LM317LCLP USB-Powered Circuit Outputting Only 3V/1.45V Instead of 3.3V/1.6V

Hello everyone! I want to design a circuit that can run off of 5V usb that will supply 3.3V and 1.6V. The IC I used is
http://www.digikey.com/product-detail/en/LM317LCLP/296-33312-5-ND/1510138

The circuit design I used is the same as in Figure 1 with C1=0.1 and C2=1uF. For R1 I used 1.2kohm resistor and R2 I used a 2kohm pot. In parallel with C2 I have two 18 ohm resistors in series as a simple voltage divider.

Unfortunately when I hooked this up to my USB port I am only getting 3V and 1.45V maximum out even when adjusting the potentiometer.

Any suggestions?

You theoretical calculation seems to be OK although notice in the datasheet it state approx 1.25V. I think there is a factor not easily shown on the datasheet that could be affecting the reference voltage. If you measure the output and the adjustment voltage and subtract the values then this will give you the most accurate result.

I did see the reference voltage in the table in the datasheet is only valid from 5 to 35V, so under 5V would be different.

It may just mean you will need to lower the value of R1. It may have been better to leave R1 at 470 ohms as well.

I could not understand this part: "In parallel with C2 I have two 18 ohm resistors in series as a simple voltage divider."

Where is the mid-point of the voltage divider connected to? The "adj" pin? If so then this could be why you are not getting the required voltage. Besides, at 3V, the total load on the IC is less than 36 ohms - about 83mA which is already pretty close to the max o/p current (100mA as per the datasheet).

Rohit-
In parallel, meaning I have one 18ohm resistor connected to Vo connected to another 18ohm resistor in series with it which is then tied to ground. Basically taking Vo=3.3V and creating a voltage divider to get approx 1.6V,

Malcolm-
I hooked the circuit up to a power supply this morning and ran it off of 7V. This time the circuit worked however the LM317 IC got hot enough to be painful to the touch. So I'm thinking you might be right trying to reduce the value of R1. Perhaps I will also go with a higher current LM317.

My only worry is when the circuit is under load I don't want the current going above 100mA at 3.3 and 1.6V, but I guess the current would be set by the total resistance through the voltage divider(i.e. the two 18ohm resistors) correct?

Looks like part of your problem might be Dropout Voltage.

The LM317 can have a fairly high dropout voltage [a value the TI datasheet seems to omit]. The ON Semi datasheet [http://www.onsemi.com/pub_link/Collateral/LM317L-D.PDF] includes a graph that shows a Drop Out Voltage of anywhere from around 1.5 and 1.9, depending on load current [@ Tamb 25ºC]. I believe the tolerence on the 5 volts supplied by a USB port is ±5%, thus the minimum expected voltage output is likely 4.75... 4.75V - 3.3V = 1.45, so you need a regulator with a Dropout Voltage at or below 1.45V. With a load of 18Ω + 18Ω = 36Ω, your quiescent load current is 3.3/36 = 92ma. Judging by the ON Semi graph (assuming the ON part is similar enough to the TI part), your dropout voltage will be around 1.8V (or even more like 1.6V, since the LM371L is probably going to get rather hot!) So, with a dropout of 1.6V and a USB port supply of 4.75, you get: 4.75 - 1.6 = 3.15V So, this could, easily be your problem. And the dropout voltage of a TI part might be greater than that depicted in the ON Semi datasheet -- try measuring the voltage across the input and output of the LM317L [that's your dropout voltage].

Now, there's another possibility. If your LM317L is, indeed, getting hot, it could be that the internal thermal limiting is kicking in and that is that is causing the output voltage to sag. It could a combination of these two influences.

Consider using an LDO regulator [and why not just use a fixed regulator?]

Also, the resistor voltage divider seems rather inefficient. Why not just use two separate regulators? Or is the real requirement that you want an _adjustable_ set of voltages where one tracks the other at half the rate?

And can you reveal to us the load requirements of these two voltage sources?

The current drawn by the two 18Ω resistors will be around 3.3/(18 + 18) = 92ma. For this circuit, alone, that will be the current through the LM317L. If you connect any other load to this, that is a different story.

Thanks for the reply. The more I started researching the more I was thinking that it could be the LM317 with people reporting a dropout voltage as high as 2 to 3 volts. Unfortunately the PCB is already made so if at all possible I'm trying to fix my mistake and make this board work. I found this regulator
http://www.digikey.com/product-detail/en/LM1086CT-ADJ/NOPB/LM1086CT-ADJ/NOPB-ND/363573
which has a typical voltage drop of 1.3V. It also appears to function in a similar manner to the LM317.

As for using the resistive voltage divider, I chose this for simplicity sake and if I could change it, I would probably go with two seperate regulators like you said.

The load I'm attaching to it is sort of a black box/unknown quantity to me. Its going to be a power supply, supplying Vcc, Vac, and GND to an evaluation board. I know with with just hooking up a multimeter to Vcc to ground I see a resistance of 618kohm and Vac to ground I see a resistance of 678kohm, however, without knowing the circuit schematic itself, I don't think these two numbers do much good.

You do realize that the 36Ω path consumes most of the available current [supplied by the LM317L] and thus leaves less than 10ma to power your "black box". So, hopefully the internal impedance truly is so high.

Also, your link goes to a 404 error.

The usual engineering cycle is prove the concept first, and then make a PCB. Proving the concept is typically done with simulation software and/or by breadboarding the circuit.

Its Texas Instruments LM1086CT I'll try posting the link to the datasheet.

http://www.ti.com/lit/ds/symlink/lm1086.pdf

The problem was it did work in Multisim, unfortunately thats without a load connected. So looks like I have a lot of changes I need to do to this circuit.

So sounds like I should:
Change out the voltage regulator to one with less of a voltage dropout
Change the voltage divider network in proportion to the load resistance measured.

I'll try to get a schematic draw up today and post it to see if you think it will work.

John - simulations will not show you the whole picture. You should read the datasheet before simulating. For example max o/p current is always specified for any voltage regulator. In simulations, you can easily approach or exceed this value without noticing much change in the o/p voltage, leading you to believe that the circuit used to simulate is indeed practicable. However in practice you will find that it overheats or does not work as expected. In the simulator, the power "probe" will show the device dissipation - but not tell you whether that dissipation is acceptable or beyond specification, that you have to decide based on the datasheet.

The importance of going through the datasheet cannot be understated - it is the most important part of designing circuits. Of course once you are conversant with a particular device and through experience one might use devices without looking at their datasheets. That said, datasheets for special-purpose ICs (like voltage regulators or IC amplifiers) have a typical application circuit - this almost always satisfies a broad range of needs and also shows what all need to be changed to suit a particular application. Straying too far from recommended values or ranges might not be desirable unless you know exactly what you're doing.

Ok thank you Rohit. Yes I am slowly learning. This is my first board design for my first internship so they've been very patient with me and said this would be a good learning experience for me from what theoretically works and what actually happens. Before I start my second design of this circuit I'm trying to get a better understanding of what this circuit is going to connect to.

I was able to hook up my "black box" to a adjustable power supply and measured the following.
Vcc=3.3VDC (set and measured)
Icc=73mA (measured)

Vac=1.6VDC(set and measured)
Iac=2.17mA

Using V=IR would it be acceptable to believe that
Rcc=~45ohm
Rac=~737ohm

I think since I'm going to redo the PCB for simplicity sake I'll do two voltage regulators. This time they will be powered off of 9VDC 700mA wall adapter.

If I use two LM317 circuits both connected to this 9VDC with one designed for 3.3V and the other 1.6V using the design equations, I should be fine, correct? The current through the load will be determined from the internal resistance of the black box?

Yes, as Rohit said, not wise to trust your design, solely to a simulator. If at all possible, it's a good idea to also breadboard it -- especially if there are features of the design that go beyond typical parameters.

Yes, that should be fine. Just ensure that the adapter is rated for the total current draw of both circuits. Also, the "black box" should not draw more current than whatever the ICs can supply.

Do you currently have access to the black box? If so, then apply those two voltages and measure the current drawn (preferably with a scope so you can see if there are any frequency components that might need filtering).

Otherwise I(we?) can't really advise you on your design. Without full exposure to the requirements, it's kind of difficult to come up with a viable design.

And, why use an LM317 for the 3.3 volts? There are fixed regulators that will supply that voltage and which don't require programming resistors. Such as the 78L33A [though it may draw more current than what you are powering--only a consideration if powered by a battery (quiescent current from 4-6ma) so you might want to look for a low current alternative, such as MCP1702 with Iq = 2µA) BUT, it current limits at 250ma, so how about the LP2950-33LP with Iq=75µA and Io=100ma -- both are LDO so be sure to read the datasheet and observe the input and output capacitor requirements].

The LM1086 looks like overkill (Io to 1.5A, Iq to 10ma [which will unnecessarily drain a battery -- if this will ever be powered by a battery]).

The LM317 has a minimum current requirement, so pay attention to that, too. Or consider the LP2951 adjustable -- 1µA min current.

The LM317L being a linear regulator has a comparatively low efficiency compared to a switching regulator, so even though you are drawing approx 100mA through the resistors there will be a comparative current draw inside the regulator, this current draw is based on the difference between the input and output voltage, this is where the heat dissipation occurs and the efficiency of the circuit is. The circuit needs to be configured to reduce the I/P and O/P difference.

Efficiency isn't always important -- depends on the requirements.

Even for a switching regulator, for a designed max o/p of 100mA (low-power), I'd reckon that the control circuitry would still draw enough current from the supply to render it inefficient. Power conversion at 98% efficiency (theoretical, practically not so sure) would mean only 2% wasted as heat. With 100W, this equates to 2W. For something of the order of 0.33W (100mA at 3.3V), 2% means 6.6mW. Would it be possible to have a low-power switching regulator having this kind of efficiency?