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'[EE]: Voltage regulation for circuits with relays,'
2001\07\09@134902 by John Waters

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Hi All,

I'm using a 78L05 to keep a stable 5V supply to a simple microcontroller
circuit, the power comes from a standard home-use 9V, 380Ma power adapter,
it works fine so far. However, I want to add a relay to my circuit, which
will easily take up to 80Ma when being turned ON, the 78L05 is now bearly
powerful enough to maintain a stable 5V. Other than that, when the relay is
drawing current, the power adapter output voltage will drop considerably,
this sudden drop and rise in input voltage to the 78L05 will affect its
regulated output as well. What is the usual way to handle this situation,
where a low power consumption circuit is driving a relay, with all power
being supplied by a domestic grade power adapter?

Thanks in advance!

John



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2001\07\09@135616 by Bob Ammerman

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I would consider driving the relay directly with the unregulated voltage.
This is easily done with an NPN transistor or logic FET and a resistor or
two.

This technique will not overload the 78L05, and the load on the wallwart
should be well under the 380ma spec leaving plenty of headroom for the
78L05.

What is the actual voltage coming out of the wallwart at low (or no)
current?

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

{Original Message removed}

2001\07\09@140404 by Spehro Pefhany

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At 01:49 PM 7/9/01 -0400, you wrote:
>Hi All,
>
>I'm using a 78L05 to keep a stable 5V supply to a simple microcontroller
>circuit, the power comes from a standard home-use 9V, 380Ma power adapter,
>it works fine so far. However, I want to add a relay to my circuit, which
>will easily take up to 80Ma when being turned ON, the 78L05 is now bearly
>powerful enough to maintain a stable 5V. Other than that, when the relay is
>drawing current, the power adapter output voltage will drop considerably,
>this sudden drop and rise in input voltage to the 78L05 will affect its
>regulated output as well. What is the usual way to handle this situation,
>where a low power consumption circuit is driving a relay, with all power
>being supplied by a domestic grade power adapter?

How stable are you expecting it to be? The 78L05 should be ok for digital
circuits, for analog you may need something better. 78M05/7805 regulators
are still very cheap and have better specs (and can dissipate more power
without a heat sink). At 80mA, I suspect you are running very hot-
considering the poor regulation on these crappy wall-warts, maybe
you need more margin.

Are you running the relay from 5V? If so, consider a second regulator for
the relay or consider running it from the 9V (but 9V relays are not as
easy to source as 12V relays).

Best regards,

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2001\07\09@143654 by John Waters

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15V at "NO" current (although marked as 9V on the wallwart).

>What is the actual voltage coming out of the wallwart at low (or no)
>current?

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2001\07\09@145312 by Peter Barick

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John,
Assuming you keep that given pwr adptr and are not running the relay
off of the regulator, have you tried this:
Have a high val C charged thru a low val R off the adptr. This is used
for the Rly and minimizes the drain at the input to the reg.

Peter, PIC Newbie

>>> EraseMEjohn_fm_watersspam_OUTspamTakeThisOuTHOTMAIL.COM 07/09/01 12:49PM >>>
Hi All,

>>I'm using a 78L05 to keep a stable 5V supply to a simple
microcontroller
>>circuit, the power comes from a standard home-use 9V, 380Ma power
adapter,
>>it works fine so far. However, I want to add a relay to my circuit

>>John



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2001\07\09@160717 by Bob Barr

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John Waters wrote:

>
>15V at "NO" current (although marked as 9V on the wallwart).
>

That's pretty typical for the unloaded output on a 9-volt wallwart.

As you increase the load, the will usually drop off fairly quickly.

Don't be surprised if the output remains above 9 volts as you crank up the
load. Many of these devices will still put out 10 or so volts even when
pretty heavily loaded. (Of course, YMMV.)

Regards, Bob

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2001\07\09@161734 by Olin Lathrop

face picon face
> I'm using a 78L05 to keep a stable 5V supply to a simple microcontroller
> circuit, the power comes from a standard home-use 9V, 380Ma power adapter,
> it works fine so far. However, I want to add a relay to my circuit, which
> will easily take up to 80Ma when being turned ON, the 78L05 is now bearly
> powerful enough to maintain a stable 5V. Other than that, when the relay
is
> drawing current, the power adapter output voltage will drop considerably,
> this sudden drop and rise in input voltage to the 78L05 will affect its
> regulated output as well. What is the usual way to handle this situation,
> where a low power consumption circuit is driving a relay, with all power
> being supplied by a domestic grade power adapter?

First, I would run the relay off the unregulated supply, although that won't
help with the overall voltage droop problem.  However, if you look closely
at the specs for your relay you will probably find a closing current and
holding current spec.  The closing current is the short term current
required to flip the relay from the off to the on state.  The holding
current is the current required to keep it in the on state once it is there.
The holding current can be several times lower.  To make use of this, tie
the unregulated supply to a resistor, then to the coil, then to the
switching transistor, then to ground.  Adjust the resistor to get the
holding current.  The closing current comes from a cap from the junction of
the resistor and coil, to the ground.  This will supply a higher current for
a short time when the relay is first switched on.  The exact value depends
on the coil resistance, the unregulated supply voltage with the relay off,
and the length of time the closing current is needed.

Oh, and don't forget the flyback diode!

********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, RemoveMEolinTakeThisOuTspamembedinc.com, http://www.embedinc.com

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2001\07\09@161810 by Lawrence Lile

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My answer: don't regulate.  Relays run fine on a wide range of voltages.  I
typically run 24 volt relays on circuits where the voltage can range from 20
volts to 35 volts, with 3 volts of ripple.  Regulation? Schmegulation.

-- Lawrence Lile


{Original Message removed}

2001\07\09@170449 by Paul Hutchinson

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face
> >15V at "NO" current (although marked as 9V on the wallwart).
> >
>
> That's pretty typical for the unloaded output on a 9-volt wallwart.
>
> As you increase the load, the will usually drop off fairly quickly.
>
> Don't be surprised if the output remains above 9 volts as you crank up the
> load. Many of these devices will still put out 10 or so volts even when
> pretty heavily loaded. (Of course, YMMV.)


Typically the output will only be at 9V when at full load.

Paul

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2001\07\09@170837 by Eisermann, Phil [Ridg/CO]

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face
I've often wondered about this, but never had time to do any testing. Do you
do this in production or personal projects? Have any data you can share?

Maybe I don't understand what goes on inside a relay. Does the coil
resistance change in steady state? Does it act like an inductor in series or
parallel with a resistor (or both)? Maybe someone can clarify this for me,
but here's the way I look at it:

Taking the original poster's 5V relay @ 80mA, the coil resistance should be
62.5 Ohm. So, if we assume the wall-wart is operating at nominal voltage (eg
9V) and the transistor saturation voltage is negligible, the coil current is
9V/62.5Ohm = 144mA. That's close to twice the specified coil current.
Realistically, the voltage will probably be higher, since the wall wart is
rated at 380mA and thus should not be all that loaded with 80mA. So the coil
current is likely to be even higher. Ok, so we need a bigger transistor, and
the saturation voltage won't quite be negligible. But still, that seems like
a lot more current than specified.

Did i get that right, or is there something else that's going on?

Looking in a DigiKey catalog, i notice some Omron relays have a %of rated
voltage for continuous operation. This goes from 110% to 200% Guess you can
do it with some relays more reliably than others?



> {Original Message removed}

2001\07\09@174745 by Bob Ammerman

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----- Original Message -----
From: "Paul Hutchinson" <RemoveMEphutchinsonspamTakeThisOuTIMTRA.COM>
To: <PICLISTEraseMEspam.....MITVMA.MIT.EDU>
Sent: Monday, July 09, 2001 5:02 PM
Subject: Re: [EE]: Voltage regulation for circuits with relays, how?


> > >15V at "NO" current (although marked as 9V on the wallwart).
> > >
> >
> > That's pretty typical for the unloaded output on a 9-volt wallwart.
> >
> > As you increase the load, the will usually drop off fairly quickly.
> >
> > Don't be surprised if the output remains above 9 volts as you crank up
the
> > load. Many of these devices will still put out 10 or so volts even when
> > pretty heavily loaded. (Of course, YMMV.)
>
>
> Typically the output will only be at 9V when at full load.
>
> Paul

....and minimum AC line voltage.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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2001\07\09@175337 by James Paul

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Regarding the wall warts, (or any power supply or transformer for
that matter), the open circuit voltage will be significantly higher
than the sepcified voltage because it is unloaded.  Just to refresh
everyones memory, the voltage/current rating on a transformer or
power supply tells you the terminal voltage at a given current draw.

For instance, a wall wart rated at 9VDC at .75 amps lets say.  This
means that the terminal voltage will be 9V when a current of 750 mA
is being drawn from it.  The open circuit voltage may be 15-20 VDC
as you are seeing here.

Just thought I'd refresh everyones memory.  I know I sometimes
forget basic things such as this, and often it drives me nuts
until I remember.  Then I wonder why I didn't thinks of that
before, or more to the point why I forgot in the first place.

                                            Regards,

                                              Jim
On Mon, 09 July 2001, Bob Barr wrote:

{Quote hidden}

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2001\07\09@182626 by Spehro Pefhany

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At 04:06 PM 7/9/01 -0500, you wrote:

>rated at 380mA and thus should not be all that loaded with 80mA. So the coil
>current is likely to be even higher. Ok, so we need a bigger transistor, and
>the saturation voltage won't quite be negligible. But still, that seems like
>a lot more current than specified.

If you have current to burn, as in the original question, you can just put
a resistor in series, reducing the nominal coil voltage (after things settle
out) to nominal under nominal input conditions. Use the next lowest relay if
the voltage isn't close enough.

Too low voltage and you can reduce the life of the relay, or it may fail to
pull in properly under low line conditions and high ambient temperature.
Too high and the coil gets too hot, meaning failures or shortened life
under high ambient temperature conditions. The coil resistance increases
with temperature, so there is a bit of self-regulation.

>Looking in a DigiKey catalog, i notice some Omron relays have a %of rated
>voltage for continuous operation. This goes from 110% to 200% Guess you can
>do it with some relays more reliably than others?

You have to look at the temperature derating and maximum Ta compared to the
actual maximum Ta for your application. More sensitive low power relays
tend to be more forgiving of high coil voltage (200% voltage = 400% power
and about 4 times the temperature rise).

Best regards,
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Embedded software/hardware/analog  Info for designers:  http://www.speff.com
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2001\07\09@182637 by Paul Hutchinson

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>
> ....and minimum AC line voltage.
>

Actually at nominal line voltage, at least with the dozen or so
manufacturers I evaluated a number of years ago.

With minimum line voltage and full current you usually get less than the
rated voltage.

Paul

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2001\07\09@183019 by Spehro Pefhany

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At 06:22 PM 7/9/01 -0400, you wrote:

>Actually at nominal line voltage, at least with the dozen or so
>manufacturers I evaluated a number of years ago.

That agrees with my experience. Average voltage at nominal input
voltage and maximum rated current. For regulation purposes you have
to take the trough voltage into account at maximum actual load
current.

Best regards,
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
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2001\07\09@210733 by Olin Lathrop

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> Maybe I don't understand what goes on inside a relay. Does the coil
> resistance change in steady state?

No, it never changes.

> Does it act like an inductor in series or
> parallel with a resistor (or both)?

For the vast majority of uses, a relay coil can be modeled very well by an
inductor in series with a resistor.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, EraseMEolinspamEraseMEembedinc.com, http://www.embedinc.com

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2001\07\10@180138 by John Waters

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I better provide more data for my original experiment:-
I used a 5V regulated supply, instead of directly using the 9V, for the
relay. From experiment, I found that my relay will close when the voltage
applied on the coil is 3.8V, the current flowing through the coil at that
moment is 82mA. If the voltage is up to 5V, the current is 98mA. So, nothing
wrong with your calculation, you just took it wrongly that I'm applying 9V
directly.

>Taking the original poster's 5V relay @ 80mA, the coil resistance should be
>62.5 Ohm. So, if we assume the wall-wart is operating at nominal voltage
>(eg
>9V) and the transistor saturation voltage is negligible, the coil current
>is
>9V/62.5Ohm = 144mA. That's close to twice the specified coil current.

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