Math don't work.

Discussion in 'The Projects Forum' started by logmode, Jun 20, 2012.

  1. logmode

    Thread Starter Member

    Sep 5, 2011
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    Hello,
    I was in electronics resources and I was thrown a LM317 regulator. I am a NEWB and have a question. I put together this circuit to study it. It doesn’t follow the rules I know. For instance parallel currents add, and series are the same. Look at the diagram, those are my meter readings, no math.ADJ has a current 57.4 UA and should be .29MA. Why is it not .29MA?
    R3 is just a load.

    Thank you
    logmode
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Possibly measurement / calculation errors.

    What measurements did you take?

    What instrument(s) did you use?

    What derived values did you calculate from your measurements? For example did you measure the voltage across R2 and use Ohm's law to work out the current based on the nominal [or measured?] resistance? Or did you measure both voltage and current?

    More information would make it easier to answer.
     
  3. logmode

    Thread Starter Member

    Sep 5, 2011
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    All measurements came from my meter, twice. no math.
    Meter- Beckman Industrial HD110
    Bought it new and it’s always given me a true answers.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    When you insert a current meter into the ADJ connection the internal resistance of the meter comes into play and changes the entire circuit values.
     
  5. logmode

    Thread Starter Member

    Sep 5, 2011
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    Even at the at the 200 U setting? If that true why do they make the meter?
     
  6. MrChips

    Moderator

    Oct 2, 2009
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    No meter is perfect.
    A voltmeter will still draw current even at 10MΩ internal resistance.
    Check the internal resistance of your ammeter. It will still have a finite resistance.
     
  7. logmode

    Thread Starter Member

    Sep 5, 2011
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    Please, I know that I don't know. But I have tested that and I found 0 A at the U setting. also my instructor has told me that there is no current.I agree with you, there must be a current but meter tell me no.
    well ok thx
     
  8. logmode

    Thread Starter Member

    Sep 5, 2011
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    So it is .29 MA. That is a lot off. Do you think it could be that off?
     
  9. takao21203

    Distinguished Member

    Apr 28, 2012
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    All current meters (amp meter) will reduce actual current that's a known fact.

    But usually this is a few % maximum.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The Adj pin current tends to be reasonably stable. It's most likely the effect of putting the meter in series with the 240 ohm resistor. A meter resistance of ~10 ohms would account pretty well for the difference in values.
     
  11. logmode

    Thread Starter Member

    Sep 5, 2011
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    In my meter manual it says:Accuracy: 200UA to 2A ranges: 0.35%. So is this BS.
    Is not the current traveling through the other resistor R2? what is R1 doing exactly to my meter? Are you saying the meter is resisting the load by 10 ohms and that throw R1?
    just trying to learn something
    thank you
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    You need to learn how to look in the manual for performance specs such as this.

    Beckman Industrial HD110 Manual

    That meter has a 250mV full-scale burden on the lower scales. So on the 200µA scale it appears to your circuit like a 1.25kΩ resistor and on the 20mA scale it looks like a 12.5Ω resistor.

    It sounds like you are saying that the ADJ pin current should be 0.29mA (and it's mA, not MA, which is megaamps - twelve orders of magnitude larger) because that's the difference between 5.27mA and 4.98mA.

    Let's assume, for the moment, that the meter has no effect on what is being measured. So what are the accuracies involved? It appears you were using the 20mA range and the manual indicates 0.75% of reading plus 1 digit. With a resolution of 10µA, that means that, at 5mA, the uncertainty is 47.5µA. So the uncertainly in each measurement is roughly equivalent to the expected ADJ pin current.

    Now, I don't know how much of this uncertainly is independent and how much is systematic.

    Now let's see how much the meter burden might be affecting the results. Let's again assume the meter is perfectly accurate, but that it does burden the circuit. You measured 125mV across the feedback resistor (and its reasonable to assume the LM317 is holding this voltage pretty constant even as the resistance changes a bit). With the meter in place you measured 4.98mA, meaning that the total resistance was 251.0Ω. When you remove the meter, that is reduced to 238.5Ω. At 1.25V, that resistor will now have 5.24mA through it.

    Notice that the difference between this and the 5.27mA is now 30µA and the apparent discrepancy is reduced to only 37µA and it is now in the other direction and it is within the uncertainty of the measurements.
     
    #12, DerStrom8, t_n_k and 1 other person like this.
  13. crutschow

    Expert

    Mar 14, 2008
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    It's simple enough to determine the internal resistance of you meter's current range. Just use another multimeter to measure the resistance across the current input. As you change the current range of the meter you will likely see the resistance change.
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This is a great example of the potential pitfalls of taking supposedly simple measurements.

    All credit also to logmode for raising the matter and being willing to look for an explanation of the dilemma.
     
  15. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Dave Jones explains the metering problem here.

    Instead of buying his (good) tool don't try to measure the current directly. Instead use your meter to measure the resistor's ohms (out of circuit) and then the voltages, and then calculate the currents.

    That's the best answer you will get without disturbing the circuit you are looking at.
     
  16. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Out of curiosity, how did you connect your meter when you were taking the measurement? I have seen some people in the past connect the probes across the points to measure current, like they would voltage, which is a no-no. I have a feeling the other guys here have already sorted everything out, but I just need to know that you're measuring it correctly :p
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    NOTE: This hadn't posted prior to me submitting my earlier post that follows it in the thread.

    Look carefully at those accuracy specs. They will almost always say some percentage of either full reading or full range and then plus one digit. Does your's say anthing like that?

    When you insert your meter into a circuit to measure current, it is like inserting a resistor in that same place. That's how the meter works, by passing the circuit current through a relatively small resistance in order to generate a measurable voltage. In most circumstances, that additional resistance has negligible effect, but that is not always the case.
     
  18. SgtWookie

    Expert

    Jul 17, 2007
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    You don't show any input or output capacitors on the regulator.

    While the LM317 doesn't necessarily require capacitors on the input or output, you should read National Semiconductors' datasheet for the LM117/LM317 on page 9 under the heading "External Capacitors" - particularly the last couple of paragraphs.

    Adding sufficient capacitance on the input and output ensures that the regulator won't oscillate. Omitting them leaves a big question mark; "Is it really stable under the conditions I'm subjecting it to?"
     
  19. logmode

    Thread Starter Member

    Sep 5, 2011
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    Thank you ladies and gentlemen for your help, especially WBahn that was nice of you.
    Ok measuring current with my meter in series with no other conductors in parallel with my meter (Derstorm8), is not a good way to do it. So I will do it Erniem way and divide the voltage drops by the resistance. R1 1.25v/240=5.21ma, and R2 23.8v/4.55k=5.23ma. Series circuits have the same currents, so ADJ must be 5.23ma- 5.21ma=20ua for ADJ.
    I measured the resistances in my meters six DC amperes settings. 200 ua = 1K ohms, 2 ma = 100K ohms, 20ma = 10K ohms, 200ma = 1.3 ohms, 2 a = .4 ohms, and 10 a = .1 ohms. So if you use your current meter for lets say a temperature controller using a RTD with a 4-20ma signal my meter is going to give me a false reading right? I have a headache that’s turning into a migraine. I have to think about this stuff.
     
  20. w2aew

    Member

    Jan 3, 2012
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    DMM ammeters basically work by measuring the voltage across a small shunt resistor inside the meter. This creates a voltage drop called a burden voltage. The lower the current range, the higher the value of the shunt resistor that is used. Burden voltages of a few hundred mV are typical, but dependent upon actual current flowing and the chosen scale. To minimize the affect of the burden voltage, always use the highest scale that will still give you the resolution you need. For low voltage circuit, sometimes the burden voltage is always a problem. In your case, I'd recommend verifying your currents a more non-intrusive way. Carefully measure the voltage drop across each resistor, then carefully measure each resistor value out of circuit, and calculate the currents. In general, the 10Mohm input impedance of the DMM measuring voltage is much less "intrusive" to circuit operation than the burden voltage of the ammeter function.
     
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