Can this transistor (2N2222) take 6 V on emitter-base? Am I reading the datasheet incorrectly?
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
Datasheet states EB maximum voltage
According to the datasheet the emitter-base maximum voltage is 6.0 V. In my circuit I am only applying 5 V. 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 40 V(?). Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V 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 5 V -- but that may have been due to being wired improperly, I'm not sure.
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 5 V on its base pin (according to the datasheet).
voltage transistors datasheet
$endgroup$
|
show 5 more comments
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
Datasheet states EB maximum voltage
According to the datasheet the emitter-base maximum voltage is 6.0 V. In my circuit I am only applying 5 V. 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 40 V(?). Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V 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 5 V -- but that may have been due to being wired improperly, I'm not sure.
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 5 V on its base pin (according to the datasheet).
voltage transistors datasheet
$endgroup$
2
$begingroup$
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
8 hours 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
8 hours 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
8 hours ago
1
$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
8 hours ago
1
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago
|
show 5 more comments
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
Datasheet states EB maximum voltage
According to the datasheet the emitter-base maximum voltage is 6.0 V. In my circuit I am only applying 5 V. 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 40 V(?). Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V 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 5 V -- but that may have been due to being wired improperly, I'm not sure.
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 5 V on its base pin (according to the datasheet).
voltage transistors datasheet
$endgroup$
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.
simulate this circuit – Schematic created using CircuitLab
Datasheet states EB maximum voltage
According to the datasheet the emitter-base maximum voltage is 6.0 V. In my circuit I am only applying 5 V. 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 40 V(?). Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V 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 5 V -- but that may have been due to being wired improperly, I'm not sure.
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 5 V on its base pin (according to the datasheet).
voltage transistors datasheet
voltage transistors datasheet
edited 17 mins ago
Peter Mortensen
1,60031422
1,60031422
asked 8 hours ago
raddevusraddevus
4721519
4721519
2
$begingroup$
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
8 hours 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
8 hours 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
8 hours ago
1
$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
8 hours ago
1
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago
|
show 5 more comments
2
$begingroup$
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
8 hours 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
8 hours 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
8 hours ago
1
$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
8 hours ago
1
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago
2
2
$begingroup$
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
8 hours 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
8 hours 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
8 hours 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
8 hours 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
8 hours 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
8 hours 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
8 hours ago
$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
8 hours ago
1
1
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$V_{EB}=V_E-V_B$.
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.
$endgroup$
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
8 hours ago
add a comment |
$begingroup$
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.)
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 $+5:text{V}$ 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 $5:text{V}$ across the base-emitter diode. Keep in mind:
- You only need from as little as $600:text{mV}$ to perhaps as much as $900:text{mV}$ (in most cases) to use the BJT as a switch.
- For each additional $60:text{mV}$ (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 $5:text{V}$!! 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.
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.
$endgroup$
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$V_{EB}=V_E-V_B$.
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.
$endgroup$
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
8 hours ago
add a comment |
$begingroup$
$V_{EB}=V_E-V_B$.
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.
$endgroup$
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
8 hours ago
add a comment |
$begingroup$
$V_{EB}=V_E-V_B$.
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.
$endgroup$
$V_{EB}=V_E-V_B$.
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.
edited 7 hours ago
Bruce Abbott
25.8k11934
25.8k11934
answered 8 hours ago
The PhotonThe Photon
86.8k398202
86.8k398202
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
8 hours ago
add a comment |
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
8 hours 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
8 hours 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
8 hours ago
add a comment |
$begingroup$
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.)
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 $+5:text{V}$ 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 $5:text{V}$ across the base-emitter diode. Keep in mind:
- You only need from as little as $600:text{mV}$ to perhaps as much as $900:text{mV}$ (in most cases) to use the BJT as a switch.
- For each additional $60:text{mV}$ (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 $5:text{V}$!! 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.
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.
$endgroup$
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
add a comment |
$begingroup$
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.)
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 $+5:text{V}$ 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 $5:text{V}$ across the base-emitter diode. Keep in mind:
- You only need from as little as $600:text{mV}$ to perhaps as much as $900:text{mV}$ (in most cases) to use the BJT as a switch.
- For each additional $60:text{mV}$ (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 $5:text{V}$!! 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.
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.
$endgroup$
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
add a comment |
$begingroup$
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.)
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 $+5:text{V}$ 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 $5:text{V}$ across the base-emitter diode. Keep in mind:
- You only need from as little as $600:text{mV}$ to perhaps as much as $900:text{mV}$ (in most cases) to use the BJT as a switch.
- For each additional $60:text{mV}$ (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 $5:text{V}$!! 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.
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.
$endgroup$
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.)
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 $+5:text{V}$ 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 $5:text{V}$ across the base-emitter diode. Keep in mind:
- You only need from as little as $600:text{mV}$ to perhaps as much as $900:text{mV}$ (in most cases) to use the BJT as a switch.
- For each additional $60:text{mV}$ (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 $5:text{V}$!! 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.
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.
edited 7 hours ago
answered 7 hours ago
jonkjonk
34.5k12875
34.5k12875
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
add a comment |
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
$begingroup$
Fantastic explanation. Very clear and helped me understand a lot. Thanks.
$endgroup$
– raddevus
7 hours ago
1
1
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
@raddevus Thanks for the kind words and I'm glad it helped out.
$endgroup$
– jonk
7 hours ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
$begingroup$
Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.)
$endgroup$
– Hearth
14 mins ago
add a comment |
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$begingroup$
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
8 hours 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
8 hours 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
8 hours ago
1
$begingroup$
No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K.
$endgroup$
– BB ON
8 hours ago
1
$begingroup$
You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again?
$endgroup$
– Peter Bennett
6 hours ago