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'[EE] Re: 5v and 3.3v power supply'
2005\08\27@105940 by Vasile Surducan

face picon face
Hi Mario,

Let me tell why no one answered you.

1. Most of the people from the list are considering this subject too easy.
2. The question was incomplete, you must tell the power or the current
you need from both supplies

However, I bet there are no more than 30 people from this 1500 people
list which knows how to dimension corect and complete a linear 5V and
3.3V power supply starting from input/output requirements. (
dimensioning and computing the transformer, rectifying bridge,
filtering capacitor, heatsinks etc.)

For your project I suggest a LM7805 and a BA033T. Both have more than
5 producer clones and various packages for 100mA, 500mA  1A and 3A.
They could be chained or not (powering the 3.3V regulator from the 5V).

cheers,
Vasile

On 8/26/05, Mario Mendes Jr. <spam_OUTmarioTakeThisOuTspammmendes.com> wrote:
{Quote hidden}

> --

2005\08\27@171906 by Mario Mendes Jr.

flavicon
face
Yes, my question was a bit unclear.  Here's the deal:

I have one of those breadboards with the power supply built into it.  It
has 2 variable outputs and a ground output.  The first variable is 0 to
+15 and the other is 0 to -15.  I currently have it setup so that it
feeds my 5v chips (pic and other logic), but needed to supply 3.3v to a
new chip I want to use along with the others, and for the prototyping
(now), I wanted to get 3.3v from the 5v output in the breadboard and
wanted to know if anyone had something similar setup, whether it was a
voltaget regulator chip or a simple zener.

Anyway, I did some searching on the internet last night and found the
National LM1086 in a TO-220 package that can supply up to 1.5A (with
heatsink), way more than I need at the moment.  The neat part of this
regulator is that it comes in both variable and fixed flavors and all
sorts of other commonly used voltages including negatives too.  Here are
some examples:

LM1086-5.0 - positive regulator LDO 1.5A 5V Current Limiting & Thermal
Protection TO-220
LM1086-ADJ - positive regulator LDO 1.5A Adjustable Current Limiting &
Thermal Protection TO-220
LM1086-3.3 - positive regulator LDA 1.5A 3.3V Current Limiting & Thermal
Portection TO-220

I ended up purchasing a few of each as they're small enough to use later
in the final board at US$2.11 each.  They're cheaper at larger
quantities but I justed needed a handful of each to keep in trays for
whenever I need one.


Thanks.


-Mario

{Original Message removed}

2005\08\27@174848 by Mark Rages

face picon face
On 8/27/05, Mario Mendes Jr. <.....marioKILLspamspam@spam@mmendes.com> wrote:
> Yes, my question was a bit unclear.  Here's the deal:
>
> I have one of those breadboards with the power supply built into it.  It
> has 2 variable outputs and a ground output.  The first variable is 0 to
> +15 and the other is 0 to -15.  I currently have it setup so that it
> feeds my 5v chips (pic and other logic), but needed to supply 3.3v to a
> new chip I want to use along with the others, and for the prototyping
> (now), I wanted to get 3.3v from the 5v output in the breadboard and
> wanted to know if anyone had something similar setup, whether it was a
> voltaget regulator chip or a simple zener.
>

What current do you need at 3.3V?  You can use a couple 1% resistors
to generate the voltage from the 5V supply.  Buffer with an op-amp for
higher current.

Also, check the datasheets to see if the other chips will run on 3.3V.
You might only need one supply.

Regards,
Mark
markrages@gmail
--
You think that it is a secret, but it never has been one.
 - fortune cookie

2005\08\27@182112 by Denny Esterline

picon face
> Yes, my question was a bit unclear.  Here's the deal:
>
> I have one of those breadboards with the power supply built into it.  It
> has 2 variable outputs and a ground output.  The first variable is 0 to
> +15 and the other is 0 to -15.  I currently have it setup so that it
> feeds my 5v chips (pic and other logic), but needed to supply 3.3v to a
> new chip I want to use along with the others, and for the prototyping
> (now), I wanted to get 3.3v from the 5v output in the breadboard and
> wanted to know if anyone had something similar setup, whether it was a
> voltaget regulator chip or a simple zener.
>

I can give you one better. Adjust your negative supply to 3.3 volts, adjust
your positive suppy to 1.7 volts. Now use the V- output as ground (0v), the
original common as +3.3v and the positive supply as 5v.

No parts requred - must be an ideal solution :-)

-Denny

2005\08\27@211932 by Mario Mendes Jr.

flavicon
face
See?!!  That is exactly why I love this group.

Thanks.

-Mario

-----Original Message-----
From: piclist-bouncesspamKILLspammit.edu [.....piclist-bouncesKILLspamspam.....mit.edu] On Behalf
Of Denny Esterline
Sent: Saturday, August 27, 2005 6:21 PM
To: Microcontroller discussion list - Public.
Subject: Re: [EE] Re: 5v and 3.3v power supply


{Quote hidden}

I can give you one better. Adjust your negative supply to 3.3 volts,
adjust your positive suppy to 1.7 volts. Now use the V- output as ground
(0v), the original common as +3.3v and the positive supply as 5v.

No parts requred - must be an ideal solution :-)

-Denny

2005\08\28@033007 by Vasile Surducan

face picon face
Much better, but still an incomplete question, which will be the
estimated load current?
This is a repetitive question, noticed?

Usualy a standard 7805 is about $0.1. Also 5V-3.3V=1.7V This don't
need either an LDO but a cheap 3.3V regulator. What you need to
dimesion carefully is the power dissipated by every regulator. This
will show you if you need or not aditional heatsinks on 5V or on 3.3V
regulators. For a low current load even a 3.3v zenner should be enough
as someone else said.

An LDO is working with very low input-output voltage. The voltage drop
is current dependant. For a low current even a standard regulator
works as an LDO. For example LM317T (dimensioned at 1A nominal
current, 1.5A short circuit current) needs between 1.5 and 2V in-out
for a 500mA load, half that it required for 1A load.
So you may guess it's a "pseudo" low drop device.

The 5V regulator must be supplied in the worst circumstances with a
voltage equal with 5V+ regulator voltage drop at the highest load
current. This means if you have a nominal mains voltage of 125V, the
regulator will must run ok even when the mains becomes 125V-15% (check
the standardised requirements for mains in your country). So what you
need to measure at the regulator input is the bottom of the ripple
voltage and not the efective DC voltage. A DVM will be unusefull for a
correct dimension of the input voltage.

cheers,
Vasile


On 8/28/05, Mario Mendes Jr. <EraseMEmariospam_OUTspamTakeThisOuTmmendes.com> wrote:
{Quote hidden}

> {Original Message removed}

2005\08\28@130425 by Mario Mendes Jr.

flavicon
face
I see what you're saying, but at this point I don't know the answer to
that question as I am still designing the thing and don't quite know
what the final 3.3v will end up looking like.

What I can tell you is that the current 5v supply can provide as much as
5A on the positive side and another 5A on the negative side.

I need to make 3.3v out of 5v to feed to an 8051 core chip.  This is the
first time I'll be playing with an 8051 chip, so initially it will only
have a few leds connected to it, but eventually I'll end up connecting
LCDs, keypads, etc.

So let's call it 3.3v at 1A max for now.

Thanks.

-Mario

{Original Message removed}

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