AC to DC Converter 35V @5A err 1.5A

Re: AC to DC Converter 35V @5A

@KlausST

The most important formula in datasheet is: Vout = 1.25V x (1 + R2/R1) + Iadj x R2

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Yes, I am aware of it.

Vout = known by you (Eshal)
R1 = 120 or 240 depending on regualtor
Iadj = given in datasheet.
--> R2 is the only one unknown value.

So you can solve the formula for R2.

This is the same way we all have to go.

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Actually I also know all these because I have read the datasheet already. Still have questions to ask. Why R1 is 120 for LM317 and 240 for LM117 to choose? Why not any other value?

I just googled for "lm317 power supply"

There are a lot of good results:
* youtube videos on how to design your own supply
* online tutorials on how to design
* schematics (you may use ist. no need to reinvent the wheel)
* PDF files with standard designs
* pdf files on how to extend on voltage and/or current.

also:
There are a lot of manufacturers for LM317. They usually have additional application notes on how to use the regulators. Go through them.

*********

You are always welcomo to ask your questions here, but please be more specific. (on where you stuck, or what parameters are unknown)

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OK

@BradtheRad
I have checked the simulation. All parameters like current through resistor, capacitor and voltages across the components are depend on the value of the components which are chosen by the designer. You agree with this, right? But the real question is how the designer know what value of the component is suitable to use in the specific condition? like there are resistors from few milliohms to thousands of megaohms but designer use just 1 from these crore resistors. How??
@schmitt trigger
My output current requirement is 1A so I will go with LM317 and thanks for praising my drawing dear

Re: AC to DC Converter 35V @5A

OK.
But the thread's name says 5 amp.:thinker:

Anyways, if you ever require 5 amps, you now know.

In post #5 Eshal said she made a mistake "5A" in the title that cannot be corrected. She wants 1.5A maximum.

Re: AC to DC Converter 35V @5A

Eshal,

Changing requirements to a variable DC out even at 1.5A brings on Mechanical Engineering challenges of cooling. But if you learn quickly that RthJA and RthJC are like Ohm's Law for thermal resistance values you can understand.

When you have a high unregulated DC input voltage and a low regulated output voltage at high current, the power dissipation is mainly lost in the LDO.

Thus if you wanted 3V @1.5A or 4.5W and your bridge was outputting 40V say , the VI drop across the LDO is 37*1.5 or 56W!!!

Now I prefer to have a Tj jcn. Temp. of 85'C max, so Total thermal resistance including jcn-case and heatsink case-ambient with forced must be around 1 deg/W above a 30'C ambient.

Since the lm317 Rjc is 5 deg/W you have to derate your current to 20% of 1.5V and now your input-output drop exceeds the absolute max of 37V with Vo=3V so it will shutdown or fail from damage.

Thus your new requirements with adjustable Vo at 1.5A and must use LM317 will not work.

You are correct, in the past I also have made mistakes in the thread's title, and it also has mislead people.

But as I said: If in the future he requires a 5 amp supply, he knows now how.
Or if he requires only 3 amps, he can use the LM350.

schmitt trigger said:

If in the future he requires a 5 amp supply, he knows now how.
Or if he requires only 3 amps, he can use the LM350.

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Hee, hee, hee. He is actually a she.
I looked up the Arabic name in Google.

@Audioguru
Yes I know every IC takes its own voltages across it to operate it. I know I need a heatsink. Actually I have gone with all these stages already I just need to confirm my all calculations here so that I could see how where I am wrong. Very good clearification sir thank you very much. but ans this.

Instead of having R2 parallel to your voltage adjustment variable resistor you should have a resistor in series with the variable resistor to limit its adjustment range to a minimum of 30V.

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I will be going to use pot in place of R2.

@SkySunnyguy
Very good explanation. But it should work on 1A output current.

@Audioguru and schmitt trigger
hahahahha.. It is very funny.. He made me he. While I am she. so funny. I am laughing again and again..

Eshal said:

@SkySunnyguy
Very good explanation. But it should work on 1A output current.

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After I pointed out reducing the current which raises the unregulated voltage well above the maximum in-out of 37V and that the power dissipation at 1.5A must be reduced to 20% even with a forced air CPU cooler means 300mA x ~45V = 13.5W for a part rated at 15-20W with diff. Suppliers makes it super hot near shutdown.

It won't work due to lack of pre-regulation and the wide excursion from no load to full load and the 37V I-O drop limit. It won't work here.

Fixed supplies are simple. higher voltage adjustable to low,and high current are not simple for linear regulators for typical LDO's

Lambda, one of the best PS makers, once used a Triac bridge to pre regulate or phase control the rectifier so maintain a small input to output LDO regulator to minimize heat loss and extreme voltage swings. Did you not try my real-time time Simulator,to adjust the design load and see what happens?

This design is impractical. SMPS is the way to go.

Eshal said:

I will be going to use pot in place of R2.

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Hi Eshal,
You said you want to adjust the output from 30V to 35V. Then without a resistor in series with the pot it will be difficult to adjust to 30.0V and any voltage in between.

Eshal; I apologize. But the anonymity of the internet makes it very hard to know the person's gender, nationality, age, religion, political leanings or other attributes.
And frankly I don't want to know, this is an electronics forum.

Therefore....Only take from my post what is relevant and important information for your electronics knowledge. Ignore the rest.

@SunnySkyguy
I understand now what you are trying to make me understand. OK got it. So my next thread will be on the SMPS but the most simplest for the beginner. So please be a good teacher for me. But I want to complete the calculation of this supply but I will not practically implement it but it will help me in future designing.

@Audioguru
What will the series resistor do in series with the pot? Does it limit the current? but a pot can also limit the current too. Isn't it?

@schmitt trigger
I can understand. By the way what a good name of your location. Very good Hexadecimal

A good teacher expect good students to learn how the basic designs work and then asks questions not spoon feed a book condensed into 1 page.

Eshal said:

@Audioguru
What will the series resistor do in series with the pot? Does it limit the current? but a pot can also limit the current too. Isn't it?

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The current in the pot for an LM317 regulator circuit is very low, 10mA. The LM317 and the 120 ohms resistor R1 limits its current.

You said that you wanted an output range of 30V to 35V but your pot by itself creates a range from 1.25V to 35V so turning it a small amount causes a large change in output voltage.
If R1 is 120 ohms then its current is 1.25V/120 ohms= 10.42mA. You want the minimum output voltage to be 30V so the value of the resistor in series with the pot should be (30V - 1.25V)/10.42mA= 2025 ohms. Use 2k ohms. Then the pot is turned to maximum then it will have a voltage that is 5V across it and have a current that is 10.42mA in it so its value is 5V/10.42mA= 352 ohms.

But nobody makes a pot that is 352 ohms so use a 500 ohms pot (or 470 ohms) and re-calculate the value of the resistor in series with it. Then the voltage range will be from around 27V to 35V.

a good teacher expect good students to learn how the basic designs work and then asks questions not spoon feed a book condensed into 1 page.

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;( ok

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Very nice explanation. This is what I actually want.
Thanks a lot. I need to ask more. Be tune to this thread. My internet connection has got problems that's why I am getting problems in login to net.