value of capacitor at the input of the LM317 regulator

Hello experts!
I hope all of you would be fine and great.

I have simple DC circuit and I am using capacitor in it parallel with 35v DC supply (this 35v DC supply is acting as a unregulated output of the DC supply) and output from capacitor is going to LM317 regulator. I am using capacitor at input stage of the regulator so that if there is any ripple in the output of the DC supply then it should be grounded. Let say we have 100Hz to 200Hz of ripple, for example. Now I want to calculate the value of capacitor to use here. Assume maximum current from the unregulated output supply is 1A.
Wordings are shown as a figure, attached.


How to calculate the capacitor value?

Thank you all.

An unregulated DC supply should already have a big main filter capacitor that reduces most of the ripple. It is easy to calculate the value required for a certain amount of ripple at a certain load current.
The LM317 needs a 0.1uF capacitor close to its pins to prevent oscillation at a high frequency.

The capacitor will provide 1A to the load during idle times lasting .01 or .02 second.

You can allow a certain amount of ripple, which is the capacitor losing a certain amount of volts during each idle time. Then you can calculate the Farad value, based on the RC time constant.

At 35 V and 1A, this calculates to an effective load of 35 ohms.

Suppose you make your smoothing capacitor 2000 uF.

Then your RC time constant is .07. That means a voltage change of 63% per .07 sec (by definition).

Or 18% per .02 sec idle time. This is a ripple value of 18% (with 2000 uF).

You may need to experiment to find out how much ripple the 317 can tolerate, and still give satisfactory operation.

Thank you all for your replies.

@Audioguru
I have already a filter capacitor in the unregulated DC supply but it is not grounding all ripples. So I decided to leave that capacitor in the unregulated supply as it is and decided to attach a capacitor at the input pin of the LM317 to the ground. Actually, I am getting 165Hz ripple from the output of the unregulated DC supply. I want to suppress this ripple nearly 0, if possible otherwise close to 0Hz.

@BradtheRad
Sir, I have already designed the unregulated supply with your help if you remember in the thread on this forum. And I am getting the ripple of 165Hz from that supply. Now, I want to regulate it but don't know how much value of capacitor is required before regulation which could make 165Hz or ripple to the nearest 0Hz, if possible. And also sir, you wrote above;

Suppose you make your smoothing capacitor 2000 uF.

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how did you calculate? Why did you suppose? Isn't there any calculation for this value?

Thank you again.

Actually, I am getting 165Hz ripple from the output of the unregulated DC supply. I want to suppress this ripple nearly 0, if possible otherwise close to 0Hz.

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The ripple frequency is defined by the input source and can't be changed by a capacitor. You can at best reduce the ripple magnitude.

But it's neither necessary nor reasonable to reduce the input ripple to zero. You need to keep a sufficient voltage difference between minimal input and regulated output voltage according to the regulator specification, e.g. 2 V.

Now, I am confused.
Why not necessary to reduce ripples in the unregulated output supply?
I want to get DC output from the LM317. So is it not necessary to provide it DC input in order to get DC output from LM317?
I know 2v is the dropout voltage for LM317 at 1A of current at 25C of temperature, approximately. But sir, I am confused with your reply now. Because those 2v you mentioned in your post#5 are the dropout voltage. These dropout voltages have not contribution to the ripple. Dropout are what the LM317 holds across it during operation.
The minimum voltage across regulator for regulation is called the dropout voltage. For the LM317T this voltage varies with the current and is typically 1.5V for currents below 200mA, rising to 1.7V at 500mA and 2V at 1A and if diode is in series with LM317 then diode's voltage i.e. 0.7V adds to the dropout voltage. For example, if unregulated power supply draws 200mA and the required output voltage is 6V, then the input voltage must be 6V plus 0.7V plus 1.5V i.e. the input voltage must be at least 2.2V higher than the output voltage. Therefore, we need to apply 8.2V minimum to the input for regulation.

Am I wrong in above Italic wording sir?

The lowest part of the input ripple must not be less than 2V more than the output voltage of the LM317 when the load current is 1A.
The regulator does not hold a voltage across it because the input ripple is causing the input voltage to vary up and down.

Eshal said:

how did you calculate? Why did you suppose? Isn't there any calculation for this value?

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We have a convenient rule of thumb for the smoothing capacitor: 1000 uF for every ampere of current.
It is not a strict formula, and you can change it to suit your purposes, and the tolerance requirements of your equipment.

I did try 1000 uF at first, but the ripple seemed too large as a result. So I raised it to 2,000.

I have already a filter capacitor in the unregulated DC supply but it is not grounding all ripples.

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You can also try connecting the smoothing capacitor to supply +. This will reference the unregulated supply to a different rail, which may help to reduce ripple.

You neither mentioned the intended regulator output voltage nor the presently observed ripple of the "unregulated 35 V supply". So no detail answer is possible.

As long as the required operation conditions for the regulator are kept, means the ripple high voltage doesn't exceed the maximum ratings and the low voltage doesn't fall below droput minimum, the ripple can be handled by the regulator. As an additional point, the finite LM317 line regulation factor allows a small fraction of the input ripple to pass the regulator and appear as output AC voltage.

@Audioguru
Very good point you picked up. New for me. But it is acceptable.

@BradtheRad
Yes, according to my experiment, I should use 2000uF. Thank you for being clear.

@FvM and all
Can you experts tell me if we bought new DC power supply from the bazaar then it is pure DC, right? If it is so then how it is pure? Why not it has any ripple content? Even in the university I use DC supply with oscilloscope then it show a straight line without any crest and trough. How company make it pure DC?

Thank you all.

A regulated power supply has usually < 100 mV (or even < 10 mV) ripple and noise. So it appears as "pure DC". You can expect similar DC quality at the LM317 output.

@FvM
Thank you very much sir for this reply. I am satisfied now. Thank you.
I am not closing this thread because if I need help then I will post it here. And ask you all to help me sir.

Thanks a lot.

Again hello to all experts and engineers

Here is what I found on the internet and exactly what I am looking for. But I will not only construct it as it is. I want to learn it. So help me step by step.
Here is a picture,


We know that the reference voltage of LM317 is typically 1.25V. You experts can see at the output of the LM317 there is a resistor of 110Ω. I want to know how to calculate the value of this resistor. Means whomever constructed or designed this circuit then how did he calculated the value of this 110Ω and 2KΩ resistors?

Thank you.

Nobody makes an LM317AH regulator. TI makes an LM317HV for high voltages.
Its datasheet says its minimum allowed load current is 12mA when its input is 60V or its output voltage might rise. So you could add a huge and hot output resistor or use the low power resistor from its output to its ADJ pin to provide the 12mA load. Its value when the input is 60V should be 1.25V/12mA= 104 ohms. Maybe 110 ohms is fine when the input is only 40V.

The pot is selected using the maximum output voltage you want and Ohm's Law. It has a current of 1.25V/110 ohms= 11.36mA in it so a 2k pot (if has an accurate resistance)m will have a voltage of 11.36mA x 2k= 22.72V across it and the maximum output voltage will be 22.72V + 1.25V= 23.97V.

If the resistor is 100 ohms then its current is 12.5mA and the maximum output voltage is 26.25V.
If the resistor is 82 ohms then you figure out the maximum voltage with a 2k pot.
75 ohms?

Nobody makes an LM317AH regulator. TI makes an LM317HV for high voltages.

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No no, sorry. Actually I make it on the multisim so it has that IC otherwise on the original where I found it, it was LM317T

Its datasheet says its minimum allowed load current is 12mA

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The word 'minimum allowed load current' you mean, the minimum current to drive the load. right?

The pot is selected using the maximum output voltage you want and Ohm's Law. It has a current of 1.25V/110 ohms= 11.36mA in it so a 2k pot (if has an accurate resistance)m will have a voltage of 11.36mA x 2k= 22.72V across it and the maximum output voltage will be 22.72V + 1.25V= 23.97V.

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Means if my requirement for maximum output voltage from the regulator is 35V then 35V+1.25V=36.25V. And then 36.25V/11.36mA=3.191kΩ or 3kΩ of pot I should use. Is this calculation correct sir?

If the resistor is 100 ohms then its current is 12.5mA and the maximum output voltage is 26.25V.
If the resistor is 82 ohms then you figure out the maximum voltage with a 2k pot.
75 ohms?

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for 82Ω, maximum output with 2kΩ pot is 30.48V.
for 75Ω, maximum output with 2kΩ pot is 33.33V. Right sir?

But Now I am little bit confused sir. For increasing the maximum output voltage you decrease the resistor from 110Ω then 100Ω then 82Ω and then 75Ω and ofcourse we get the increment in the maximum output voltage with that 2kΩ pot. But initially the selected resistor or 110Ω was calculated with the minimum load current from the datasheet. This parameter is fixed for LM317 so every time 1.25V dividing by minimum load current gives approximately the 110Ω of resistor. So if we use 100Ω, 82Ω or 75Ω, will it affect the minimum load current? Because we don't get these three values by this calculation 1.25V(fixed and defined in datasheet)/12mA(fixed and defined in datasheet) = 110Ω. Are you getting what am I trying to ask you sir?

Thank you for your great help.

Eshal said:

No no, sorry. Actually I make it on the multisim so it has that IC otherwise on the original where I found it, it was LM317T

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Multisim is stupid when it shows a part number that is WRONG.
If you use an ordinary LM317 then show it on the schematic but you should use the LM317HV because it is made for a high input voltage like you have.

The word 'minimum allowed load current' you mean, the minimum current to drive the load. right?

Click to expand...

The output of an LM317, LM338 and LM350 adjustable regulators have all of their operating current sent to the output pins so the ADJ pin has very low current. But ordinary fixed regulators like the 78xx has all of their ouperating current sent to ground. The adjustable regulators do not have a ground pin. The maximum operating current is 10mA when the supply voltage is high. If the adjustable regulator does not have a load that draws at least 10mA then some of them with the maximum operating current will have their output voltage pulled up too high (which might destroy some circuits it powers).
You have a choice of loading the output so the current is always at least 10mA (a resistor to ground at the output might need to be big and hot) or using a low power resistor from the output to the ADJ pin that uses 10mA. This resistor helps the pot set the output voltage.

Means if my requirement for maximum output voltage from the regulator is 35V then 35V+1.25V=36.25V. And then 36.25V/11.36mA=3.191kΩ or 3kΩ of pot I should use. Is this calculation correct sir?

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No.
If you want an output of 35V then the 110 ohm resistor has 1.25V across it and the pot has 35V - 1.25V= 33.75V across it because these resistors are in series from the output to ground.
The current in the resistors is 1.25V/110 ohms= 11.36mA. Then the pot is 33.75V/11.36mA= 2971 ohms. But you cannot buy a pot with that resistance. If you use a lower value resistor then you can use a lower value pot.
82 ohms produces 1.25V/82= 15.2mA then the pot can be 33.75V/15.2mA= 2220 ohms.
If you use 75 ohms then the current is 1.25V /75= 16.7mA. The pot value is 33.75V/16.7mA= 2021 ohms so a 2k pot can be used if its value is not less than 2021 ohms.

Multisim is stupid when it shows a part number that is WRONG.
If you use an ordinary LM317 then show it on the schematic but you should use the LM317HV because it is made for a high input voltage like you have.

Click to expand...

OK sir, when I will buy the IC from the bazaar then I will ask for LM317HV to the shopkeeper.

The output of an LM317, LM338 and LM350 adjustable regulators have all of their operating current sent to the output pins so the ADJ pin has very low current. But ordinary fixed regulators like the 78xx has all of their ouperating current sent to ground. The adjustable regulators do not have a ground pin. The maximum operating current is 10mA when the supply voltage is high. If the adjustable regulator does not have a load that draws at least 10mA then some of them with the maximum operating current will have their output voltage pulled up too high (which might destroy some circuits it powers).
You have a choice of loading the output so the current is always at least 10mA (a resistor to ground at the output might need to be big and hot) or using a low power resistor from the output to the ADJ pin that uses 10mA. This resistor helps the pot set the output voltage.

Click to expand...

Got it sir.

If you want an output of 35V then the 110 ohm resistor has 1.25V across it and the pot has 35V - 1.25V= 33.75V across it because these resistors are in series from the output to ground.
The current in the resistors is 1.25V/110 ohms= 11.36mA. Then the pot is 33.75V/11.36mA= 2971 ohms. But you cannot buy a pot with that resistance. If you use a lower value resistor then you can use a lower value pot.
82 ohms produces 1.25V/82= 15.2mA then the pot can be 33.75V/15.2mA= 2220 ohms.
If you use 75 ohms then the current is 1.25V /75= 16.7mA. The pot value is 33.75V/16.7mA= 2021 ohms so a 2k pot can be used if its value is not less than 2021 ohms.

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I understood. I did wrong calculation. Sorry sir.
OK so if I want to change my output requirement then I should change 110Ω resistor as well as 2kΩ pot. Right? For example, if I want output 40V instead of 35V then 40V - 1.25V = 38.75V is the voltage across the pot. And 38.75V/11.36mA = 3.411kΩ. Am I right this time sir?

Thank you for your help

Regards,
Princess

With a 40V output then the input voltage must be at least 42V. Then if the output voltage is turned down the poor little regulator will get too hot and shutdown if the current is only a few hundred mA or more.

The resistor must be 110 ohms or less. Since you cannot buy a 3411 ohms pot then the resistor can be 68 ohms and the pot can be 2k ohms but the pot will get very hot.

EDIT: Princess??
Oh, you are a ****** girl. Do you ... never mind.

With a 40V output then the input voltage must be at least 42V. Then if the output voltage is turned down the poor little regulator will get too hot and shutdown if the current is only a few hundred mA or more.

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I will use heat sink then.

The resistor must be 110 ohms or less. Since you cannot buy a 3411 ohms pot then the resistor can be 68 ohms and the pot can be 2k ohms but the pot will get very hot.

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Now, what I understand. Whatever is my output voltage requirement. What I just need to do is to use 110Ω resistor or less but 2kΩ fix. Right?

EDIT: Princess??
Oh, you are a ****** girl. Do you ... never mind.

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I am what ???
And by the way I am a girl. Don't you believe? hehehehehe

With such a high voltage and not too much current, if the output voltage was set low like maybe 5V then the LM317 on a huge heatsink will still be too hot, maybe then you need a fan blowing on the huge heatsink.

Your name is connected with a religion in the middle east. Not many girls from there come here.