Power Circuit Problem

Started by Scott Lingerfelt June 25, 2004

Anyone,

I just created a PCB which I have put the following components for my 5V
power circuit. I do not get 5V out I am getting around 4.2V. Any
suggestions? I do not have anything connected but a ULN2803A driver.
Should I upsize my 10uF cap? Do I need to load the circuit.

Wall jack connector
diode
10uF cap
5V regulator
.1uF cap

----Diode-----Regulator-------------VCC
| | |
C | C
Wall Jack A 10uF | A .1uF Output
P | P
| | |
| | |
GND

Thanks in advance, Scott
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I do not see any mention of input voltage but because the rectifier will only be a half wave rectifier the input voltage should not be too low.  I would say 9v at least and 12 would be better.  The more current that is drawn the more clipping would result and as I remember the 2803 is an output driver so check what current you draw. More than a couple of tenths of milli’s and rather use a full wave bridge.  Preferably there should be a 100u cap before and after the regulator.

 

To sum it up the input voltage and current drawn is the limiting factors to the clipping that you’re seeing that produces a saw wave and is not too healthy for electronic circuitry.

 

Regards

Martin

 

From: Scott Lingerfelt [mailto:s...@covcable.com]
Sent: Saturday, 26 June 2004 9:22
To: Piclist
Subject: [piclist] Power Circuit Problem

 


      Anyone,

I just created a PCB which I have put the following components for my 5V
power circuit.  I do not get 5V out I am getting around 4.2V. Any
suggestions?  I do not have anything connected but a ULN2803A driver.
Should I upsize my 10uF cap?  Do I need to load the circuit.

      Wall jack connector
      diode
      10uF cap
      5V regulator
      .1uF cap

           ----Diode-----Regulator-------------VCC
                      |           |      |
                      C           |      C
  Wall Jack                A 10uF |      A .1uF        Output
                      P           |      P
                      |           |      |
                      |           |      |
           GND

      Thanks in advance, Scott
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In a message dated 6/25/2004 6:02:33 PM Eastern Daylight Time, s...@covcable.com writes:


I just created a PCB which I have put the following components for my 5V
power circuit.  I do not get 5V out I am getting around 4.2V. Any
suggestions?  I do not have anything connected but a ULN2803A driver.
Should I upsize my 10uF cap?  Do I need to load the circuit.



Why the diode - that will cost you .7VDC.  Take it out - you dont need it.

Sid


I assume you are using a 7805 regulator? You must not have any lower
than 8 volts to the input to the regulator in order to keep enough
headroom for the regulator to operate properly and stable. Your wall
transformer may or may not have a filtering capacitor. If it doesn't,
the 10uf is not nearly enough capacitance to keep the ripple from
dipping less than 8 volts on the dip. Especially if the diode in the
transformer is only a half wave, which they usually are. If you are sure
it is a DC output, you don't need the extra diode. If you are preventing
someone from using an AC output transformer, you are wise to leave it in
there. If you are using a 7805, you need a 0.1uf cap on the input of the
regulator too. No need for one on the output. I can't tell by your
schematic because the text may have jumped on the line showing your
values.

Check out my website for some hints on designing simple regulator
circuits. It's at:
http://www.pic101.com/why_is_my_regulator_getting_hot.htm

Rick

Scott Lingerfelt wrote:

>
> Anyone,
>
> I just created a PCB which I have put the following components for my
> 5V
> power circuit. I do not get 5V out I am getting around 4.2V. Any
> suggestions? I do not have anything connected but a ULN2803A driver.
> Should I upsize my 10uF cap? Do I need to load the circuit.
>
> Wall jack connector
> diode
> 10uF cap
> 5V regulator
> .1uF cap
>
> ----Diode-----Regulator-------------VCC
> | | |
> C | C
> Wall Jack A 10uF | A .1uF Output
> P | P
> | | |
> | | |
> GND
>
> Thanks in advance, Scott
> ---
> Outgoing mail is certified Virus Free.
> Checked by AVG anti-virus system (http://www.grisoft.com).
> Version: 6.0.707 / Virus Database: 463 - Release Date: 6/15/2004
>




I'm sure the diode is for reverse voltage protection, so probably a good idea.  However, that will cost you around .7v
 
The wall power supply need to supply at least 9v under load for this to be able to deliver 5v.
-----Original Message-----
From: Scott Lingerfelt [mailto:s...@covcable.com]
Sent: Friday, June 25, 2004 4:22 PM
To: Piclist
Subject: [piclist] Power Circuit Problem


      Anyone,

I just created a PCB which I have put the following components for my 5V
power circuit.  I do not get 5V out I am getting around 4.2V. Any
suggestions?  I do not have anything connected but a ULN2803A driver.
Should I upsize my 10uF cap?  Do I need to load the circuit.

      Wall jack connector
      diode
      10uF cap
      5V regulator
      .1uF cap

           ----Diode-----Regulator-------------VCC
                      |           |      |
                      C           |      C
  Wall Jack                A 10uF |      A .1uF        Output
                      P           |      P
                      |           |      |
                      |           |      |
           GND

      Thanks in advance, Scott
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
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So far I think I've seen 3 different "min input voltage" claims.
None of which jibe with the 7805 datasheet that claims 2V drop out at
1A load. So, according to the datasheet (always a good place to
look), you need 7 Volts input. However, at 500 mA output load, the
dropout is about 1.5V and at 0 mA load, the dropout is about 1 V.

Phil



The datasheet specifies the "minumim" voltage required. Are you sure you
want to design your circuit so it runs at your minimum? Bad idea. The
different "claims" are legitimate depending upon ones parameters in
their own design. Running at the mimimum doesn't allow for dips or sags
in the line voltage and load variations, and as long as your circuit can
function with the 5 volt supply dropping to the 4 volt region, go for
it.
Rick

Phil wrote:

> So far I think I've seen 3 different "min input voltage" claims.
> None of which jibe with the 7805 datasheet that claims 2V drop out at
> 1A load. So, according to the datasheet (always a good place to
> look), you need 7 Volts input. However, at 500 mA output load, the
> dropout is about 1.5V and at 0 mA load, the dropout is about 1 V.
>
> Phil





Check your voltmeter.
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This is a somewhat theoretical exercise in that I'm not designing
anything here. But if I was building a 1A supply then I probably
wouldn't use a 7805 since its at the limit of the specs and would
take a pretty good heatsink to keep it within temperature limits. If
it was designed for a 500 mA load, then 7V would be getting close to
a reasonable over-margin. From the sounds of the original post, the
load is probably less than 100 mA. One should also be aware that
extra voltage on the 78xx series is disipated as heat though a couple
of volts extra won't fry an egg.

To your general point, yes, I agree one should not engineer to the
edge of the specifications.

My point was not that others are wrong but rather one should read the
data sheet and understand their design requirements. Saying things
like "needs X volts" without understanding design requirements is a
bit hasty, IMO.

--- In , rixy04 <rixy04@v...> wrote:
> The datasheet specifies the "minumim" voltage required. Are you
sure you
> want to design your circuit so it runs at your minimum? Bad idea.
The
> different "claims" are legitimate depending upon ones parameters in
> their own design. Running at the mimimum doesn't allow for dips or
sags
> in the line voltage and load variations, and as long as your
circuit can
> function with the 5 volt supply dropping to the 4 volt region, go
for
> it.
> Rick
>
> Phil wrote:
>
> > So far I think I've seen 3 different "min input voltage" claims.
> > None of which jibe with the 7805 datasheet that claims 2V drop
out at
> > 1A load. So, according to the datasheet (always a good place to
> > look), you need 7 Volts input. However, at 500 mA output load,
the
> > dropout is about 1.5V and at 0 mA load, the dropout is about 1 V.
> >
> > Phil
> >
> >