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Is there a way to bypass a component in series in a circuit if that component fails?



The Next CEO of Stack OverflowWhat does this component do in this series pass regulatorSeries and Parallel CircuitSimplifying Series/ Parallel RC circuitled wont turn on in series circuitHow to determine total circuit voltage and individual component resistance for a series circuitSeries parallel resistor. Circuit redrawCircuit redraw. Series and parallel resistors240v Split phase series circuitSeries RL Circuit Transient AnalysisIf the order of any component is altered in a series circuit, will the total voltage be affected?










1












$begingroup$


Imagine a couple of speakers connected in series. Could there be a way to build such circuit with the ability to close the circuit if one component isn't working anymore? Like a different path that ignores the malfunctioning component, maybe using a voltmeter to trigger a switch in the circuit when it reads zero.



(just a question I'd given my electronics teacher in class. He failed to give me an answer)










share|improve this question









$endgroup$
















    1












    $begingroup$


    Imagine a couple of speakers connected in series. Could there be a way to build such circuit with the ability to close the circuit if one component isn't working anymore? Like a different path that ignores the malfunctioning component, maybe using a voltmeter to trigger a switch in the circuit when it reads zero.



    (just a question I'd given my electronics teacher in class. He failed to give me an answer)










    share|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      Imagine a couple of speakers connected in series. Could there be a way to build such circuit with the ability to close the circuit if one component isn't working anymore? Like a different path that ignores the malfunctioning component, maybe using a voltmeter to trigger a switch in the circuit when it reads zero.



      (just a question I'd given my electronics teacher in class. He failed to give me an answer)










      share|improve this question









      $endgroup$




      Imagine a couple of speakers connected in series. Could there be a way to build such circuit with the ability to close the circuit if one component isn't working anymore? Like a different path that ignores the malfunctioning component, maybe using a voltmeter to trigger a switch in the circuit when it reads zero.



      (just a question I'd given my electronics teacher in class. He failed to give me an answer)







      series






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 days ago









      InTheMoodForNowInTheMoodForNow

      203




      203




















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$


          Is there a way to bypass a component in series in a circuit if that component fails?




          This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.



          There exist some examples of this already where it is feasible.



          • mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out.

          • battery management system (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure.

          • some LED drivers have string open detection to detect and/or bypass individual without excess current.

          • "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure

          • AC grid fault detection systems have re-routing switch methods for some fault conditions

          • The internet was designed with redundant paths where a "switch" is just one unit in a system.

          Yes/No/Maybe



          That depends on a lot of unstated assumptions, which one can specify by questions.



          What kind of component or failure? Open? , short? or in between?

          Will it reduce reliability? e.g. bypass a fuse

          Is it worth it? Cost/benefit for an added detector, multiplexer to bypass

          Is it better to choose a more reliable part, design or process in the 1st place?

          What if bypassing that component damages the previous or next?

          What if the circuit detecting a failure is less reliable, fails and bypasses in error?

          What is the expected mean time between failures MTBF? and to repair MTTR?






          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you for this!
            $endgroup$
            – InTheMoodForNow
            2 days ago


















          3












          $begingroup$

          It's possible in certain circumstances but generally not. In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit.



          Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. Truebeard's Stumper explains this mechanism:




          [Inside the bulb there is a shunt resistor.] It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights.




          Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures will (eventually) follow.



          Another example you can research is runway lighting. Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. You can research this yourself to see how faults are handled.



          In general your scheme isn't going to work. Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. In addition there is the problem of how to energise the switch to reset the device on power-on. If the device has shorted itself out there is no way for it to power itself back on.






          share|improve this answer









          $endgroup$












          • $begingroup$
            In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
            $endgroup$
            – HandyHowie
            2 days ago










          • $begingroup$
            That's an option I hadn't considered. You might have an answer there.
            $endgroup$
            – Transistor
            2 days ago


















          1












          $begingroup$

          Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. But how you would do this in concrete depends on the kind of device we're talking about. In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker.
          As stated, in generally you only need to detect a failure and be able to switch off the device.






          share|improve this answer











          $endgroup$













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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$


            Is there a way to bypass a component in series in a circuit if that component fails?




            This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.



            There exist some examples of this already where it is feasible.



            • mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out.

            • battery management system (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure.

            • some LED drivers have string open detection to detect and/or bypass individual without excess current.

            • "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure

            • AC grid fault detection systems have re-routing switch methods for some fault conditions

            • The internet was designed with redundant paths where a "switch" is just one unit in a system.

            Yes/No/Maybe



            That depends on a lot of unstated assumptions, which one can specify by questions.



            What kind of component or failure? Open? , short? or in between?

            Will it reduce reliability? e.g. bypass a fuse

            Is it worth it? Cost/benefit for an added detector, multiplexer to bypass

            Is it better to choose a more reliable part, design or process in the 1st place?

            What if bypassing that component damages the previous or next?

            What if the circuit detecting a failure is less reliable, fails and bypasses in error?

            What is the expected mean time between failures MTBF? and to repair MTTR?






            share|improve this answer









            $endgroup$












            • $begingroup$
              Thank you for this!
              $endgroup$
              – InTheMoodForNow
              2 days ago















            3












            $begingroup$


            Is there a way to bypass a component in series in a circuit if that component fails?




            This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.



            There exist some examples of this already where it is feasible.



            • mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out.

            • battery management system (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure.

            • some LED drivers have string open detection to detect and/or bypass individual without excess current.

            • "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure

            • AC grid fault detection systems have re-routing switch methods for some fault conditions

            • The internet was designed with redundant paths where a "switch" is just one unit in a system.

            Yes/No/Maybe



            That depends on a lot of unstated assumptions, which one can specify by questions.



            What kind of component or failure? Open? , short? or in between?

            Will it reduce reliability? e.g. bypass a fuse

            Is it worth it? Cost/benefit for an added detector, multiplexer to bypass

            Is it better to choose a more reliable part, design or process in the 1st place?

            What if bypassing that component damages the previous or next?

            What if the circuit detecting a failure is less reliable, fails and bypasses in error?

            What is the expected mean time between failures MTBF? and to repair MTTR?






            share|improve this answer









            $endgroup$












            • $begingroup$
              Thank you for this!
              $endgroup$
              – InTheMoodForNow
              2 days ago













            3












            3








            3





            $begingroup$


            Is there a way to bypass a component in series in a circuit if that component fails?




            This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.



            There exist some examples of this already where it is feasible.



            • mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out.

            • battery management system (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure.

            • some LED drivers have string open detection to detect and/or bypass individual without excess current.

            • "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure

            • AC grid fault detection systems have re-routing switch methods for some fault conditions

            • The internet was designed with redundant paths where a "switch" is just one unit in a system.

            Yes/No/Maybe



            That depends on a lot of unstated assumptions, which one can specify by questions.



            What kind of component or failure? Open? , short? or in between?

            Will it reduce reliability? e.g. bypass a fuse

            Is it worth it? Cost/benefit for an added detector, multiplexer to bypass

            Is it better to choose a more reliable part, design or process in the 1st place?

            What if bypassing that component damages the previous or next?

            What if the circuit detecting a failure is less reliable, fails and bypasses in error?

            What is the expected mean time between failures MTBF? and to repair MTTR?






            share|improve this answer









            $endgroup$




            Is there a way to bypass a component in series in a circuit if that component fails?




            This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.



            There exist some examples of this already where it is feasible.



            • mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out.

            • battery management system (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure.

            • some LED drivers have string open detection to detect and/or bypass individual without excess current.

            • "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure

            • AC grid fault detection systems have re-routing switch methods for some fault conditions

            • The internet was designed with redundant paths where a "switch" is just one unit in a system.

            Yes/No/Maybe



            That depends on a lot of unstated assumptions, which one can specify by questions.



            What kind of component or failure? Open? , short? or in between?

            Will it reduce reliability? e.g. bypass a fuse

            Is it worth it? Cost/benefit for an added detector, multiplexer to bypass

            Is it better to choose a more reliable part, design or process in the 1st place?

            What if bypassing that component damages the previous or next?

            What if the circuit detecting a failure is less reliable, fails and bypasses in error?

            What is the expected mean time between failures MTBF? and to repair MTTR?







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 days ago









            Sunnyskyguy EE75Sunnyskyguy EE75

            70k225101




            70k225101











            • $begingroup$
              Thank you for this!
              $endgroup$
              – InTheMoodForNow
              2 days ago
















            • $begingroup$
              Thank you for this!
              $endgroup$
              – InTheMoodForNow
              2 days ago















            $begingroup$
            Thank you for this!
            $endgroup$
            – InTheMoodForNow
            2 days ago




            $begingroup$
            Thank you for this!
            $endgroup$
            – InTheMoodForNow
            2 days ago













            3












            $begingroup$

            It's possible in certain circumstances but generally not. In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit.



            Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. Truebeard's Stumper explains this mechanism:




            [Inside the bulb there is a shunt resistor.] It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights.




            Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures will (eventually) follow.



            Another example you can research is runway lighting. Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. You can research this yourself to see how faults are handled.



            In general your scheme isn't going to work. Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. In addition there is the problem of how to energise the switch to reset the device on power-on. If the device has shorted itself out there is no way for it to power itself back on.






            share|improve this answer









            $endgroup$












            • $begingroup$
              In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
              $endgroup$
              – HandyHowie
              2 days ago










            • $begingroup$
              That's an option I hadn't considered. You might have an answer there.
              $endgroup$
              – Transistor
              2 days ago















            3












            $begingroup$

            It's possible in certain circumstances but generally not. In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit.



            Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. Truebeard's Stumper explains this mechanism:




            [Inside the bulb there is a shunt resistor.] It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights.




            Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures will (eventually) follow.



            Another example you can research is runway lighting. Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. You can research this yourself to see how faults are handled.



            In general your scheme isn't going to work. Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. In addition there is the problem of how to energise the switch to reset the device on power-on. If the device has shorted itself out there is no way for it to power itself back on.






            share|improve this answer









            $endgroup$












            • $begingroup$
              In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
              $endgroup$
              – HandyHowie
              2 days ago










            • $begingroup$
              That's an option I hadn't considered. You might have an answer there.
              $endgroup$
              – Transistor
              2 days ago













            3












            3








            3





            $begingroup$

            It's possible in certain circumstances but generally not. In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit.



            Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. Truebeard's Stumper explains this mechanism:




            [Inside the bulb there is a shunt resistor.] It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights.




            Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures will (eventually) follow.



            Another example you can research is runway lighting. Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. You can research this yourself to see how faults are handled.



            In general your scheme isn't going to work. Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. In addition there is the problem of how to energise the switch to reset the device on power-on. If the device has shorted itself out there is no way for it to power itself back on.






            share|improve this answer









            $endgroup$



            It's possible in certain circumstances but generally not. In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit.



            Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. Truebeard's Stumper explains this mechanism:




            [Inside the bulb there is a shunt resistor.] It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. (This is exactly the same effect as twisting the lamp to short the wires together!) This completes the circuit, and the set lights.




            Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures will (eventually) follow.



            Another example you can research is runway lighting. Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. You can research this yourself to see how faults are handled.



            In general your scheme isn't going to work. Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. In addition there is the problem of how to energise the switch to reset the device on power-on. If the device has shorted itself out there is no way for it to power itself back on.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 2 days ago









            TransistorTransistor

            88k785189




            88k785189











            • $begingroup$
              In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
              $endgroup$
              – HandyHowie
              2 days ago










            • $begingroup$
              That's an option I hadn't considered. You might have an answer there.
              $endgroup$
              – Transistor
              2 days ago
















            • $begingroup$
              In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
              $endgroup$
              – HandyHowie
              2 days ago










            • $begingroup$
              That's an option I hadn't considered. You might have an answer there.
              $endgroup$
              – Transistor
              2 days ago















            $begingroup$
            In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
            $endgroup$
            – HandyHowie
            2 days ago




            $begingroup$
            In the loudspeaker example, wouldn't the switch replace the faulty speaker with a resistor of the same value as the resistance of the speaker?
            $endgroup$
            – HandyHowie
            2 days ago












            $begingroup$
            That's an option I hadn't considered. You might have an answer there.
            $endgroup$
            – Transistor
            2 days ago




            $begingroup$
            That's an option I hadn't considered. You might have an answer there.
            $endgroup$
            – Transistor
            2 days ago











            1












            $begingroup$

            Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. But how you would do this in concrete depends on the kind of device we're talking about. In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker.
            As stated, in generally you only need to detect a failure and be able to switch off the device.






            share|improve this answer











            $endgroup$

















              1












              $begingroup$

              Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. But how you would do this in concrete depends on the kind of device we're talking about. In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker.
              As stated, in generally you only need to detect a failure and be able to switch off the device.






              share|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. But how you would do this in concrete depends on the kind of device we're talking about. In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker.
                As stated, in generally you only need to detect a failure and be able to switch off the device.






                share|improve this answer











                $endgroup$



                Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. But how you would do this in concrete depends on the kind of device we're talking about. In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker.
                As stated, in generally you only need to detect a failure and be able to switch off the device.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Sim SonSim Son

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