value of capacitor at the input of the LM317 regulator

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OK I will use 35V for maximum output with 1A of current. I will use 75Ω and 2k pot for this requirement according to this calculation 35V-1.25V=33.75V and 1.25V/75Ω=16.7mA then 33.75V/16.7mA=2021Ω i.e. 2kΩ of pot.
And C1=2000uF.

Now sir tell me how to calculate the value for the C2 capacitor? Refer to the fig in the post#13. In that figure, C2=10uF. How did designer calculate this value?

Your name is connected with a religion in the middle east. Not many girls from there come here.
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Yes, you guessed right. I am from Asia. And agree that girls from these regions don't often use forum, mostly these technical forum like electronics, electrical, mechanical, etc. I agree with you sir.

Thank you a lot
 

Hello sir
Thank you for your help again.
How to know whether we should use 1uF or 25uF? But the designer used 10uF.

Thanks again
 

The datasheet shows schematics with 1uF and 10uF but does not say if they are tantalum or aluminum electrolytic.
The text in the datasheet says that a 1uF tantalum is good at high frequencies and a 25uF electrolytic is also good at high frequencies so they recommend using either one at the output. A 10uF aluminum electrolytic can also be used at the output.
 
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OK. And what is the purpose of this capacitor? Is it acting as the filter capacitor?
 

Eshal said:
OK. And what is the purpose of this capacitor? Is it acting as the filter capacitor?
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The text in the datasheet explains about the capacitors. You should read the datasheet.
The small input capacitor is important to prevent the regulator from oscillating at a high frequency since the huge main filter capacitor is a poor filter for high frequencies.
The output capacitor improves transient response and reduces the output impedance at high frequencies.
A capacitor can be added to the ADJ pin to ground to improve ripple rejection.
 
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The text in the datasheet explains about the capacitors. You should read the datasheet.
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Yes, I read and get what you are trying to point out.

The small input capacitor is important to prevent the regulator from oscillating at a high frequency since the huge main filter capacitor is a poor filter for high frequencies.
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You are saying small input capacitor at high frequency. Your statement is based on this fact C=1/2πfXc². This equation implies that Capacitor value is inversely proportional to the frequency. That's why if frequency is higher then capacitance would be low. Right sir? And same for the statement, "the huge main filter capacitor is a poor filter for high frequencies"

A capacitor can be added to the ADJ pin to ground to improve ripple rejection.
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What value should I use to bypass the ADJ pin? And if I use bypass capacitor then what about 2k pot? Is bypass capacitor parallel to the 2k pot?

Thank you sir.
 

The datasheet has text that talks about the capacitors and what different types and different values do. It has graphs that show ripple rejection, output impedance and line transient response with and without the ADJ pin capacitor. It shows a schematic with the ADJ pin capacitor connected.

A ceramic disc or solid tantalum capacitor is a very low impedance at high frequencies. A large value electrolytic capacitor is made with its foil wrapped around and around which causes it to be a high impedance inductor at high frequencies.
 
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OK sir, got it. Thank you
Now final question related with the figure which I posted previously. There is a C3=100uF capacitor across which we take output of the regulated supply. What is its purpose?
 

Eshal said:
Now final question related with the figure which I posted previously. There is a C3=100uF capacitor across which we take output of the regulated supply. What is its purpose?
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I explained what the datasheet says about the output capacitor in post #26. It can be 1uF to 1000uF but the graphs show that 10uF works well.
 
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Yes yes, it responses to the transient response. Now I will design further then tell you sir.

Thank you very much.
 

Please look at this paragraph
Look at it this way: the 317's reference voltage is based on holding an accurate 1.25v across R1. This is multiplied by (R2/R1) to get an output voltage of 1.25*(1+(R2/R1)). In effect, R2 adds positive feedback from the sum of R1 current and the internal voltage reference current! What C2 does is bypass (reduce) this positive feedback: effectively it reduces 'noise gain' to the internal amplifier. At DC, the voltage gain remains R2/R1, but to AC signals the voltage gain tends towards (1+ 0/R1) = 1 because C2 'looks like' a very small impedance at high frequencies. This lowered noise gain is where the much better ripple rejection shown in the LM317 (and similar parts) datasheet comes from. So, a big hint: make Z (the impedance) of C2 much smaller than R2 at twice your AC Mains frequency to suppress ripple better; 10uF will be more effective than 1uF for example, and 100uF may be better still.
in this given link
**broken link removed**

I want to know about this theory. How R2 adds positive feedback? How positive feedback concept comes up here?
 

Your link does not work. It should be http://www.acoustica.org.uk/t/3pin_reg_notes1.html
There is no "positive feedback".

The datasheet says that "the capacitor from the ADJ pin to ground prevents ripple from being amplified as the output voltage is increased. With a 10 μF bypass capacitor 80dB ripple rejection is obtainable at any output level. Increases over 10 μF do not appreciably improve the ripple rejection at frequencies above 120Hz".

Without the capacitor 1V of 120Hz input ripple is reduced -60dB down to 1mV.
With the capacitor 1V of 120Hz input ripple is reduced -80dB down to 100uV.
 
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Transformers have VA ratings which is simply the secondary voltage multiplied by secondary current -- this is strictly true only if the attached load is purely resistive (i.e has a power factor of 1.0). A reactive load containing capacitors or inductors (which one would expect for such a power supply) has a low power factor (i.e. less that 1.0) and thus de-rates the transformer's power capacity to the stated VA multiplied by the power factor because it draws more current than a purely resistive load. So, when choosing a transformer for a reactive load, one needs to divide the load in watts by the load's power factor to arrive at the VA needed which has sufficient "headroom" to accommodate the low power factor.

I found this paragraph from the site. Can anyone elaborate the bold sentence with example?

@audioguru
 

I have never used a transformer with an inductive load like a motor so I do not look at the power factor of a load. The sentence in BOLD explains that a transformer with extra power is needed when its load has a poor power factor. "Headroom" is the extra amount of power required from the transformer.

An AC-DC power supply for an audio amplifier fully charges its main filter capacitor only one time at the beginning when it is turned on. Then the capacitor is only "topped up" to keep it charged. Most of the time the power transformer for an audio amplifier provides 10% of the full amplifier power (plus some extra power making heat) then music or vocal peaks make full power plus more heat occasionally. But Acid Rock "music" makes continuous full blast noise that might strain an amplifier and its power transformer.
 
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