OT: Integrated Circuit Power Supply

Archiver

I've got an integrated circuit which consists of a PIC16F819, an LCD and
couple of buttons. I want to power the IC with a 12 volt battery I have
because it is the correct physical size. So here is my problem.



1. If I use a 6 volt radio shack power supply everything runs fine.
2. If I use a 6 volt camera battery everything works as well.
3. If I use a 12 volt radio shack power supply with a resistor to
reduce the voltage to 6 volts the IC initializes fine but if I press a
button the chip resets.
4. If I use a 12 volt garage remote battery with a resister to reduce
the voltage to 6 volts the system initializes but resets when a button is
pressed.



I've tried the resister on both the positive and negative leads with same
results.



Any ideas?



The ultimate goal is to power the IC with the small 12 volt battery.



Thanks a bunch!



Scott



I probably need not remind everyone that I am an electronics novice.







[noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

Scott,

Does pressing the button cause some solenoid to turn on or activate some
lights or any devices that will cause the current consumed by the system to
increase?

What is likely happening is that when you press the button, the current
required by the system increases. Since you have a resistor in series with the
power supply line, a current increase through that resistor will cause more
voltage to be dropped across that resistor, resulting in not enough voltage to
keep
your processor running.

Do you have a voltmeter (scope is better). What happens to your "6volts" when
you press the button? It probably drops.

What voltage does your system require? 5 volts? (six seems like an odd value)
What kind of circuit is this? Does the circuit have an on-board 5 volt
regulator? If yes, you can run 12 volts to it as long as this circuit is not
consuming gobs of current.

If there is no on board 5 volt regulator, and the system runs off of 5 volts,
get yourself a 7805 regulator. You can then run 12 volts to the regulator
input and the output will be at 5 volts. You will also need some de-coupling and

filter caps on the input and output of the regulator.

Write back if this does not make sense.

Ken

---

1. If I use a 6 volt radio shack power supply everything runs fine.
2. If I use a 6 volt camera battery everything works as well.
3. If I use a 12 volt radio shack power supply with a resistor to
reduce the voltage to 6 volts the IC initializes fine but if I press a
button the chip resets.
4. If I use a 12 volt garage remote battery with a resister to reduce
the voltage to 6 volts the system initializes but resets when a button is
pressed.



I've tried the resister on both the positive and negative leads with same
results.


[noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

From: "Scott Wilkinson" <smwilkinson@e...>

> I've got an integrated circuit which consists of a PIC16F819, an LCD and
> couple of buttons. I want to power the IC with a 12 volt battery I have
> because it is the correct physical size. So here is my problem [noparse][[/noparse]...]
>
You need a voltage regulator in that circuit, Scott. Look up an LM7805.
Costs 50¢ at Digikey and solves exactly the problems you're having. Here's
the spec: http://makeashorterlink.com/?Q45B26A46

Gary

Archiver

Ken,

Thanks for the response.

The buttons are active low from the pins on the chip to ground. The
buttons count a single increment per press and write the total to
EEPROM memory on the chip and refresh the display. When the buttons
are pressed the voltage drops from 6.1 volts to 5.8 volts. The
voltages and drops are the same with the power supplies that do work
(6.1 Volt no resistor) and those that don't (12 volts de-regulated
to 6.1 volts with resistor).

Its kind of a long story as to why I'm using 6 Volts. The chip I'm
using wants 5 volts and the LCD wants 5 volts as well. While using
a 7805 regulator and capacitor I was ACCIDENTLY sending 7.5 volts to
my project all through the program development stage. Still,
everything ran fine. When it came time to switch to battery power I
noticed that at 5 volts the LCD was a little dim and slow. When I
bumped up the power to 6 Volts(through lower resistors) the display
looked better. However, the chip would reset when I pressed a
button. I tried raising the voltage all the way up to 8 volts but
it didn't help with the reset. When I ran 6 volts through the radio
shack power supply (it doesn't have a 5 volts setting) it runs
fine. I got a 6 volt battery and the project works well also but
the battery is too big.

Basically, any time I use any power supply with a resistor a button
press causes a chip reset.

Since the project is so small I'd rather try not to use the 7805 if
possible.

Let me know if you need more info.

Thanks Again!

Scott

--- In basicstamps@yahoogroups.com, smartdim@a... wrote:
> Scott,
>
> Does pressing the button cause some solenoid to turn on or
activate some
> lights or any devices that will cause the current consumed by the
system to
> increase?
>
> What is likely happening is that when you press the button, the
current
> required by the system increases. Since you have a resistor in
series with the
> power supply line, a current increase through that resistor will
cause more
> voltage to be dropped across that resistor, resulting in not
enough voltage to keep
> your processor running.
>
> Do you have a voltmeter (scope is better). What happens to
your "6volts" when
> you press the button? It probably drops.
>
> What voltage does your system require? 5 volts? (six seems like
an odd value)
> What kind of circuit is this? Does the circuit have an on-board 5
volt
> regulator? If yes, you can run 12 volts to it as long as this
circuit is not
> consuming gobs of current.
>
> If there is no on board 5 volt regulator, and the system runs off
of 5 volts,
> get yourself a 7805 regulator. You can then run 12 volts to the
regulator
> input and the output will be at 5 volts. You will also need some
de-coupling and
> filter caps on the input and output of the regulator.
>
> Write back if this does not make sense.
>
> Ken
>
>

---

>
>
>
>
> 1. If I use a 6 volt radio shack power supply everything runs
fine.
> 2. If I use a 6 volt camera battery everything works as well.
> 3. If I use a 12 volt radio shack power supply with a resistor
to
> reduce the voltage to 6 volts the IC initializes fine but if I
press a
> button the chip resets.
> 4. If I use a 12 volt garage remote battery with a resister to
reduce
> the voltage to 6 volts the system initializes but resets when a
button is
> pressed.
>
>
>
> I've tried the resister on both the positive and negative leads
with same
> results.
>
>
> [noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

Gary,

Thanks for the reply.

I just tried the 7805 with the 12 volt battery and you are correct
it works. I guess I may have to try to cram it into the project
somewhere.

Is there any other smaller solution you can think of?

Thank you!

Scott

--- In basicstamps@yahoogroups.com, "Gary W. Sims" <simsgw@c...>
wrote:
> From: "Scott Wilkinson" <smwilkinson@e...>
>
> > I've got an integrated circuit which consists of a PIC16F819, an
LCD and
> > couple of buttons. I want to power the IC with a 12 volt
battery I have
> > because it is the correct physical size. So here is my problem
[noparse][[/noparse]...]
> >
> You need a voltage regulator in that circuit, Scott. Look up an
LM7805.
> Costs 50¢ at Digikey and solves exactly the problems you're
having. Here's
> the spec: http://makeashorterlink.com/?Q45B26A46
>
> Gary

Archiver

From: "lcg30339" <smwilkinson@e...>

> I just tried the 7805 with the 12 volt battery and you are correct
> it works. I guess I may have to try to cram it into the project
> somewhere.
>
> Is there any other smaller solution you can think of?

Well, a TO-220 3-pin configuration is not huge, but we can probably get you
something smaller since your current requirements are probably quite low.
What are we looking at? 50ma total? I assume you want pin leads, not a
surface mount component, but would it help to have a lower part? That is a
part that lies close to the board? At very low current levels, I think we
can find something like an 8-pin DIP package which would take up a lot less
volume than a TO-220, especially if the 7805 you tried has a heat sink
attached.

Let me know,

Gary

Archiver

You MIGHT be able to get away with putting the regulator at the battery, and
run the +5 and ground to your board. Do you know what the maximum current
consumption of the circuit is? There is another 5 v regulator that is a T0-92
package which is the size of a small signal transitor, but is only good for
about
100 ma. You can go on-line to mouser.com or digikey and see package sizes and
output current/voltages.

Ken

---

---

> I just tried the 7805 with the 12 volt battery and you are correct
> it works. I guess I may have to try to cram it into the project
> somewhere.
>
> Is there any other smaller solution you can think of?


[noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

http://www.mouser.com//index.cfm?handler=search.processsearchb&supplierpage=%2
0511-L78L05ABZ


5 volt regulator, T0-92 package, but only good for 100 mA

Ken

---

From: "Scott Wilkinson" <smwilkinson@e...>

> I've got an integrated circuit which consists of a PIC16F819, an LCD and
> couple of buttons. I want to power the IC with a 12 volt battery I have
> because it is the correct physical size


[noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

From: <smartdim@a...>
>
>
> 5 volt regulator, T0-92 package, but only good for 100 mA
>
That's good, Ken. I forgot about TO-92. At Digikey, I found three models in
TO-92 good for 150ma from Sharp, TI, and Linear Technologies. Here's the
last one for an example, Scott: http://makeashorterlink.com/?S5AE16A46. It's
an LT1121-5 and it offers a low drop-out voltage, which is surely not an
issue since your 12V battery at full discharge won't approach the drop-out
voltage for a 5V regulator. On the other hand, the most expensive of these
units won't cost as much quantity one as the shipping charges<g>, so
over-spec'd features aren't terribly important.

Gary

Archiver

Gary and Ken,

Thanks a bunch. I'd love to tell you how many mA I'm drawing but I
can't figure out how to properly setup and read my meter. I told
you I was a novice. My gut tells me I'm not drawing much amperage.
I'm about to go lookup each of your suggested solution but they
sound plausible. My entire project size is 2in x 1.25in x .6in.
That has to hold 4 buttons, an 18 Pin PIC, LCD, battery, and a
motion switch that takes up almost 1/4 of the space.

Just as a side note. I find it strange that the LCD is not as crisp
and is slower with the 4708 vs when I used the straight resistor.
However, the chip doesn't reset with the 4708 and it does with the
resistor.

Thanks for your time.

Have a great night!

Scott

--- In basicstamps@yahoogroups.com, "Gary W. Sims" <simsgw@c...>
wrote:
> From: <smartdim@a...>
> >
> >
> > 5 volt regulator, T0-92 package, but only good for 100 mA
> >
> That's good, Ken. I forgot about TO-92. At Digikey, I found three
models in
> TO-92 good for 150ma from Sharp, TI, and Linear Technologies.
Here's the
> last one for an example, Scott: http://makeashorterlink.com/?
S5AE16A46. It's
> an LT1121-5 and it offers a low drop-out voltage, which is surely
not an
> issue since your 12V battery at full discharge won't approach the
drop-out
> voltage for a 5V regulator. On the other hand, the most expensive
of these
> units won't cost as much quantity one as the shipping charges<g>,
so
> over-spec'd features aren't terribly important.
>
> Gary

Archiver

If you are not comfortable measuring current with your meter, you can
calculate the current by measuring voltage........

Find the smallest value resistor you have.......1 ohm or less (this procedure
might "smoke" your resistor so wear safety glasses and be ready to disconnect
power........

Measure the resistance of this low ohm resistor. Then connect this resistor
in series with your 6 volt power supply the same way you did it earlier (you
should also notice that the voltage will probably be well above 6 volts,
compared to earlier)

Now with your meter connected across the resistor, power your circuit and do
what it takes to consume max current (button press etc). Record the voltage
(if you get a minus reading voltage, just note the absolute value).

The current used by your circuit = the voltage reading divided by the
measured resistance. This will get you pretty close to knowing what your circuit

consumes.

Now......if your circuit still resets with the 1 ohm or less resistor, your
reading is not valid......but if no reset you will have a ballpark value.
ken

---

Gary and Ken,

Thanks a bunch. I'd love to tell you how many mA I'm drawing but I
can't figure out how to properly setup and read my meter. I told
you I was a novice. My gut tells me I'm not drawing much amperage.
I'm about to go lookup each of your suggested solution but they
sound plausible. My entire project size is 2in x 1.25in x .6in.
That has to hold 4 buttons, an 18 Pin PIC, LCD, battery, and a
motion switch that takes up almost 1/4 of the space.

Just as a side note. I find it strange that the LCD is not as crisp
and is slower with the 4708 vs when I used the straight resistor.
However, the chip doesn't reset with the 4708 and it does with the
resistor.

Thanks for your time.

Have a great night!

Scott


[noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

Ken,

I appreciate the attempt to explain how to measure the current. I
read and reread your post 5 or 6 times and I still don't think I am
doing it correctly. With a 6 volt battery and a 1 ohm resister I
measure 5.9 volts to the circuit. When I press a button I get a .3
Volt power drop. The problem is I get measure 5.9 volts on both
sides of the 1 ohm resistor. I'm not using a digital meter so their
cold be a very small difference across the resistor.

By the way the 6 volt battery and the 1 ohm resistor did not cause a
reset. The desplay looks alot more crisp as well. its a shame I
cant fit that batter in my project.

Thanks anyway!

Scott

--- In basicstamps@yahoogroups.com, smartdim@a... wrote:
> If you are not comfortable measuring current with your meter, you
can
> calculate the current by measuring voltage........
>
> Find the smallest value resistor you have.......1 ohm or less
(this procedure
> might "smoke" your resistor so wear safety glasses and be ready to
disconnect
> power........
>
> Measure the resistance of this low ohm resistor. Then connect this
resistor
> in series with your 6 volt power supply the same way you did it
earlier (you
> should also notice that the voltage will probably be well above 6
volts,
> compared to earlier)
>
> Now with your meter connected across the resistor, power your
circuit and do
> what it takes to consume max current (button press etc). Record
the voltage
> (if you get a minus reading voltage, just note the absolute value).
>
> The current used by your circuit = the voltage reading divided by
the
> measured resistance. This will get you pretty close to knowing
what your circuit
> consumes.
>
> Now......if your circuit still resets with the 1 ohm or less
resistor, your
> reading is not valid......but if no reset you will have a ballpark
value.
> ken
>

---

> Gary and Ken,
>
> Thanks a bunch. I'd love to tell you how many mA I'm drawing but
I
> can't figure out how to properly setup and read my meter. I told
> you I was a novice. My gut tells me I'm not drawing much
amperage.
> I'm about to go lookup each of your suggested solution but they
> sound plausible. My entire project size is 2in x 1.25in x .6in.
> That has to hold 4 buttons, an 18 Pin PIC, LCD, battery, and a
> motion switch that takes up almost 1/4 of the space.
>
> Just as a side note. I find it strange that the LCD is not as
crisp
> and is slower with the 4708 vs when I used the straight resistor.
> However, the chip doesn't reset with the 4708 and it does with the
> resistor.
>
> Thanks for your time.
>
> Have a great night!
>
> Scott
>
>
> [noparse][[/noparse]Non-text portions of this message have been removed]

Archiver

What I should have explained is this is simply ohms law.......V = I*R

With some simple algebra, I = V/R

The voltmeter is connected across the resistor...that is one lead of the
meter to the battery positive, other lead of the voltmeter to the "other" side
of
the resistor......
This might be very hard to do with an analog meter.....In any event, based on
what you desribed, the circuit does not appear to use much current.



I appreciate the attempt to explain how to measure the current. I
read and reread your post 5 or 6 times and I still don't think I am
doing it correctly. With a 6 volt battery and a 1 ohm resister I
measure 5.9 volts to the circuit. When I press a button I get a .3
Volt power drop. The problem is I get measure 5.9 volts on both
sides of the 1 ohm resistor. I'm not using a digital meter so their
cold be a very small difference across the resistor.

By the way the 6 volt battery and the 1 ohm resistor did not cause a
reset. The desplay looks alot more crisp as well. its a shame I
cant fit that batter in my project.

Thanks anyway!


[noparse][[/noparse]Non-text portions of this message have been removed]