Is it possible to use a NPN BJT as switch, from single power source? The Next CEO of Stack...
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Is it possible to use a NPN BJT as switch, from single power source?
The Next CEO of Stack OverflowSimple transistor switching example should show LED offProblems Getting NPN Bipolar Transistor to Switch OnUsing NPN transistor as switchWhat type of transistor would be required?Inverting a push buttonNPN transistor not switching 12V from microcontrollerUsing a single power source for both the gate and source of my transistor?Newbie Help - Trouble with NPN resistorUse current from SMD LED to switch larger currentHow to remove leakage current from nRES transistor switch?
$begingroup$
I'm attempting to build the following circuit to better understand how to use a NPN BJT as a switch.
What I've Tried
I'm calculating the current across the LED (in order to light it up) as:
5.2V* - 1.7V (LED drop) = 3.5V
3.5V / 17mA = 200Ohms
*NOTE - The power source is 5.2V because I'm using 4 AA rechargeables in series (at 1.3V each).
The Problem
The problem I see when I push the button to turn the circuit on is that the transistor becomes very hot. I noticed this the first time because I could smell something. Then I touched it. Ouch! :) I'm assuming I'm providing too much current on the be (base to emitter) circuit.
Things I've Tried / Additional Problem
However, when I attempt to add resistance into the be circuit then the LED doesn't light up, even when my resistor gets down to a value as low as 47Ohms.
Questions
- Is it possible (due to ratios of current needed) to even power both sides of the circuit from the same power source? Or is it ridiculously difficult or something and not done?
- Can you help me understand the additional calculation(s) I should be using to power the circuit so my LED will light when I push the button?
simulate this circuit – Schematic created using CircuitLab
transistors bjt switching
$endgroup$
add a comment |
$begingroup$
I'm attempting to build the following circuit to better understand how to use a NPN BJT as a switch.
What I've Tried
I'm calculating the current across the LED (in order to light it up) as:
5.2V* - 1.7V (LED drop) = 3.5V
3.5V / 17mA = 200Ohms
*NOTE - The power source is 5.2V because I'm using 4 AA rechargeables in series (at 1.3V each).
The Problem
The problem I see when I push the button to turn the circuit on is that the transistor becomes very hot. I noticed this the first time because I could smell something. Then I touched it. Ouch! :) I'm assuming I'm providing too much current on the be (base to emitter) circuit.
Things I've Tried / Additional Problem
However, when I attempt to add resistance into the be circuit then the LED doesn't light up, even when my resistor gets down to a value as low as 47Ohms.
Questions
- Is it possible (due to ratios of current needed) to even power both sides of the circuit from the same power source? Or is it ridiculously difficult or something and not done?
- Can you help me understand the additional calculation(s) I should be using to power the circuit so my LED will light when I push the button?
simulate this circuit – Schematic created using CircuitLab
transistors bjt switching
$endgroup$
add a comment |
$begingroup$
I'm attempting to build the following circuit to better understand how to use a NPN BJT as a switch.
What I've Tried
I'm calculating the current across the LED (in order to light it up) as:
5.2V* - 1.7V (LED drop) = 3.5V
3.5V / 17mA = 200Ohms
*NOTE - The power source is 5.2V because I'm using 4 AA rechargeables in series (at 1.3V each).
The Problem
The problem I see when I push the button to turn the circuit on is that the transistor becomes very hot. I noticed this the first time because I could smell something. Then I touched it. Ouch! :) I'm assuming I'm providing too much current on the be (base to emitter) circuit.
Things I've Tried / Additional Problem
However, when I attempt to add resistance into the be circuit then the LED doesn't light up, even when my resistor gets down to a value as low as 47Ohms.
Questions
- Is it possible (due to ratios of current needed) to even power both sides of the circuit from the same power source? Or is it ridiculously difficult or something and not done?
- Can you help me understand the additional calculation(s) I should be using to power the circuit so my LED will light when I push the button?
simulate this circuit – Schematic created using CircuitLab
transistors bjt switching
$endgroup$
I'm attempting to build the following circuit to better understand how to use a NPN BJT as a switch.
What I've Tried
I'm calculating the current across the LED (in order to light it up) as:
5.2V* - 1.7V (LED drop) = 3.5V
3.5V / 17mA = 200Ohms
*NOTE - The power source is 5.2V because I'm using 4 AA rechargeables in series (at 1.3V each).
The Problem
The problem I see when I push the button to turn the circuit on is that the transistor becomes very hot. I noticed this the first time because I could smell something. Then I touched it. Ouch! :) I'm assuming I'm providing too much current on the be (base to emitter) circuit.
Things I've Tried / Additional Problem
However, when I attempt to add resistance into the be circuit then the LED doesn't light up, even when my resistor gets down to a value as low as 47Ohms.
Questions
- Is it possible (due to ratios of current needed) to even power both sides of the circuit from the same power source? Or is it ridiculously difficult or something and not done?
- Can you help me understand the additional calculation(s) I should be using to power the circuit so my LED will light when I push the button?
simulate this circuit – Schematic created using CircuitLab
transistors bjt switching
transistors bjt switching
asked 3 hours ago
raddevusraddevus
4501519
4501519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.
One way to use a switch and transistor to contol an LED is:
simulate this circuit – Schematic created using CircuitLab
R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.
Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.
$endgroup$
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
add a comment |
$begingroup$
Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.
I think this is closer to what you're looking for:
simulate this circuit – Schematic created using CircuitLab
Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.
I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).
$endgroup$
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.
One way to use a switch and transistor to contol an LED is:
simulate this circuit – Schematic created using CircuitLab
R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.
Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.
$endgroup$
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
add a comment |
$begingroup$
When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.
One way to use a switch and transistor to contol an LED is:
simulate this circuit – Schematic created using CircuitLab
R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.
Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.
$endgroup$
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
add a comment |
$begingroup$
When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.
One way to use a switch and transistor to contol an LED is:
simulate this circuit – Schematic created using CircuitLab
R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.
Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.
$endgroup$
When you close the switch, you are applying 5.2 volts across the base/emitter junction, which normally doesn't like more than 0.7 volts - this will destroy the transistor.
One way to use a switch and transistor to contol an LED is:
simulate this circuit – Schematic created using CircuitLab
R2 will limit the base current when thet switch is closed. R3 pulls the base low when the switch is open, to ensure the transistor is not conducting.
Pressing the switch will provide base current through R2, allowing the transistor to conduct, drawing current through the LED and R1. R1 limits the LED current to a safe value.
answered 2 hours ago
Peter BennettPeter Bennett
37.9k13068
37.9k13068
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
add a comment |
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
$begingroup$
Fantastic! Thanks very much for explaining that so clearly. I will try it out and mark this as answer later today or tomorrow.
$endgroup$
– raddevus
2 hours ago
add a comment |
$begingroup$
Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.
I think this is closer to what you're looking for:
simulate this circuit – Schematic created using CircuitLab
Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.
I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).
$endgroup$
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
add a comment |
$begingroup$
Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.
I think this is closer to what you're looking for:
simulate this circuit – Schematic created using CircuitLab
Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.
I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).
$endgroup$
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
add a comment |
$begingroup$
Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.
I think this is closer to what you're looking for:
simulate this circuit – Schematic created using CircuitLab
Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.
I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).
$endgroup$
Your circuit is not designed correctly for what you're trying to do. You currently have the BJT connected in parallel with the LED ciruit, instead of being in series with it. Because of this, in you circuit, when the BJT is turned on, all of the current is flowing directly from +5.2V to GND through the BJT (basically like a short circuit), which is why the BJT is getting so hot. Since there is 0 resistance between +5.2V and GND through the BJT, none of the current is going through the LED (thus staying off). Additionally, the switch SW1 is not connected to be able to turn on the BJT, rather it is connected to provide power to everything all at once.
I think this is closer to what you're looking for:
simulate this circuit – Schematic created using CircuitLab
Note how this allows the switch SW1 to control the BJT via its base pin, which in turn allows the BJT (which is now in series with the LED circuit) to pass current through it from 5.2V, through the D1, through R1, through Q1 and finally to GND.
I tried to make this as similar-looking as possible to your original circuit to make it clearer what was incorrect. However, note that you'll still want to include a pull-down resistor and series-resistor on the Q1 base pin (similar to Peter Bennett's post).
edited 2 hours ago
answered 2 hours ago
mithmith
25915
25915
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
add a comment |
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
3
3
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Applying 5.2 volts directly across the base/emitter junction will kill the transistor. You need a 1K or more resistor between the switch and the transistor base, and possibly a pull-down resistor to ensure the transistor turns off when the switch is open.
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
2 hours ago
add a comment |
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