Help: Need DC Voltage Reducer

[Greg]

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

 

[Old Mac User]

Well, yes, that circuit will draw current from the battery all the
time. Depending on the values of the resistors, this may degrade the
battery quickly.

Another way is to set up circuit with the battery, a resistor and the
timer in series.
The appropriate value for the resistor depends upon the resistance of
the timer.
The idea is to pick a resistor that absorbs 7.5 v and leaves 1.5 v for
the timer.
This, too, is not the most efficient possible circuit since most of the
energy from
the battery will be consumed by the resistor.


Greg wrote:

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

 

[Chris]

Greg wrote:

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

Hi, Greg. Congratulations on getting the Forrest Mims book -- it's
probably the best total newbie start to learning electronics. You can
easily do what you're suggesting, and it won't short out the battery
(view in fixed font or cut&paste to M$ Notepad):
|
| .--------. .
| | |
| | .-. |
| | 750| | |I
| | | | |
| | '-' V
|9V | |
| +| o-----------.
| --- | |
| - .-. |
| | 150| | | _ |
| | | | |I1 / \ |I2
| | '-' | ( T ) |
| | | V \_/ V
| | | |
| | | |
| '--------o-----------'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

This will provide you with a steady 1.5V at the timer -- as long as it
doesn't use very much current.

Look at the circuit above. I = I1 + I2. If I2 is near zero, then
there's about 900 ohms load across your battery, and I = 10mA (Ohms Law
-- 9V / 900 ohms = 0.01 amps = 10mA). If you've got one of those small
LCD watch battery clocks, they generally use about 1mA of current.
That won't affect the voltage divider too much -- a 1mA timer load will
bring the divider voltage down to 1.35V or so. You should still be OK
here.

From a practical standpoint, I'd assume the current draw from the
little timer isn't quite steady -- it probably requires more current in

very short pulses for the timer crystal and the display IC. This is
probably going to interfere with the timer operation. If you want to
use this circuit, you really should go to Radio Shack and purchase a
10uF (microfarad) cap, and put it in parallel with the 150 ohm resistor
like this:

|
| .--------. .
| | |
| | .-.
| | 750| |
| | | |
| | '-'
|9V | |
| +| o-----o-----.
| --- | | |
| - .-. | |
| | 150| | +| _
| | | | --- / \
| | '-' --- ( T )
| | | 10uF| \_/
| | | | |
| | | | |
| '--------o-----o-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Be sure to attach the lead of the cap marked "+" on the side of the
circuit where the 750 ohm and 150 ohm resistors meet. This will work
for a fleapower load. But there are two problems here.

First, if your little clock has a piezo alarm, current draw on the 1.5V
is going to go way up when the alarm goes off. That will bring the
voltage way down, and probably stop your clock. Even worse, if you've
got one of those quartz clocks, their average current draw is more like
10mA, which is completely unacceptable, alarm or not. Not only that,
but the quartz clock draws current in spikes at the moment the hands
move, which will ruin things for sure.

The second problem is also a biggie. For the above circuit, you're
burning 10mA to get a useful 1mA at 1.5V with 10% regulation. In other
words, you're using 90 milliwatts of power (9V * 10mA) of power to get
a useful 1.5 milliwatt (1.5V * 1mA). Your batteries aren't going to
last nearly as long as they could.

If this is just an experiment, you should be OK here. If you're
planning on running this permanently and don't own stock in Ray O Vac,
you can get an IC from Radio Shack that will make your voltage
regulator more efficient. This IC is required if you've got a quartz
clock -- you might be able to make the ohms of the resistors a lot
lower to get better regulation for a larger load, but the 9V battery
can't crank that much current.

If you'd like, please feel free to post again for more advice. Be sure
to describe what kind of timer you're using, how much current it needs
if you know, and also if it has an alarm output and what type.

Good luck
Chris

 

[John Popelish]

Greg wrote:

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

You might use a small 3 terminal voltage regulator, like the LM317LZ,
with the output voltage set with a couple resistors.

http://cache.national.com/ds/LM/LM317L.pdf

If you can live with 1.25 volts out, instead of 1.5, you need only a
minimum load resistor. The data sheet shows a 240 ohm resistor from
output to the ADJ pin, to handle the absolute worst case for minimum
load current, but it will probably work fine with a 560 to 1000 ohm
resistor there (for lower battery load). The data sheet says that the
typical minimum load current is about 1.5 mA with a 3 to 15 volt
input. That would be 1.25/.0015 = 833 ohms, especially true if this
thing will never see its full rated temperature.

 

[Chris]

John Popelish wrote:

Greg wrote:
I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

You might use a small 3 terminal voltage regulator, like the LM317LZ,
with the output voltage set with a couple resistors.

http://cache.national.com/ds/LM/LM317L.pdf

If you can live with 1.25 volts out, instead of 1.5, you need only a
minimum load resistor. The data sheet shows a 240 ohm resistor from
output to the ADJ pin, to handle the absolute worst case for minimum
load current, but it will probably work fine with a 560 to 1000 ohm
resistor there (for lower battery load). The data sheet says that the
typical minimum load current is about 1.5 mA with a 3 to 15 volt
input. That would be 1.25/.0015 = 833 ohms, especially true if this
thing will never see its full rated temperature.

Hi, John. If the OP wants to go to Radio shack, he'll have to settle
for an LM317T:

http://www.national.com/ds.cgi/LM/LM117.pdf

Radio Shack page:
http://tinyurl.com/ze5q9
Catalog #: 276-1778
$2.29 USD

The LM317 IC is a 3-pin linear voltage regulator available at Radio
Shack, that will maintain a steady output voltage for a varying load.
Not only that, but it will only use 3.5mA extra, instead of the 10mA we
were talking about for your voltage divider. Not quite as good as the
LM317L, but still OK for battery use.

If the OP gets one of these, he can follow the pinout on the package
after reading the data sheets, and put this fairly simple circuit
together:

|
| _____
| IN | |OUT
| .-------o-----|LM317|-o-----o-----.
| | +| |_____| | +| |
| | --- | .-. --- |
| +| --- ADJ| | | --- |
| --- 10uF| |360| |10uF| |
| - | | '-' | |
|9V | | | | | |
| | | o----' | / \
| | | | | ( T )
| | | | | \_/
| | | .-. | |
| | | 68| | | |
| | | | | | |
| | | '-' | |
| | | | | |
| '-------o--------o----------o-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

He'll need a 360 ohm resistor, a 68 ohm resistor, and two 10uF caps
(all of which he should be able to get at Radio Shack). And as John
says, if your timer can live with 1.23V, you can forget the 68 ohm
resistor, and just connect the ADJ pin and the other side of the 360
ohm resistor to GND. But whether the load is an LCD clock or a quartz
clock, you really should include the 360 ohm resistor to GND -- either
load is pulsed current, and you should always make sure there's at
least 3.5mA going through the LM317.

The great thing about this circuit is, he will be able to get a steady
1.5V output over the entire current range of his transistor battery
(from zero to 100mA). Also, he only "wastes" 3.5mA. This would mean
that, for a 1mA load, he'll only need 4.5mA from his battery rather
than 90mA.

Such a deal.

Thanks for the spot, John.

Cheers
Chris

 

[Chris]

Chris wrote:
..

Hi, John. If the OP wants to go to Radio shack, he'll have to settle
for an LM317T:

http://www.national.com/ds.cgi/LM/LM117.pdf

Radio Shack page:
http://tinyurl.com/ze5q9
Catalog #: 276-1778
$2.29 USD

The LM317 IC is a 3-pin linear voltage regulator available at Radio
Shack, that will maintain a steady output voltage for a varying load.
Not only that, but it will only use 3.5mA extra, instead of the 10mA we
were talking about for your voltage divider. Not quite as good as the
LM317L, but still OK for battery use.

If the OP gets one of these, he can follow the pinout on the package
after reading the data sheets, and put this fairly simple circuit
together:

|
| _____
| IN | |OUT
| .-------o-----|LM317|-o-----o-----.
| | +| |_____| | +| |
| | --- | .-. --- |
| +| --- ADJ| | | --- |
| --- 10uF| |360| |10uF| |
| - | | '-' | |
|9V | | | | | |
| | | o----' | / \
| | | | | ( T )
| | | | | \_/
| | | .-. | |
| | | 68| | | |
| | | | | | |
| | | '-' | |
| | | | | |
| '-------o--------o----------o-----'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

He'll need a 360 ohm resistor, a 68 ohm resistor, and two 10uF caps
(all of which he should be able to get at Radio Shack). And as John
says, if your timer can live with 1.23V, you can forget the 68 ohm
resistor, and just connect the ADJ pin and the other side of the 360
ohm resistor to GND. But whether the load is an LCD clock or a quartz
clock, you really should include the 360 ohm resistor to GND -- either
load is pulsed current, and you should always make sure there's at
least 3.5mA going through the LM317.

The great thing about this circuit is, he will be able to get a steady
1.5V output over the entire current range of his transistor battery
(from zero to 100mA). Also, he only "wastes" 3.5mA. This would mean
that, for a 1mA load, he'll only need 4.5mA from his battery rather
than 90mA.

Such a deal.

Thanks for the spot, John.

Cheers
Chris

Post in haste, repent at leisure. ;-)

This would mean
that, for a 1mA load, he'll only need 4.5mA from his battery rather
than 90mA.

Rather than 10mA.

But whether the load is an LCD clock or a quartz
clock, you really should include the 360 ohm resistor to GND -- either
load is pulsed current, and you should always make sure there's at
least 3.5mA going through the LM317.

The OP will always need the load resistor of 360 ohms for low output
current. He needs a minimum current from the LM317 OUT pin of 3.5mA,
or the output voltage will rise above its resistor-programmed point.
If his load never goes below 3.5mA, he can use higher value resistors
to set the programmed output voltage. And of course, if he only needs
1.2v, and his load current is greater than 3.5mA, he can forget both
resistors entirely. But connecting it directly to GND is only if he
only needs 1.2V.

Assuming the OP is using an LCD timer clock without alarm, he may only
need 1mA. Using an LM317 may only then double his battery life, which
may or may not be a good investment. But if he's using a quartz clock,
or if he's got an alarm function, he probably won't have a choice, if
he wants his battery to last a while. Getting good regulation for even
a 10mA load with a voltage divider will mean cranking 100mA from the
battery, which should kill it quickly.

Possibly the OP could describe his timer a little better.

Cheers
Chris

 

[Dorian McIntire]

"Greg" <gfight@yourpantstampabay.rr.com> wrote in message
news:gfight-A818AC.08471431052006@news-server1.tampabay.rr.com...

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

Since your source is a battery and doesn't require filtering and if your
current requirements are a only a few milliamps I would simply use a 7.5
volt Zener diode in series with the timer.



Unless your current requirements are low all the solutions presented here,
including mine, are very wasteful of battery power. An efficient switcher
design however would be a complicated solution to your problem.



Dorian

 

[Greg]

In article <1149082364.036179.315810@y43g2000cwc.googlegroups.com>,
"Old Mac User" <chendrixstats@juno.com> wrote:

Well, yes, that circuit will draw current from the battery all the
time. Depending on the values of the resistors, this may degrade the
battery quickly.

Another way is to set up circuit with the battery, a resistor and the
timer in series.
The appropriate value for the resistor depends upon the resistance of
the timer.
The idea is to pick a resistor that absorbs 7.5 v and leaves 1.5 v for
the timer.
This, too, is not the most efficient possible circuit since most of the
energy from
the battery will be consumed by the resistor.


Greg wrote:
I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

Thanks folks for a very informative set of solutions. It looks like I'm
making this more complicated than need be. It would be much simpler to
add an AA battery in a holder for the timer.

The timer is not a quartz device. It's an inexpensive pocket timer with
LCD display and a piezo buzzer alarm. I am going to disconnect the
buzzer and the leads will go to turn on another circuit (powered by the
9v batt) via an SCR.

I don't know the current draw of the timer. I'm still trying to get my
hands on some of them from an outfit called OzoneLabs in Canada. I
can't get them to respond to my emails or phone messages. Little timers
like these are hard to find (countdown from 99 hours).

Greg

 

在 2006年5月31日星期三 UTC+8下午8:47:15，Greg写道：

I have a DC circuit powered by a 9v battery and I need to power a 1.5v
timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book
from Radio Shack: Two resistors in series are connected to the two
terminals of a battery. V+ out is taken from between the two resistors,
V- from the neg. battery contact. V out is determined by the values of
the resistors. I have no problem doing the math, but wouldn't that
circuit short out the battery?

Thanks,
Greg

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Number one, check your line feed settings, or learn to make paragraphs.

Number two, you are responding to a post that is old enough to have a driver's license in many states.

As far as the OP goes, anyone want to sit there and type for a while go ahead. But there is such a thing as books.

 

[Jasen Betts]

On 2015-09-16, jurb6006@gmail.com <jurb6006@gmail.com> wrote:

Number one, check your line feed settings, or learn to make paragraphs.

Number two, you are responding to a post that is old enough to have a driver's license in many states.

As far as the OP goes, anyone want to sit there and type for a while go ahead. But there is such a thing as books.

what sort of drivers licence can 9 year olds get?

--
\_(ツ)_