Knee voltage in pn junction diode
Added on 2021-04-17
8 Pages1289 Words109 Views
Problem 1:Solution:Condition 1: Given data: R1=12KΩ, R2=3.3KΩ, RC=1.2KΩ, RE=330KΩ, β=300, VCC=10VAssume VBE, VBC =0.7V, VCE(SAT)=0.2VTo Find: VC,VB, VESituation 1: The resistor R1 is openedCircuit diagram:VCC –ICRC - VBC – IBRB =0 (Apply Kirchhoff's voltage law)VCC –βIBRC - VBC – IBRB =0 Hence, IB =( VCC - VBC ) / (RB + βRC ) ------ Equation 1(Here RB is R2)Applying the values to Equation 1 we get, IB = 2.55×10-5 AHence, IC = βIB = 7.65 mA IE ≈ICAns: VE = IERE =2.508 V, VB = VE + VBE =3.208V, VC = VCC – ICRC = 0.88V
Situation 2: The resistor R2 is openedCircuit diagram: VCC - VBE – IBRB – IERE = 0 (Apply Kirchhoff's voltage law) ------------------------- Equation 2IE = IB + IC = IB + βIB = (β +1) IB (Since IC = βIB) ------------------------------------- Equation 3Substitute Equation 3 in Equation 2 we get,VCC - VBE – IBRB - (β +1) IBRE = 0 (Here RB is R1)Hence IB = (VCC - VBE) / (RB + (β +1) IBRE) ----------------------------------------------------------- Equation 4From Equation 4 we get IB = 8.35 × 10-5 A IC = βIB = 0.025A IE ≈ICAns: VE = IERE = 8.25 V, VB = VE + VBE = 8.95V, VC = VCC – ICRC = -20V.Situation 3: The resistor RC is openedCircuit diagram:
The thevenin voltage is given by the formulaVth = (VCC × R2) / (R1 + R2)Hence Vth = 7.84VRth = (R1×R2) / (R1+R2)Hence Rth = 2588 ΩNow, from the thevenin’s circuit we can getVth – IBRth –VBE –IERE = 0(Apply Kirchhoff's voltage law) ------------------------- Equation 5IE =IB +IC (Here IC = 0) Therefore Equation 5 becomesVth – IBRth –VBE – IBRE = 0IB = (Vth –VBE) / (Rth + RE) ---------------------------------------------------------------- Equation 6Substituting the given values in Equation 6 we getIB = 2.4mAIE = IC + IB = 2.4 mA( since IC = 0)Ans: VE = IERE = 0.8074 V, VB = VE + VBE = 1.507V, VC = 0V.
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