2
\$\begingroup\$

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 ?

\$\endgroup\$
12
\$\begingroup\$

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.

\$\endgroup\$
  • 2
    \$\begingroup\$ 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 ! \$\endgroup\$ – Maverick Sep 25 '18 at 7:33
  • 1
    \$\begingroup\$ Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)? \$\endgroup\$ – Maverick Sep 25 '18 at 7:42
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Bimpelrekkie Sep 25 '18 at 8:22
  • \$\begingroup\$ Base(+) to Emitter(-) I get 5.85V thats ok. But Base(-) to Emitter(+) I get ~1.2 (reverse biased). Is that ok to get 1.2V in reverse biased ? Cos in the last case we dont have two BE junctions is series if I am correct. \$\endgroup\$ – Maverick Jan 10 at 20:16
  • \$\begingroup\$ @Maverick When probing B(+) and E(-) you get 2 diodes in forward so 1.2 V. When probing E(+) and B(-) there will be 8120 ohms (both resistors in series, the BE diodes will be off) so the voltage you get will depend on what current your meter uses for the diode test. If you're sure you probed E(+) and B(-) and got 1.2 V then you have a weird diode test on your meter. \$\endgroup\$ – Bimpelrekkie Jan 10 at 20:59
0
\$\begingroup\$

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

\$\endgroup\$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.