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Voltage Regulator Fundamentals

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I would like to use an LM317L 3 terminal adjustable regulator to supply(say) 2.4v output from a stable 24v supply.
The anticipated output current should be no higher than 25 ma. Well within the 100ma max of the regulator.

SO, what's the question I hear you all asking.......

What will the current be in the 24 volt input??

Is it simply output current *(v.out/v.in) ie.. 25*(2.4/ 24) = 2.5ma ??
OR do I miss the point somewhere??

Best Wishes.

Tony Hunt
 
The current at the output is determined by the load and the voltage applied to the load. If it's 2.4 volts then you need to know the load. If the load, as you understand it, will draw 25 ma at 2.4 volts then the "output" is 25 ma. You are missing something but work on the problem and you'll catch on then wonder how it could not have been obvious. Don't be concerned and your lack of understanding. Everyone starts at the beginning and it takes time for the concepts to sink in. Good luck.
 
The load is simply a 96ohm resistor.

What I wanted to know was what the current in the 24v input would be.

As nothing is 100% efficient, it would be nice to know how to determine it exactly.

Best Wishes

Tony Hunt
 
The load is simply a 96ohm resistor.

What I wanted to know was what the current in the 24v input would be.

As nothing is 100% efficient, it would be nice to know how to determine it exactly.

Best Wishes

Tony Hunt

Tony; here is a hint. The output current from the 317 goes to two parallel branches; one is the 96Ω load resistor, and the other is the voltage divider that makes 1.25V out of 2.4V. Since you do not show those, you will have to calculate the current that flows through the divider. It should be ~5mA, so the total current coming out of the pin is 30mA.

Now if you were to look at the 317's data sheet, it would tell you that the Adujustment Pin current that flows out of that pin is ~50uA, which is tiny compared to the load current

If you consider Kirchoff's current law, it basically says, sum of current in = current out, which tells us that the current flowing into the 317 is 30mA.

Now consider this: if we think about the power supply, it is delivering 24V*30mA = 0.72W.

Where is the Power going? Well, the power dissipated in the load resistor is only 2.4V*25mA = 0.06W. Where is the rest (0.72-0.06 = 0.66W) going?

You guessed it, into the regulator, and the voltage divider resistors. The voltage divider resistors dissipate 2.4V*5mA = 0.012W (insignificant), leaving 0.648W for the regulator itself.

Now, I'm here to tell you that if you ask a LM317 in a TO220 metal tab package (without an add-on heatsink) to to dissipate 0.65W, it will get hot to the touch, but be short of burning your fingers. If you ask a TO-92 style 317 to dissipate 0.65W, it will get stinking hot to the point where it will shut itself down, reducing the output voltage to near zero, thereby reducing the current through it, to protect itself, and your next post to the forum would be "how come my 317 doesn't put out the voltage I expected"
 
Thanks a lot. I KNEW I should have not dozed off in the Kirchoffs Law classes 40 years ago.

"how come my 317 doesn't put out the voltage I expected"

OH no.... It was actually going to be "Why does my 317 leave Scorch Marks in my thumb??"
but now I know.............. An SOT23 case can't handle 0.65W <BG>

SO, the next question IS obvious.
A 317 linear voltage regulator is apparently the wrong thing.
How DO I get 2.4V from 24 V without all the smoke and flames??
0.06W seems such a small amount of energy to handle.

Price not important......(Not often you hear that said.)
I'd rather pay £5.00 for a component that worked rather than 75p for one that didn't..

Best Wishes

Tony Hunt
 
If I needed 2.4V @ 25mA starting from 24V, I would first answer these questions:

It the 25mA load always present and constant?

Is overall efficiency important?

Answer me those, and we'll go to the next step
 
Thank you Mike.

The load will vary between 10 and 25ma.

Overall Efficiency is very important.
Best Wishes

Tony Hunt
 
If you want high efficiency than a modified version of the Black Regulator may work for you.
 
Since you didn't put in a location when you registered,where can you buy parts?
 
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It's difficult to get a high efficiency with such a low voltage output and a high input voltage. About the best I could do, according to my simulations, is 60% (which, of course, is still a lot better than the 10% of a linear regulator for these input and output voltages).

For my simulations of this circuit, I changed the zener (ZD1) to 3.3V, which gives an output of 2.6V. I changed R2 to 10kΩ, and Rz to 25kΩ. I used an inductor resistance of 1Ω. That gave me a simulated efficiency of 60% with an output of 26mA at 2.6V, and an input of 4.7mA at 24V.
 
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@Mike. I deal with Farnell in U.K. and Digikey in U.S. but I'm fairly adventurous, I'll buy from anybody.

@Carl. 60% sounds SO much better than the 10% of a linear regulator. Is this a linear relationship?? E.G. would an output of 2.6mA have an input of 0.47mA??

I was hoping for a single chip solution but am now far more confident to try anything.

Best Wishes

Tony Hunt
 
Go to this page.

I arbitrarily picked the third one down; the LT3470. Not a friendly package for homebrew, but oh well...

I opened the LT3470 test jig that comes with LTSpice, and diddled the load and feedback resistors and the input voltage to produce this sim:

Efficiency is not too bad; 73mW in for 59.3mW out.
 

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@Carl. 60% sounds SO much better than the 10% of a linear regulator. Is this a linear relationship?? E.G. would an output of 2.6mA have an input of 0.47mA??
Tony Hunt
It's not linear since there are some fixed loses (like the transistor bias current and the current through the zener) so the efficiency goes down as the current goes down (look at the efficiency curves for some of the Black Regulator designs).

Also in my simulation the circuit became erratic when the output current goes below about 4mA at which point the input current was 1.8mA. The circuit requires a certain minimum load for proper operation. Not sure if there's a way to get it to operate properly at such low output current.
 
Thank you both very much indeed. My overall understanding has improved DRAMATICALLY in the last 24hrs.

I've ordered an LT3470 from Farnell AND enough components to build the Black Regulator..
Looks like 80% efficient to me.... That's good.

SO, it's like this.. My wife looked over my shoulder and said "It's obviously that bright red light that's making it hot.."
(OK It's an LED but she is SO right)

SO, I replace the 220r resistor with a 2.2k one. (as per my circuit I must blushingly say. Must be going colour blind.)
The current drops from 25mA to 5 mA, the bright red 'Light' goes dimmer but visible enough, the 317 is only slightly warm, my thumb has healed and life is GOOD...

Moral of the story..... If you think it's too hot..... It probably is..

If either of you is ever in the U.K. contact me and I'll treat you here.... **broken link removed**

I'm downloading LTSpice as we speak.........

Stand by for more silly questions.......

Best Wishes

Tony Hunt
 
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