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Can I write my BE KVL with Rb and Rx in parallel? Also, I am not sure where to include Vbe in my KVL.

BJT circuit

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  • \$\begingroup\$ It appears your \$i_B\$ will be split off into two separate branches: Across \$R_b\$ and across \$R_x\$. \$V_{be}\$ will be the voltage after \$R_b\$ minus the voltage seen at \$V_E\$, are both things you haven't initiated calculations yet. \$\endgroup\$ – KingDuken Oct 27 '19 at 2:32
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Perhaps if the circuit is drawn differently, you may see a different solution.

enter image description here

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  • \$\begingroup\$ So I have 2 KVL equations (1) $$Vi-Vd = IbRb + Vbe$$ (2)$$Vi-Vd = IxRx$$ How are they related to the CE KVL? \$\endgroup\$ – Hector Oct 27 '19 at 17:42
  • \$\begingroup\$ That's correct, notice the first equation is independent of Rx? Using a simple voltage source model of a forward biased diode the voltage input to Rb is independent of the value of Rx. As we know the voltage across a voltage source is a constant value. \$\endgroup\$ – sstobbe Oct 27 '19 at 18:05
  • \$\begingroup\$ Had the diode of been say a resistor, you would find your first equation would have 2 unknowns and you would need both equations for a solution. \$\endgroup\$ – sstobbe Oct 27 '19 at 18:06
  • \$\begingroup\$ Does this mean that Rx will not affect the input voltage, Vi, at which the LED will turn on? \$\endgroup\$ – Hector Oct 27 '19 at 21:18
  • \$\begingroup\$ @Hector Yes, that is correct. With a basic voltage source model of the forward biased diode, Rx has no affect on the Vi turn on voltage. If you were to apply the full exponential Shockley equation to the diode forward voltage you, would find a solution that requires a slightly larger Vi. \$\endgroup\$ – sstobbe Oct 27 '19 at 21:26

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