Test a transistor with a multimeter

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I used the guide at https://vetco.net/blog/test-a-transistor-with-a-multimeter/2017-05-04-12-25-37-07 to test a transistor with my DMM. The NPN transistor I used to test is the "mospec tip130 No11D" and it was off from the circuit.



My question is that, STEP 3 (Emitter to Base) failed to show “OL” (Over Limit). Instead I read about ~1.2V. I used another same transistor that I am sure that it works on the circuit and I still get the same reading. Is the transistor a) bad b) wrong test c) good for some reason ?










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    up vote
    2
    down vote

    favorite












    I used the guide at https://vetco.net/blog/test-a-transistor-with-a-multimeter/2017-05-04-12-25-37-07 to test a transistor with my DMM. The NPN transistor I used to test is the "mospec tip130 No11D" and it was off from the circuit.



    My question is that, STEP 3 (Emitter to Base) failed to show “OL” (Over Limit). Instead I read about ~1.2V. I used another same transistor that I am sure that it works on the circuit and I still get the same reading. Is the transistor a) bad b) wrong test c) good for some reason ?










    share|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I used the guide at https://vetco.net/blog/test-a-transistor-with-a-multimeter/2017-05-04-12-25-37-07 to test a transistor with my DMM. The NPN transistor I used to test is the "mospec tip130 No11D" and it was off from the circuit.



      My question is that, STEP 3 (Emitter to Base) failed to show “OL” (Over Limit). Instead I read about ~1.2V. I used another same transistor that I am sure that it works on the circuit and I still get the same reading. Is the transistor a) bad b) wrong test c) good for some reason ?










      share|improve this question















      I used the guide at https://vetco.net/blog/test-a-transistor-with-a-multimeter/2017-05-04-12-25-37-07 to test a transistor with my DMM. The NPN transistor I used to test is the "mospec tip130 No11D" and it was off from the circuit.



      My question is that, STEP 3 (Emitter to Base) failed to show “OL” (Over Limit). Instead I read about ~1.2V. I used another same transistor that I am sure that it works on the circuit and I still get the same reading. Is the transistor a) bad b) wrong test c) good for some reason ?







      transistors pcb multimeter npn test






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      edited Sep 25 at 7:33

























      asked Sep 25 at 7:04









      Maverick

      1287




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          2 Answers
          2






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          up vote
          12
          down vote



          accepted










          Instead I can read about ~1.2V



          That makes perfect sense as the TIP130 is a Darlington transistor.



          It has an internal schematic like:



          enter image description here



          Note how between base and emitter there are actually two BE junctions in series, added up those two would have a forward voltage of around 1.2 V.



          Also note the additional diode between collector and emitter, it is only present in some Darlington transistors. Most "single" bipolar transistors don't have this diode.



          Bonus sidenote:



          Why does this type of transistor exist?



          Because is has a very high current amplification! A single transistor will usually have a current amplification (beta) of around a factor 100 to 500. But power transistors needed to control large currents (1 A or more) often have quite a low beta, often less than 30. Now by adding a (low power but high beta) transistor we can multiply the betas so we get a beta of (for the TIP130) of between 500 and 15000. So a lot less current is needed to control a large current.






          share|improve this answer


















          • 2




            I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
            – Maverick
            Sep 25 at 7:33






          • 1




            Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
            – Maverick
            Sep 25 at 7:42







          • 1




            Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
            – Bimpelrekkie
            Sep 25 at 8:22


















          up vote
          0
          down vote













          I think "Emitter to Base" may mean the PN junctions are reverse biased. The meter should show an open circuit. "Base to Emitter" measurement should be about 1.2V
          jlp






          share|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            12
            down vote



            accepted










            Instead I can read about ~1.2V



            That makes perfect sense as the TIP130 is a Darlington transistor.



            It has an internal schematic like:



            enter image description here



            Note how between base and emitter there are actually two BE junctions in series, added up those two would have a forward voltage of around 1.2 V.



            Also note the additional diode between collector and emitter, it is only present in some Darlington transistors. Most "single" bipolar transistors don't have this diode.



            Bonus sidenote:



            Why does this type of transistor exist?



            Because is has a very high current amplification! A single transistor will usually have a current amplification (beta) of around a factor 100 to 500. But power transistors needed to control large currents (1 A or more) often have quite a low beta, often less than 30. Now by adding a (low power but high beta) transistor we can multiply the betas so we get a beta of (for the TIP130) of between 500 and 15000. So a lot less current is needed to control a large current.






            share|improve this answer


















            • 2




              I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
              – Maverick
              Sep 25 at 7:33






            • 1




              Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
              – Maverick
              Sep 25 at 7:42







            • 1




              Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
              – Bimpelrekkie
              Sep 25 at 8:22















            up vote
            12
            down vote



            accepted










            Instead I can read about ~1.2V



            That makes perfect sense as the TIP130 is a Darlington transistor.



            It has an internal schematic like:



            enter image description here



            Note how between base and emitter there are actually two BE junctions in series, added up those two would have a forward voltage of around 1.2 V.



            Also note the additional diode between collector and emitter, it is only present in some Darlington transistors. Most "single" bipolar transistors don't have this diode.



            Bonus sidenote:



            Why does this type of transistor exist?



            Because is has a very high current amplification! A single transistor will usually have a current amplification (beta) of around a factor 100 to 500. But power transistors needed to control large currents (1 A or more) often have quite a low beta, often less than 30. Now by adding a (low power but high beta) transistor we can multiply the betas so we get a beta of (for the TIP130) of between 500 and 15000. So a lot less current is needed to control a large current.






            share|improve this answer


















            • 2




              I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
              – Maverick
              Sep 25 at 7:33






            • 1




              Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
              – Maverick
              Sep 25 at 7:42







            • 1




              Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
              – Bimpelrekkie
              Sep 25 at 8:22













            up vote
            12
            down vote



            accepted







            up vote
            12
            down vote



            accepted






            Instead I can read about ~1.2V



            That makes perfect sense as the TIP130 is a Darlington transistor.



            It has an internal schematic like:



            enter image description here



            Note how between base and emitter there are actually two BE junctions in series, added up those two would have a forward voltage of around 1.2 V.



            Also note the additional diode between collector and emitter, it is only present in some Darlington transistors. Most "single" bipolar transistors don't have this diode.



            Bonus sidenote:



            Why does this type of transistor exist?



            Because is has a very high current amplification! A single transistor will usually have a current amplification (beta) of around a factor 100 to 500. But power transistors needed to control large currents (1 A or more) often have quite a low beta, often less than 30. Now by adding a (low power but high beta) transistor we can multiply the betas so we get a beta of (for the TIP130) of between 500 and 15000. So a lot less current is needed to control a large current.






            share|improve this answer














            Instead I can read about ~1.2V



            That makes perfect sense as the TIP130 is a Darlington transistor.



            It has an internal schematic like:



            enter image description here



            Note how between base and emitter there are actually two BE junctions in series, added up those two would have a forward voltage of around 1.2 V.



            Also note the additional diode between collector and emitter, it is only present in some Darlington transistors. Most "single" bipolar transistors don't have this diode.



            Bonus sidenote:



            Why does this type of transistor exist?



            Because is has a very high current amplification! A single transistor will usually have a current amplification (beta) of around a factor 100 to 500. But power transistors needed to control large currents (1 A or more) often have quite a low beta, often less than 30. Now by adding a (low power but high beta) transistor we can multiply the betas so we get a beta of (for the TIP130) of between 500 and 15000. So a lot less current is needed to control a large current.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Sep 25 at 19:01

























            answered Sep 25 at 7:20









            Bimpelrekkie

            43.4k23996




            43.4k23996







            • 2




              I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
              – Maverick
              Sep 25 at 7:33






            • 1




              Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
              – Maverick
              Sep 25 at 7:42







            • 1




              Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
              – Bimpelrekkie
              Sep 25 at 8:22













            • 2




              I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
              – Maverick
              Sep 25 at 7:33






            • 1




              Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
              – Maverick
              Sep 25 at 7:42







            • 1




              Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
              – Bimpelrekkie
              Sep 25 at 8:22








            2




            2




            I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
            – Maverick
            Sep 25 at 7:33




            I suspected the darlington connection that I read in the datasheet, but I dont have the experience to understand if that matters in this case. Thanx !
            – Maverick
            Sep 25 at 7:33




            1




            1




            Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
            – Maverick
            Sep 25 at 7:42





            Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
            – Maverick
            Sep 25 at 7:42





            1




            1




            Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
            – Bimpelrekkie
            Sep 25 at 8:22





            Yes, that's the diode. If you probe + at emitter and - at collector you should measure around 0.6 V. If you probe the other way round (- at emitter and + at collector) you should get "OL" as then there should be no conduction at all.
            – Bimpelrekkie
            Sep 25 at 8:22













            up vote
            0
            down vote













            I think "Emitter to Base" may mean the PN junctions are reverse biased. The meter should show an open circuit. "Base to Emitter" measurement should be about 1.2V
            jlp






            share|improve this answer
























              up vote
              0
              down vote













              I think "Emitter to Base" may mean the PN junctions are reverse biased. The meter should show an open circuit. "Base to Emitter" measurement should be about 1.2V
              jlp






              share|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                I think "Emitter to Base" may mean the PN junctions are reverse biased. The meter should show an open circuit. "Base to Emitter" measurement should be about 1.2V
                jlp






                share|improve this answer












                I think "Emitter to Base" may mean the PN junctions are reverse biased. The meter should show an open circuit. "Base to Emitter" measurement should be about 1.2V
                jlp







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Sep 25 at 17:21









                jlpayton

                1




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