voltage regulator question adjustable voltage needed

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.

OP crossposted excessively, I have taken the liberty of setting
followups to sci.electronics.basics *ONLY*
panfilero wrote:
> 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.

panfilero wrote:
> 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

> much thanks!
> J.

In article
<ccc61329-7213-490c-95e4-(E-Mail Removed)>,
panfilero <(E-Mail Removed)> wrote:


> 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.

In article
<ccc61329-7213-490c-95e4-(E-Mail Removed)>,
panfilero <(E-Mail Removed)> wrote:

> 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

--
Private Profit; Public Poop! Avoid collateral windfall!

On Dec 22, 11:30*pm, Hugh Gibbons <hughgibb...@noemail.bitbucket.net>
wrote:
> In article
> <ccc61329-7213-490c-95e4-60b4deb11...@r37g2000prr.googlegroups.com>,
>
> *panfilero <panfil...@gmail.com> wrote:
> > 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.


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!

In article
<c8a3f6df-fba5-45d4-86fc-(E-Mail Removed)>,
panfilero <(E-Mail Removed)> wrote:

> 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

--
Private Profit; Public Poop! Avoid collateral windfall!

panfilero wrote:
> On Dec 22, 11:30 pm, Hugh Gibbons <hughgibb...@noemail.bitbucket.net>
> wrote:
>> In article
>> <ccc61329-7213-490c-95e4-60b4deb11...@r37g2000prr.googlegroups.com>,
>>
>> panfilero <panfil...@gmail.com> wrote:
>>> 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.
>
>
> 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.

--
Sue

panfilero <(E-Mail Removed)> writes:
>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.

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