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Is it possible to use a NPN BJT as switch, from single power source?
The Next CEO of Stack OverflowCan this transistor (2n2222) take 6V on emitter-base? Am I reading datasheet incorrectly?Simple 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
asked yesterday
raddevusraddevus
4621519
$endgroup$
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
asked yesterday
raddevusraddevus
4621519
$endgroup$
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
asked yesterday
raddevusraddevus
4621519
$endgroup$
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 yesterday
raddevusraddevus
4621519
asked yesterday
raddevusraddevus
4621519
edited 18 hours ago
raddevus
asked yesterday
raddevusraddevus
4621519
asked yesterday
raddevusraddevus
4621519
asked yesterday
raddevusraddevus
4621519
4621519
add a comment |
add a comment |
1 Answer
1
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.
answered yesterday
Peter BennettPeter Bennett
38k13068
$endgroup$
$begingroup$
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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
answered yesterday
Peter BennettPeter Bennett
38k13068
$endgroup$
$begingroup$
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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 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.
answered yesterday
Peter BennettPeter Bennett
38k13068
$endgroup$
$begingroup$
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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 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.
answered yesterday
Peter BennettPeter Bennett
38k13068
$endgroup$
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 yesterday
Peter BennettPeter Bennett
38k13068
answered yesterday
Peter BennettPeter Bennett
38k13068
answered yesterday
Peter BennettPeter Bennett
38k13068
answered yesterday
Peter BennettPeter Bennett
38k13068
38k13068
$begingroup$
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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 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
yesterday
$begingroup$
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
yesterday
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 hours ago
$begingroup$
I built the circuit and examined it with multimeter also and it works great. The transistor no longer becomes hot at all either of course (because of the voltage divider). Great help, thanks again.
$endgroup$
– raddevus
23 hours ago
add a comment |
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