I cannot make out the POT value from the diagram.

It's a "10K POT", but I measured the resistance to be 11.56KΩ to 2.5Ω

Data says 5K.

How did you get to that? Sorry, still learning to make sense of datasheets and how to use the information.

That's fine, but my point was a 1K will give the OP better resolution because he has a 12V input voltage.

What's resolution?

R1 always has 1.249 volts across it. That is a constant for this chip.

So, I should have used the Reference Voltage? Which would also be the minimum output voltage?

 

What's resolution?

How fine you have to twiddle the knob to hit the right spot.

, I should have used the Reference Voltage? Which would also be the minimum output voltage?

Yep.

 

BTW... You guys are awesome. You're way more helpful, understanding, and patient than the last forum I was on. Thank you.

 

You're way more helpful, understanding, and patient than the last forum I was on.

We're probably just dumber than the other guys. That makes it easier to work with beginners.

 

It's good that you're aware of the issue with counterfeit parts, but testing is probably beyond the capabilities of most hobbyists.

Basically you'd need to test all of the parameters that the manufacturers check and it's unlikely you'll have the appropriate equipment. In addition to reference accuracy, you'd need to check: max current, thermal protection, load regulation, line regulation, ripple rejection, drift, ....

If you're determined to do this, you'll need to build some test circuits and have an oscilloscope.

I'm down with that... But, I gotta get this variable power supply box working first.

 

What's resolution?

If you use a 5K pot, only the first 20% of wiper movement will give you any voltage change. If you use a 1K pot, the voltage change will be spread over a wider wiper movement; giving you better resolution.

 

I'm down with that...

It'll be less work to just buy from a reputable source.

 

If you use a 5K pot, only the first 20% of wiper movement will give you any voltage change. If you use a 1K pot, the voltage change will be spread over a wider wiper movement; giving you better resolution.

So, I'm assuming that by lowering the resistance; from 10K => 5K => 1K, I'm also reducing the scope of VOUT. That would imply a relationship between resistance and voltage. And that by using a higher POT would result in more VOUT range but less resolution. And by using a lower POT I increase resolution but reduce my VOUT range.

Is there some formula to express the change to Resistance to Voltage?

Would chaining multiple 1K POTs in serial give me both a larger VOUT scope and resolution?

 

It'll be less work to just buy from a reputable source.

That comes with experience that I don't yet have.

 

Didn't you read the datasheet

 

That comes with experience that I don't yet have.

http://forum.allaboutcircuits.com/t...ronics-parts-where-you-are.88279/#post-637121

So, I'm assuming that by lowering the resistance; from 10K => 5K => 1K, I'm also reducing the scope of VOUT.

By using the right pot for the job, you are achieving the Vout you want without 80% of the pot doing nothing.

 

So, I'm assuming that by lowering the resistance; from 10K => 5K => 1K, I'm also reducing the scope of VOUT. That would imply a relationship between resistance and voltage. And that by using a higher POT would result in more VOUT range but less resolution. And by using a lower POT I increase resolution but reduce my VOUT range.

Using a 5K or 10K pot won't allow the output voltage to increase above Vin - Vdropout. In your last schematic, that would be about 9V.

Is there some formula to express the change to Resistance to Voltage?

Being inexperienced doesn't excuse you from reading the datasheet.

With I1 about 10mA, you can ignore the effect of Iadj*R2; and delta Iadj.

Would chaining multiple 1K POTs in serial give me both a larger VOUT scope and resolution?

Not based on your last schematic.

 

Didn't you read the datasheet

Being inexperienced doesn't excuse you from reading the datasheet.

Read, yes.... Comprehend 100%, no. Thus, is my primary problem.

Take for example Io(min) in the table below. Io(min) is the lowest amperage required for the element, yet in the Max column 10 mA is listed. As per a previous comment, 10mA is the minimum amperage required. So, why is it in the Max column and not the Min column?

Under section 6 of the datasheet in Application Information, they give VO = VREF (1 + R2/R1) + IADJ R2 to calculate the output voltage. In that, I don't know what to use for VREF(Reference Voltage) I know they list 1.25V. Does VREF change as resistance or voltage change? Same for IADJ(Adjustment Pin Current) that has 50µA.

What does "Vi-Vo" mean exactly, other than the obvious Voltage Output subtracted from Voltage Input.

 

Hello,

You are correct with " What does "Vi-Vo" mean exactly, other than the obvious Voltage Output subtracted from Voltage Input."
You might want to have a look at the attached PDF on how to read a datasheet.

Bertus

 

Take for example Io(min) in the table below. Io(min) is the lowest amperage required for the element, yet in the Max column 10 mA is listed. As per a previous comment, 10mA is the minimum amperage required. So, why is it in the Max column and not the Min column?

The datasheet didn't give a minimum value. It gave typical and maximum. Typical is the value that parts typically exhibit and maximum gives you the worst case value.

If you're designing a circuit that needs to work all of the time, you use the, in this case, maximum value given by the manufacturer.

Once you get your power supply working, you can see what happens if you lower the minimum output current to less than typical (e.g. 1mA) and see what happens.

FWIW, I've had LM371T that did not work at a load current of 10mA. The device was out of spec, but what recourse does a low volume user have? It would cost me more time, money, and effort to return the out-of-spec device than to deal with it by increasing the current (which I did) or discarding it as defective.

 

Simply buy your parts from Mouser, or Digikey, or any other reputable USA distributor that sells in small quantities. The prices of a 317 from Mouser is under a buck for qty 1, but the gotcha is that their minimum shipping is around 6 bucks, so order enough stuff so that the shipping will not be too much of an irritation. Ebay is the most likely place to be stuck with counterfeit parts, especially from foreign market sellers. Stick with Mouser and you will not be screwed.

 

Hello,

You are correct with " What does "Vi-Vo" mean exactly, other than the obvious Voltage Output subtracted from Voltage Input."
You might want to have a look at the attached PDF on how to read a datasheet.

Bertus

Informative, but doesn't quite help with the vernacular and relating data to application. Which I presume comes from noobish "does this work in this matter" type of questions. At least until I'm able to draw the lines between the dots. Someone should really write an electronics dictionary for noobs.

 

So, I've just been introduced to the concept of a Voltage Divider. (Which likely will lead to another thread.)

Is that essentially what the resistors are doing? Except that Z2 is a potentiometer, and Vout from the LM317T goes into Vin, and Vout goes to the ADJ pin?

 

Is that essentially what the resistors are doing? Except that Z2 is a potentiometer, and Vout from the LM317T goes into Vin, and Vout goes to the ADJ pin?

Sort of...

A better way to look at it is that Z1 between Vout and Adj creates a current source who's value is determined by 1.25V/Z1. The output voltage then becomes 1.25V + I1*Z2.

From the datasheet:

 

Sort of...

A better way to look at it is that Z1 between Vout and Adj creates a current source who's value is determined by 1.25V/Z1. The output voltage then becomes 1.25V + I1*Z2.

From the datasheet:
View attachment 112666

I see... So, how does the LM317T avoid the Ouroboros effect resulting in exponential change? What's fed into ADJ affects Vout which is then changed by R1 & R2 and fed back into ADJ.

Being inexperienced doesn't excuse you from reading the datasheet.

With I1 about 10mA, you can ignore the effect of Iadj*R2; and delta Iadj.

Where'd you get that schematic from? I can't find it in the one I have from Mouser. Why can I ignore the effect? "Delta Iadj"? What's with that second Vout after R2 in your schematic? Am I supposed to join the two Vouts?