unebonnevie 
ok, I skimmed over the spec and see lots of LM317 calculators (like http://www.whatcircuits.com/lm31... ) on the net that have one resistor held constant (R2) and varies the Vout and the adjustable resistor (R1). What I don't get is what the value of Vin to have the desired Vout.
The Vout equation does not express in terms of Vin at all.
Vout = 1.25(1 + R2/R1)
There has to be a Vin to R1 adjusted to get the desired Vout, no? Or as long as my Vin is within the LM317's spec, I am good?
Thanks
valusoft 
The 317 is a linear voltage regulator and so long as the Vin is greater than the Vout + the drop out voltage (3 volts according to my datasheet) all will be good. Except (isn't there always an except?) you are limited by the ability of the 317 to dissipate the heat generated across the 317 due to the current and voltage across it (Vin-Vout). There is also a maximum Vin rating for the 317 which is defined as being 40 volts greater than Vout. So, you can use the 317 in high voltage regulator designs so long as you obey these rules.
Cheers,
Ross
unebonnevie The 317 is a linear voltage regulator and so long as the Vin is greater than the Vout + the drop out voltage (3 volts according to my datasheet) all will be good.
My Vin = 5V
My Vout = 2.8V
My LM317 is from STMicroelectronic (http://www.st.com/stonline/books... ). Drop out voltage is 2.5V max, based on the spec. (I am not sure if I got that right.) Then for my case:
Vin < (Vout + drop out voltage) is
5V < (2.8V + 2.5V) = 5.3V, that is, 5V < 5.3V.
Ok, what other alternative for me (besides using two 1.5V batteries)? Looks like the LM317 is not good to use in my case.
Maybe a couple diodes in series to reduce 5V to 3V?
Thanks.
leon_heller 
Use an LDO regulator, like the LM1117.
unebonnevie valusoft wrote:The 317 is a linear voltage regulator and so long as the Vin is greater than the Vout + the drop out voltage (3 volts according to my datasheet) all will be good.My Vin = 5V
My Vout = 2.8V
My LM317 is from STMicroelectronic (http://www.st.com/stonline/books... ). Drop out voltage is 2.5V max, based on the spec. (I am not sure if I got that right.) Then for my case:Vin < (Vout + drop out voltage) is
5V < (2.8V + 2.5V) = 5.3V, that is, 5V < 5.3V.
Ok, what other alternative for me (besides using two 1.5V batteries)? Looks like the LM317 is not good to use in my case.
Maybe a couple diodes in series to reduce 5V to 3V?
Thanks.
Wait a minute...How come Sparkfun has a product that use the LM317 and able produce 3.3V out of the 5V supply from USB? This seems to violate the rule that Vin is definitely less than 3.3V + 3V drop out.
See http://www.sparkfun.com/datashee... from product http://www.sparkfun.com/commerce...
From http://www.electronics-lab.com/a... it says: "Note: The input voltage to the LM317 must be at least 1.5v greater than the output voltage. "
So, I should be OK.
valusoft@: Please explain how you come about the statement "The 317 is a linear voltage regulator and so long as the Vin is greater than the Vout + the drop out voltage (3 volts according to my datasheet) all will be good."
leon_heller It should be OK. See the dropout graph in the data sheet.
unebonnevie
Wait a minute...How come Sparkfun has a product that use the LM317 and able produce 3.3V out of the 5V supply from USB? This seems to violate the rule that Vin is definitely less than 3.3V + 3V drop out.See http://www.sparkfun.com/datashee... from product http://www.sparkfun.com/commerce...
From http://www.electronics-lab.com/a... it says: "Note: The input voltage to the LM317 must be at least 1.5v greater than the output voltage. "
So, I should be OK.
So, I wired up like the above schematic from Sparkfun, but with the adj resistor to be 295ohm to get a 2.79V, according to the LM317 calculator http://www.electronics-lab.com/a... .
But...I only get 2.02V. My VCC is the USB power supply, which is not exactly 5V but 4.88V.
With adj resistor at 360ohm, I am supposed to get 3.13V but getting 3.09V solid.
I use 5% tolerance resistors. Not sure if using 1% would make a difference.
uhmm... :D
leon_heller Set the voltage to 2.8V with the variable resistor.
valusoft Maybe my spec sheet is out of date .... but this is what it "says".
Cheers,
Ross
Jepael
Wait a minute...How come Sparkfun has a product that use the LM317 and able produce 3.3V out of the 5V supply from USB? This seems to violate the rule that Vin is definitely less than 3.3V + 3V drop out.See http://www.sparkfun.com/datashee... ... Supply.pdf from product http://www.sparkfun.com/commerce... ... ts_id=8376
Do you actually have that Sparkfun device and have verified the output is 3.3V? Sure it might be 3.3V but only for very light load, or then it is not 3.3V but something less in reality. They just advertise it does the job, but they do not say if it does it in reality or in what conditions.
From http://www.electronics-lab.com/a... it says: "Note: The input voltage to the LM317 must be at least 1.5v greater than the output voltage. "So, I should be OK.
No you are not OK. You should not believe everything you read on the Internet. That comment about 1.5V dropout voltage is valid if you don't draw more than 20mA of current at room temperature of 25C or greater. When temperature is over 75C, you can draw up to 200mA.
I bet that LM317 page was done quickly by some hobbyist, and now he is distributing his opinions and measurements based on 1-hour experiment on single LM317 unit, and that may not apply to all LM317 chips from different manufacturers or different manufacturing dates.
Ross, that figure just says the specifications are measured in conditions where Vin is at least 3V more than Vout.
It does not mean you always have to have at least 3V more on the input, it just always works in all temperature and all current conditions when the voltage difference is at least 3V. There is a Dropout Voltage figure somewhere in the LM317 datasheet.
But 5-3.3=1.7V which is pushing it. Or if it is measured to be 4.88, 4.88-3.3=1.58.
I think LM317 is just wrong component for this type of thing, I have succesfully done 5V->3.3V with LM1117/LD1117 too, or just use any other LDO.
valusoft Jepael, yes I understand about the spec conditions ... also that marketing will ALWAYS describe their products under the best "light". So that is probably why I used that figure (mea culpa). I also agree that the 317 is not the sharpest chisel in the toolbox.
I guess using the graph would have been more helpful. Apologies.
Ross
Jepael So that is probably why I used that figure (mea culpa).
Have no worries, using the 3V as the first rule of thumb is excellent for LM317 and 78xx regulators and I do that too, because then they will always work. 2V, not enough in all cases, 2.5V might usually work, 3V is perfect.
ahh yet another case where more is better.. and size does matter
But if the 5V in is regulated why not just use 3 1N4007 diodes and a small load resistor (any regulator have a leak aswell) so the drop will be about 3*0.6 V or 5-1.8=3.2V dropping to about 3V at 1A load that is ok for most parts.
unebonnevie
No you are not OK. You should not believe everything you read on the Internet.
Uhmm...I am reading the above on the internet now. :P
That comment about 1.5V dropout voltage is valid if you don't draw more than 20mA of current at room temperature of 25C or greater. When temperature is over 75C, you can draw up to 200mA.I bet that LM317 page was done quickly by some hobbyist, and now he is distributing his opinions and measurements based on 1-hour experiment on single LM317 unit, and that may not apply to all LM317 chips from different manufacturers or different manufacturing dates.
All the posts have been very helpful to me, yours, Ross's, the "hobbyist" above. They help me to understand!
Thanks, all!
unebonnevie Measuring the current of the LM317 Vout. Out of curiosity I want to measure the current of the Vout of the LM317.
How would I do that safely? My Vcc is 5V and the LM317's Vout is 3.09V based on the adjust resistor I used.
I know that I need to set my multi-meter to DCA and the RED cable to the 10ADC socket. Black cable to COM socket. I set the dial to 20m
Jepael Vout current? To where, your own circuit or the Vadj setting resistors?
If you want you will measure the current from the lead that goes to your 3V circuitry. What do you have there?
unebonnevie Vout current? To where, your own circuit or the Vadj setting resistors?If you want you will measure the current from the lead that goes to your 3V circuitry. What do you have there?
Basically I want to use the Vout, which is a little less than 3.09V, from the LM317 as the power supply for the Nokia 110 LCD. I want to make sure that there is not too much current go through the LCD. Obviously, I need to put resistor to limit the current, but how many ohms, since I need to know the current.
The Nokia 1100 takes about 2.8V for its VCC.
Brutte LCDs consume currents of ~uA and you are using an LM317 from STM, which is rated at 2A. Aren't you exaggerating a little bit (about a million times)? Use a pot of several kohms instead and you will be fine. The LCD from Nokia 1100 has leds onboard (totally independent supply pins) and you can power them from any voltage as long as you do not exceed current (use resistor).
unebonnevie LCDs consume currents of ~uA and you are using an LM317 from STM, which is rated at 2A. Aren't you exaggerating a little bit (about a million times)?
Not at all.
The LM317 is what I have to use. If I had a voltage regulator with less amperage rating, I certainly would use, unless, of course, the 2A rating would cause my circuit to smoke. Then that exaggeration is bad. My circuit is powered by the USB power from the PC. 5V and max of 500mA.
Use a pot of several kohms instead and you will be fine. The LCD from Nokia 1100 has leds onboard (totally independent supply pins) and you can power them from any voltage as long as you do not exceed current (use resistor).
Sure that limits the current. How about the voltage regulating from VCC = 5V to 2.8-3.0V, which is this thread is all about, thus, my using the LM317 [for now] and other approaches that people have suggested.
clawson 
You seem to have missed Brutte's point. Remember that current is drawn, not injected. (Last time I said "current sucks, not blows" is caused some confusion. Just because the regulator is rated at 2A it doesn't mean it'll be injecting a full 2A into the circuit. The design of the circuit will draw a certain current and by the sounds of it this is in microamp territory so the 2A the regulator may provide is millions (certainly hundreds of thousands of times) more than will actually be drawn (well unless it short-circuits through virtually 0 Ohms that is!)
Brutte You use a linear regulator when the output regulation (voltage or current or any linear combination of that) must be very precise with low ripple. This is not required for digital circuits.
I do not have a datasheet of the display, but lets suppose it operates at 2.7V, current ~uA.
USB voltage can be within 4.75V to 5.25V at a receptacle (according to bus powered USB device I guess).
Set the voltage divider current to 1mA and that is all you need. Take a 2k7 and (5k-2k7)=2k3 resistors and connect them in series to USB power(if it is 2.3V you got, swap them). You will have 2.7V in between.
Jepael If I had a voltage regulator with less amperage rating, I certainly would use, unless, of course, the 2A rating would cause my circuit to smoke. Then that exaggeration is bad. My circuit is powered by the USB power from the PC. 5V and max of 500mA.
No you don't insert any resistors anywhere.
Let's quickly think what linear regulators do. They regulate voltage, so the voltage stays constant where you set it independent of current taken by what is powered with it (your LCD).
Of course the currents and voltages have to be within the maximum real-world limits of both power supply and the regulator itself, i.e. you cannot draw 3A from 1A regulator or 500mA supply, and your output voltage is always less than input voltage.
Your 2A regulator is okay, even if your power supply can only supply 500mA and your LCD will only take 1mA.
In my opinion, you cannot break that regulator, but you should protect frying your USB connector. Maybe a fuse would be in order, as you cannot put any series resistor between 5V and regulator because the voltage is already so low.
And with that voltage drop, seen from the datasheet curves, you can only draw something around 100mA max anyway before the regulator output voltage is going to drop.
LM317 requires a MINIMUM of 20ma load current. If you want micopower, this is NOT the regulator.
The suggested feedback network takes 10ma. So, you need to guarantee a minimum additional load of 10ma.
So, with an LM317, why turn the micro down to the microamp level?
Jim
Jepael LM317 requires a MINIMUM of 20ma load current.
Some manufacturers can do less, my National datasheet says 10mA minimum load current. The voltage adjustment network is usually made to provide all of that.
My bad. I meant 10ma. The National Semi spec sheet suggests a feedback net that draws 5ma (1.25V/240 ohms). So, you have to supply at least 5 ma additional.
Jim
ArnoldB Still not exactly right. The minimum current is a function of Vout - Vin. The smaller the difference the smaller can be the minimum current (until you reach the dropout voltage). See the "Minimum Operating Current" diagram in the datasheet.
The diagram shows the typical current. To be on the safe side you better double the values in the diagram. Then you end up at e.g. 3 mA if you manage to keep the difference at 5 V.
