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?
*NO* or is there a beinifit to folowing the example on
*YES*
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.
My supply voltage is around 24V, and I want a 10V output... so I was thinking about putting a voltage divider at the output of the regulator and using a pot to dial in 10V out... the load will have a draw of around 300mA.
Could you elaborate a little more on please on what they are trying to do on that datasheet, and on what would be wrong with my approach?
PS - I thought a 12V regulator outputs 12V... but you said you can use it to adjust between 12-15V... is this true?
"> N.B LM340-12 is absolutely the *wrong* choice unless you only want to
You should *ONLY* post to groups where your question is ON TOPIC and relevant. If you aren't sure if its an appropriate group, you may ask
*briefly* if it is. E.g. Subject "LM340-12: Can I a beginner's question?", Body "I need an adjustable voltage output of around 10V. I have read the LM340-12 data sheet. I need help with designing a circuit. Can I ask here or is there a better group?" would have got you *polite* directions to sci.electronics.basics or maybe even the comment "Well this isn't really the right group but have you looked at ......... site where there is a circuit that will help you?". Before posting to a group *always* read about a week's worth of recent messages to get a feel for the group. This reading without posting is called 'lurking' and is a *GOOD* thing.
Some of the regulars on S.E.D (common abbreviation to first letters
*within* a group) have designed chips that sold millions. As an ordinary tech with many years in the repair trade who's been into electronics all my life, I know enough to follow their arguments and can understand their circuits if I study them hard, but there is an elegance in the designs they knock out in an hour that I couldn't duplicate without years of work. There are also some, mostly of lesser accomplishment who delight in misleading newcomers.
I don't read alt.engineering.electrical, but there is precious little connection between most of electrical engineering and low power electronics. If you want to design a custom mains transformer, try there.
S.E.misc is low traffic, for stuff that really doesn't fit elsewhere.
This group is the 'tutorial' group.
*MANY* users killfile (automatically throw away unread) *all* messages to more than 3 different groups. These are usually the users that have been using the groups the longest and have the most experience. They are exactly the users you should want to interest in your question.
Due to inappropriate posts and spam from Google Groups users, there in an increasing move towards killfiling all of you on the technical groups. There is not a lot you can do about that except to get proper newsreader software (which works like an email program that runs on
*YOUR* computer, not on a webpage inside your browser window) and sign up with a News (or NNTP) Server. Some of us still read posts coming from Google. Few of us use Google to post from.
As I *dont* use google, and I quoted the whole of your message, You
*MAY* get some answers from those who don't normally see posts from Google.
Thats enough education about USENET, I'll try to give technical help in the other fork of your thread.
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The ground pin isn\'t connected to ground, so it\'ll no longer have 0V on
it, it\'ll have on it whatever the voltage is at the junction of R1 and
R2.
The regulator will use that voltage as a reference and will output a
voltage 12V higher than that.
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
My supply voltage is around 24V, and I want a 10V output... so I was thinking about putting a voltage divider at the output of the regulator and using a pot to dial in 10V out... the load will have a draw of around 300mA.
Could you elaborate a little more on please on what they are trying to do on that datasheet, and on what would be wrong with my approach?
PS - I thought a 12V regulator outputs 12V... but you said you can use it to adjust between 12-15V... is this true?
"> N.B LM340-12 is absolutely the *wrong* choice unless you only want to
A LM317 might be a better choice but at 0.3A you'll be dissipating at least
3.6W and more as you adjust for lower voltage (higher volt drop in the regulator), so look into how big a heat sink you'll need.
You can reduce the dissipation by finding a lower voltage source.
The adjustable regulator circuit raises the "ground" pin of the regulator above ground, but the regulator still tries to keep 12 volts between its output pin and its ground pin. The circuit shown, when used with a 12 volt regulator, will allow you to adjust the output voltage from 12 volts to about 22 volts, with your 24 volt input. Using the LM340-5 5 volt regulator in this circuit will let you adjust the voltage from 5 to 22 volts, with your 24 volt input.
A better regulator choice would be an LM317, which is designed to be used as an adjustable regulator. The same circuit would allow you to adjust the output voltage from 1.2 volts to 22 volts.
As another poster mentioned, you _will_ have to put the regulator on a good heatsink as it will have to dissipate (24V - output voltage) * output current watts.
For the 300 mA at 10 volts you want, that's 4.2 watts.
--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
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To get acceptable regulation the current in the pot should be around ten times the load current. You need 300mA at 10V. The pot you are using as the voltage divider would need to be a 4 ohm one rated at over 36 watts and the regulator wouldn't be able to supply enough current and would either shut down or burn up.
If you try to use a higher value lower wattage pot, you will have crap regulation. Pot's wattage ratings actually define the maximum current the track can take. You need 300 ma out at 10V so if one reduces that
10:1 divider current to load current ratio to 1:1 and tolerate crap regulation, you will have 600mA flowing in the top part of the track which by definition *must* be dropping 2V. The remainder of the track has 10V across it and is passing 300mA. This would need a 36 ohm pot, a value that is not going to be easy to find.
OK, try a 50 ohm pot. The regulation will now be truly awful. The top part of the track is carrying 516mA, the lower part, 216mA and all should work. One small problem, the pot needs to be rated at over 13.5 watts or it will overheat and catch fire. A 50 ohm loudspeaker control pot is good for about 3.5 watts, you need 5 times better.
I hope you can now see why a resistive dropper, especially one based on a pot is impracticable at more than a few tens of milliamps output current.
On the data sheet they are showing you how to adjust the output voltage of the regulator upwards by 'lifting' the voltage at the common terminal as it *always* acts to keep its designed output voltage 12V above its common. They can do this because the current in the common leg is about
6mA, though it would be advisable to design the potential divider to pass around 10 times that. You need to start with a LM340-05 regulator to be able to get 10V out though.
Personally, I'd put a small 6.2V Zener diode in series with the common leg, cathode towards the regulator which would give 10.2V (+- 0.1V) output and forget the resistors which would work for *most* 10V loads.
N.B. the capacitors shown in the data sheet are *essential* or it can start oscillating at a frequency too high to hear, get very hot and burn up, possibly shorting and taking your load with it.
You can go a lot higher than 15, but it would be preferable to start with a higher voltage regulator. You *cant* go below 12 (without some trickery with a negative supply to the common terminal). See data sheet for examples.
You've had another suggestion in S.E.D to use a LM317T. Its fully adjustable with a minimum output of 1.2V, and the data sheet is much more understandable.
Basically, its designed to keep its output 1.2V above its Adj terminal (the equivalent of common on the 78xx or LM340-xx series regulators). It also works with 100 times less current through that pin. This means that it works the same way by lifting the voltage on the Adj terminal but you can use pots of normal (well under 1 watt) ratings and values of several kiloohms. Much easier. It still needs the caps for the same reason.
If you are going to need various voltages in the future, get a few LM317T regulators (you can get any voltage if you have a selection of resistors handy), but the 5V and 12V fixed regulators are useful too. If its a one off need for a fixed 10V, the 5V regulator with the 6.2V Zener is the simplest circuit and is reliable.
What is your 24V supply and what is your 10V 300mA load? There may be other problems.
N.B. If the supply is from a vehicle, BEWARE that whenever the starter motor or other high current devices operate there are often transients up to several hundred volts which *will* blow the regulator and your load. If its only plugged in with the engine off and cold it will be fine, but sooner or later you know it will get plugged in with the engine running :-(
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....
IanM, I appreciate you taking the time to give me such a detailed answer, and I am really trying to understand it:
When you say "To get acceptable regulation the current in the pot should be around ten times the load current."
ok, I believe you are referring to my voltage divider solution with the reference of the voltage regulator hooked up to ground. So, I'm curious where you got the '10 times the load current' rule... why is that?
comparing what you are saying to john field's answer above he says: PD =3D (Vs - Vl) Il =3D (12V - 10V) * 0.3A =3D 0.6 watts
which doesn't sound bad... but he's not following the same conventions your pointing out... his way seems like the voltage divider method aint bad....
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.
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.
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That\'s the power being dissipated by the 680 ohm resistor, not the
regulator.
The power being dissipated by the regulator will be the difference
between the supply voltage and the output voltage multiplied by the load
current:
Pd = (24V - 12V) * 0.3A = 3.6 watts.
>
>which doesn\'t sound bad... but he\'s not following the same conventions
>your pointing out... his way seems like the voltage divider method
>aint bad....
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