More Current out of 7805
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Posts: 46,084
Hi,
My project requires a peak current of about 1.5 amps. I need a regulated
source that I can put together quickly out of parts I have. Can I put to
7805s in parallel t get more current? How well do they share or is there
some other way of doing it?
Thanks
Mike
My project requires a peak current of about 1.5 amps. I need a regulated
source that I can put together quickly out of parts I have. Can I put to
7805s in parallel t get more current? How well do they share or is there
some other way of doing it?
Thanks
Mike
Comments
> My project requires a peak current of about 1.5 amps. I need a regulated
> source that I can put together quickly out of parts I have. Can I put to
> 7805s in parallel t get more current? How well do they share or is there
> some other way of doing it?
I may be incorrect, but I don't think you can parallel 7805's. How
long is your peak current spike? How much current does your project need
in quiescent mode? 1 Amp current (rated value) for a 7805 will require
a good heatsink and good air circulation to avoid letting the smoke out.
You may want to go with a TO3 version 5V regulator, even still. If your
1.5Amp spike is a transient, a good cap to buffer that spike will handle
it. It just depends.
DLC
--
============================================================================
* Dennis Clark Aristocrat at heart dlc@v... www.verinet.com/~dlc *
* Be well, do good work, and stay in touch -- Garrison Keillor *
============================================================================
but I'm sure he won't mind if I put it here:
<<I agree, don't use two devices in parallel.
Here is a simple circuit that will give you all the
current you will ever want.
Use a PNP transistor, of sufficient current to meet
your needs, For a cool 2 amps use a TIP-30.
connect the emitter (pin3) to B+. connect the
collector to "load" (pin2). Connect the base to your
7805 regulator IC (pin1 of TIP-30)to (pin1 of 7805).
Connect the output of the regulator(pin3)to collector
of TIP-30 (pin2). Ground the 7805 (pin2). Add .1 caps
to emitter and collector of transistor. Place a 470uf
cap to Collector (pin2) of transistor. This is clean
and cool, with minimum parts.
A schematics would have been better, but hope this
helps.>>
Duncan
On paper, parallel 7805s don't share the load well at all; the one with the
higher output voltage takes most of the load and the other one loafs. But
do you really care? As long as your output voltage is reasonably close, it
really doesn't matter which device is carrying the load. I ran a set of
three 7805s like this one time specifically to see if I could induce a
thermal shutdown oscillation (The overachiever that takes most of the load
gets too hot, shuts down, passes the load to the next guy until he shuts
down etc.) I never did get it to misbehave; it worked perfectly every time.
I'd have to think it would be a better idea for production to go to a
larger regulator like the LM138 or LM196, but that violates the premise of
putting it together quickly out of parts you have.
Whatever you do, you're going to have to deal with the power dissipation
question, unless the 1.5 Amp peak current is short and/or low duty cycle.
(if it IS short and/or low duty cycle, you might get by with just a big
capacitor on the output). If you need more help, try to give us a little
more information as to the duration and frequency of the current peak as
well as any quiescent load you might have.
It sounds like you're not proposing to put this into production, so I give
you permission to try paralleling the 7805s; it's not a pretty solution,
but I think it will work, and pretty sure it won't hurt anything.
Bob U.
At 02:15 PM 3/28/00 -0500, you wrote:
>Hi,
>
>My project requires a peak current of about 1.5 amps. I need a regulated
>source that I can put together quickly out of parts I have. Can I put to
>7805s in parallel t get more current? How well do they share or is there
>some other way of doing it?
>
>Thanks
>Mike
>
>
>
>eGroups.com Home: http://www.egroups.com/group/basicstamps/
>http://www.egroups.com - Simplifying group communications
regulated
> > source that I can put together quickly out of parts I have. Can I put
to
> > 7805s in parallel t get more current? How well do they share or is
there
> > some other way of doing it?
snip
I don't think its a great idea to parallel 7805's, its been done, most times
it works,
some times it doesn't, and Dennis is right about a large heat sink, and
plenty of air.
try the lm323k, its a to-3 case, its preset at 5 volts, the case is ground,
and its rated
at 3 amps continuous.
they are available from most sources.
funny one Newark electronics $ 0.36 each while remaining stock lasts.
regular
jdr $ 2.99
jameco $ 2.95
strange one digi-key $ 10.50 each.
just goes to show, shopping around can save you some money.
I got these prices by using www.findchips.com
norm
components. I tried to parallel them and the voltage was about 4.9v but was
unsure if it could slip at anytime. It will probably be at about 1.5 amps
for 8 - 10 seconds until it spins up then drop to about .5 - .8
Thanks
Mike
Original Message
From: "Bob Underwood" <bobu@n...>
To: <basicstamps@egroups.com>
Sent: Tuesday, March 28, 2000 4:23 PM
Subject: [noparse][[/noparse]basicstamps] Re: More Current out of 7805
> Mike
>
> On paper, parallel 7805s don't share the load well at all; the one with
the
> higher output voltage takes most of the load and the other one loafs. But
> do you really care? As long as your output voltage is reasonably close, it
> really doesn't matter which device is carrying the load. I ran a set of
> three 7805s like this one time specifically to see if I could induce a
> thermal shutdown oscillation (The overachiever that takes most of the load
> gets too hot, shuts down, passes the load to the next guy until he shuts
> down etc.) I never did get it to misbehave; it worked perfectly every
time.
>
> I'd have to think it would be a better idea for production to go to a
> larger regulator like the LM138 or LM196, but that violates the premise of
> putting it together quickly out of parts you have.
>
> Whatever you do, you're going to have to deal with the power dissipation
> question, unless the 1.5 Amp peak current is short and/or low duty cycle.
> (if it IS short and/or low duty cycle, you might get by with just a big
> capacitor on the output). If you need more help, try to give us a little
> more information as to the duration and frequency of the current peak as
> well as any quiescent load you might have.
>
> It sounds like you're not proposing to put this into production, so I give
> you permission to try paralleling the 7805s; it's not a pretty solution,
> but I think it will work, and pretty sure it won't hurt anything.
>
> Bob U.
>
> At 02:15 PM 3/28/00 -0500, you wrote:
> >Hi,
> >
> >My project requires a peak current of about 1.5 amps. I need a regulated
> >source that I can put together quickly out of parts I have. Can I put to
> >7805s in parallel t get more current? How well do they share or is there
> >some other way of doing it?
> >
> >Thanks
> >Mike
> >
> >
> >
> >eGroups.com Home: http://www.egroups.com/group/basicstamps/
> >http://www.egroups.com - Simplifying group communications
>
>
>
>
> eGroups.com Home: http://www.egroups.com/group/basicstamps
> www.egroups.com - Simplifying group communications
>
>
a regulator. If you have a PNP laying around, I would go with it.
Good luck,
Ray McArthur
Original Message
From: <orthner@s...>
To: <basicstamps@egroups.com>
Sent: Tuesday, March 28, 2000 3:35 PM
Subject: [noparse][[/noparse]basicstamps] Re: More Current out of 7805
> This was posted to another list by Tom Umble <ctech_pro@y...>
> but I'm sure he won't mind if I put it here:
>
> <<I agree, don't use two devices in parallel.
> Here is a simple circuit that will give you all the
> current you will ever want.
> Use a PNP transistor, of sufficient current to meet
> your needs, For a cool 2 amps use a TIP-30.
> connect the emitter (pin3) to B+. connect the
> collector to "load" (pin2). Connect the base to your
> 7805 regulator IC (pin1 of TIP-30)to (pin1 of 7805).
> Connect the output of the regulator(pin3)to collector
> of TIP-30 (pin2). Ground the 7805 (pin2). Add .1 caps
> to emitter and collector of transistor. Place a 470uf
> cap to Collector (pin2) of transistor. This is clean
> and cool, with minimum parts.
> A schematics would have been better, but hope this
> helps.>>
>
> Duncan
>
>
>
>
>
> -- 20 megs of disk space in your group's Document Vault
> -- http://www.egroups.com/docvault/basicstamps/?m=1
>
>
help equalize the current flow (and heat) ?
richard
Original Message
From: Bob Underwood <bobu@n...>
To: <basicstamps@egroups.com>
Sent: Tuesday, March 28, 2000 3:23 PM
Subject: [noparse][[/noparse]basicstamps] Re: More Current out of 7805
> Mike
>
> On paper, parallel 7805s don't share the load well at all; the one with
the
> higher output voltage takes most of the load and the other one loafs. But
> do you really care? As long as your output voltage is reasonably close, it
> really doesn't matter which device is carrying the load. I ran a set of
> three 7805s like this one time specifically to see if I could induce a
> thermal shutdown oscillation (The overachiever that takes most of the load
> gets too hot, shuts down, passes the load to the next guy until he shuts
> down etc.) I never did get it to misbehave; it worked perfectly every
time.
>
> I'd have to think it would be a better idea for production to go to a
> larger regulator like the LM138 or LM196, but that violates the premise of
> putting it together quickly out of parts you have.
>
> Whatever you do, you're going to have to deal with the power dissipation
> question, unless the 1.5 Amp peak current is short and/or low duty cycle.
> (if it IS short and/or low duty cycle, you might get by with just a big
> capacitor on the output). If you need more help, try to give us a little
> more information as to the duration and frequency of the current peak as
> well as any quiescent load you might have.
>
> It sounds like you're not proposing to put this into production, so I give
> you permission to try paralleling the 7805s; it's not a pretty solution,
> but I think it will work, and pretty sure it won't hurt anything.
>
> Bob U.
>
> At 02:15 PM 3/28/00 -0500, you wrote:
> >Hi,
> >
> >My project requires a peak current of about 1.5 amps. I need a regulated
> >source that I can put together quickly out of parts I have. Can I put to
> >7805s in parallel t get more current? How well do they share or is there
> >some other way of doing it?
> >
> >Thanks
> >Mike
> >
> >
> >
> >eGroups.com Home: http://www.egroups.com/group/basicstamps/
> >http://www.egroups.com - Simplifying group communications
>
>
>
> -- Check out your group's private Chat room
> -- http://www.egroups.com/ChatPage?listName=basicstamps&m=1
>
>
> if you put a small resistance on each output leg (like 0.33 ohm) would that
> help equalize the current flow (and heat) ?
Load sharing does not work with current sensitive devices (transistors
for example) because of a "feature" called thermal runaway. As current
increases, temperature increases, increased temperature causes lower
resistance, which increases current... I think you see where this is going.
What always happens is that one of the paralleled components will heat up
a bit more than the others, that one will start passing more current, ...
The others will not do anything, the vast majority will go through one of
the parts. Only FETs do not share this problem, with increased current the
voltage drop across the FET increases, this causes heat (energy dissipation)
which increases the FET's resistance, which chokes back current allowing
other paralleled parts to pass more current. That wasn't a very techy
description, but it gets the point across I guess. 8^)
DLC
>
Original Message
> From: Bob Underwood <bobu@n...>
> To: <basicstamps@egroups.com>
> Sent: Tuesday, March 28, 2000 3:23 PM
> Subject: [noparse][[/noparse]basicstamps] Re: More Current out of 7805
>
>
> > Mike
> >
> > On paper, parallel 7805s don't share the load well at all; the one with
> the
> > higher output voltage takes most of the load and the other one loafs. But
> > do you really care? As long as your output voltage is reasonably close, it
> > really doesn't matter which device is carrying the load. I ran a set of
> > three 7805s like this one time specifically to see if I could induce a
> > thermal shutdown oscillation (The overachiever that takes most of the load
> > gets too hot, shuts down, passes the load to the next guy until he shuts
> > down etc.) I never did get it to misbehave; it worked perfectly every
> time.
> >
> > I'd have to think it would be a better idea for production to go to a
> > larger regulator like the LM138 or LM196, but that violates the premise of
> > putting it together quickly out of parts you have.
> >
> > Whatever you do, you're going to have to deal with the power dissipation
> > question, unless the 1.5 Amp peak current is short and/or low duty cycle.
> > (if it IS short and/or low duty cycle, you might get by with just a big
> > capacitor on the output). If you need more help, try to give us a little
> > more information as to the duration and frequency of the current peak as
> > well as any quiescent load you might have.
> >
> > It sounds like you're not proposing to put this into production, so I give
> > you permission to try paralleling the 7805s; it's not a pretty solution,
> > but I think it will work, and pretty sure it won't hurt anything.
> >
> > Bob U.
> >
> > At 02:15 PM 3/28/00 -0500, you wrote:
> > >Hi,
> > >
> > >My project requires a peak current of about 1.5 amps. I need a regulated
> > >source that I can put together quickly out of parts I have. Can I put to
> > >7805s in parallel t get more current? How well do they share or is there
> > >some other way of doing it?
> > >
> > >Thanks
> > >Mike
> > >
> > >
> > >
> > >eGroups.com Home: http://www.egroups.com/group/basicstamps/
> > >http://www.egroups.com - Simplifying group communications
> >
> >
> >
> > -- Check out your group's private Chat room
> > -- http://www.egroups.com/ChatPage?listName=basicstamps&m=1
> >
> >
>
>
>
>
> eGroups.com home: http://www.egroups.com/group/basicstamps
> http://www.egroups.com - Simplifying group communications
>
>
>
>
--
============================================================================
* Dennis Clark Aristocrat at heart dlc@v... www.verinet.com/~dlc *
* Be well, do good work, and stay in touch -- Garrison Keillor *
============================================================================
Good idea; yes, that solves the problem of current sharing, but it costs
you a couple hundred millivolts of load regulation. The degeneration
resistance idea works well with parallel transistors where you can put the
current sharing resistors inside the loop and the feedback keeps the output
voltage accurate, but there's no (easy) way to do that with a 7805.
The PNP booster transistor idea can work very well with one caution: you
lose most of the protection features of the 7805. The current is limited
only by the beta of the boost transistor and the thermal shutdown will
protect the 7805, but NOT the PNP.
In Michael's second message, he says he wants to use this to spin up a hard
drive and only needs the extra current during the time the motor is coming
up to speed. That will be several seconds, so a large capacitor is probably
not practical.
I don't understand why the voltage would be 4.9V with two 7805s in
parallel; in my experience, the output goes to higher of the two voltages
at light to moderate loads, then falls to the voltage of the second one
when the first is overloaded. But modern 7805s are much more accurate than
the data sheet implies, so you would probably be hard pressed to see that
happening. I'd have to guess that the 7805s are putting out 5.00V and you
are losing it in the wiring.
Instead of paralleling 7805s, how about using two of them separately, one
to power the motor and another to power your logic? That way, you don't
care if the motor one isn't very accurate during spin-up, your logic should
be completely happy at all times and both regulators should be OK.
Bob U.
At 07:04 AM 3/29/00 -0600, you wrote:
>if you put a small resistance on each output leg (like 0.33 ohm) would that
>help equalize the current flow (and heat) ?
>
>richard
>I am trying to spin up a hard drive and power a breadboard packed with
>components. I tried to parallel them and the voltage was about 4.9v but was
>unsure if it could slip at anytime. It will probably be at about 1.5 amps
>for 8 - 10 seconds until it spins up then drop to about .5 - .8
>Thanks
>Mike
>Richard/Mike
>
>Good idea; yes, that solves the problem of current sharing, but it costs
>you a couple hundred millivolts of load regulation. The degeneration
>resistance idea works well with parallel transistors where you can put the
>current sharing resistors inside the loop and the feedback keeps the output
>voltage accurate, but there's no (easy) way to do that with a 7805.
>
>The PNP booster transistor idea can work very well with one caution: you
>lose most of the protection features of the 7805. The current is limited
>only by the beta of the boost transistor and the thermal shutdown will
>protect the 7805, but NOT the PNP.
>
>In Michael's second message, he says he wants to use this to spin up a hard
>drive and only needs the extra current during the time the motor is coming
>up to speed. That will be several seconds, so a large capacitor is probably
>not practical.
>
>I don't understand why the voltage would be 4.9V with two 7805s in
>parallel; in my experience, the output goes to higher of the two voltages
>at light to moderate loads, then falls to the voltage of the second one
>when the first is overloaded. But modern 7805s are much more accurate than
>the data sheet implies, so you would probably be hard pressed to see that
>happening. I'd have to guess that the 7805s are putting out 5.00V and you
>are losing it in the wiring.
Might depend on the accuracy of the test instrument as well, I'd think.
>
>Instead of paralleling 7805s, how about using two of them separately, one
>to power the motor and another to power your logic? That way, you don't
>care if the motor one isn't very accurate during spin-up, your logic should
>be completely happy at all times and both regulators should be OK.
>
>Bob U.
>
>At 07:04 AM 3/29/00 -0600, you wrote:
>>if you put a small resistance on each output leg (like 0.33 ohm) would that
>>help equalize the current flow (and heat) ?
>>
>>richard
> >I am trying to spin up a hard drive and power a breadboard packed with
> >components. I tried to parallel them and the voltage was about 4.9v but was
> >unsure if it could slip at anytime. It will probably be at about 1.5 amps
> >for 8 - 10 seconds until it spins up then drop to about .5 - .8
> >Thanks
> >Mike
>
>
>
>
>-- Check out your group's private Chat room
>-- http://www.egroups.com/ChatPage?listName=basicstamps&m=1
>
>
>