Finding necessary voltage for diode conduction

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In the following, I need to find $V_a(v_in)$ such that $D_3$ is on. Assume a forward voltage drop of $0.7V$. From past computations, it turned out to be around $10.75V$ (as you can check from the sim, whilst putting $10V$ would turn it off); however, I can't find the issue in my computation.



Since $$i_1=i_2+i_3,quad fracv_i-0.7-v_aR_1=fracv_aR_2+fracv_a-0.7+6R_3$$
In order for the diode to turn on, I then set $v_a=6.7$ and solve for $v_in$; however, I find a bigger value ($14.75V$) than it should be. Where's that flaw?










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    schematic





    simulate this circuit – Schematic created using CircuitLab



    In the following, I need to find $V_a(v_in)$ such that $D_3$ is on. Assume a forward voltage drop of $0.7V$. From past computations, it turned out to be around $10.75V$ (as you can check from the sim, whilst putting $10V$ would turn it off); however, I can't find the issue in my computation.



    Since $$i_1=i_2+i_3,quad fracv_i-0.7-v_aR_1=fracv_aR_2+fracv_a-0.7+6R_3$$
    In order for the diode to turn on, I then set $v_a=6.7$ and solve for $v_in$; however, I find a bigger value ($14.75V$) than it should be. Where's that flaw?










    share|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite













      schematic





      simulate this circuit – Schematic created using CircuitLab



      In the following, I need to find $V_a(v_in)$ such that $D_3$ is on. Assume a forward voltage drop of $0.7V$. From past computations, it turned out to be around $10.75V$ (as you can check from the sim, whilst putting $10V$ would turn it off); however, I can't find the issue in my computation.



      Since $$i_1=i_2+i_3,quad fracv_i-0.7-v_aR_1=fracv_aR_2+fracv_a-0.7+6R_3$$
      In order for the diode to turn on, I then set $v_a=6.7$ and solve for $v_in$; however, I find a bigger value ($14.75V$) than it should be. Where's that flaw?










      share|improve this question















      schematic





      simulate this circuit – Schematic created using CircuitLab



      In the following, I need to find $V_a(v_in)$ such that $D_3$ is on. Assume a forward voltage drop of $0.7V$. From past computations, it turned out to be around $10.75V$ (as you can check from the sim, whilst putting $10V$ would turn it off); however, I can't find the issue in my computation.



      Since $$i_1=i_2+i_3,quad fracv_i-0.7-v_aR_1=fracv_aR_2+fracv_a-0.7+6R_3$$
      In order for the diode to turn on, I then set $v_a=6.7$ and solve for $v_in$; however, I find a bigger value ($14.75V$) than it should be. Where's that flaw?







      circuit-analysis diodes






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      share|improve this question











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      asked 5 hours ago









      edmz

      1284




      1284




















          3 Answers
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          up vote
          2
          down vote



          accepted










          If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.



          Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?






          share|improve this answer




















          • It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago

















          up vote
          2
          down vote













          Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,



          $$i_3 = fracv_a-(0.7+6)R_3 $$



          Note the brackets.






          share|improve this answer




















          • The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago










          • @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
            – sstobbe
            4 hours ago


















          up vote
          2
          down vote













          If $V_a = 6.7V$ and $D_3$ diode threshold voltage is $0.7V$ the $I_3$ current is $0A$. Therefore we can find $V_IN$ using voltage divider equation.



          $$V_IN = V_a cdot (1+ fracR_1R_2) + 0.7V = 6.7V cdot 1.5 + 0.7V = 10.75V$$






          share|improve this answer






















          • You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
            – edmz
            4 hours ago











          • Yep. You are right.
            – G36
            4 hours ago










          Your Answer




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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.



          Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?






          share|improve this answer




















          • It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago














          up vote
          2
          down vote



          accepted










          If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.



          Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?






          share|improve this answer




















          • It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.



          Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?






          share|improve this answer












          If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.



          Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 5 hours ago









          Andy aka

          233k10172396




          233k10172396











          • It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago
















          • It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago















          It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
          – edmz
          5 hours ago




          It definitely is. The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
          – edmz
          5 hours ago












          up vote
          2
          down vote













          Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,



          $$i_3 = fracv_a-(0.7+6)R_3 $$



          Note the brackets.






          share|improve this answer




















          • The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago










          • @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
            – sstobbe
            4 hours ago















          up vote
          2
          down vote













          Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,



          $$i_3 = fracv_a-(0.7+6)R_3 $$



          Note the brackets.






          share|improve this answer




















          • The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago










          • @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
            – sstobbe
            4 hours ago













          up vote
          2
          down vote










          up vote
          2
          down vote









          Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,



          $$i_3 = fracv_a-(0.7+6)R_3 $$



          Note the brackets.






          share|improve this answer












          Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,



          $$i_3 = fracv_a-(0.7+6)R_3 $$



          Note the brackets.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 5 hours ago









          sstobbe

          1,82528




          1,82528











          • The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago










          • @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
            – sstobbe
            4 hours ago

















          • The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
            – edmz
            5 hours ago










          • @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
            – sstobbe
            4 hours ago
















          The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
          – edmz
          5 hours ago




          The thing that made me exclude that was because then $i_3=0$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on.
          – edmz
          5 hours ago












          @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
          – sstobbe
          4 hours ago





          @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as $eta V_T ln( I_3/ I_s)$ but the solution becomes very difficult.
          – sstobbe
          4 hours ago











          up vote
          2
          down vote













          If $V_a = 6.7V$ and $D_3$ diode threshold voltage is $0.7V$ the $I_3$ current is $0A$. Therefore we can find $V_IN$ using voltage divider equation.



          $$V_IN = V_a cdot (1+ fracR_1R_2) + 0.7V = 6.7V cdot 1.5 + 0.7V = 10.75V$$






          share|improve this answer






















          • You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
            – edmz
            4 hours ago











          • Yep. You are right.
            – G36
            4 hours ago














          up vote
          2
          down vote













          If $V_a = 6.7V$ and $D_3$ diode threshold voltage is $0.7V$ the $I_3$ current is $0A$. Therefore we can find $V_IN$ using voltage divider equation.



          $$V_IN = V_a cdot (1+ fracR_1R_2) + 0.7V = 6.7V cdot 1.5 + 0.7V = 10.75V$$






          share|improve this answer






















          • You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
            – edmz
            4 hours ago











          • Yep. You are right.
            – G36
            4 hours ago












          up vote
          2
          down vote










          up vote
          2
          down vote









          If $V_a = 6.7V$ and $D_3$ diode threshold voltage is $0.7V$ the $I_3$ current is $0A$. Therefore we can find $V_IN$ using voltage divider equation.



          $$V_IN = V_a cdot (1+ fracR_1R_2) + 0.7V = 6.7V cdot 1.5 + 0.7V = 10.75V$$






          share|improve this answer














          If $V_a = 6.7V$ and $D_3$ diode threshold voltage is $0.7V$ the $I_3$ current is $0A$. Therefore we can find $V_IN$ using voltage divider equation.



          $$V_IN = V_a cdot (1+ fracR_1R_2) + 0.7V = 6.7V cdot 1.5 + 0.7V = 10.75V$$







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 5 hours ago









          G36

          4,6951411




          4,6951411











          • You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
            – edmz
            4 hours ago











          • Yep. You are right.
            – G36
            4 hours ago
















          • You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
            – edmz
            4 hours ago











          • Yep. You are right.
            – G36
            4 hours ago















          You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
          – edmz
          4 hours ago





          You surely meant $6.colorred7Vcdot 1.5$. Still, thanks for your approach.
          – edmz
          4 hours ago













          Yep. You are right.
          – G36
          4 hours ago




          Yep. You are right.
          – G36
          4 hours ago

















           

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