Newbie circuit clarifications

Discussion in 'General Electronics Chat' started by doug3460, Apr 20, 2009.

  1. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    I've attached the schematics for a segment of the adjustable voltage regulator I've been working on. My drawing is really a simple composite from several similar designs available on the net & in AAC. My question deals with the component layout in the schematic & the general operation of the circuit.

    QUESTION 1: Is the order relevant for component placement on the OUT line of the IC? I am aware that sometimes component placement is very relevant. In other circumstance, it is not. The top two drawings are from datasheets. They differ slightly. The one on the left is an application hint for diode protection. The one on the right is for improved ripple rejection. Notice the OUT line resistor & diode are reversed between the two drawings. If my understanding is correct, their order in this circumstance is not relevant. I am aware that the 1μF tantalum cap on the OUT line needs to be physically placed as close to the pin as possible, but I see that as a board issue.

    QUESTION 2: How does current distinguish between resistors in parallel & series? In my PLANNED APPLICATION drawing, I added a resistor in parallel to the 5k adjustable resistor. The math gives me a voltage limit of 15.00V max out (which I desire), although by adjusting the value of the new resistor I can increase (or decrease) this limitation. However, let's suppose I swap the position of the 10μF cap & the added resistor (if position in the circuit is not relevant). That resistor could then be placed directly below the 240Ω resistor. So even though it's parallel to the 5k linear resistor, it's in series with the 240Ω one. If I treat these two in series the math changes (albiet slightly, i.e., 15.28V out). But I don't want 15.28V, I want 15.00V, lol.

    Appreciate any help so I can get a firm grasp on how the circuit is operating & honing my understanding for the math.
     
    Last edited: Apr 21, 2009
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    That 5.6K resistor R5 is going to have a deleterious effect on the adjustment pot. Just leave it out. It will adjust to 15 volts just fine (assuming the input voltage is 17.5 volts or better). Better yet, put the circuit together on a breadboard and try it out to see how it works.

    Is your app really so critical that you can't just stuff in a 7815 and be within 10 mv?
     
  3. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    The value of 240Ω is for LM117 only.

    For LM317, which has a minimum output loading requirement of 10mA max. instead of 5mA for LM117, one should really use a 120Ω resistor instead.
     
  4. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    Thanks for the replies. Apologies for the delayed response. Couple things.

    beenthere wrote:
    Bear with me please. This isn't about the app or substituting a 7815. Eventually I'll get some pictures & schematics up in the projects section of the forum & then it'll probably make more sense. I'm working on the second half of my PSU for my bench. But, due to my work schedule & chaotic life :)D), I have limited bench time in the man cave. So the questions I posted were the result of computations & research conducted away from there. The circuit will be bread boarded of course (there's more to it than this one section). I'm just trying to get a solid grip on the math & general circuit function. Heck, I spent a couple hours the other night playing with product over sum resistor combinations in the 317M V\small_{OUT} formula because I wanted to see what the effects were & to get the math down pat. :rolleyes: That's when it dawned on me that the component placement could be an issue & so I asked the question.

    I am trying to educate myself so when I finish my PSU I know & can reasonably defend every choice I made regarding the components, their function & placement, & the math that goes into all that stuff. Obviously, I'm having some trouble since I can't intuitively grasp why two parallel resistors arranged in series with a single resistor are computed as:

    \frac{R_1R_2}{R_1+R_2}+{R_3} and not as \frac{R_1(R_2+R_3)}{R_1+(R_2+R_3)}

    As I noted in my original post, the difference is small, but that's not my point is it? lol.​

    eblc1388 wrote:


    Apologies. It slipped back in. You had noted the same thing in an earlier post I had made regarding controlling circuit current. I researched it at that time (not that I don't believe you, mind you, it's just that of the dozens of 317 circuits I've examined, the 240Ω or close (Tony VanRoon uses a 220Ω in his) is consistently in them ;)). Even in many of the Application Hints specifically labelled for the 317 on the datasheets from the different manufacturers a 240 is used. But I understand what you're getting at. It's Ohms Law. The regulator needs a load to function. The datasheets list these as "Typ(ical)" & "Max" current. The regulator puts out a nominal 1.25V so the load between the Adj & Out pins is compared based on that resistor. The math tells me that any resistor between 390Ω (416Ω actually) & 120Ω should function based on the range listed by the datasheet for the 317M.​


    Appreciate you both taking the time to respond. Will follow-up as I progress.:)
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    And does that automatically make them correct? :mad:

    Most likely they will work because 240Ω is suitable for "typical" LM317. This means 80% or 90% of LM317 or whatever.

    You have to use a resistor, either 240Ω or 120Ω. There is no extra component cost involved. Using 240Ω have the possible chance that the LM317 becoming unstable while using 120Ω will not.

    Take your pick.​
     
  6. doug3460

    Thread Starter Active Member

    Oct 19, 2008
    87
    0
    Thanks eblc1388 for the reply. I see I omitted in my last post my intention to use your recommendation for a lower value resistor (please see below) when I make the breadboard workup.:)

    eblc1388 wrote (in reference to datasheets):
    :confused: Not sure of your point or the angry face. For a fledgling hobbiest such as myself, there'd be no reason to suspect such a widespead error had crept into all the schematics. One "error" in one example on one datasheet from one manufacturer would be a mistake. The same error on multiple examples on multiple datasheets from different manufacturers is something altogether different. Plus it's occurring in several reference publications, in addition to numerous sample schematics of working circuits. Additionally, this series of adjustable regulators isn't new or even a recent development. There's been plenty of time to work out the bugs & issue revisions, tech notes, etc..

    Regardless, as you point out, the datasheet does show the typical & maximum current values available (I can only hope those values are accurate! ;)), so there is logically a workable range of values (which I have empirically confirmed).

    Some additional thoughts along this line...

    The engineers appear to have decided to use the max current value of 5mA to determine the load resistor for the 117 chip. The math indicates that the load on the 117 with a 240Ω resistor is slightly above the max current listed (5.208mA is the computed value), so there must be a safety margin.

    Using the same logic of setting the load toward the maximum for the chip, your recommendation of 120Ω for the 317 & it's 10mA limit makes sense.

    However, I wonder how far outside the max range one can go before there's a problem. For the 117, we know at least ~0.21mA above the max is not only acceptable, but apparently a designed tolerance which can go even higher. Given no other specifications, that 240Ω resistor at 5% tolerance could actually place the current at ~0.48mA above the max if the particular resistor went down to the low end of the 5% limit (228Ω). Using a 120Ω on the 317 & ignoring the 5% is going to put the current ~0.42mA above the max. Worse case scenario it can go to ~0.97mA. Now that would be getting really close to 11mA total, which to me would imply it's now too far from the limit. These numbers are probably not significant, but there must be an upper end to that designed tolerance/safety margin.

    Consequently, I was thinking 130Ω may be a more appropriate choice since ideally that would put the load ~0.39mA below the listed maximum. Now the worse case scenario @ 5% would place the current only ~0.12mA above the max for the chip & below the known tolerance of 0.21mA for the series.

    Anyway, just musing. I have a long drive to & from work, so these things I ponder. lol ;)
     
  7. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The worst case LM317 needs to have a load current of 10mA only when its input to output voltage is 40V[/quote] and it does not have any other load. The curves on the datasheet show that a typical LM317 needs a minimum load current of only 2.1mA when it has an input to output voltage of only 20V.

    Many people use the 240 ohms (for the more expensive LM117) shown on the first page of the datasheet for a single project and it will probably work but some will not work when an LM317 is used.

    I always use 120 ohms with an LM317.
     
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