# Adjustable voltage regulator

1. Aug 5, 2016

### Enochfoul

1. The problem statement, all variables and given/known data

Calculate the output voltage range for the adjustable voltage regulator circuit shown in FIGURE 1.

I have read an older post and have understood how the formula works but I am thrown off by the inclusion of the 50uA current coming from the Adjust Pin. How would I cater for this current in the equation?

2. Relevant equations

$V_{OUT}=V_{REF}\times(1+ \frac{R_2}{R_1})$
3. The attempt at a solution

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2. Aug 5, 2016

### Staff: Mentor

Is this line how you intend?

The relation you are probably looking for is:
current through the 5kΩ = current through the 400Ω + 50μA.

3. Aug 5, 2016

### Enochfoul

I messed up with the LaTeX there I apologise. Am I to ignore the vref of 1.25V in this circuit or does it have influence on the circuit? Thanks

4. Aug 5, 2016

### Enochfoul

Vout = vref(1+R2/R1)

5. Aug 5, 2016

### Staff: Mentor

I think Vref is the voltage at the junction of the voltage-divider resistors.
Vref is the voltage across the upper resistor, here 400Ω.

Last edited: Aug 5, 2016
6. Aug 5, 2016

### Enochfoul

I've had a go at a solution how does this look?
Current through R2 =1.25/5000= 0.25mA
Current through R1 =1.25/400= 3.12mA
Then do I need to +50μA to R1 or R2 as well?

7. Aug 5, 2016

### Staff: Mentor

Right.
This isn't right. It's the upper resistor that has the fixed 1.25V across it, then its current goes through the lower resistor.
Add it into the current through the lower resistor. (Are you told that the 50uA is fixed for all outputs?)

Last edited: Aug 5, 2016
8. Aug 5, 2016

### Enochfoul

1.25*(1+5000/400)=16.87V

9. Aug 5, 2016

### Staff: Mentor

That looks right, when the 50μA is ignored.

Now, can you recalculate taking into account that 50μA.

Last edited: Aug 5, 2016
10. Aug 5, 2016

### Merlin3189

I may be wrong here, but that doesn't sound right to me. I thought the 1.25 V was the difference between the Output and the Adj. Since the regulator has no other ground reference, how can it know what the potential on the Adj is with respect to ground?

My thought for this problem was that there is a constant 1.25V across the 400Ω resistor. The minimum output is therefore trivial and the maximum output is when the other resistor is 5k and Adj is still at 1.25V below the output. Just a V=IR calculation.

This is the LM317. If you look at the lefthand comparator, I think it switches at 1.25V between Adj and Output.
And

Last edited: Aug 5, 2016
11. Aug 5, 2016

### Staff: Mentor

You're definitely right. I'll go back and fix that.

Are you able to find specs saying whether the current marked 50μA is approximately constant? If so it could be included in the calculation to give a more precise answer, as I understand OP to be asking.

12. Aug 5, 2016

### Merlin3189

Yes. This data sheet shows LM317 as typical 50μA, max 100μA.
It's not the clearest data sheet I found, but the others were pdf, so I couldn't copy the image I wanted.

Yes. If there's fixed 1.25V across the 400Ω, that's a standing current of xxx.
When control is 0, neither current matters. Adj is at 0V, so output is at yyy
When control is 5k, both currents flow through the 5k, putting Adj at zzz and output at yyy more.

13. Aug 6, 2016

### Enochfoul

I know if I just have to account for the 1.25v I would use the following formulas to get the range but where would the 50uF figure fit into the equation?
VOUT=1.25∗(1+0/400)
VOUT=1.25∗(1+5000/400)

14. Aug 6, 2016

### Merlin3189

Your equations are based on a potential divider with no current flowing into or out of the junction, Vtap = Vtop x Rtap/Rtotal, which would be the case if no current flowed into or out of the Adj terminal.
But here we are told that there is a small current, 50μA, so the potentiometer formula is not exactly right.
The difference is small, so people normally disregard it and use the formulae you quote, but it's easy enough to calculate things using common sense rather than a formula.

If you look at the diagram of the LM317 adjustable regulator, you can see first that the 50μA arises because they use a zener with a 50μA constant current through it as the voltage reference. This zener current has to flow out of the Adj terminal to ground. Secondly there is a comparator (or op amp) which controls the output current to maintain the difference in voltage between the Adj terminal and Output terminal at 1.25 V

The 400Ω resistor is between Output and Adj, so there is always 1.25 V across it. So there is always 1.25 V / 400 Ω = 3.125 mA flowing in it.
The 5k variable also has this current flowing in it and another 50μA from the Adj terminal = 3.125 mA + 0.050 mA = 3.175 mA
Now you can calculate the voltage at Adj for any setting of the potentiometer, because these currents are fixed. And the voltage at the Output will always be 1.25V more than at Adj.

You ask how you can modify your formulae to take account of the zener current. I'm sure I could do that with a few minutes maths, but I wouldn't bother. IMO Once you understand what the IC does, you don't need formulae (unless you count the basic V=IR as a formula.)

15. Aug 6, 2016

### Enochfoul

Hi thanks for your patience so far which will probably run out after this:

Min Voltage V=IR 3.125*400=1.25V
Max Voltage V=IR 3.175*5000=15.875V

16. Aug 6, 2016

### Staff: Mentor

Yes, apart from the missing $\require{color}\colorbox{yellow}{\times10^{-3}}$ to make your equation correct!

(Had the problem statement referred you to use the IC's spec sheet then I think you would take into consideration the max/min values of the zener current, meaning for max possible o/p voltage you'd calculate using the 100μA figure, rather than the typical figure.)

17. Aug 6, 2016

### Enochfoul

After reading the spec sheet it referred to the current from the Adj pin as negligible and that it normally can be ignored. I am wondering something though, the spec sheet has a dropout min and max listed. It states the minimum is 1.3V. With this in mind would my calculation of 1.25V be an issue
Yes of course I converted my answer. Thanks for the help

18. Aug 6, 2016

### Staff: Mentor

That's true. Normally. But I can picture a situation where efficiency is paramount, and the current drained by that pair of resistors represents waste. So it may be decided to scale them both up by a factor of x10 or x20. In such a case, that 50uA/100uA may no longer be insignificant when selecting fixed resistor values.. So, in practice it may normally be ignored, but in principle it should not be forgotten.

The manufacturer would be specifying that minimum dropout to still allow the 1.25V reference.

19. Aug 6, 2016

### Merlin3189

I think you're still a bit low with your max voltage:
Rvar = resistance of variable resistor ...............0 Ω................................... 5kΩ
VAdj = Rvar x 3.175mA ...........................0Ω x 3.175mA= 0V ........5kx3.175mA = 15.875V
VOut = VAdj + 1.25V ...............................0V+1.25V=1.25V ..........15.875 + 1.25 =17.125V
The Output is always 1.25 V above Adj.

Yes, for the LM317, 1.25V is the "typical" value, but it can vary from 1.2 to 1.3, and the 50μA is also typical, but could be up to 100μA (they don't mention a minimum for this on my datasheet: perhaps, since it is ideally zero, no one worries if it is smaller than expected?)
I think these values are given, not so much so that you can calculate the output exactly rather than approximately (you can't !), but so that you can estimate the possible error.

Say you were designing a power supply to give 12V, you could not be certain that all the circuits you produced would be exactly 12V. You would have to allow some tolerance (or permissible margin of error.) Apart from the regulator chip even the resistors are not exactly what they say. A cheap resistor might be 5% tolerance, or you could go upmarket a bit for 2% or 1% tolerance. Better than that and the price rises rapidly. (I'm not currently buying these, so my figures may be dated - I even used to buy cheap resistors with 20% tolerance! - but the principle is going to be the same.) The variable resistor is even more difficult to manufacture accurately. Just looking at a cheap (is 2 pounds cheap for a pot these days?) potentiometer I see they quote ±20%. So if you used that for your 5k variable, your maximum resistance could be anything from 4k to 6k. That'll make more difference than 1.2 or 1.3V or the 50μA. And most components change their value with temperature.

At the end of the day you add up all the possible errors to find the worst case above and below your target, then you can specify that your power supply will give 12V ± 1V say. If that is not good enough, you need to redesign it. Maybe find a better chip, use closer tolerance components (at least for the critical component), put in a variable component and adjust it to get the output closer to your target, or get clever with the design in some way (that's one reason why many real circuits don't look like the simple circuits you see in textbooks. )