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'[EE]: ~ 3.3V from 7805'
2001\11\09@182147 by Chetan Bhargava

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Hello list,

Is it possible to achieve near about 3.3 volts out of a 7805 plus a couple
of IN4007 (1 volt drop) on the output pin?

Thanks

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2001\11\09@204953 by Byron A Jeff

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On Fri, Nov 09, 2001 at 02:59:15PM -0800, Chetan Bhargava wrote:
> Hello list,
>
> Is it possible to achieve near about 3.3 volts out of a 7805 plus a couple
> of IN4007 (1 volt drop) on the output pin?

It'll have very poor regulation characteristics dependant on the diodes,
temperature, and current draw.

A much better idea is to replace the 7805 with a LM317 and a couple of
precision resistors.

BAJ

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2001\11\10@024211 by Richard Sloan

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Unless you really need the 5V!

I just did exactly this on a project and for me it worked out fine.

My 3.3V is only at 50ma

R.

>>  On Fri, Nov 09, 2001 at 02:59:15PM -0800, Chetan Bhargava wrote:
>>  > Hello list,
>>  >
>>  > Is it possible to achieve near about 3.3 volts out of a 7805 plus a
>>  couple
>>  > of IN4007 (1 volt drop) on the output pin?

>>  It'll have very poor regulation characteristics dependant on the diodes,
>>  temperature, and current draw.

>>  A much better idea is to replace the 7805 with a LM317 and a couple of
>>  precision resistors.

>>  BAJ

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>>  (like ads or off topics) for you. See http://www.piclist.com/#topics








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2001\11\10@065941 by Russell McMahon

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> Is it possible to achieve near about 3.3 volts out of a 7805 plus a couple
> of IN4007 (1 volt drop) on the output pin?


A slightly better result than using diodes at not too much more cost is to
use a single transistor and 2 resistors.

BC337 or similar small TO92 NPN transistor
Collector to 5 volts
Emitter is the 3V3 output
1K2 resistor to base from 5 volts.
3K9 resistor from base to ground.

The two resistors provide about 3.8V to the transistor base.
The emitter is about 0.6v below this at about 3.2V.
Juggle the resistor values to provide different output voltages.
(eg 1K, 3K9 = about 3.4V)
The resistor values can be multiplied by 10 for small output currents (say
12K & 39K)

A transistor and 2 resistors doesn't cost much more than 2 diodes
Adding an output capacitor (10uF to 100 uF)  would be a very good idea in
either case.



regards



                Russell McMahon

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2001\11\10@085617 by Byron A Jeff

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On Sat, Nov 10, 2001 at 12:30:53AM -0500, Richard Sloan wrote:
> Unless you really need the 5V!
>
> I just did exactly this on a project and for me it worked out fine.
>
> My 3.3V is only at 50ma

While I agree it'll work OK with a small current range and constant temperature
I'd still have concerns about using a device that's not designed for voltage
regulation for regulation. That's just me I guess.

If you need dual voltages, simply leave the 7805 and feed the 5V into the
LM317.

But the original question was phased it's unclear as to whether or not the
5V is required.

But now rereading the question is also seems that an exact 3.3 wasn't
required. I'm now interested in what that 3.3V is going to be used for.

Personally I'd use the voltage regulator and be done with it. But I find that
I approach everything from a hobby perspective, so cost may be a significant
but undiscussed issue.

BAJ
{Quote hidden}

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2001\11\12@024716 by Vasile Surducan

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Well, if you don't want only 50mA why don't you use a simple 3.3V zenner ?
regards, Vasile



On Sat, 10 Nov 2001, Richard Sloan wrote:

{Quote hidden}

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2001\11\12@180107 by Chetan Bhargava

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I just need to power an EZ-USB chip that uses 3.3v. Would the zener approach
be ok?


----- Original Message -----
From: "Vasile Surducan" <EraseMEvasilespam_OUTspamTakeThisOuTL30.ITIM-CJ.RO>
To: <PICLISTspamspam_OUTMITVMA.MIT.EDU>
Sent: Sunday, November 11, 2001 11:42 PM
Subject: Re: [EE]: ~ 3.3V from 7805


{Quote hidden}

diodes,
> >  >>  temperature, and current draw.
> >
> >  >>  A much better idea is to replace the 7805 with a LM317 and a couple
of
> >  >>  precision resistors.
> >
> >  >>  BAJ
> >
> >  >>  --
> >  >>  http://www.piclist.com hint: The list server can filter out
subtopics
> >  >>  (like ads or off topics) for you. See
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2001\11\12@220531 by artstar

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Crude, but nevertheless OK in this application.

Adios,
LarZ

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-----Original Message-----
From: pic microcontroller discussion list
[spamBeGonePICLISTspamBeGonespamMITVMA.MIT.EDU] On Behalf Of Chetan Bhargava
Sent: Tuesday, 13 November 2001 10:00
To: TakeThisOuTPICLISTEraseMEspamspam_OUTMITVMA.MIT.EDU
Subject: Re: [EE]: ~ 3.3V from 7805

I just need to power an EZ-USB chip that uses 3.3v. Would the zener
approach
be ok?

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2001\11\13@044719 by Vasile Surducan

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what load current do you have and from what supply  ?
Example: in  5V, out 3.3V supply current 100mA.
Answer: yes  R = ( 5 - 3.3 )/ 0.1 = 17 ohm, choose R=15ohm/0.5W
dissipation on zenner without load : 0.3W choose a 1W zenner
don't forget 10uF(tantal) or 10uF (paper dielectric) + 100nF parallel with
zenner, probably you have to choose a zenner to have the right 3.3V value
( read data spec about absolute supply values for your chip )
Vasile

On Mon, 12 Nov 2001, Chetan Bhargava wrote:

> I just need to power an EZ-USB chip that uses 3.3v. Would the zener approach
> be ok?
>
>
> {Original Message removed}

2001\11\13@064038 by Byron A Jeff

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On Tue, Nov 13, 2001 at 11:26:53AM +0200, Vasile Surducan wrote:
> what load current do you have and from what supply  ?
> Example: in  5V, out 3.3V supply current 100mA.
> Answer: yes  R = ( 5 - 3.3 )/ 0.1 = 17 ohm, choose R=15ohm/0.5W
> dissipation on zenner without load : 0.3W choose a 1W zenner
> don't forget 10uF(tantal) or 10uF (paper dielectric) + 100nF parallel with
> zenner, probably you have to choose a zenner to have the right 3.3V value
> ( read data spec about absolute supply values for your chip )
> Vasile
>

I can see very little advantage of the zener over a true regulator. I read
the chip's specs, an it requires a +-9% regulated supply:

1) What's the max overshoot with the zener?
2) What's the ripple voltage?
3) What's the regulation tolerance?
4) What's the current draw in the quiescent state?

All of these items are listed in the LM317 data sheet, because it's a self
contained regulator. It consists of exactly one part more than the zener
solution and can make some guarantees.

In this instance I'm just wonder why the zener is a better choice. Even in
singles the price difference is only 14 cents US (0.50 vs. 0.36 on DigiKey).

I'm just wondering....

BAJ

> On Mon, 12 Nov 2001, Chetan Bhargava wrote:
>
> > I just need to power an EZ-USB chip that uses 3.3v. Would the zener approach
> > be ok?
> >
> >
> > {Original Message removed}

2001\11\13@073025 by Vasile Surducan

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On Tue, 13 Nov 2001, Byron A Jeff wrote:

> On Tue, Nov 13, 2001 at 11:26:53AM +0200, Vasile Surducan wrote:
> > what load current do you have and from what supply  ?
> > Example: in  5V, out 3.3V supply current 100mA.
> > Answer: yes  R = ( 5 - 3.3 )/ 0.1 = 17 ohm, choose R=15ohm/0.5W
> > dissipation on zenner without load : 0.3W choose a 1W zenner
> > don't forget 10uF(tantal) or 10uF (paper dielectric) + 100nF parallel with
> > zenner, probably you have to choose a zenner to have the right 3.3V value
> > ( read data spec about absolute supply values for your chip )
> > Vasile
> >
>
> I can see very little advantage of the zener over a true regulator. I read
> the chip's specs, an it requires a +-9% regulated supply:
>
> 1) What's the max overshoot with the zener?
 ???

> 2) What's the ripple voltage?

 does not matter, the input is stabilised +5V

> 3) What's the regulation tolerance?

 idem, you said +/- 9% , that means approx 3.0 to 3.6V, perfectly for a
3v3 zenner

> 4) What's the current draw in the quiescent state?

 almost all current flow through zenner, no problem.
>
> All of these items are listed in the LM317 data sheet, because it's a self
> contained regulator. It consists of exactly one part more than the zener
> solution and can make some guarantees.
>
guarantees same as a zenner...


> In this instance I'm just wonder why the zener is a better choice. Even in
> singles the price difference is only 14 cents US (0.50 vs. 0.36 on DigiKey).
>
 no one say that, but is possible...and there was someone who said it
isn't. Sometime classic devices are still working. That's all.

> I'm just wondering....

 fell free...

regards, Vasile
{Quote hidden}

> > > {Original Message removed}

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