Test a transistor with a multimeter
Clash Royale CLAN TAG#URR8PPP
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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 âÂÂOLâ (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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up vote
2
down vote
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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 âÂÂOLâ (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
add a comment |Â
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 âÂÂOLâ (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
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 âÂÂOLâ (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
transistors pcb multimeter npn test
edited Sep 25 at 7:33
asked Sep 25 at 7:04
Maverick
1287
1287
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2 Answers
2
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oldest
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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:
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.
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 !
â Maverick
Sep 25 at 7:33
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â Bimpelrekkie
Sep 25 at 8:22
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0
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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
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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:
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.
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 !
â Maverick
Sep 25 at 7:33
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â Bimpelrekkie
Sep 25 at 8:22
add a comment |Â
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:
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.
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 !
â Maverick
Sep 25 at 7:33
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â Bimpelrekkie
Sep 25 at 8:22
add a comment |Â
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:
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.
Instead I can read about ~1.2V
That makes perfect sense as the TIP130 is a Darlington transistor.
It has an internal schematic like:
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.
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 !
â Maverick
Sep 25 at 7:33
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â Bimpelrekkie
Sep 25 at 8:22
add a comment |Â
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 !
â Maverick
Sep 25 at 7:33
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â 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 !
â 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 !
â Maverick
Sep 25 at 7:33
1
1
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â Maverick
Sep 25 at 7:42
Looking the schematic.... shouldn't I have also some voltage between Emmiter (+ proble) and Collector (- proble)?
â 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.
â 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.
â Bimpelrekkie
Sep 25 at 8:22
add a comment |Â
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
add a comment |Â
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
add a comment |Â
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
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
answered Sep 25 at 17:21
jlpayton
1
1
add a comment |Â
add a comment |Â
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