Could someone explain how a LT3080 works

[MoighonFweeman]

From what I have seen, the LT3080 is one of the most popular LDO regulators out there. Less than half a volt of dropout, as opposed to the three volts of the popular LM317. The catch is that there seems to be way less information on how to use it. The only half-useful thing I can find is the attached picture. So, I'd be grateful if someone could explain a few things to me:

1. What is the circle with the arrow in it supposed to be? I know it's some sort of voltage reference, but what exactly does it do?
2. What is the voltage at the noninverting input of the opamp?
3. How can I adjust the output voltage using a voltage rather than a resistor? I'm aiming for 0 to 12 volts, and I only have a 1K pot.

If someone gives me the answer to the first question, I may or may not be able to figure out the other things on my own. Anyways, thanks!

[Alex Eisenhut]

From what I have seen, the LT3080 is one of the most popular LDO regulators out there. Less than half a volt of dropout, as opposed to the three volts of the popular LM317. The catch is that there seems to be way less information on how to use it. The only half-useful thing I can find is the attached picture. So, I'd be grateful if someone could explain a few things to me:

1. What is the circle with the arrow in it supposed to be? I know it's some sort of voltage reference, but what exactly does it do?
2. What is the voltage at the noninverting input of the opamp?
3. How can I adjust the output voltage using a voltage rather than a resistor? I'm aiming for 0 to 12 volts, and I only have a 1K pot.

If someone gives me the answer to the first question, I may or may not be able to figure out the other things on my own. Anyways, thanks!

All the information you could ever need is right there. That circle with the arrow is the symbol for a current source. The regulator is sending a fixed current out the pin and the voltage drop across your Rset sets the operating point.

So, 2 is R*I. So 3 depends on the device.

[Phaedrus]

1.) from the datasheet:

"A precision '0' TC 10uA internal current source is connected to the non-inverting input of a power operational amplifier. The power operational amplifier provides a low impedance buffered output to the voltage on the non-inverting input. A single resistor from the non-inverting input to ground sets the output voltage and if this resistor is set to zero, zero output results...
...
A true current source allows the regulator to have gain and frequency response independent of the impedance on the positive input. Older adjustable regulators, such as the LT1086 have a change in loop gain with output voltage as well as bandwidth changes when the adjustment pin is bypassed to ground. For the LT3080-1, the loop gain is unchanged by changing the output voltage or bypassing. Output regulation is not fixed at a percentage of the output voltage but is a fixed fraction of millivolts. Use of a true current source allows all the gain in the buffer amplifier to provide regulation and none of that gain is needed to amplify up the reference to a higher output voltage."

http://cds.linear.com/docs/en/datasheet/30801fb.pdf

2.) The voltage at the non-inverting input is set by your set resistor.

3.) It would probably be best to simply use your 1k pot as the set resistor, in series with a fixed resistor if desired.

[Len]

Guess who did a video about this chip

[Marco]

1. What is the circle with the arrow in it supposed to be? I know it's some sort of voltage reference, but what exactly does it do?

It's the symbol of a current source.

2. What is the voltage at the noninverting input of the opamp?

The same voltage as the SET pin, seeing as they are connected ... with a Rset it's 10 uA*Rset, like the picture says.

3. How can I adjust the output voltage using a voltage rather than a resistor? I'm aiming for 0 to 12 volts, and I only have a 1K pot.

What's your input voltage? In principle you can just connect the pot between the input voltage rails and the wiper of the potentiometer to SET. It will set the voltage on the non inverting input almost linearly because the resistance is so small compared to 10 uA (if the current was larger it would still work, but the effect of turning the pot would become nonlinear). BUT at 12V input voltage the power dissipated in the pot would already be 144 mW ... more with higher input voltage obviously. Also the input voltage has to be steady obviously.