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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?










2












$begingroup$


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



  1. 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?

  2. 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?

schematic





simulate this circuit – Schematic created using CircuitLab










share|improve this question









$endgroup$
















    2












    $begingroup$


    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



    1. 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?

    2. 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?

    schematic





    simulate this circuit – Schematic created using CircuitLab










    share|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      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



      1. 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?

      2. 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?

      schematic





      simulate this circuit – Schematic created using CircuitLab










      share|improve this question









      $endgroup$




      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



      1. 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?

      2. 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?

      schematic





      simulate this circuit – Schematic created using CircuitLab







      transistors bjt switching






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 3 hours ago









      raddevusraddevus

      4501519




      4501519




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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:





          schematic





          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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


















          0












          $begingroup$

          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:





          schematic





          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).






          share|improve this answer











          $endgroup$








          • 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











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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:





          schematic





          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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















          2












          $begingroup$

          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:





          schematic





          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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













          2












          2








          2





          $begingroup$

          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:





          schematic





          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.






          share|improve this answer









          $endgroup$



          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:





          schematic





          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.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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
















          • $begingroup$
            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













          0












          $begingroup$

          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:





          schematic





          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).






          share|improve this answer











          $endgroup$








          • 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















          0












          $begingroup$

          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:





          schematic





          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).






          share|improve this answer











          $endgroup$








          • 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













          0












          0








          0





          $begingroup$

          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:





          schematic





          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).






          share|improve this answer











          $endgroup$



          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:





          schematic





          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).







          share|improve this answer














          share|improve this answer



          share|improve this answer








          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












          • 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

















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