voltage regulator question adjustable voltage needed

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

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!
Reply to
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OP crossposted excessively, I have taken the liberty of setting followups to sci.electr> Hello,
*NO* or is there a beinifit to folowing the example on
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.
Reply to
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?
Reply to
In article , panfilero wrote:
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.
Reply to
Hugh Gibbons
In article , panfilero wrote:
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
Reply to
Salmon Egg
Seems to me his best route is to get himself an LM317
Reply to
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....
Reply to
In article , panfilero wrote:
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.
Reply to
Salmon Egg
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
Reply to
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.
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.
Reply to
Dave Martindale

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