What is an input?

Discussion in 'General Electronics Chat' started by sureshparanjape, Apr 22, 2013.

  1. sureshparanjape

    Thread Starter Member

    Feb 10, 2012
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    Consider a dc circuit, in which ac is rectified and fed to LM317 through a 50 V capacitor. DC voltage before the capacitor is 12 v. What should be the value of Vin in the calculations of the restriction Vin - Vo < 3 in using LM317? Vin= 12?
    sureshparanjape
     
  2. TheComet

    Member

    Mar 11, 2013
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    Didn't you just answer your own question?

    I don't understand what you're asking to be honest, sorry.
     
  3. Mike33

    AAC Fanatic!

    Feb 4, 2005
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    You need to provide a more detailed description or picture (schematic) of the circuit you are trying to work on.

    And more information, such as What is the value of the AC power source you're using? What capacitors are in the path for filtering? How are you setting the adjustment for the LM317? What is your desired output voltage/current?

    I can make an assumption that if you are getting 12V at the cap before the LM317, you are within the 317's ratings. You may then set up a condition to get somewhat less than 10V at the output, or lower. Voltage going "through" a cap to the 317 doesn't sound right, however. There should be one going to GROUND just before it, for additional filtration, but you don't have it in the path. DC voltage would be blocked by it, and all that would pass would be ripple voltage.
     
  4. JohnInTX

    Moderator

    Jun 26, 2012
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    The Vo-Vin >3V restriction means that your input voltage must be at least 3V higher than your output voltage when using an LM317. So 12V in can only get up to 9V out.
     
    Last edited: Mar 1, 2016
    Evanguy, killivolt and #12 like this.
  5. sureshparanjape

    Thread Starter Member

    Feb 10, 2012
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    You seniors, by knowledge(age very unlikely - I am 75 years!), you would have guessed that I am, if at all, just a beginner. The reason I asked the question is that I have read, "the capacitor smooths out ripples in the input voltage" but not any reason for having capacitor of say 50 v instead of say 10 V.
    If the input voltage weren't there in the circuit, the capacitor, theoretically, would be the source of LM317.If LM 317 is to control the output to 15 v, LM317 would keep drawing voltage 15 from the capacitor till it can(of course, it would be damaged some time!). Now we assume there is an input of 12 v to the capacitor. Now, even if the input voltage is 12 V (less than accepted by LM317)and the capacitor's capacity is 50 V, for LM317 input voltage is bigger than output voltage, till it drops below 18V. Am I making any sense? Again if the above logic is not wrong, then the system would break down, if the rate by which the capacitor gets charged is less than the rate by which it gets discharged.
    Again does it make any sense?
    Many thanks to all the quick replies to my query.
    sureshparanjape
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    This seems backwards. The resitriction should be along the lines of

    Vin - Vo > 3V

    meaning that Vin must be at least 3V higher than the output voltage.

    But it depends on context. It may be related to a part of the spec where they are specifically talking about the performance near the dropout point.

    The 3V seems a bit high, but not by much (and I haven't looked at an LM317 data sheet lately).
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    No, it doesn't. It's pretty much babble. And I don't mean that as a put down in any way. It just means that you are struggling to understand something that you haven't quite got a handle on yet, which is why you came here in the first place, so everything is still in its proper place.

    Part of the problem is that you haven't yet developed a feel for the scale of things, so the notion that if you remove Vin that the cap is going to power things for some amount of time is understandable, but not practical. The input capacitor on a regulator like this is often on the order of 0.1μF. Let's say that Vin is charged up to 43V and the output is adjusted to deliver 10V, meaning that there is 30V of overhead that can be used up before the circuit stops functioning properly. Now let's say that you have a single LED that consumes 10mA of current. Once you disconnect Vin, how long will the capacitor be able to power the circuit? Since Q=CV and average current is I=Q/T, you have

    T = Q/I = CV/I = 0.1μF*30V/10mA = 10μs

    That doesn't mean that capacitors are never used as power sources, but those applications are typified by very small currents, very large capacitors, or very short time durations (and preferably two or more of these).

    The capacitor rating (the 50V instead of 10V question) is a measure of how much voltage the capacitor can withstand before something bad happens (and the bad varies from capacitor type to capacitor type and includes everything up to an including exploding and/or bursting into flame).

    If you apply 12V to a 10V-rated cap, you run the risk of letting the magic smoke out.

    Think of it like the weight rating on a chair. If you are having a party and expect a lot of big guys to attend, you want chairs that are rated for 350lb and not chairs that are rated for 120lb.

    So let's say that you put 200lb of water in a chair rated for 350lb. Just because the chair is rated for 350lb does not mean that, somehow, you can take 350lb of water from it -- you only have 200lb of water available.

    So if you have a capacitor rated for 50V and you have in input of 12V, the capacitor only has 12V across it and, if your output is set for 15V, you have a regulator that can't do it's job and you will probably see something around 9V at the output.
     
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  8. Metalmann

    Active Member

    Dec 8, 2012
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    Thank you WBahn, there is still something about 555s I don't understand. Not quite sure what is so troublesome.....:confused:
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Okay.... that's quite the non sequitur.

    Whatever....
     
  10. Mike33

    AAC Fanatic!

    Feb 4, 2005
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    It's a basic misunderstanding, is all.
    The input cap is in parallel with the input signal (the power source), and isn't really involved in 'supplying' voltage. It just removes any ripple that may be on the input voltage. Some call this a "smoothing" cap. It's not like a battery, as WBahn explained very well above.

    In the OP, the man said his voltage is "supplied" by a capacitor of 50V rating. So he probably was thinking 'series'.

    More learning about the fundamentals of electronics is necessary! I do remember well when I could only get an LED to blink when I'd remove the (DC) power, as I had a cap in series with it...and I couldn't understand the difference between DC and AC :eek:) More reading caused a light to finally go on about how these components behave in different applications!
     
  11. sureshparanjape

    Thread Starter Member

    Feb 10, 2012
    64
    2
    Dear Mr.WBahn, I highly appreciate your comments.Particularly the part that tells how much time it would take to drain a capacitor of its capacity, even by a simpleton LED!
    Thanks a lot.
    sureshparanjape
     
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