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'[EE]:Question about L7805 Voltage regulator'
2002\11\11@093129 by Luberth Dijkman

picon face
Hello,

Could someone tell me if it is allowed to use +24VDC as input for L7805
Voltage regulator

Some kits say input voltage 8 to 15 Volt
What i can make up from the L7800.pdf maximum for L7805 = 35V
or am i wrong
can i use 24VDC as input???


Luberth
Dutch is my mothers language ;-)
Http://luberth.com

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2002\11\11@094133 by Sid Weaver

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In a message dated 11/11/2002 09:31:53 Eastern Standard Time,
.....LuberthKILLspamspam@spam@GEOCITIES.COM writes:


> Could someone tell me if it is allowed to use +24VDC as input for L7805
> Voltage regulator
>
>

Luberth, the 7805 will take 24VDC.  It might heat up a little and you may
have to use a heat sink.

Sid

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2002\11\11@094340 by Olin Lathrop

face picon face
> Could someone tell me if it is allowed to use +24VDC as input for L7805
> Voltage regulator

Read the spec sheet.


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2002\11\11@103600 by 4HAZ

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----- From: "Luberth Dijkman" <Luberth@

> Hello,
>
> Could someone tell me if it is allowed to use +24VDC as input for L7805
> Voltage regulator
>
> Some kits say input voltage 8 to 15 Volt
> What i can make up from the L7800.pdf maximum for L7805 = 35V
> or am i wrong
> can i use 24VDC as input???

That depends on the total heat you can dissipate vs. generated heat.
If you are trying for 50ma. for example:
24v - 5v = 19v across the regulator
19v * .05a = .95watts
The little 78L05 in the TO-92 package is rated 175 degrees Celsius per watt  Junction-to-Ambient so that puts the junction temp 175 * .95 or 166.25 degrees C above the ambient temp (also in Celsius), this is obviously too much as the max junction temp is specked at 125c.
However same circuit with a 7805 in a TO-220 package rated 65c/W J-A works because
65 * .95 puts the junction temp 61.75 above the ambient so little or no heatsink required.
The Junction-to-Case for the 7805 in the TO-220 package is 5c/W add the heatsinks Sink-to-Ambient and you can calculate how hot the junction temp will be with a heatsink.

Lonnie - KF4HAZ -

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2002\11\11@104219 by Tony Harris

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Yes, you can feed it 24V (per the spec sheets max) - I am using one right
now tied off a 24 volt @ CT transformer (need a pic for certain controls)
and it works beautifully, but it does get very warm / borderline hot - so if
you are feeding it that much be sure to have a heatsink.

-Tony
{Original Message removed}

2002\11\11@105054 by Spehro Pefhany

picon face
At 09:46 AM 11/11/02 -0600, you wrote:
>Yes, you can feed it 24V (per the spec sheets max) - I am using one right
>now tied off a 24 volt @ CT transformer (need a pic for certain controls)
>and it works beautifully, but it does get very warm / borderline hot - so if
>you are feeding it that much be sure to have a heatsink.

I presume you are using the CT to give you around 17VDC rather than a FW
rectifier; 24VAC FW rectified and filtered is not an acceptable input
for a 7805 (too close to the 35VDC Abs. max input voltage), regardless of
power dissipation considerations.

Best regards,

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2002\11\11@112836 by Tony Harris

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I should have asked for more specification from the guy in terms of his
current requirements - I really didn't think about it at the time.  with a
24 volt input, without any heatsink considerations (see below) the guy who
originally asked would be limited to a max of 100mA (1.9 watts *65 deg C =
123.5 - not much breathing room).

But, in MY situation....

After filtering, rectification, and everything, the actual feed to it is
~22VDC

Using ~100mA, I'm only at ~110 Deg Cel (1.7 watts * 65 Deg C) - which is
below the max operating temp of the 7805 (125 deg cel), although I did add a
heatsink to it as I was concerned regardless, especially if my design
changed at all that required any more current, or a variety of other
possibilities while it is in use.  I am using the TO220 package.

But, there is one thing I don't understand and am hoping that perhaps you,
or Lonnie might be able to answer...

By adding a heatsink, how does that change the base formula for calculating
temps?   It's one of those things that I've always added one when it seemed
too hot, but I'd really like to have a better understanding of  how it will
change my formulas - especially adjusting max current and such.  Most of my
projects on that front have been low current projects, but I do have one
coming up that from everything I want, it looks like I'll be up at 500mA,
which could cause a serious problem if I don't have the right numbers to
make it work .  Any help on that front (in terms of how adding a heat sink
affects the numbers) would be most appreciated.

It's one of those things - I know it needs to be done, but I don't know how
to factor for it :).

-Tony

{Original Message removed}

2002\11\11@114658 by Olin Lathrop

face picon face
> By adding a heatsink, how does that change the base formula for
calculating
> temps?

A heatsink changes (hopefully lowers) the device's thermal resistance to
the ambient temperature.  The device temperature above ambient is the
power being dissipated times the thermal resistance to ambient.

For example, let's assume a bare device had a thermal reistance of
100deg/watt, is dissipating 800mW, and ambient temperature is 20deg.  The
device will be 80deg warmer than ambient, which puts it a 100deg.  If a
heatsink changes the thermal resistance to 30deg/watt, then the device
temperature will be 800mW * 30deg/W = 24deg.  Its temperature will be
ambient + rise = 20deg + 24deg = 48deg.

The orientation of heatsink fins can have big effect on its thermal
resistance because hot air will try to rise.  Therefore vertically
oriented fins will increase airflow due to heating of the air by the
heatsink.  Forced air cooling (a fan) has a major effect on overall
thermal resistance.  See a heatsink datasheet.


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2002\11\11@120328 by Alan B. Pearce

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>But, there is one thing I don't understand and am
>hoping that perhaps you, or Lonnie might be able to answer...

Think of each stage of getting the heat from the chip to the air as a
"thermal resistance". This is just like a normal resistor is an "electrical
resistance".

In the case of an electrical resistance it is quoted as "ohms" which is
calculated as volts/amp.

Now thermal resistance is calculated as degrees C/watt. You will find this
figure in the data sheet, often using the symbol Rth, given for several
possible cases.

There will be Rth to the air (Rth junction-air). There will be Rth to the
case (Rth junction-case), which will be a much lower figure. In each case
this is the total thermal resistance from the chip junction generating the
heat to the item mentioned. The difference between the two figures is the
thermal resistance of the case to the air. It is this figure that you can
vary by adding a heatsink. All manufacturers and/or distributors (AFAIK)sell
heatsinks, showing a thermal resistance in the catalogue against each
heatsink type. You can add this thermal resistance to the junction to case
thermal resistance that the manufacturer gives for the chip. You can then
compare this to the junction to air thermal resistance that they give for
the chip. If the ratio of the figures is 2:1, then you will be able to run
twice as much current through the chip, to reach the same temperature.


You can work everything out without needing a thermometer (read wet finger)
and a running system, if you know the current draw you expect. Note that the
125 C rating is the temperature of the chip inside the case, not the
temperature of the case. The temperature of the case will need to be less
than 125 C by a value that you can calculate using the Rth(junction-case)
given by the chip manufacturer. In practice it is not good form to run the
chip at anything like 125C.

I would be looking to keep it below 100C at worst case ambient temperature
(probably 50-60C ambient), which will mean that for most things you will
need a heatsink. In practice this will typically mean you can keep your
finger on the chip at room temperature without getting a burn on your skin.

However do be careful, as many on this list (as well as many others) will
testify, there have been many occasions when things are that hot, they have
left their fingerprint on the chip ;)))))

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2002\11\11@121341 by Tony Harris

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>
> For example, let's assume a bare device had a thermal reistance of
> 100deg/watt, is dissipating 800mW, and ambient temperature is 20deg.  The
> device will be 80deg warmer than ambient, which puts it a 100deg.  If a
> heatsink changes the thermal resistance to 30deg/watt, then the device
> temperature will be 800mW * 30deg/W = 24deg.  Its temperature will be
> ambient + rise = 20deg + 24deg = 48deg.
>

Ok, so let me make sure I understand this.

To use as an example...

http://www.digikey.com/scripts/us/dksus.dll?Detail?Ref=73546&Row=159293

it's Thermal resistivity is 15.6deg/W, heat dissipation 6W (I assume max?)

Does this mean that if I use this particular heat sink that it will drop my
calcs from mA * 65deg/watt (unsinked 7805  TO220) to  mA*15.6deg/Watt ?

-Tony

ps - there is a fan too, but I want to figure base numbers should the fan
fail.

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2002\11\11@121539 by fred jones

picon face
Hi,
I am putting 24volts in to a 7905 to get a -5volts out.  The 7905 does get
warm with a large heat sink...no it gets hot.  Is there something like the
Black regulator that I could use to cool things down?  Keeping ripple to a
minimum is very important, thats why I used a linear regulator.  I've always
had a difficult time reducing ripple.  Any help is appreciated.
Thanks,
Fred

>Luberth, the 7805 will take 24VDC.  It might heat up a little and you may
>have to use a heat sink.
>
>Sid


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2002\11\11@123243 by Olin Lathrop

face picon face
> it's Thermal resistivity is 15.6deg/W, heat dissipation 6W (I assume
max?)
>
> Does this mean that if I use this particular heat sink that it will drop
my
> calcs from mA * 65deg/watt (unsinked 7805  TO220) to  mA*15.6deg/Watt ?

I didn't follow the link, but it sounds like the 15.6deg/W figure is for
the bare heat sink.  You need to add the junction to case thermal
resistance to that to get the overall junction to ambient thermal
resistance.


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2002\11\11@123900 by Olin Lathrop

face picon face
> I am putting 24volts in to a 7905 to get a -5volts out.  The 7905 does
get
> warm with a large heat sink...no it gets hot.  Is there something like
the
> Black regulator that I could use to cool things down?  Keeping ripple to
a
> minimum is very important, thats why I used a linear regulator.  I've
always
> had a difficult time reducing ripple.  Any help is appreciated.

Try a buck regulator that produces an efficient -7 or -8 volts, then let
the 7905 drop that to a clean -5.

How clean do you really need it?  It shouldn't be hard to end up with less
than 200mV p-p ripple on the 5V output of a bare buck regulator.  That
should be no problem powering logic circuits or a PIC.  You could also try
a series inductor followed by a capacitor to ground after the buck
regulator output.  That should greatly attenuate the ripple at a small
increase in supply impedence.


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2002\11\11@124914 by Tim McDonough

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On Mon, 11 Nov 2002 11:15:33 -0600, fred jones wrote:
>Hi, I am putting 24volts in to a 7905 to get a -5volts out.  The
>7905 does get warm with a large heat sink...no it gets hot.  Is
>there something like the Black regulator that I could use to cool
>things down?  Keeping ripple to a minimum is very important, thats
>why I used a linear regulator.  I've always had a difficult time
>reducing ripple.  Any help is appreciated.

You don't say what the current is but you're dropping 19 volts so heat loss will add up in a big hurry. You might try adding a dropping resistor in series with the input of the regulator or using another regulator ahead of the 7905 to share the load. Maybe drop it to -12 volts first.

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2002\11\11@125546 by Tony Harris

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Ahhhhhhh - ok - so I add the junction to case 5degC/Watt to the
15.6degC/Watt of the heatsink to get a total of 20.6degC/Watt total...  Am I
on the right track?

So if I were to throw that heatsing on my 7805, I could move up to 250mA and
only have a temp of 87.55degC + ambient? and assuming my ambient temp is
~21degC, my total temp would be at ~108DegC, which would be less then the
max of 125DegC....(assuming total volts to 7805 is 22Vdc)....

Do I have my calculations correct?

-Tony


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2002\11\11@142251 by Peter L. Peres

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On Mon, 11 Nov 2002, Luberth Dijkman wrote:

*>Hello,
*>
*>Could someone tell me if it is allowed to use +24VDC as input for L7805
*>Voltage regulator
*>
*>Some kits say input voltage 8 to 15 Volt
*>What i can make up from the L7800.pdf maximum for L7805 = 35V
*>or am i wrong
*>can i use 24VDC as input???

You can use 24V input but watch out for dissipated power.

Peter

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2002\11\11@142301 by Peter L. Peres

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On Mon, 11 Nov 2002, fred jones wrote:

*>Hi,
*>I am putting 24volts in to a 7905 to get a -5volts out.  The 7905 does get
*>warm with a large heat sink...no it gets hot.  Is there something like the
*>Black regulator that I could use to cool things down?  Keeping ripple to a
*>minimum is very important, thats why I used a linear regulator.  I've always
*>had a difficult time reducing ripple.  Any help is appreciated.

Add a 7912 regulator before the 7905. This will reduce the ripple and
spread the heat. You can put it on the same heatsink or on another (beware
the tab on 79xx is not ground).

Peter

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2002\11\11@142907 by Olin Lathrop

face picon face
> Ahhhhhhh - ok - so I add the junction to case 5degC/Watt to the
> 15.6degC/Watt of the heatsink to get a total of 20.6degC/Watt total...
Am I
> on the right track?
>
> So if I were to throw that heatsing on my 7805, I could move up to 250mA
and
> only have a temp of 87.55degC + ambient? and assuming my ambient temp is
> ~21degC, my total temp would be at ~108DegC, which would be less then
the
> max of 125DegC....(assuming total volts to 7805 is 22Vdc)....
>
> Do I have my calculations correct?

I didn't check your arithmetic, but it seems you understand the concept.


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2002\11\11@164445 by Russell McMahon

face
flavicon
face
> I am putting 24volts in to a 7905 to get a -5volts out.  The 7905 does get
> warm with a large heat sink...no it gets hot.  Is there something like the
> Black regulator that I could use to cool things down?  Keeping ripple to a
> minimum is very important, thats why I used a linear regulator.  I've
always
> had a difficult time reducing ripple.  Any help is appreciated.

You could use something EXACTLY lime the Black Regulator as a pre-regulator
provided it meets your power requirements.

For a negative input and output voltage you need to "invert" every active
component.
Replace any NPN with a PNP, and PNP with an NPN, reverse the "direction of
all diodes and zeners, feed input negative below ground and expect negative
output below ground (ground connection stays the same in old and new
versions)  and voila - a negative input & output black regulator. Using this
to feed you existing 7905 with about 8 volts will allow it to operate at
much lower temperature while still providing the good regulation of a linear
regulator.

As you are dissipating highish power now you are probably working at the
better part of 1 amp load. If so you will need to design the black
pre-regulator accordingly. Suitable output transistor and an inductor that
doesn't saturate at the required current levels are probably the main
requirements.

Roman may be prepared to suggest component values to get you started.



       Russell McMahon

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2002\11\11@165442 by Russell McMahon

face
flavicon
face
> Add a 7912 regulator before the 7905. This will reduce the ripple and
> spread the heat. You can put it on the same heatsink or on another (beware
> the tab on 79xx is not ground).

It will spread the heat to another component (the 7912) BUT if they are on
an identical heatsink as before the heatsink temperature will be the same as
before. This may be acceptable but be aware of this.

If the input voltage is always about the same (24 volts) then a series
resistor is probably the simplest solution (but not the most elegant).

eg for

I mA load,
Vo output volts,
Vin supply input volts,
Vhr regulator "headroom" (minimum allowable drop across regulator) .
Vri regulator  input volts ( = Vin-Vo-Vrh)

Leave the regulator a little more headroom than the databook says and ALWAYS
design for worst case load. AFAIR the 7905 needs about 2v headroom at full
load.

For a linear (non switching regulator)

   Power total = Vin * I
   Power in load = Vo * I
   Power lost in regulator)(s) = (Vin - Vo) * I

R = (Vin - Vo - Vrh) * 1000 / I

Power in R is (I/1000)^2 * R

For example for 24 volts in, 5 volts out, 3 volts regulator headroom, 500 mA
MAX load current.

R = (24 - 5 - 3) * 1000 / 500 = 32 ohms
Power in R = 0.5^2 * 32 = 8 watts

Resistor should be substantially higher powered than power dissipated.
Say in this case 2 x 10 watt resistors.
Note that at 500 mA total power = 24 x 0.500 = 12 watts so there is quite
some heat to dissipate.
A switching preregulator looks quite attractive unless you have power to
burn and don't mind the heat.

If you told us what the application was and why you had to use 24v it may
assist with a better solution.


       Russell McMahon

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2002\11\11@221015 by Luberth Dijkman

picon face
>>If you told us what the application was and why you had to use 24v it may
>>assist with a better solution.

yee what have i started a lot off messages on my question
i work as a mechanic in a large bread factory
there is a bread slicing/packing machine
a PLC has 5 outputs connected to a display on the other side off machine
in case of an error (door open, safety switch etcetera) the PLC ouputs a binary
value
the display connects a message to this valeu so the operator nows whats wrong

the display is more broken then that its working and repair is expensive

so i thougt i use a pic 16f84 with 4bit LCD connected to portb  and connect the
5 outputs from plc to porta (optocouplers)
there are 30 error messages and i think the pic16f84 can handle this
most machinery and PLC's i have seen and also this machine use 24VDC
so i wanted to use the 24VDC and power the pic/lcd(16x2) with it via L7805

i was not sure if i could use +24VDC for L7805

Dutch is my language => Sorry if i wrote something stupid

Luberth




Russell McMahon wrote:

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2002\11\11@231741 by fred jones

picon face
Thanks to all of you for so many good suggestions.  I appreciate everyone
taking the trouble to answer.  What I'm doing is building a metal detector.
The 24volts is being supplied by 2 12volt gel cells in series.  The 24 volts
is used to pulse the coil (the more, the better the depth) this is used to
supply a 7905 for -5v to the negative supply on the analog components such
as op amps.  The negative supply is then stepped up with a couple of charge
pumps to provide the positive 5v supply for powering the PIC16f877 as well
as the analog electronics such as the positive rail for the op amps.  I had
heck trying to get a clean level out of the charge pumps.  That is why I'm a
little gun shy on another switcher.  The heat coming off of the 7905 may
make it worth the effort though, let alone the wasted energy.
Thanks again,
Fred






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2002\11\12@014455 by Roman Black

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fred jones wrote:
>
> Thanks to all of you for so many good suggestions.  I appreciate everyone
> taking the trouble to answer.  What I'm doing is building a metal detector.
> The 24volts is being supplied by 2 12volt gel cells in series.  The 24 volts
> is used to pulse the coil (the more, the better the depth) this is used to
> supply a 7905 for -5v to the negative supply on the analog components such
> as op amps.  The negative supply is then stepped up with a couple of charge
> pumps to provide the positive 5v supply for powering the PIC16f877 as well
> as the analog electronics such as the positive rail for the op amps.  I had
> heck trying to get a clean level out of the charge pumps.  That is why I'm a
> little gun shy on another switcher.  The heat coming off of the 7905 may
> make it worth the effort though, let alone the wasted energy.


Maybe a re-design is best? If using dual rail
opamps why not put the Gnd at mid point between the
2 batteries, then generate the +5v and -5v with a
7805 and 7905, easier and less heat for the regulators.
Your only problem is to drive the 24v "pulser"
transistor but that should be easily fixed.
-Roman

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2002\11\12@015345 by Roman Black

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Luberth Dijkman wrote:

> there is a bread slicing/packing machine
> a PLC has 5 outputs connected to a display on the other side off machine
> in case of an error (door open, safety switch etcetera) the PLC ouputs a binary
> value
> the display connects a message to this valeu so the operator nows whats wrong
>
> the display is more broken then that its working and repair is expensive
>
> so i thougt i use a pic 16f84 with 4bit LCD connected to portb  and connect the
> 5 outputs from plc to porta (optocouplers)
> there are 30 error messages and i think the pic16f84 can handle this
> most machinery and PLC's i have seen and also this machine use 24VDC
> so i wanted to use the 24VDC and power the pic/lcd(16x2) with it via L7805

The PIC and LCD only draw 50mA or less, even
with LCD backlight so a 7805 should work with
a small heatsink since you want reliability.
You could put a 12v 3W zener in series with the
7805, so the zener drops the input voltage to
12v and the 7805 runs nice and cool (no heatsink).
Put a 47uF cap across the zener too.

Alternatively if you want to fiddle with a couple
of transistors you can use my Black regulator
which will be quite efficient for 24v -> 5v use
and won't need the 7805 regulator. See;
http://www.romanblack.com/smps.htm
-Roman

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2002\11\12@022652 by Roman Black

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Russell McMahon wrote:
{Quote hidden}

Sure, given the data for max load current and
expected input voltage range. :o)
-Roman

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2002\11\12@024612 by Roman Black

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Tony Harris wrote:
>
> Ahhhhhhh - ok - so I add the junction to case 5degC/Watt to the
> 15.6degC/Watt of the heatsink to get a total of 20.6degC/Watt total...  Am I
> on the right track?
>
> So if I were to throw that heatsing on my 7805, I could move up to 250mA and
> only have a temp of 87.55degC + ambient? and assuming my ambient temp is
> ~21degC, my total temp would be at ~108DegC, which would be less then the
> max of 125DegC....(assuming total volts to 7805 is 22Vdc)....
>
> Do I have my calculations correct?


Hi Tony, your calculations sound ok but your
design philosophy might need revising. ;o)
Any soldered component run over about 50'C
(that's about 25'C over ambient) REALLY reduces
reliability. The PCB will darken, the solder
will be exposed to expansion and heat stresses
and it will soon get dry joints (disaster on
a voltage regulator!)
-Roman

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2002\11\12@042000 by Alan B. Pearce

face picon face
> there is a bread slicing/packing machine
> a PLC has 5 outputs connected to a display on the other side off machine
> in case of an error (door open, safety switch etcetera) the PLC ouputs a
binary
> value
> the display connects a message to this valeu so the operator nows whats
wrong
>
> the display is more broken then that its working and repair is expensive
>
> so i thougt i use a pic 16f84 with 4bit LCD connected to portb  and
connect the
> 5 outputs from plc to porta (optocouplers)

If the original display is LED, then I would suggest using LED's again
rather then LCD. There is also the advantage that LED is easier to see in a
factory type environment.

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2002\11\12@074314 by Tony Harris

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Well, I did find some mongo heat sinks that would be at ~5degC/Watt, which
would bring it into the range you are talking about - I'll have to look into
those more.

-Tony
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2002\11\12@142942 by Peter L. Peres

picon face
On Tue, 12 Nov 2002, Russell McMahon wrote:

*>> Add a 7912 regulator before the 7905. This will reduce the ripple and
*>> spread the heat. You can put it on the same heatsink or on another (beware
*>> the tab on 79xx is not ground).
*>
*>It will spread the heat to another component (the 7912) BUT if they are on
*>an identical heatsink as before the heatsink temperature will be the same as
*>before. This may be acceptable but be aware of this.

Yes but the temperature inside each device won't be as before. He said it
works but it's a little too hot. Now it will be less hot for each device
but equally hot for the whole ;-)

Peter

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