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voltage regulator question adjustable voltage needed

P

panfilero

Jan 1, 1970
0
Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

much thanks!
J.
 
I

IanM

Jan 1, 1970
0
OP crossposted excessively, I have taken the liberty of setting
followups to sci.electronics.basics *ONLY*
panfilero said:
Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok?
*NO*
or is there a beinifit to folowing the example on
the datasheet? *YES*

much thanks!
J.
You looked up the data sheet so are probably smart enough to learn basic
electronics without too much difficulty. You need a much better grasp
of circuit theory and instruction on asking intelligent questions before
you can expect the professional users of groups other than
sci.electronics.basics to be friendly.

You can start learning by answering the questions (in the group
sci.electronics.basics *ONLY*):

What is your load and what is its minimum and maximum current requirements?

What is your supply voltage and what is it from?

N.B LM340-12 is absolutely the *wrong* choice unless you only want to
adjust from 12V to lets say 15V.

That's my good deed for the day done, Merry Christmas all.
 
P

panfilero

Jan 1, 1970
0
That is basically a 78XX, 5V, 12V ,and 15V output.
The output of teh 5V can be boosted to 12V by lifting ground reference with resistors.
If you only need 12V, use a 7812!!!!!!!!!!!

If you need a programmable voltage or current, why not use a LM317?
Easier to get.
The LM317 goes all the way down to about 1.22 V.
Diagram is the same.

Thanks for the reply.... I'm confused now, I thought the LM340-xx was
a fixed output voltage regulator... and that if I order LM340-12 I
will be getting a 12V voltage regulator (and if I order LM340-15 it
would be the 15v regulator... and the LM340-05...etc etc). So you are
saying that if I get a LM340-05 (5V regulator) and lift the ground
(I'm not sure what lift the ground means) that I can turn it into a
12V regulator?

I do only need a 12V regulator so I will look into the 7812... looks
like I was looking at the wrong stuff... much thanks, appreciate your
help!
 
P

panfilero

Jan 1, 1970
0
That is basically a 78XX, 5V, 12V ,and 15V output.
The output of teh 5V can be boosted to 12V by lifting ground reference with resistors.
If you only need 12V, use a 7812!!!!!!!!!!!

If you need a programmable voltage or current, why not use a LM317?
Easier to get.
The LM317 goes all the way down to about 1.22 V.
Diagram is the same.

ok.... you recommended the 7812 over the LM340-12.... but I can't tell
the difference (besides the LM340 series sounding a little more
robust) check out these two descriptions for each series:

The LM78XX series of three terminal positive regulators
are available in the TO-220 package and with several
fixed output voltages, making them useful in a wide
range of applications. Each type employs internal current
limiting, thermal shut down and safe operating area protection,
making it essentially indestructible. If adequate
heat sinking is provided, they can deliver over 1A output
current. Although designed primarily as fixed voltage
regulators, these devices can be used with external components
to obtain adjustable voltages and currents.

The LM140/LM340A/LM340/LM78XXC monolithic 3-terminal positive voltage
regulators employ internal current-limiting, thermal shutdown and safe-
area compensation, making them essentially indestructible. If adequate
heat sinking is provided, they can deliver over 1.0A output current.
They are intended as fixed voltage regulators in a wide range of
applications including local (on-card) regulation for elimination of
noise and distribution problems associated with single-point
regulation. In addition to use as fixed voltage regulators, these
devices can be used with external components to obtain adjustable
output voltages and currents.


they kinda sound basically the same to me... is the 7812 really a
better more obvious choice for a fixed voltage regulator?
 
S

Spehro Pefhany

Jan 1, 1970
0
Yes, sometimes if a bit more voltage is required then say 5V,
they add a diode in series with the ground connection,
then, for a Si diode, you get 5.7V.
I am not sure if it will work very well to do make a 7805 output 12V.

You must have see the little formula on page one of the pdf datasheet,
where its says what Vout is in relation to those 2 resistors, and its
effect on load regulation.
The load regulation clearly gets worse.

You can put a 6.8V zener in the ground circuit and get around 11.8V
out. The current out of the GND pin is fairly constant.

This sometimes comes in handy if you need +/- supplies, can float the
"ground". You do have to make sure not to draw too much current from
the "minus" supply to ground (but to the + supply is okay).

But in general, if you need a 12V regulator, just buy one.
 
D

daestrom

Jan 1, 1970
0
panfilero said:
Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

Use the example in the datasheet.

The 'guts' of that thing works by passing current through from the supply
until the output pin rises to 12 V above the ground pin.

So by connecting the ground pin to the center of the voltage divider, you
trick it into passing current from the input to the output pin until just
that part of the pot has 12V across it. If the pot is set to the mid-point,
that means 12V across half the pot's resistance, or 24V across the total pot
from end to end. So the output pin is 24V above the return line and just
12V above the ground pin of the chip.

Does that help?

daestrom
 
E

ehsjr

Jan 1, 1970
0
panfilero said:
ok.... you recommended the 7812 over the LM340-12.... but I can't tell
the difference (besides the LM340 series sounding a little more
robust) check out these two descriptions for each series:

The LM78XX series of three terminal positive regulators
are available in the TO-220 package and with several
fixed output voltages, making them useful in a wide
range of applications. Each type employs internal current
limiting, thermal shut down and safe operating area protection,
making it essentially indestructible. If adequate
heat sinking is provided, they can deliver over 1A output
current. Although designed primarily as fixed voltage
regulators, these devices can be used with external components
to obtain adjustable voltages and currents.

The LM140/LM340A/LM340/LM78XXC monolithic 3-terminal positive voltage
regulators employ internal current-limiting, thermal shutdown and safe-
area compensation, making them essentially indestructible. If adequate
heat sinking is provided, they can deliver over 1.0A output current.
They are intended as fixed voltage regulators in a wide range of
applications including local (on-card) regulation for elimination of
noise and distribution problems associated with single-point
regulation. In addition to use as fixed voltage regulators, these
devices can be used with external components to obtain adjustable
output voltages and currents.


they kinda sound basically the same to me... is the 7812 really a
better more obvious choice for a fixed voltage regulator?

The 7812 costs less, and may be more widely available.

Ed
 
H

Hugh Gibbons

Jan 1, 1970
0
Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

Yes, a huge advantage. The output pin can put out an amp of current,
but used that way, it only produces output voltages greater than the
nominal output. So you can turn a 12V regulator into a 13V or 15V
regulator if you want.

What you seem to want is a lower voltage regulator. The best way to
get that is to use a lower voltage regulator.

Starting from +12V and dividing down to a lower level won't give you a
regulated voltage. It'll give you the equivalent of a regulated voltage
with a large source impedance.
 
S

Salmon Egg

Jan 1, 1970
0
panfilero said:
Hello,

ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

http://www.national.com/ds/LM/LM340.pdf

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...

would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

much thanks!
J.

Even simple regulator integrated circuits are complex devices. The
equation indicates that the output voltage will be 5V greater than
whatever the divider supplies.

I have not studied this data sheet in detail. It is not part of my
retirement description. So with the clue here and those given by others,
you have some work to do on your own

Bill
 
J

Jasen Betts

Jan 1, 1970
0
The 7812 costs less, and may be more widely available.

the LM340T-12 is a 7812. The price will be the same because it is
exactly the same part.
 
P

panfilero

Jan 1, 1970
0
Yes, a huge advantage.  The output pin can put out an amp of current,
but used that way, it only produces output voltages greater than the
nominal output.  So you can turn a 12V regulator into a 13V or 15V
regulator if you want.  

What you seem to want is a lower voltage regulator.   The best way to
get that is to use a lower voltage regulator.  

Starting from +12V and dividing down to a lower level won't give you a
regulated voltage.  It'll give you the equivalent of a regulated voltage
with a large source impedance.


Hmmmm.... well since I don't need 1A of current... I can't see the
huge advantage... and when you say it will give me the equivalent of a
regulated voltage with a large source impedance.... is that a bad
thing? The equivalent of a regualted votlage sounds good... and so
does a large source impedance... I'm sorry, I'm not seeing the
drawback here... I'm probally just not understanding what you are
trying to say there....

thanks!
 
S

Salmon Egg

Jan 1, 1970
0
panfilero said:
Hmmmm.... well since I don't need 1A of current... I can't see the
huge advantage... and when you say it will give me the equivalent of a
regulated voltage with a large source impedance.... is that a bad
thing? The equivalent of a regualted votlage sounds good... and so
does a large source impedance... I'm sorry, I'm not seeing the
drawback here... I'm probally just not understanding what you are
trying to say there....

Again, I am not fully reading all the background.

One of the reasons one uses a voltage regulator is to have a LOW
IMPEDANCE SOURCE. To have a high impedance source is to have a constant
current source.

Bill
 
P

Palindrome

Jan 1, 1970
0
panfilero said:
Hmmmm.... well since I don't need 1A of current... I can't see the
huge advantage... and when you say it will give me the equivalent of a
regulated voltage with a large source impedance.... is that a bad
thing? The equivalent of a regualted votlage sounds good... and so
does a large source impedance... I'm sorry, I'm not seeing the
drawback here... I'm probally just not understanding what you are
trying to say there....

The large source impedance means that the output voltage will drop (a
lot) as the load current increases. So, even though the source voltage
feeding the divider is well regulated, the output voltage from the
divider chain won't be.

This may, or may not, be a problem. If your load is extremely light,
then its effect on the voltage divider will be small.

If your load is extremely constant, a simple series resistor (to drop
the excess voltage) may be all that is needed - the combination of load
and series resistor will form the voltage divider.

However, using a design that produces a regulated output voltage with a
low source impedance has the advantage that it can be used with a wide
variety of loads, without the output voltage being affected.
 
D

Dave Martindale

Jan 1, 1970
0
panfilero said:
ok, I'm looking at the datasheet for a 12V voltage regulator...
LM340-12.... here's the datasheet:

On the first page under Typical Applications theres a schematic
titled: adjustable output regulator... and I can't figure it out,
there's an equation there for Vout, but to me it looks like Vout would
be 12V because it's from the output pin to ground... there's a voltage
divider with a pot, but I don't see how the pot does anythign since
Vout is taken from Vout pin with respect to ground... Woooo but the
ground pin of the voltage regulator is in the middle of the votlage
divider.... so there is 12V across R1.... I'm confused.

Just about the only sane way to use this particular voltage regulator is
as a fixed-voltage regulator at the voltage it's designed for (12 V), or
possibly slightly higher. If you want to adjust the voltage, there are
much better options - e.g. the LM317.
Basically I want to use this voltage regulator as an adjustable
voltage out... I was originally just planning to put in a voltage
divider to ground at the Vout pin and have one of the resistors be a
pot and take my voltage out from the middle of the divider...
would my way be ok? or is there a beinifit to folowing the example on
the datasheet?

Your approach will work fine if the voltage out of the pot has
essentially zero load on it - e.g. if it's connected to one input of a
DMM. But if it's supplying any significant amount of current to
anything, then the "adjusted" voltage will be pulled down by the extra
load that's effectively in parallel with the lower half of the pot.
Worse, if the load varies, the voltage will vary too. You can reduce
the amount of variation by using a lower-resistance pot, but then it
wastes more power in the pot itself.

Plus you need to watch the maximum rating of the pot. A 1/4 W pot
means it can dissipate 1/4 W over the entire resistance element. But if
you have the shaft set to the 90% position, all the load current is
passing through only 10% of the pot element, and it can handle only 25
mW. In practice, this means there's a maximum current it can handle no
matter the shaft position. For example, a 1/4 W pot across 12 V can
handle about 144 mA - and that's the sum of the load current *plus* the
current dumped through the pot itself.

In contrast, if you use a LM317, you can also use a pot to set the
output voltage, but the pot wiper is connected to the regulator's
adjustment terminal, not the load. The pot handles a few mA of current
in the voltage divider, so there's no heating to worry about. You can
easily adjust the output voltage from about 1.25 V up to 30 V or so, and
the voltage will remain constant under a wide range of load current.
And that's what you want a regulated power supply to do!

Lookup up the LM317. It requires at most a few more parts around it
than a fixed regulator, but it's much more flexible.

Dave
 
E

ehsjr

Jan 1, 1970
0
Jasen said:
the LM340T-12 is a 7812. The price will be the same because it is
exactly the same part.

Wrong. Same part, *different* price. Check Digikey, Mouser, Newark.
I did the work for you. For a TO-220 package qty1:

***** *COST* *******
LM7812 LM340T12
Digikey | $0.45 | $1.74 |
Mouser | $0.26 | $0.50* |
Newark | $0.375** | $1.65 |

* = " Cross Reference Match For: LM340T12" and the
cross-reference pn given is 511-L7812CV
** Newark has a promotional price of $0.329

The part numbers found by the search and used in the table
above:

******** *Part Numbers* ********
LM7812 LM340T12
Digikey | LM7812ACT-ND | LM340T-12-ND |
Mouser | 512-LM7812ACT | 511-L7812CV |
Newark | 86K1600 | 41K4771 |

In each case, the 7812 costs less.

Ed
 
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