BS2p power usage
Archiver
Posts: 46,084
I've built a circuit that uses a BS2p 40 pin stamp. It's powered by an
unregulated wall wart that provides about 14 volts dc. I run that voltage
through a 7805 to give the stamp 5v. When I plug my circuit in the 7805
gets hot and the circuit as a whole draws about 50ma. The rest of the
circuit is relays and leds, but they're all driven by the stamp and the
stamp is currently programmed to just send a debug message.
If I plug in the circuit without the stamp it draws 5ma, which I assume is
being used by the 7805
My questions are:
is a 7805 supposed to run hot? how much amperage should a BS2p be drawing
if it's not using it's I/O ports at all?
Any advice welcome!
unregulated wall wart that provides about 14 volts dc. I run that voltage
through a 7805 to give the stamp 5v. When I plug my circuit in the 7805
gets hot and the circuit as a whole draws about 50ma. The rest of the
circuit is relays and leds, but they're all driven by the stamp and the
stamp is currently programmed to just send a debug message.
If I plug in the circuit without the stamp it draws 5ma, which I assume is
being used by the 7805
My questions are:
is a 7805 supposed to run hot? how much amperage should a BS2p be drawing
if it's not using it's I/O ports at all?
Any advice welcome!
Comments
What kind of bypassing caps do you have befor and after the '05? also adding
just a small bolt on type heat sink helps alot with heat disapation.
Lincoln Worsham
Slap Jack Sound
Original Message
From: "raoul vaneigem" <raoul@c...>
To: <basicstamps@yahoogroups.com>
Sent: Monday, July 08, 2002 10:33 AM
Subject: [noparse][[/noparse]basicstamps] BS2p power usage
> I've built a circuit that uses a BS2p 40 pin stamp. It's powered by an
> unregulated wall wart that provides about 14 volts dc. I run that voltage
> through a 7805 to give the stamp 5v. When I plug my circuit in the 7805
> gets hot and the circuit as a whole draws about 50ma. The rest of the
> circuit is relays and leds, but they're all driven by the stamp and the
> stamp is currently programmed to just send a debug message.
>
> If I plug in the circuit without the stamp it draws 5ma, which I assume is
> being used by the 7805
>
> My questions are:
>
> is a 7805 supposed to run hot? how much amperage should a BS2p be drawing
> if it's not using it's I/O ports at all?
>
> Any advice welcome!
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>I've built a circuit that uses a BS2p 40 pin stamp. It's powered by an
>unregulated wall wart that provides about 14 volts dc. I run that voltage
>through a 7805 to give the stamp 5v. When I plug my circuit in the 7805
>gets hot and the circuit as a whole draws about 50ma. The rest of the
>circuit is relays and leds, but they're all driven by the stamp and the
>stamp is currently programmed to just send a debug message.
>
>If I plug in the circuit without the stamp it draws 5ma, which I assume is
>being used by the 7805
>
>My questions are:
>
>is a 7805 supposed to run hot? how much amperage should a BS2p be drawing
>if it's not using it's I/O ports at all?
>
>Any advice welcome!
Some useful information:
http://www.chipcenter.com/circuitcellar/askus/feb00/Q1_00_41.htm
-Check out "Maximum Average Power Dissipation" under "Typical Performance
Characteristics" for a TO-220 case style.
-Also look at "Peak Output Current" under "Typical Performance
Characteristics",
noticing the input to output voltage differential.
http://www.national.com/ds/LM/LM7805C.pdf
5mA at 25C sounds about right from just the regulator.
By the sound of your post, it seems as though you are pulling much more
than 50mA.
How are you measuring current, and when you say "the circuit as a whole draws
about 50ma" what exactly do you mean?
Assuming that the BS2p draws 50mA just sitting there, and the regulator
draws 5mA,
then dropping from 14V to 5V creates 9V that needs to be dissipated through
the
regulator in the form of HEAT. 9V x 55mA = 0.450 Watts This should be ok for
a 7805 which should be able to handle about 1A without a Heatsink unless
you have
a 78L05 which can only handle about 100mA, but suppose you have a couple of
led's
on at 20mA a piece and a relay that draws 50mA which would typically be on
the low
side. That's an additional 90mA to your existing 55mA making your
regulator sweat
1.3Watts (9V x 145mA) ...POP!
...Add a heatsink!
As one person suggested, using a bolt helps greatly. If you find you need
more heat
dissipation that that you can cut 1 inch squares of aluminum with a hole
drilled in
the center(s) or large washer(s), each aluminum square/large washer
separated by a
smaller washer to provide a small air gap and 'kabob' them onto the bolt.
One last question, What if any capacitors do you have on the
regulator? Both input
and output?
Beau Schwabe Mask Designer IV - ATL
National Semiconductor Enterprise Networking Business Unit
500 Pinnacle Court, Suite 525 Wired Communications Division
Mail Stop GA1 Norcross, GA 30071
>5mA at 25C sounds about right from just the regulator.
>
>By the sound of your post, it seems as though you are pulling much more
>than 50mA.
>
>How are you measuring current, and when you say "the circuit as a whole draws
>about 50ma" what exactly do you mean?
I plug the circuit into a wall wart, then measure the current used with my
multimeter, by disconnecting the power and then placing the leads of the
multimeter in series with the power supply. That way all the current flows
through the multimeter and can be measured. Is this correct?
>One last question, What if any capacitors do you have on the
>regulator? Both input
>and output?
I have a 1uf on the 12v input and a 100 uf on the output
> >
> >5mA at 25C sounds about right from just the regulator.
> >
> >By the sound of your post, it seems as though you are pulling much more
> >than 50mA.
> >
> >How are you measuring current, and when you say "the circuit as a whole
> draws
> >about 50ma" what exactly do you mean?
>
>I plug the circuit into a wall wart, then measure the current used with my
>multimeter, by disconnecting the power and then placing the leads of the
>multimeter in series with the power supply. That way all the current flows
>through the multimeter and can be measured. Is this correct?
That's correct.
> >One last question, What if any capacitors do you have on the
> >regulator? Both input
> >and output?
>
>I have a 1uf on the 12v input and a 100 uf on the output
http://www.national.com/ds/LM/LM78XX.pdf
The 100uf is a bit excessive, all you need is a .1uF on the output and a .22uF
on the input. Refer to the PDF under "Typical Applications-->Fixed Output
Regulator"
Other places to look in the PDF;
"Application Hints-->Transient Voltages"
and
"Typical Applications-->High Input Voltage Circuits"
R = (Vsupply - Vreg) / ( 2 x Imax )
Taking your values from earlier...
R = (14V - 5V) / ( 2 x 50mA)
R = 9V / 100mA
R = 90 Ohms (or closest approximation 100 Ohm)
...However if you expect larger currents as I suggested of 145mA
or more...
R = (14V - 5V) / ( 2 x 145mA)
R = 9V / 290mA
R = 31 Ohms (or closest approximation 33 Ohm)
Beau Schwabe Mask Designer IV - ATL
National Semiconductor Enterprise Networking Business Unit
500 Pinnacle Court, Suite 525 Wired Communications Division
Mail Stop GA1 Norcross, GA 30071
Thanks for the info. One more question for you- are the excessive
capacitors causing power drain? Do I need to unsolder them and replace
them, or can I just leave them in and do it right next time? [noparse];)[/noparse]
Incidentally, I added a heat sink which helped with the 7805 overheating
> > >One last question, What if any capacitors do you have on the
> > >regulator? Both input
> > >and output?
> >
> >I have a 1uf on the 12v input and a 100 uf on the output
>
>
>http://www.national.com/ds/LM/LM78XX.pdf
>
>The 100uf is a bit excessive, all you need is a .1uF on the output and a .22uF
>on the input. Refer to the PDF under "Typical Applications-->Fixed Output
>Regulator"
>
>
>Other places to look in the PDF;
>
>"Application Hints-->Transient Voltages"
>
>and
>
>"Typical Applications-->High Input Voltage Circuits"
> R = (Vsupply - Vreg) / ( 2 x Imax )
>
> Taking your values from earlier...
>
> R = (14V - 5V) / ( 2 x 50mA)
>
> R = 9V / 100mA
>
> R = 90 Ohms (or closest approximation 100 Ohm)
>
> ...However if you expect larger currents as I suggested of 145mA
>or more...
>
> R = (14V - 5V) / ( 2 x 145mA)
>
> R = 9V / 290mA
>
> R = 31 Ohms (or closest approximation 33 Ohm)
>
>
>
>
>
>
>
>Beau Schwabe Mask Designer IV - ATL
>National Semiconductor Enterprise Networking Business Unit
>500 Pinnacle Court, Suite 525 Wired Communications Division
>Mail Stop GA1 Norcross, GA 30071
>
>
>
>
>To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
>from the same email address that you subscribed. Text in the Subject and
>Body of the message will be ignored.
>
>
>Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>Beau,
>
>Thanks for the info. One more question for you- are the excessive
>capacitors causing power drain? Do I need to unsolder them and replace
>them, or can I just leave them in and do it right next time? [noparse];)[/noparse]
This might be ok... the .1uF is mainly to stabalize the high frequency
ripple generated from the regulator.... a 100uF is not fast enough to
respond to the ripple.
>Incidentally, I added a heat sink which helped with the 7805 overheating
>
>
> > > >One last question, What if any capacitors do you have on the
> > > >regulator? Both input
> > > >and output?
> > >
> > >I have a 1uf on the 12v input and a 100 uf on the output
> >
> >
> >http://www.national.com/ds/LM/LM78XX.pdf
> >
> >The 100uf is a bit excessive, all you need is a .1uF on the output and a
> .22uF
> >on the input. Refer to the PDF under "Typical Applications-->Fixed Output
> >Regulator"
> >
> >
> >Other places to look in the PDF;
> >
> >"Application Hints-->Transient Voltages"
> >
> >and
> >
> >"Typical Applications-->High Input Voltage Circuits"
> > R = (Vsupply - Vreg) / ( 2 x Imax )
> >
> > Taking your values from earlier...
> >
> > R = (14V - 5V) / ( 2 x 50mA)
> >
> > R = 9V / 100mA
> >
> > R = 90 Ohms (or closest approximation 100 Ohm)
> >
> > ...However if you expect larger currents as I suggested of 145mA
> >or more...
> >
> > R = (14V - 5V) / ( 2 x 145mA)
> >
> > R = 9V / 290mA
> >
> > R = 31 Ohms (or closest approximation 33 Ohm)
> >
> >
> >
> >
> >
> >
> >
> >Beau Schwabe Mask Designer IV - ATL
> >National Semiconductor Enterprise Networking Business Unit
> >500 Pinnacle Court, Suite 525 Wired Communications Division
> >Mail Stop GA1 Norcross, GA 30071
> >
> >
> >
> >
> >To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> >from the same email address that you subscribed. Text in the Subject and
> >Body of the message will be ignored.
> >
> >
> >Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
>from the same email address that you subscribed. Text in the Subject and
>Body of the message will be ignored.
>
>
>Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
Beau Schwabe Mask Designer IV - ATL
National Semiconductor Enterprise Networking Business Unit
500 Pinnacle Court, Suite 525 Wired Communications Division
Mail Stop GA1 Norcross, GA 30071