.
 
VictorPS said:
Why? because of 5V-1.25V drop= 3.75V? further drop to 3.3V will not stable?

The dropout voltage of an LM317 is not 1.25 V. The reference voltage is 1.25 V.

https://www.electro-tech-online.com/custompdfs/2007/11/LM117.pdf, 4th graph on page 6 shows the drop out voltage.

The dropout voltage depends on current and temperature. At 500 mA, at 25 °C, it is about 1.8 V, and that is measuered when there is a 100 mV reduction of the output voltage.

You need a low dropout regulator to produce 3.3 V from 5 V.
 
You are correct.

Also, the regulator's characteristics are specified with the input 3V higher than the output.

That graph is the typical dropout voltage and doesn't give the worst case scenario.

You could probably run it at 500mA with the source 1.8V higher than the output but it won't regulate very well.
 
With all said, perhaps we can derive 1.5V from 5V and for few mA. even if 10mA min load can be managed by a red LED on the load side. Incidentally these Cell chargers on car output at least 5.! volts and there should not be a problem.

thre is no second thought about one going to modern ICs which work more efficiently.


Sarma
 
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Hero999 said:
You are correct.

Also, the regulator's characteristics are specified with the input 3V higher than the output.

Since this is the case, you could incorporate a 7808 into your PSU. To get your 5V back, run your 8V output through a 7805.

To get 3.3V, tap the 8V output of the 7808 and input it into your 317. To get your 1.8V tap the o/p of the 7805 to a second LM 317, since tapping it from the 7808 may cause heating problems (i.e the regulator would need to drop about 8-1.8 or 6.2V which I fear may cause excessive heating).

Just curious is this a microprocessor project you're embarking on?
 
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Where did you get the idea that he has voltage available to power a 7808? And if he did, what would be the point?
 

there is a confusion on my part -

the best approach appears to be --use MC/LM7833 (5 to 3.3 LDO regulator)and LM317L to work with 5V input and be done with it. the 317L can be programmed for 1.8V with 5V input .
 
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Roff said:
Where did you get the idea that he has voltage available to power a 7808? And if he did, what would be the point?

Perhaps he does, perhaps he doesn't - he hasn't educated us on that. My suggestion may or may not help him. If it has-my pleasure. If it further meddles things up, there are smarter guys out there to ensure he doesn't go by a wrong advice.

And as I can see he's got himself out of the hole. All's well!
 
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You certainly see the world differently than I do. The OP has never posted again anywhere on these forums.
 
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