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'[EE]: can I use 7805 to step down 24V to 5V?'
2005\06\22@135954 by John Waters

picon face
Hi All,

My circuit needs both 24V and 5V, but I have only a single 24V supply. Is
there any way to step down 24V to 5V? Can I use a 7805? If this is not a
good way, any better alternative?

Thanks in advance!

John


2005\06\22@142432 by phil B

picon face
Probably.  The input voltage within spec but you don't
say current requirements.  You will dissipate the
different in heat so I'd be conservative on picking
the package and be prepared to use a heat sink.  
Linears like the 78xx family are pretty inefficient
but if that's not an issue you will be ok.  Also,
calculate the additional drain on the 24V supply.


--- John Waters <spam_OUTjohn_fm_watersTakeThisOuTspamhotmail.com> wrote:

{Quote hidden}

> --

2005\06\22@143413 by Spehro Pefhany

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At 10:59 AM 6/22/2005 -0700, you wrote:
>Hi All,
>
>My circuit needs both 24V and 5V, but I have only a single 24V supply. Is
>there any way to step down 24V to 5V? Can I use a 7805? If this is not a
>good way, any better alternative?
>
>Thanks in advance!
>
>John

How much current, and is the 24V supply regulated? If it's regulated,
(or doesn't go much higher than 30V peak under any conditions) and the
current is relatively small (say 10mA or 25mA) then the 7805
is fine, and is certainly cheap.

If the current is more like 200mA or 400mA, you should consider using a
switching regulator.  If it's a one-off there are prepackaged regulators
for a price, or if you've got quantity, there are some nice chips with or
without the power switch etc. built-in.

Note that you *could* use the 7805 at higher currents, but you'd need
a heat sink and you'd be wasting a lot of power. The power you have to
dispose of is about

Pd = (Vin - 5V)*(Iout) + Vin*4mA

Let's take an example:

Vin = 24V, Iout = 30mA

Pd = 570mW + 24mA = 593mW (0.6W)

A TO-252 with no copper tab or pour would have a theta-JA of    103°C/W
according to National's AN 1028. So there would be a rise of 62°C above
the ambient inside the enclosure. If that's 15°C above Ta, and you find
a Tj of 100°C acceptable, then the maximum ambient temperature would
be about normal room temperature (23°C). Not very practical.

Stick a square inch of 2-oz copper on the top layer attached to the tab
and you can better than halve the thermal resistance (to 47°C/W), so you
could have an ambient of 43°C and still keep Tj to 100°C. Abs max Tj is
125°C, so short excursions to much higher temperatures would likely
be tolerated.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
.....speffKILLspamspam@spam@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff

2005\06\22@145554 by peiserma

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face
piclist-bounces@mit.edu wrote:
> Hi All,
>
> My circuit needs both 24V and 5V, but I have only a single 24V
> supply. Is there any way to step down 24V to 5V? Can I use a 7805? If
> this is not a good way, any better alternative?

In theory, yes. Primary thing to watch out for is power dissipation.
24V input and 5V output means the regulator is dropping 19V. You
will need to figure out your circuit's maximum current draw.

quick explanation: Ohms Law says P=IV. So multiply the voltage drop
by the current required by your circuit. Then find the thermal
resistance on the datasheet (specified in Celsius per Watt or
sometimes Kelvin per Watt) and see if the junction temperature
remains within specifications.

say your circuit draws 100mA under worst case conditions. Calculate
power by P = 19V*0.1A = 1.9W. A LM7805C (TO-220 case) has a thermal
resistance from junction to ambient of 54 C/W. So 54C/W*1.9W, or
102.6 Celsius heat rise. Note that this is the temperature increase,
above ambient. If ambient is 25C, then the junction temperature
would be 127.6C, which is takes the device outside specs. So you
need a heatsink.... There will be thermal resistance from junction
to case, case to heatsink, and heatsink to ambient. Add them all
up to get the junction to ambient thermal resistance.

You can't do anything about the junction to case, or really the
case to sink (not much anyway), leaving the case to ambient. So
you have to work the equation backwards based on maximum current
draw of the circuit, and then figure out the maximum heatsink to
ambient thermal resistance permissible. Oh, and make sure the
heatsink is capable of the power dissipation you require.

Does that help get you started?

2005\06\22@145701 by John J. McDonough

flavicon
face
----- Original Message ----- From: "Spehro Pefhany" <speffspamKILLspaminterlog.com>
Subject: Re: [EE]: can I use 7805 to step down 24V to 5V?


> How much current, and is the 24V supply regulated? If it's regulated,

Going from 24 volts to 5, that 7805 is going to get mighty hot, even with relatively little current.  The good news is that it is really hard to let the smoke out of a 7805.  But you might keep it away from other components that might not like being in the oven.  And be very careful touching the thing. It really is incredible how much heat a 7805 can survive.

--McD

{Quote hidden}

> -

2005\06\22@155355 by Richard Prosser

picon face
If you know the current & it's reasonably constant, you can put a
fixed resistor in series with the supply to drop the volts down to
something a linear such as a 7805 can handle easier (9volts or so. The
resistor will need to be able to handle the power of course. You may
even be able to find a small light bulb to do the job (at reduced
reliability).
Or do a search in the archives or google for the "Black Regulator" for
a simple switcher.

Richard P

On 23/06/05, John J. McDonough <EraseMEmcdspam_OUTspamTakeThisOuTis-sixsigma.com> wrote:
> {Original Message removed}

2005\06\22@164224 by Spehro Pefhany

picon face
At 07:53 AM 6/23/2005 +1200, you wrote:
>  You may even be able to find a small light bulb to do the job (at reduced
>reliability).

A light bulb is a (sort of) constant current device. The 7805 with a
constant load on the output looks like a fairly good constant current sink
(looking into the input).

What will happen when you put two constant current circuits in series?

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speffspamspam_OUTinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff


2005\06\22@164637 by fred jones

picon face
Yes they can survive a lot of heat but you have to be careful because they
will go into thermal shutdown to protect themselves.  If it gets too hot or
current consumption is a priority, then a switcher might be in order.
FJ
============================================================

Going from 24 volts to 5, that 7805 is going to get mighty hot, even with
relatively little current.  The good news is that it is really hard to let
the smoke out of a 7805.  But you might keep it away from other components
that might not like being in the oven.  And be very careful touching the
thing. It really is incredible how much heat a 7805 can survive.

--McD


2005\06\22@164757 by Charles Linquist

flavicon
face

If cost isn't a huge factor, I think a better solution would be to use a Power Trends 78ST105HC (Power Trends
is now owned by TI)

This regulator is a switching type that has the same pinout as a 7805.  If I remember correctly, it can take up to
36V on the input, and since it is a switcher, generates virtually NO heat.  The only downside is that the footprint is
slighly larger than a 7805.

Charles Linquist



Richard Prosser wrote:

{Quote hidden}

>>{Original Message removed}

2005\06\22@170222 by Wouter van Ooijen

face picon face
> What will happen when you put two constant current circuits in series?

Is that a rethorical question? :)

If they are ideal current sources of unequal currents something (one of
the sources, the wiring, the isolation, reality, whatever) will have to
give, just like when you connect two voltage sources of different
voltages.

With practical (series regulating, not real source) circuits the higher
current one will 'saturate' to its lowest series resistance and the
lower current one will determine the current.

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu


2005\06\22@170405 by Richard Prosser

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The thing is that the light bulb is only a "sort of" constant current.
Initially the 7805 will be dissipating a bit more, but as the bulb
heats up, the bulb will start to take the load.
Provided it is sized correctly, which is the trick.
It just can work out a lot cheaper to use an automotive bulb (for
example) than a 20W 10ohm resistor.

RP

On 23/06/05, Spehro Pefhany <KILLspamspeffKILLspamspaminterlog.com> wrote:
{Quote hidden}

> -

2005\06\22@170837 by Jinx

face picon face
> Note that you *could* use the 7805 at higher currents, but you'd
> need a heat sink and you'd be wasting a lot of power. The power
> you have to dispose of is about

I'd drop the 24V down to about 8-9V with a resistor, then regulate

If you haven't power to spare as lost heat doing this, switchers would
be more attractive

2005\06\22@171800 by Charles Craft

picon face
Ouch!
(there was a disclaimer about cost but wow)

>From Digikey:
Price Break        Unit Price
1                   17.03000
25                 15.89000
100               14.75500
500               14.18750        


{Original Message removed}

2005\06\22@203635 by Tony Smith

picon face

>> How much current, and is the 24V supply regulated? If it's regulated,
>
> Going from 24 volts to 5, that 7805 is going to get mighty hot, even with
> relatively little current.  The good news is that it is really hard to let
> the smoke out of a 7805.  But you might keep it away from other components
> that might not like being in the oven.  And be very careful touching the
> thing. It really is incredible how much heat a 7805 can survive.
>
> --McD


I've seen people put a 7812 between the 24v and 7805 to split up the load.
You still create as much heat, but it's shared between the 2 devices.

Tony

2005\06\22@204143 by olin piclist

face picon face
John J. McDonough wrote:
> Going from 24 volts to 5, that 7805 is going to get mighty hot, even
> with relatively little current.  The good news is that it is really
> hard to let the smoke out of a 7805.  But you might keep it away from
> other components that might not like being in the oven.  And be very
> careful touching the thing. It really is incredible how much heat a
> 7805 can survive.

Actually the 7805 will protect itself.  The circuit may not be happy though
when it's power gets suddenly shut down without notice.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

2005\06\22@205924 by David Minkler

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face
...which isn't bad at all under the right circumstances.  If it's a one-up, size constrained, time constrained, skills constrained, lots of current required from 5V and/or same current, amp for amp, unavailable from existing 24V supply or many subsets of same, it's a very reasonable choice.  Roman's "Black converter" (see archives for link) isn't a bad choice if you really want cheap and efficient and don't mind the unconventional.  Plan on spending a little time optimizing (give it a little board space, spend time, need some skill, probably not worth it for a one-up).

Dave

Charles Craft wrote:

{Quote hidden}

>{Original Message removed}

2005\06\22@224130 by Russell McMahon

face
flavicon
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>>Richard Prosser wrote:
>>>If you know the current & it's reasonably constant, you can put a
>>>fixed resistor in series with the supply to drop the volts down to
>>>something a linear such as a 7805 can handle easier.

Resistor is a good solution if you MUST use 24v and a 7805.
This way you can get most of the energy which must be dissipated in
the resistor.

Specify I max.

1.    Vdrop_max = Vin - Vdropoutmin - Vout = (24-3-5) = 16 volts

2.    R = V/Imax = 16/1 = 16r at 1 amp.
Larger at less max amps.

3.    Resistor power rating = Vdrop_max x Imax.
At 1 amp that's 16 watts.

Choose a resistor that's free air rated at at least twice the wattage
you get from 3. above.

You can series and/or parallel connect resistors as required.

       RM

2005\06\23@040230 by Alan B. Pearce

face picon face
Charles Craft wrote:

>Ouch!
>(there was a disclaimer about cost but wow)
>
>>From Digikey:
>Price Break Unit Price
>1                   17.03000
>25                 15.89000
>100               14.75500
>500               14.18750
>


>If cost isn't a huge factor, I think a better solution would be to use a
>Power Trends 78ST105HC (Power Trends

Actually, when you look at that cost, if one needs a switching regulator,
then the cost will be about the same, by the time you purchase all the parts
and put them on a PCB. Chip, inductor, probably two tantalum caps and about
3 ceramics, anywhere between two and 5 resistors, seem to make up a fairly
typical switcher requirement. Sure you can get ones with fixed outputs which
do away with a number of the parts, but they also tend to be restricted in
input voltage, so for 24v in I suspect a more generic chip will be required,
which will require more parts.

2005\06\23@041801 by Buehler, Martin

picon face
using a step down converter would be more efficient for sure, but if 24v
is within the spec of the 7805, this should work.
if you're getting into problems with power dissipation (or if 24v is too
much for a 7805), you could use a 7815 and a 7805 in series, to spread
heat production over two devices. so you might eventually be able to get
rid of heat sinks.
tino

************************************************************************
******************************


>{Original Message removed}

2005\06\23@045754 by William Chops Westfield

face picon face

--Apple-Mail-4--524047711
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On Jun 23, 2005, at 1:02 AM, Alan B. Pearce wrote:

>> If cost isn't a huge factor, I think a better solution would be to
>> use a
>> Power Trends 78ST105HC [at $17 in ones!]
>
> Actually, when you look at that cost, if one needs a switching
> regulator,
> then the cost will be about the same...

I dunno.  Here's a tentative (untested) PCB layout for the low-cost
"Roman Black" switching regulator, in a 3-terminal configuration...


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2005\06\23@053017 by David P Harris

picon face
William Chops Westfield wrote:

>
> On Jun 23, 2005, at 1:02 AM, Alan B. Pearce wrote:
>
>>> If cost isn't a huge factor, I think a better solution would be to
>>> use a
>>> Power Trends 78ST105HC [at $17 in ones!]
>>
>>
>> Actually, when you look at that cost, if one needs a switching
>> regulator,
>> then the cost will be about the same...
>
>
> I dunno.  Here's a tentative (untested) PCB layout for the low-cost
> "Roman Black" switching regulator, in a 3-terminal configuration...

How big is this circuit board?  Looks about 0.8 x 1 inch.  Would a smd
one approach the size of a TO220?
David


2005\06\23@160731 by William Chops Westfield

face picon face

On Jun 23, 2005, at 2:30 AM, David P Harris wrote:

>> ack" switching regulator, in a 3-terminal configuration...
>
> How big is this circuit board?  Looks about 0.8 x 1 inch.  Would a smd
> one approach the size of a TO220?
>
Yeah, about .7x.9 inch :-)  I have an SMD layout, but it's not much
smaller,
though that becomes highly dependent on which packages you think you can
manage to solder.  It turns out that an SOT23 transistor has close to
the
same footprint a TO92, and a 1206 resistor about the same footprint
as a vertical TH resistor (though of course they're much THINNER.)
And inductor/capacitor size in SMT is pretty variable depending on how
much you're willing to pay.  (The black regulator isn't designed for the
low inductances of vert small SMT parts.)  You get some space
back going to double sided as well.

OTOH, I figure the board as shown is already pretty close the the
average
TO220 plus smallish heatsink...

BillW

2005\06\23@214341 by David P Harris

picon face
William Chops Westfield wrote:

{Quote hidden}

Looks like that board has components on both sides.

You could make some pcbs through http://www.sparkfun.com for $5 each
($5/sq. inch).

David


2005\06\24@042724 by Alan B. Pearce

face picon face
>You could make some pcbs through http://www.sparkfun.com
>for $5 each ($5/sq. inch).

Better still, put them in Olimex for some free ones in their competition.

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