Can this transistor (2n2222) take 6V on emitter-base? Am I reading datasheet incorrectly? The Next CEO of Stack OverflowIs it possible to use a NPN BJT as switch, from single power source?How to read the datasheet of a Digital transistor?Is this transistor datasheet correct?NPN transistor high resistance across collector and emitterCalculating base current and resistor required from transistor datasheet? Do I need a different transistor?Is it possible to overcome a transistor's current limit by running it at lower voltage?H bridge's voltage, breaksdown when connected to a motorBJT Transistors - On CharacteristicsSelecting a transistor - Reading transistor datasheetsDo BJT (Darlington) transistors have a minimum collector current?ESD diode with zener configuration is equivalent to a circuit containing diode, transistor and zener diode, so how this could be equivalent

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Can this transistor (2n2222) take 6V on emitter-base? Am I reading datasheet incorrectly?



The Next CEO of Stack OverflowIs it possible to use a NPN BJT as switch, from single power source?How to read the datasheet of a Digital transistor?Is this transistor datasheet correct?NPN transistor high resistance across collector and emitterCalculating base current and resistor required from transistor datasheet? Do I need a different transistor?Is it possible to overcome a transistor's current limit by running it at lower voltage?H bridge's voltage, breaksdown when connected to a motorBJT Transistors - On CharacteristicsSelecting a transistor - Reading transistor datasheetsDo BJT (Darlington) transistors have a minimum collector current?ESD diode with zener configuration is equivalent to a circuit containing diode, transistor and zener diode, so how this could be equivalent










1












$begingroup$


I have created the following circuit to better understand how to use a transistor as a switch*.



I've been struggling with understanding datasheets for transistors.





schematic





simulate this circuit – Schematic created using CircuitLab



DataSheet States EB Max Voltage



According to the datasheet the Emitter-Base voltage max is 6.0V.
In my circuit I am only applying 5V.

From reading the datasheet (below) would you expect that to be too much?



Also, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40V? Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40V on it?



Am I reading the datasheet properly?



I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5V -- but that may have been due to being wired improperly, I'm not sure.



2n2222a max ratings



You can see the complete datasheet at:



https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF



* Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source? However, I had the original circuit built improperly there and I'm now wondering if the 2N2222 can take 5V on it's base pin (according to datasheet).










share|improve this question









$endgroup$







  • 1




    $begingroup$
    Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
    $endgroup$
    – BB ON
    51 mins ago











  • $begingroup$
    @BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
    $endgroup$
    – raddevus
    47 mins ago







  • 1




    $begingroup$
    Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
    $endgroup$
    – BB ON
    41 mins ago











  • $begingroup$
    @BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
    $endgroup$
    – raddevus
    37 mins ago






  • 1




    $begingroup$
    No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
    $endgroup$
    – BB ON
    29 mins ago















1












$begingroup$


I have created the following circuit to better understand how to use a transistor as a switch*.



I've been struggling with understanding datasheets for transistors.





schematic





simulate this circuit – Schematic created using CircuitLab



DataSheet States EB Max Voltage



According to the datasheet the Emitter-Base voltage max is 6.0V.
In my circuit I am only applying 5V.

From reading the datasheet (below) would you expect that to be too much?



Also, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40V? Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40V on it?



Am I reading the datasheet properly?



I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5V -- but that may have been due to being wired improperly, I'm not sure.



2n2222a max ratings



You can see the complete datasheet at:



https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF



* Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source? However, I had the original circuit built improperly there and I'm now wondering if the 2N2222 can take 5V on it's base pin (according to datasheet).










share|improve this question









$endgroup$







  • 1




    $begingroup$
    Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
    $endgroup$
    – BB ON
    51 mins ago











  • $begingroup$
    @BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
    $endgroup$
    – raddevus
    47 mins ago







  • 1




    $begingroup$
    Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
    $endgroup$
    – BB ON
    41 mins ago











  • $begingroup$
    @BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
    $endgroup$
    – raddevus
    37 mins ago






  • 1




    $begingroup$
    No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
    $endgroup$
    – BB ON
    29 mins ago













1












1








1





$begingroup$


I have created the following circuit to better understand how to use a transistor as a switch*.



I've been struggling with understanding datasheets for transistors.





schematic





simulate this circuit – Schematic created using CircuitLab



DataSheet States EB Max Voltage



According to the datasheet the Emitter-Base voltage max is 6.0V.
In my circuit I am only applying 5V.

From reading the datasheet (below) would you expect that to be too much?



Also, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40V? Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40V on it?



Am I reading the datasheet properly?



I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5V -- but that may have been due to being wired improperly, I'm not sure.



2n2222a max ratings



You can see the complete datasheet at:



https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF



* Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source? However, I had the original circuit built improperly there and I'm now wondering if the 2N2222 can take 5V on it's base pin (according to datasheet).










share|improve this question









$endgroup$




I have created the following circuit to better understand how to use a transistor as a switch*.



I've been struggling with understanding datasheets for transistors.





schematic





simulate this circuit – Schematic created using CircuitLab



DataSheet States EB Max Voltage



According to the datasheet the Emitter-Base voltage max is 6.0V.
In my circuit I am only applying 5V.

From reading the datasheet (below) would you expect that to be too much?



Also, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40V? Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40V on it?



Am I reading the datasheet properly?



I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5V -- but that may have been due to being wired improperly, I'm not sure.



2n2222a max ratings



You can see the complete datasheet at:



https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF



* Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source? However, I had the original circuit built improperly there and I'm now wondering if the 2N2222 can take 5V on it's base pin (according to datasheet).







voltage transistors datasheet






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 1 hour ago









raddevusraddevus

4621519




4621519







  • 1




    $begingroup$
    Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
    $endgroup$
    – BB ON
    51 mins ago











  • $begingroup$
    @BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
    $endgroup$
    – raddevus
    47 mins ago







  • 1




    $begingroup$
    Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
    $endgroup$
    – BB ON
    41 mins ago











  • $begingroup$
    @BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
    $endgroup$
    – raddevus
    37 mins ago






  • 1




    $begingroup$
    No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
    $endgroup$
    – BB ON
    29 mins ago












  • 1




    $begingroup$
    Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
    $endgroup$
    – BB ON
    51 mins ago











  • $begingroup$
    @BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
    $endgroup$
    – raddevus
    47 mins ago







  • 1




    $begingroup$
    Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
    $endgroup$
    – BB ON
    41 mins ago











  • $begingroup$
    @BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
    $endgroup$
    – raddevus
    37 mins ago






  • 1




    $begingroup$
    No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
    $endgroup$
    – BB ON
    29 mins ago







1




1




$begingroup$
Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
$endgroup$
– BB ON
51 mins ago





$begingroup$
Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other.
$endgroup$
– BB ON
51 mins ago













$begingroup$
@BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
$endgroup$
– raddevus
47 mins ago





$begingroup$
@BBON So it's a ratio between base and emitter side? And I need to increase the emitter side since the base side is so high? If I had 4.5 volts on the emitter side then maybe the base side would be ok? I'm guessing here. Still not sure. I'm sorry, I know I'm missing something foundational. EDIT - Also everyone always mentions 0.7V and I'm not sure where that comes from. Part of my missing knowledge. I don't ever see that in the datasheet. thx
$endgroup$
– raddevus
47 mins ago





1




1




$begingroup$
Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
$endgroup$
– BB ON
41 mins ago





$begingroup$
Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory.
$endgroup$
– BB ON
41 mins ago













$begingroup$
@BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
$endgroup$
– raddevus
37 mins ago




$begingroup$
@BBON Very good explanation of "forward biased at that voltage...applying 5V doesn't make it conduct more". That helped a lot. I will read more on BJT theory. I've read things like learn.sparkfun.com/tutorials/transistors/all but trying to then read datasheet was difficult for me. Thanks again.
$endgroup$
– raddevus
37 mins ago




1




1




$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
29 mins ago




$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
29 mins ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

$V_EB=V_E-V_B$.



The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.



With $V_BE$ ($V_B-V_E$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.






share|improve this answer











$endgroup$








  • 2




    $begingroup$
    @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
    $endgroup$
    – Transistor
    1 hour ago



















0












$begingroup$

This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)





schematic





simulate this circuit – Schematic created using CircuitLab



The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.



But the idea of directly tying $+5:textV$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $5:textV$ across the base-emitter diode. Keep in mind:



  1. You only need from as little as $600:textmV$ to perhaps as much as $900:textmV$ (in most cases) to use the BJT as a switch.

  2. For each additional $60:textmV$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

You were applying $5:textV$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)



This is why a resistor is often applied to the base circuit.





schematic





simulate this circuit



The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.



By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.



As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.






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









    5












    $begingroup$

    $V_EB=V_E-V_B$.



    The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.



    With $V_BE$ ($V_B-V_E$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.






    share|improve this answer











    $endgroup$








    • 2




      $begingroup$
      @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
      $endgroup$
      – Transistor
      1 hour ago
















    5












    $begingroup$

    $V_EB=V_E-V_B$.



    The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.



    With $V_BE$ ($V_B-V_E$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.






    share|improve this answer











    $endgroup$








    • 2




      $begingroup$
      @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
      $endgroup$
      – Transistor
      1 hour ago














    5












    5








    5





    $begingroup$

    $V_EB=V_E-V_B$.



    The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.



    With $V_BE$ ($V_B-V_E$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.






    share|improve this answer











    $endgroup$



    $V_EB=V_E-V_B$.



    The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.



    With $V_BE$ ($V_B-V_E$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 21 mins ago









    Bruce Abbott

    25.8k11934




    25.8k11934










    answered 1 hour ago









    The PhotonThe Photon

    86.8k398201




    86.8k398201







    • 2




      $begingroup$
      @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
      $endgroup$
      – Transistor
      1 hour ago













    • 2




      $begingroup$
      @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
      $endgroup$
      – Transistor
      1 hour ago








    2




    2




    $begingroup$
    @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
    $endgroup$
    – Transistor
    1 hour ago





    $begingroup$
    @raddevus: To be clear - $ V_EB $ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction).
    $endgroup$
    – Transistor
    1 hour ago














    0












    $begingroup$

    This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)





    schematic





    simulate this circuit – Schematic created using CircuitLab



    The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.



    But the idea of directly tying $+5:textV$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $5:textV$ across the base-emitter diode. Keep in mind:



    1. You only need from as little as $600:textmV$ to perhaps as much as $900:textmV$ (in most cases) to use the BJT as a switch.

    2. For each additional $60:textmV$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

    You were applying $5:textV$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)



    This is why a resistor is often applied to the base circuit.





    schematic





    simulate this circuit



    The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.



    By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.



    As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.






    share|improve this answer











    $endgroup$

















      0












      $begingroup$

      This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)





      schematic





      simulate this circuit – Schematic created using CircuitLab



      The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.



      But the idea of directly tying $+5:textV$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $5:textV$ across the base-emitter diode. Keep in mind:



      1. You only need from as little as $600:textmV$ to perhaps as much as $900:textmV$ (in most cases) to use the BJT as a switch.

      2. For each additional $60:textmV$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

      You were applying $5:textV$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)



      This is why a resistor is often applied to the base circuit.





      schematic





      simulate this circuit



      The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.



      By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.



      As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.






      share|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)





        schematic





        simulate this circuit – Schematic created using CircuitLab



        The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.



        But the idea of directly tying $+5:textV$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $5:textV$ across the base-emitter diode. Keep in mind:



        1. You only need from as little as $600:textmV$ to perhaps as much as $900:textmV$ (in most cases) to use the BJT as a switch.

        2. For each additional $60:textmV$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

        You were applying $5:textV$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)



        This is why a resistor is often applied to the base circuit.





        schematic





        simulate this circuit



        The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.



        By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.



        As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.






        share|improve this answer











        $endgroup$



        This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)





        schematic





        simulate this circuit – Schematic created using CircuitLab



        The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.



        But the idea of directly tying $+5:textV$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $5:textV$ across the base-emitter diode. Keep in mind:



        1. You only need from as little as $600:textmV$ to perhaps as much as $900:textmV$ (in most cases) to use the BJT as a switch.

        2. For each additional $60:textmV$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

        You were applying $5:textV$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)



        This is why a resistor is often applied to the base circuit.





        schematic





        simulate this circuit



        The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.



        By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.



        As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 8 mins ago

























        answered 15 mins ago









        jonkjonk

        34.4k12874




        34.4k12874



























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