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'[EE]: 7805 regulator current increase'
2002\07\13@111006 by Jim

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Hi I was wondering is is possible to parallel 2 of them together to double the current out or do I have to use
transistors to increase current?
Thanks
Jim

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2002\07\13@113933 by Patrick J

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You can parallel them. Use a small resistor (like 0.33 ohm) in series with
each of them so they deliver half the current each.
I have a design here somewhere that use 10x 7824 in parallel...

----- Original Message -----
I was wondering is is possible to parallel 2 of them together to double the
current out or do I have to usetransistors to increase current?
Thanks
Jim

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2002\07\13@114809 by Thomas C. Sefranek

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I prefer the PNP power transistor fix.
I use a 1 ohm 1 watt current sense/bias resistor.

Patrick J wrote:

> You can parallel them. Use a small resistor (like 0.33 ohm) in series with
> each of them so they deliver half the current each.
> I have a design here somewhere that use 10x 7824 in parallel...
>
> {Original Message removed}

2002\07\13@121146 by Pic Dude

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Yes, I'm sure you can parallel them, but I'm not sure what
the effect would be if they aren't perfectly matched.  One
of the experts can chime in here.

In the meanwhile...

Was doing some 5V regulator research recently and I remember
that there is a 1.5A alternative to the 7805 in a TO-220
package, which is called the LM1086.  It's even an LDO.  And
digikey has them.  Only difference I noticed is that metal
tab is not connected to ground, but to Vout.

Also, there is a 3A alternative in a TO-220 package, called
the LT323AT.

So perhaps one of these will work for you.

Cheers,
-Neil.




{Original Message removed}

2002\07\13@122643 by Spehro Pefhany

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At 11:08 AM 7/13/02 -0500, you wrote:
>Yes, I'm sure you can parallel them, but I'm not sure what
>the effect would be if they aren't perfectly matched.  One
>of the experts can chime in here.

It will probably "sort of" work, with one regulator hovering
on the edge of over-current or thermal shutdown. Not good.

<snippage>

>So perhaps one of these will work for you.

There are also some nice ones from Sanken, some in the
thermally superior TO-218 package.

Best regards,

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2002\07\13@123848 by Jim

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Thanks for the fast response.
The reason I ask was I just got a bunch of 78L05's 100ma from ebay.
But then again I also got a bunch of smt pnp's @ 600ma so maybe I'll go the
transistor
route. I tryed a pnp with an lm317 a year or so ago and had some problems
with it
like smoking transistors or somthing I don't remember. With a schematic that
I found on the web
on several sites.

Thanks
Jim

{Original Message removed}

2002\07\13@164703 by Matt Pobursky

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Just remember to look carefully at your steady-state power dissipation
in your pass transistor or regulator. This is probably the most common
problem I see with new or inexperienced designer's power circuits.
"Gee, the transistor is rated for 5A, but it gets HOT!". Basic laws of
physics are always at work and you can't dissipate 1W in a SOT-23, SOT
-89 or TO-92 package without it getting very hot and (probably)
destroying it.

Also remember the hotter you run a particular semiconductor the sooner
it will fail. Cool running silicon works virtually forever (all other
possible stresses aside). Most of my designs are conservative and
generally, no semiconductor runs hotter than you can continuously touch
with your finger (but use a temp. probe to check if you're not sure!).
Many of those designs are still running in commercial and industrial
applications after 15-20 of years continuous use too.  

Just my thoughts on the subject.

Matt Pobursky
Maximum Performance Systems

On Fri, 12 Jul 2002 21:40:58 -0700, Jim wrote:
>Thanks for the fast response.
>The reason I ask was I just got a bunch of 78L05's 100ma from ebay.
>But then again I also got a bunch of smt pnp's @ 600ma so maybe I'll
>go the transistor route. I tryed a pnp with an lm317 a year or so ago
>and had some problems with it like smoking transistors or somthing I
>don't remember. With a schematic that I found on the web on several
>sites.
>
>Thanks Jim

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2002\07\14@175621 by Russell McMahon

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> I was wondering is is possible to parallel 2 of them
> together to double the current out or do I have
> to use transistors to increase current?

As Patrick noted, you can add a SMALL input resistor to each one to
encourage them to share the load. The reason this is needed is that each
regulator is liable to have very slightly different characteristics. As
noted below, this may not work as well as expected. An alternative is to add
a VERY small output resistor to each one (see discussion for why).

If you just parallel them then one may want to provide say 4.99 volts while
the other wants to supply 4.98 volts (example voltages only). Once the
output gets to 4.98 volts one regulator will be 'happy" and shut down while
the higher voltage unit will keep driving the output. If there is a lower
resistance load than it can drive at 4.99v it will go all the way up to its
rated capacity before its output voltage starts to sag. At that stage its
internal current limit will operate and the voltage will drop until it is
low enough for the other regulator to start supplying. If there are more
than 2 then there will tend to be a "pecking order". In practice there will
probably be some overlap as the regulators will not have a perfectly sharp
regulating voltage.

Note that the above result is not necessarily terrible - it is the internal
current limit causing the load sharing and NOT the internal thermal limiter.
As long as the regulator is not THERMALLY limiting this is not an
excessively high stress situation. For a TO220 package without a heatsink
you can dissipate about 2 watts (depends on ambient temperature) so this
means that with a 7805 regulator you WOULD almost certainly run into thermal
limiting as well. This is because the 7805 has about a 1.8v dropout (input
to output) voltage at full load and when running hot. To dissipate 2 watts
at 1 amp you need a 2 volt drop which is JUST above this minimum dropout
voltage.
CONCLUSION: If running TO220 package 7805 regulators in parallel without
input resistors the maximum input voltage without a heatsink is 8 volts IF
thermal limiting is not to occur.
Once you thermal limit you start to reduce your power handling so it is
undesirable, quite apart from reliability issues.

If you add a small series input resistor to each regulator, the individual
regulator's input voltage will drop as its own current drain rises.  While
the 7805 (and any regulator) is relatively immune to input voltage
regulations (after all, it is meant to be a regulator). Spec sheet says that
line regulation is typically 5 mV (120 mV max) for 17v change in input
voltage. You are not going to see MUCH output change with the drop in input
voltage caused by a small input resistor. As long as the regulator are
closely matched for output voltage it will work and if they are not it will
still help the current limiting sharing mentioned above.

Now imagine that each regulator has a 0.01 ohm (10 milliohm) output resistor
to the common output point. If one regulator supplied 1 amp and the other
supplied nothing then the output voltage would be I x R = 10 millivolts
below the regulator output voltage. If a 0.1 ohm resistor was used then it
would be 0.1 volt below. Even at 10 millivolts we are getting a change
equivalent to an input voltage swing of 17 volts. The second (or many
other ) regulators are liable to start helping with even 10 milliohms output
resistance. Obviously this is not ideal as it adds a degree of noise
coupling in your supply line but 10 milliohms is in the order of PCB track
resistance anyway (and track resistance will be FAR more than this in some
designs.).
Worth thinking about.(maybe :-) ).

Using a parallel transistor can work well BUT the transistor lacks current
and thermal limiting and is therefore much less "fail-safe" than a regulator
only solution.



           Russell McMahon

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2002\07\14@234746 by Jim

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Very informative Russell Thanks

Jim

-----Original Message-----
From: Russell McMahon <apptechspamKILLspamPARADISE.NET.NZ>
To: .....PICLISTKILLspamspam.....MITVMA.MIT.EDU <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU>
Date: Sunday, July 14, 2002 2:57 PM
Subject: Re: [EE]: 7805 regulator current increase


>> I was wondering is is possible to parallel 2 of them
>> together to double the current out or do I have
>> to use transistors to increase current?
>
>As Patrick noted, you can add a SMALL input resistor to each one to
>encourage them to share the load. The reason this is needed is that each
>regulator is liable to have very slightly different characteristics. As
>noted below, this may not work as well as expected. An alternative is to
add
{Quote hidden}

limiter.
>As long as the regulator is not THERMALLY limiting this is not an
>excessively high stress situation. For a TO220 package without a heatsink
>you can dissipate about 2 watts (depends on ambient temperature) so this
>means that with a 7805 regulator you WOULD almost certainly run into
thermal
{Quote hidden}

that
>line regulation is typically 5 mV (120 mV max) for 17v change in input
>voltage. You are not going to see MUCH output change with the drop in input
>voltage caused by a small input resistor. As long as the regulator are
>closely matched for output voltage it will work and if they are not it will
>still help the current limiting sharing mentioned above.
>
>Now imagine that each regulator has a 0.01 ohm (10 milliohm) output
resistor
>to the common output point. If one regulator supplied 1 amp and the other
>supplied nothing then the output voltage would be I x R = 10 millivolts
>below the regulator output voltage. If a 0.1 ohm resistor was used then it
>would be 0.1 volt below. Even at 10 millivolts we are getting a change
>equivalent to an input voltage swing of 17 volts. The second (or many
>other ) regulators are liable to start helping with even 10 milliohms
output
>resistance. Obviously this is not ideal as it adds a degree of noise
>coupling in your supply line but 10 milliohms is in the order of PCB track
>resistance anyway (and track resistance will be FAR more than this in some
>designs.).
>Worth thinking about.(maybe :-) ).
>
>Using a parallel transistor can work well BUT the transistor lacks current
>and thermal limiting and is therefore much less "fail-safe" than a
regulator
{Quote hidden}

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2002\07\15@102802 by Roman Black

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Jim wrote:

> Subject: Re: [EE]: 7805 regulator current increase
>
> >> I was wondering is is possible to parallel 2 of them
> >> together to double the current out or do I have
> >> to use transistors to increase current?


Where regulators have tight regulation (within 10mV from
0A to 1A), like 7805 etc, you can parallel them with a small
resistor on the OUTPUT. Drop about 30mV on the output at
full current, so about 0.03 ohms is enough. 0.1 ohm
resistors will do if they are all you can get.

Russell McMahon wrote:
>As Patrick noted, you can add a SMALL input resistor to each one to
>encourage them to share the load.

I don't think you can do it with input resistors, they
have to be on the output, as the output is closely
regulated by the opamp in the chip regardless of input
volts. ie, connect one 7805 to 12v and one to 8v, both
will output the same volts, at least within a couple mV
chip tolerance. Resistors must be on the output, and YES
it will work fine.
-Roman

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2002\07\15@103627 by john

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A while back, when we still bought/got given Linear data books(yes, in paper
format), i recall seeing a circuit incorporating a high power pnp transistor
in a boost configuration, to alleviate the regulator's load.

The PNP derived its drive from a resistor in the input lead of the regulator
and was calculated so that the resistor was just biasing the PNP on when the
regulator was approaching the maximum current available. The output of the
pnp and the regulator were tied together. This bias point i assume could be
reduced to a suitable level by increasing the input resistance to the
regulator.

Hope that helps.

John

On Monday 15 July 2002 04:26 pm, you wrote:
{Quote hidden}

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Thank-you for your time.

John Ward

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2002\07\15@110330 by Russell McMahon

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> Russell McMahon wrote:
> >As Patrick noted, you can add a SMALL input resistor to each one to
> >encourage them to share the load.
>
> I don't think you can do it with input resistors, they
> have to be on the output, as the output is closely
> regulated by the opamp in the chip regardless of input
> volts. ie, connect one 7805 to 12v and one to 8v, both
> will output the same volts, at least within a couple mV
> chip tolerance. Resistors must be on the output, and YES
> it will work fine.

Yes. I started with the input resistor case as it had been mentioned
elsewhere. Further down I pointed out that the regulator action fought
against you and that output resistors gave far better results. In fact our
recommended resistor values and the compromise values we suggested are both
very similar.

Then again, buying a higher current regulator ma be easier :-)
(But I think Jim has a lot of 78L05s to use up (based on a prior post) so
this may be a good reason to get out the resistor box).




       RM

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2002\07\15@112143 by Spehro Pefhany

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At 12:26 AM 7/16/02 +1000, you wrote:

>Where regulators have tight regulation (within 10mV from
>0A to 1A), like 7805 etc, you can parallel them with a small
>resistor on the OUTPUT. Drop about 30mV on the output at
>full current, so about 0.03 ohms is enough. 0.1 ohm
>resistors will do if they are all you can get.

Besides the regulation, there's a tolerance on the output
voltage- something like +/-200mV at 25'C for the LM7805.
If one regulator has a nominal output of 5.2 and the other
4.8, they won't try to share current at all, *due* to the
tight regulation- that is the output current will go
from 5mA to 1.5A with typically only 9mV of output change.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2002\07\15@200001 by Patrick J

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> > Russell McMahon wrote:
> > >As Patrick noted, you can add a SMALL input resistor to each one to
> > >encourage them to share the load.

Actually Patrick MEANT that the resistor(s) should be on the OUTPUT.
What he wrote seems to have been missunderstood/missquoted ;-)

<mutter>
darn, can't sleep.. i've been missquoted on the piclist !
</mutter>

/Patrick

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2002\07\15@220717 by Russell McMahon

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Sorry - I was the misquoter :-)
Misunderstood your intention.

I went to some length to explain why output resistors would be better. If
I'd read your post more closely I could have skipped that part.


{Quote hidden}

Happens to me all the time I might add :-)



       Russell

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