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?
Chain wire methods together in Lightning Web Components
How to install OpenCV on Raspbian Stretch?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
A Man With a Stainless Steel Endoskeleton (like The Terminator) Fighting Cloaked Aliens Only He Can See
Why does standard notation not preserve intervals (visually)
How a 64-bit process virtual address space is divided in Linux?
The exact meaning of 'Mom made me a sandwich'
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
Newlines in BSD sed vs gsed
Is it possible to use a NPN BJT as switch, from single power source?
Why is my new battery behaving weirdly?
Won the lottery - how do I keep the money?
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
What happened in Rome, when the western empire "fell"?
Make solar eclipses exceedingly rare, but still have new moons
"misplaced omit" error when >centering columns
How do I align (1) and (2)?
Why is quantifier elimination desirable for a given theory?
Can MTA send mail via a relay without being told so?
When you upcast Blindness/Deafness, do all targets suffer the same effect?
Is wanting to ask what to write an indication that you need to change your story?
What was the first Unix version to run on a microcomputer?
Why didn't Khan get resurrected in the Genesis Explosion?
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 2 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
1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");
StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f429894%2fis-it-possible-to-use-a-npn-bjt-as-switch-from-single-power-source%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
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.
$endgroup$
$begingroup$
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
1 hour 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
1 hour 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 1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour 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 1 hour ago
answered 1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour ago
$begingroup$
Thanks, I edited the post to make that clarification
$endgroup$
– mith
1 hour ago
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f429894%2fis-it-possible-to-use-a-npn-bjt-as-switch-from-single-power-source%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown