HELLO !

Im new here and i need help calculating the values of the two resistors(R1+R2) in my voltage regulator circuit. The chip im using is a 5 pin low dropout voltage regulator called EZ1582CM. Im plan on using 3 volts and 5 volts like the chip states for the input. The output im trying to set will be 1.80 volts and id like to be able to adjust it up or down a little. The problem im having is i dont understand the formula or how to calculate the formula for the resistor values.(R1+R2) (Im not that great at algebra but am sure i can understand if someone will explain how the formula works)


Heres the formula i copied and pasted....

....................R1
Vout = Vref (1+-- ) + I Adj x R2
....................R2

Heres a picture of the screenshot of the formula. ( in case i pasted wrong)


And i've also attached the datasheet for more info if needed.

Thanks For the help.
Chris Oborski
.

 

Hi Chris,

If you change your ic to a LM317 you need only one input and you can still adjust the voltage between 1.2v and about 30v. all depend on the transformer you use. So if you're willing to do so, i will then show you how to work out the resistors. I can also give you a complete diagram if you want one.

SODA

 

The formula:
Vout = Vref(1+ R2/R1)+ Iadj*R2
OK, so Vref is typically 1.250V, but may range from 1.243v to 1.257v.
(See the table on page 3, to the right of "Reference Voltage".)
Minimum load current to guarantee regulation is typically 5mA, but could be as high as 10mA. (this comes from the same table, a bit further down)
To ensure guaranteed regulation with no load on the output, use a maximum of 124 Ohms for R1.

Further down in that same table, to the right of "Adjust Pin Current", you'll see Iadj; typically 50uA but may be as high as 120uA.

With the typical values you have:
Vout = 1.250(1+ R2/124 Ohms)+ 50uA*R2
You already know what Vref is; that's the difference between the OUT and ADJ pins.
So if Vref is 1.25v and R1 = 124 Ohms, the current through R1 is roughly 10.08mA.
Add 50uA Iadj to that to get 10.13mA.
You want a 1.8v output for starters. Subtract Vref (1.25v) from 1.8v = 0.55v
Since R=E/I and you know that I = 10.13mA and E = 0.55v, R2 needs to be 54.3 Ohms (rounded off)

Consulting a table of standard resistance values: http://www.logwell.com/tech/components/resistor_values.html (bookmark this)
we find that 54.3 Ohms is not a standard value of resistance.

So, we can use an online calculator for getting something close:
http://www.qsl.net/in3otd/parallr.html (bookmark this page too!)

You can see that there are a number of series and parallel combinations that will come very close to what you need.

 

Soda

I appreciate your help but i wanted to use a regulator with a higher amp rating than the LM317.

SgtWookie

Vout = Vref(1+ R2/R1)+ Iadj*R2
OK, so Vref is typically 1.250V, but may range from 1.243v to 1.257v.
(See the table on page 3, to the right of "Reference Voltage".)

I understood what some of the values meant on the datasheet but i didnt realise that i had to stay within some of the reference voltages that it stated. That makes more sense.

(See the table on page 3, to the right of "Reference Voltage".)
Minimum load current to guarantee regulation is typically 5mA, but could be as high as 10mA. (this comes from the same table, a bit further down)
To ensure guaranteed regulation with no load on the output, use a maximum of 124 Ohms for R1.

124 ohms for R1 did you get this by using 10mA Max...I think thats what you meant. That would make sense anyway.

Add 50uA Iadj to that to get 10.13mA.
You want a 1.8v output for starters. Subtract Vref (1.25v) from 1.8v = 0.55v
Since R=E/I and you know that I = 10.13mA and E = 0.55v, R2 needs to be 54.3 Ohms (rounded off)

Im still trying to understand how you worked out the formula. I wasnt very good at algebra in high school but i want to learn how to do this myself. I know you had to reverse the the equation and break it down. can you show me ...or is there a reference somewhere to show how to break the equation down? I hope im not asking to much but thank you

 

I understood what some of the values meant on the datasheet but i didnt realise that i had to stay within some of the reference voltages that it stated. That makes more sense.

Yes. They give typical, and frequently minimums and maximums.
This regulator has a Vref that has much tighter specifications than an LM317. An LM317's Vref is nominally 1.25v, but can be anywhere from 1.2v to 1.3v, and still be within manufacturer's specifications.

124 ohms for R1 did you get this by using 10mA Max...I think thats what you meant. That would make sense anyway.

Yes.
The specifications state that to ensure guaranteed regulation, the regulator needs to have a load of 10mA on it.
The minimum Vref (voltage from OUTPUT to ADJ) specification is 1.243v.
So to ensure a minimum 10mA load, 1.243v/ 10mA = 124.3 Ohms. Since fractional value resistors aren't readily available, I just rounded down to 124 Ohms. 124 Ohms is actually a standard E96 value (look it up in the link to the standard resistor value table I posted earlier).

The regulator will source current from OUTPUT in order to keep the Vref stable at somewhere between 1.243v and 1.257v.

Im still trying to understand how you worked out the formula. I wasnt very good at algebra in high school but i want to learn how to do this myself. I know you had to reverse the the equation and break it down. can you show me ...or is there a reference somewhere to show how to break the equation down? I hope im not asking to much but thank you

Frankly, I wasn't very good with algebra either - and time really hasn't helped!

But anyway, back to here, which is how I work through it.
If Vref is 1.25v and R1 is 124 Ohms, I = E/R (Current = Voltage/Resistance) = 1.25v/124 Ohms = 10.08mA flowing through R1 from the OUTPUT terminal.

Add 50uA for Iadj (flowing from the ADJ terminal) to that 10.08mA to get 10.13mA total current that needs to be sunk from the ADJ terminal through R2.

You want a 1.8v output.
Subtract Vref (1.25v) from 1.8v = 0.55v, because the ADJ terminal will be 1.25v (nominally) less than the OUTPUT terminal; so 0.55v is the voltage you need to drop across R2.

Since R=E/I (Resistance = Voltage / Current) and you know that I = 10.13mA and E = 0.55v, R2 needs to be 54.3 Ohms (rounded off)

Now if you go back to the original equation:
Vout = Vref(1+ R2/R1)+ Iadj*R2
and plug in the numbers obtained from the datasheet and derived so far:
? = 1.25(1+ 54.3/124) + 50uA * 54.3
= 1.25(1+ 0.4379) + 0.002715
= 1.25(1.4379) + 0.002715
= 1.797375 + 0.002715
= 1.80009
I've done some rounding here and there, but even with that rounding I'm only 90 microvolts off from your specified 1.8v.

Does this help explain it a bit more?

Note that Iadj has very little effect when R2 is a low value. However, it can be quite significant when R1 is large, resulting in a large R2; that 50uA to 120uA across several KOhms can "bite" you if you don't account for it.

 

YES!!
Thank you!

Im going to copy and paste this post so i can use for future Reference! It totally makes sense.I see how you broke it down.Now i see how everything calculates out. I know all the ohms equations but from the begining the "BIG" formula confused he crap out of me and couldn't figure any way to take a swing at it. The other thing is i wasnt putting two and two together when reading the datasheet or that i had to stay within the maximium guidelines or use the typical values when figuring out the circuit on the second page. I just figured that i had to stay within the "Maximum Absolute Values" on the first page.

Thanks for clarifying

Im wicked happy!!
Thanks Again..

 

Glad that helped

Now if you want to be really "sneaky", you'll notice that with the 10.13mA current through R1 when it's 124 Ohms, then for every Ohm of R2, you get roughly 10mV + Vref output.

So if you wanted 5v out, 5v-1.25v=3.75v, so 375 Ohms of resistance will get you very close.
If you actually calculate it out with the 10.13mA figure, you'll see that the actual requirement is 370 Ohms. The error is only a couple percent.