Series-type ohmmeter

  1. in determining resistance R1 & R2 in an ohmmeter... the resistance at half scale deflection is used...why particularly this resistance is being used.. why not the resistances at deflection of other current values...?? Is this for convenience??
    Also , Rh (resistance at half scale deflection ) is regarded as internal resistance of ohmmeter looking into terminals A & B.. how..?? Is thevenin's is being used over here.that's why such calculations are there..( Rh=R1+R2Rm/R2+Rm) where Rm=meter resistance
     

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  2. jcsd
  3. jim hardy

    jim hardy 4,336
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    R1's value has to be whatever is midscale on the meter's face, and that is an artistic or aesthetic choice not an electronic one.
    I'll explain that odd sounding statement below but i hope you're patient.

    1. That schematic does not look to me correct. Plus it's not quite legible. More later.

    2. To understand the answer to your ohmmeter question, i really recommend you learn the principle of ohmmeter operation ,
    so to that end do this thought experiment:

    ....... Start thought experiment.....................................................................

    Just for a moment consider the meter ideal, that is it has infinite resistance , it indicates the voltage applied to it.
    And for the same moment let the "zero adjust" resistor (R2?) be not there, or zero ohms.

    Now --- You have two resistances in series: Runknown and R1 which is the internal halfscale resistor.

    Battery voltage divides between Runknown and R1 in accordance with their values.
    Here's the important point: The meter actually measures the voltage across internal resistor R1.
    If Runknown is zero, voltage across R1 is full battery voltage. Current is Vbattery/R1.
    If Runknown is infinite, voltage across R1 is zero because no current can flow.
    If Runknown is equal to R1, voltage across meter is half battery voltage. Current is Vbattery/2R1 .

    And that's why i think your schematic is wrong - it shows meter measuring voltage across zero adjust resistor not R1, which won't work.

    So redraw your schematic and work the circuit in your mind until it becomes obvious how the ohmmeter is really a voltmeter that's reporting the voltage across a known resistor.

    Now we only figured three points:

    Runknown = infinite,
    where no current flows so the meter does not deflect. That's why the left hand end of the ohms scale is marked "infinity".

    Runknown = zero,
    where the meter reports full battery voltage. That's why the right hand full deflection end of the ohms scale is marked "zero".
    That's also why there's an R2(?) zero adjust , it lets you adjust for fullscale deflection even if the batteries are slightly run down AND it lets you adjust out the resistance of your test leads.

    Runknown = R1,
    where the meter indicates half battery voltage, that is it goes to midscale indication (provided we remembered to short the test leads and adjust R2 for full scale first).

    So for the ohms scale, what number would you put midscale? The value of R1, of course.
    However - a midscale value that's more than about ten doesn't give good resolution of low ohms(and i think ten is pushing it),
    and one that's less than about four gives an ugly scale that looks unbalanced, the big numbers are all scrunched up on the left
    so your art guys should lay out the scale to make it have best combination of readability and appearance
    and the electronics guys should match R1 to those aesthetics.

    For every value of Runknown you can calculate what voltage the meter will see hence how much it will deflect,
    so you know where on the ohms scale to place that number.
    Observe that voltage divider action gives voltage out that is not linear and that's why the ohms scale is non-linear. Play with it on a spreadsheet.... or a slide rule in my case.

    So finally the answer to your question:
    When you have decided what number you want to appear in the middle of your ohms scale you have decided the value of R1 - not the other way round.

    That's the end of the thought experiment.

    ..........................End thought experiment..............................................................


    Now to the practical side:
    you'll have to re-calculate R1's exact value to correct for the presence of Rmeter in parallel with it...
    and you'll have to decide how much adjustment you want to give the user with R2 to accomodate aging batteries and an ohm or so of test lead..

    It is important to have the above understanding of how your ohm meter works.
    That way you can use it to do other stuff like check semiconductor junctions.
    Since it's a voltmeter indicating the difference between battery voltage and voltage across your unknown,,,,,
    and battery voltage is probably 1.5 volts,
    when you forward bias a silicon diode you expect it to drop 0.6 volts...
    So your meter should indicate ~0.9 volts which will be close to 2/3 deflection.
    If meter shows zero deflection your junction is not conducting. It may be open or reverse biased.
    If meter shows full scale deflection your junction is shorted.


    It is also important to recognize this :
    remember above we said current into Runknown = Vbattery/R1 ?
    My Triplett 630 meter has 4.3 ohms midscale.
    That means on RX1 it will send about 1.5V/4.3Ω = around 350 milliamps into a zero ohm Runknown, and midscale would be about 1.5/8.6 = ~174 milliamps.
    That's enough to wreck a small signal transistor
    so ALWAYS check semiconductors on RX10 or RX100 scale where the meter applies less current.
    Fortunately cheap meters don't have a RX1 scale.


    Last point:

    In your schematic it isn't shown whether the red or the black test lead goes to battery negative.
    In US made meters (Simpson 260, Triplett 630) the red test lead will be positive when reading ohms.
    Every Japanese meter i've encountered is the opposite , its red lead is negative when ohms is selected.
    That'll confuse you when checking semiconductors with different meters. whenever i pick up a new ohm-meter I always use a voltmeter to check whether its red lead is positive or negative.

    Whew ! Long winded, am i not ?
    But:
    If you make yourself aware of these details about ohm-meters you'll not blow up so many of them as most students do.
    AND you'll find it a very useful tool. In my career i did 95% of my troubleshooting with an analog meter and an analog oscilloscope.

    .... the humble analog ohm meter has a lot to teach us. It is a good friend to have.

    old jim
     
    Last edited: Aug 14, 2014
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  4. The diagram is correct according to my reference book..in it.. the meter is in series with R1 & Rx and parallel with the shunting resistor R2..
     
  5. jim hardy

    jim hardy 4,336
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    Okay - i thought about this last night and wondered if that's what your author was up to.

    That scheme would work provided R2 is a shunt which has really low resistance compared to R1 the halfscale resistor.
    That way you still have a reasonable voltage divider , and current is determined by R1 and Runknown since R1//Rmeter is small in comparison. You can measure the current and get an equally satisfactory measurement.

    So i owe you an apology for doubting your author - what he drew should work - and i humbly admit my mistake.

    What's important is that you grasp the clever trick of comparing Runknown to a known Rmidscale and calculating Runknown's value by laws of Ohm and Kirchoff, even if that calculation is just an analog one by a simple non-linear meter scale.

    In my description i described the ohm-meters to which i am accustomed.
    Remember i said you'd have to re-calculate R1 for effect of meter in parallel with it.

    Your textbook instead describes a meter in series to measure current instead of in parallel to measure voltage. Nothing at all is wrong with that concept. You'd have to recalculate R1 for the effect of the meter in series, that's all.


    Okay .... I agree, that's very close to what Rh will be.
    Recall that at full scale deflection the voltage across R1 is full battery voltage minus the drop across R2//Rmeter .
    If R2 is a shunt for the meter movement it should be generating only about 0.05 to 0.1 volts for full deflection, and half that for half scale deflection.


    It's pretty common to use a raw meter movement across a shunt, but i'm not accustomed to seeing them across an adjustable shunt.

    Okay let's think for a minute here........
    Meters for measuring current are typically 50 or 100 millivolts for full scale deflection.
    which is a small fraction of battery voltage.
    At this point putting some real world numbers to your example will demonstrate things better than juggling alphabets.. at least that's how my alleged brain works best.

    ----------------------------------Thought experiment for your shunt based ohm-meter ---------
    Let us assume midscale is 5 ohms.
    Also that Vbattery is 1.5 volts.
    And that we have a meter movement that takes 1 milliamp for full deflection and has Rmeter of 100ohm, so drops 100 millivolts. That is not an unusual movement to find in cheap meters. Better ones are more like 50 microamps 50 millivolts.

    In your mind, short meter leads together.

    Vbattery divides across R1 and R2//Rmeter.
    We know that R2//Rmeter sees 0.1 volt
    so R1 sees [STRIKE]1.49[/STRIKE] 1.40 volts

    meaning R1 = [STRIKE]14.9[/STRIKE] 14.0 X R2//Rmeter
    or R2//Rmeter = R1/14
    We could write one equation but it'll have two unknowns.
    Here it is:
    1.5 = I X (R1 + R2//Rmeter)

    Now connect meter leads to a 5 ohm resistor (Rmidscale) and write another equation.
    Note that midscale indication means we have half the current.
    1.5 = 0.5I X (R1 + R2//Rmeter + 5)

    Now divide one by the other and finish the algebra:

    1= (R1 + R2//Rmeter)/ 0.5(R1 + R2//Rmeter + 5)
    0.5(R1 + R2//Rmeter + 5) = (R1 + R2//Rmeter)
    2.5 = 0.5(R1 + R2//Rmeter)
    R1 + R2//Rmeter = 5 ( how 'bout that)

    R1 + R1/14.9 R1/14 = 5
    R1(1 + 1/14) = 5
    R1 = 5/[STRIKE]1.067114[/STRIKE] 1.0714 ... = [STRIKE]4.6855[/STRIKE] R1 = 4.6667 ohms
    and R2//Rmeter = 0.3333 ohms

    SO the R2 is indeed a low ohm shunt.

    Remember Rmeter was 100ohm

    so R2 = [STRIKE].3146[/STRIKE][STRIKE] oops, .3155 ohms[/STRIKE] oops2 R2 = 0.3345 ohms

    ------------------------------------------end thought experiment --------------------

    You can build an ohmmeter that way. Your author has proposed it.
    A lesson for me - times they are a'changin.

    If you build one that way, connect the wiper to one end of your rheostat. That way if the wiper opens, which is not an unusual occurrence , it'll be less of an overload to your meter.

    I really suggest you study the circuit. Calculate what how far off your halfscale indication will be when battery runs down to 1.25 volts and you've re-adjust R2 for full scale indication with leads shorted,
    and again when you have a fresh battery of perhaps 1.75 volts.

    Lastly
    Most of the meters i run across are of voltage measuring type not current as in your example.
    When somebody accidentally connects that type ohmmeter across 120 vac housepower while selected to RX1 , it smokes the internal halfscale resistor . Manufacturer often intends it as a fuse, they're usually metal film.
    You can buy such meters at thrift stores very cheap . The telltale is they read full scale when connected to even a 100 ohm resistor or flashlight bulb. That's why they were junked. It also tellls you the meter movement at least moves and isn't burnt open. Open them up and look on the circuit board, you'll see a charred Rhalfscale.
    It'll look something like this, top right:
    [​IMG]

    Now you know how to figure out what resistor to buy to fix a bargain junkshop meter.

    Thanks, i learned something and hope you did too.
    Polish up that algebra above and show it to teacher..

    old jim
     
    Last edited: Aug 15, 2014
  6. see..now I have many doubts to ask... as this topic is really new to me and I am very confused about everything and I even forgetting the basics.. !!
    first of all..if voltage across R2//Rm is 0.1 then voltage across R1 should be 1.5-0.1 i.e., 1.4 volt but here you wrote 1.49 Volt..??
    secondly how this expression .. R1=14.9* (R2//Rm) came up..??

    are you trying to say here that the net resistance is 5 ohm which is the mid scale resisitance..?? what are we getting from this.. what actually should be clear by now to me.. ?? it may sound ridiculous but I am actually confused...!!
    I got your calculations , explanations but were you trying to explain me..??
    please..help..its really required..!!
     
  7. jim hardy

    jim hardy 4,336
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    Oh how embarassing

    you are exactly right

    that's what i get for rushing to finish a post before dinner, and dinner was already out of the oven

    But how pleased i am that you caught my arithmetic mistake ! I found three others before my final click on "save changes"..

    I think you have "got it"

    I'll fix that post now. Check it again in about a half hour.

    old jim

    old jim
     
    Last edited: Aug 15, 2014
  8. jim hardy

    jim hardy 4,336
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    okay, i think it's fixed now.

    Yes, net meaning R1 + R2//Rmeter not including the external 5 ohm Runknown...

    No it's not ridiculous. My explanations were off-course because i've never encountered a series ohm-meter as in your textbook. My prejudice is to blame for your confusion.

    What should be clear to you is simply this -
    an analog meter type ohm-meter applies a voltage to an unknown resistance through a known internal resistance ,
    and calculates the unknown either of two ways:
    a. It measures the current and calculates the unknown by simple Ohm's law,

    Imeasured = Vbattery/(Rinternal + Runknown) ,,, where Imeasured , Vbattery and Rinternal are all known;

    or,

    b. It measures the voltage across its internal resistance and calculates the unknown by
    solving the voltage divider formula ,

    Vmeasured = Vbattery X Rinternal/(Rinternal + Runknown) ,,, where Vmeasured , Vbattery and Rinternal are all known.

    a. would be your series type current measuring meter

    b. would be my parallel type voltage measuring meter

    They're really not very different.


    I was just trying to get you up to the point you really understand the ohm meter.
    And some of its not-so-obvious uses. You seemed to have genuine interest.

    If you can calculate the values of R1 and R2 for an ohm meter with these parameters:
    Vbattery = 1.5
    Rmid = 10 ohms
    Meter = 0.1 ma full scale, Rmeter = 500 ohms
    then you've got it and you can design one from scratch.

    I hope i haven't made anymore mistakes.
    getting late here - g'night all.

    old jim
     
    Last edited: Aug 16, 2014
  9. what is R internal here ..?? Is it R2//Rm..?? and why haven't you considered R1 over here..??
     
  10. ohkk... I got tht..its R1+R2//Rm....
     
  11. and the Imeasured is full deflection current ..right???
     
  12. jim hardy

    jim hardy 4,336
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    yes !
     
  13. jim hardy

    jim hardy 4,336
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    Imeasured is whatever is measured
    when Runknown = zero it's Vbat/Rinternal which would be full deflection

    when Runknown = Rinternal it's Vbat/2Rinternal = half deflection
    if Runknown = 2Rinternal then Imeasured = Vbat/(Rinternal + 2Rinternal) = Vbat/3rinternal = 1/3 deflection
    and so on... that's how you figure out where on your scale to write the Ohm numbers !

    Glad to see your progress. Soon you won't be able to remember when it wasn't intuitive ..

    old jim
     
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  14. thanks a lot sir... its been really helpful...
     
  15. well I have gone through a ques.. in which we have to design a ohmmeter..Vbattery = 3V , current for full scale deflection = 0.5mA , , Rm=50 ohm , half-scale resistance =3000ohm..we have to find R1 & R2..!!
    so in 1st case we can find Rinternal as Rinternal= Vbattery / I (full-scale )
    but how will we find R1 & R2..??
    even if we form 2 equ... how we'll got R1 & R2..as those terms will be cancelled...
    see its not a h.w ques..the only thing is I am trying to sorting out this ohmmeter problem..!!
     
  16. jim hardy

    jim hardy 4,336
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    Ohh , you are so close now !

    go back and take a look at the thought experiment

    the key is you know the ratio of R1 to R2//Rmeter
    At full scale deflection what is voltage drop across R2//Rmeter?

    we wrote our second equation for current with Rhalfscale connected to the terminals which gives us current of I/2

    so when you divide the two equations, I cancels out but the other terms do not
    leaving you a not too bad looking equation with terms R1, Rhalfscale, R2//Rmeter, and [itex]1/2[/itex]

    since you know ratio of R1 to R2//Rmeter you know either one in terms of the other, so plug it in and you have one equation with only one unknown, .....



    be rigorous in your algebra, and write neatly in horizontal lines one step at a time
    else you'll make silly little mistakes, like i do...

    old jim
     
  17. after seeing the thought process another doubt is there which I asked also..but I guess u forgot to ans n so I..
    how the expression : meaning R1 = 14.0 X R2//Rmeter came up???
     
  18. jim hardy

    jim hardy 4,336
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    From voltage across the two resistances at full scale current when Runknown was zero.
    Vbattery = 1.5
    V across R2//Rmeter is in the thought experiment 0.1 volt, 100 millivolts. That was assumed:
    At full deflection, R2//Rm sees 0.1 volts and R1 sees the rest of Vbattery, 1.4 volts.
    The ratio of those two voltages, 1.4/0.1, is 14.0

    Think of it as a resistive voltage divider. Voltage divides in proportion to the resistances as you know from your basic circuit class.


    In your new example where Vbattery = 3 volts
    Rmeter is 50 ohms, and it's a 0.5milliamp movement
    so voltage across meter at full indication will be 25 millivolts , 0.025 volts

    so R1 will see 2.975 volts
    and R2//Rmeter will see 0.025 volts.
    What is that ratio ? Is that not the ratio of the two resistances, by laws of Ohm and Kirchoff?

    Write the two equations for current,
    Write first one with zero ohms for Runknown gives just I, and I is not known yet.
    Write second one with Rhalfscale ohms for Runknown , that one gives I/2

    divide one by the other to remove the I term because that eliminates one unknown.
    You're left with an equation having just terms R1 and R2//Rmeter and Rhalfscale and a consant of 0.5
    since you know R2 in terms of R1//Rmeter (their ratio from above), plug that in and you get an equation with only one unknown.

    You can trust your elementary algebra provided you are meticulous with it..
    Have more faith in your basics. Talk through them one step at a time.
    We all want to rush straight to an answer; to slow down our thinking and take "one step at a time" is contrary to the overstimulated age in which we live.
    glad to see you are persisting . That's how you become a master !


    old jim

    EDIT: PS you're doing a good job of phrasing your questions.
    You appear to have the electrical theory down just fine, it's a mere algebra problem now.
     
  19. but we are knowing the value of I.. since its full scale deflection and we are knowing the full scale deflection current i.e., 0.5mA so we should put it over here..!!
     
  20. jim hardy

    jim hardy 4,336
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    No we don't know the current I . 0.5 ma through the meter movement is just part of it.
    We do know that 0.5 is the current through the meter movement.
    We do not know yet how much more current flows through R2 in parallel with the meter movement.

    The current through R1 and Runknown is the sum of those two currents.
    That's I. It's the SUM of current through the meter movement AND the current through R2.

    That's why you must divide the two equations to get that unknown I out of them.
    After you've solved for R1 and R2//Rmeter, you can go back and solve for current I with any value of Runknown.

    I think you'll find full scale current I is one ma, half of it through R2 and the other half through your meter movement. Halfscale current will be 0.5 ma, again half through R2 and half through your meter.

    Try your algebra and see if it takes you there.

    old jim
     
    Last edited: Aug 16, 2014
  21. there's something wrong in your previous expression R1=14.0*R2//Rm...
    because according to voltage divider formula.. it'll be R1 = 14 *(R1+R2//Rm)..& not tht you'hv written..!!
    VR1=V(R2//Rm) * (R1/R1+R2//Rm)...
     
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