The choice of PNP or NPN will depend on the Second-last Digit of your Matriculation Number....

  

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The choice of PNP or NPN will depend on the Second-last Digit of your Matriculation Number. If it is odd, use PNP as in Fig. 3. If it is even, use NPN as in Fig. 4. For example: U2212334A Design the PNP (2N3906) emitter follower in Fig. 3 U2212345B Design the NPN (2N3904) emitter follower in Fig. 4 Note that for the convenience of working with positive DC voltages, DC calculations from 3.1-3.3 must again reference the most negative supply in the circuit as GND (OV), and hence the total dual-supply potential is referred to as Vccxx = 2xVcc V where Vcc 2.xx V and xx the last two digits of your Matriculation Number. Note in Fig. 3 and Fig. 4 we represented the OPA344 output as a Thévenin-equivalent source with voltage source vo and RTH =-02 which reflects an ideal opamp output. 3.1) We must design a DC operating point to satisfy the requirement of AvT > 0.95. a) Write out the equation for Avr. b) What is the minimum value for gmRL' such that Avr > 0.95? c) Observe from R = Re//Rt that Re' will attain a maximum value no matter how much Re increases. Prove this maximum value of RL = RL. d) Assuming the maximum value of RL', find the minimum value of gm required to achieve the minimum value for gmRL' in (b). e) Hence derive the minimum value of Ic. In practice, R.' will be <RL due to finite Re hence: gm must be larger than the value found in (d), and Ic must be larger than the value found in (e). Note that the value of Ic is set by your design choice of Ve and Re. For PNP: le le (Vecz - Ve)/Re For NPN: Ic le Ve/Re Which are in turn set by Ve and Veel = 0.75V For PNP: Ve=Vs+ Ves For NPN: VE V-VSE Hence you can maximize gm by adjusting the value of Va (using the Rr and Rz voltage divider), and by reducing Re. However, observe that reducing Re also reduces RL. f) Hence find a value of Va and then select Re from the Resistor Choices which is low enough to produce high g, but not so low as to kill the value of R', and thereby meet the minimum gRL found in (b). (Note Rr and Ra will be chosen later, in 3.2) Ensure that your final design choice of g and Re is compatible with physical values for both Va and Ve that must lie between your Vcca supply rail and GND Summarize your design choices below, assuming [Va] = 0.75V. Va (V)= VE (V)- Ic (mA) = gm (1/2)= BJT PNP/NPN RE (0) R₁(0)= g-R= g) Calculate the maximum AC swing on vo given mic 2uApk. Use the same values of Ruc and Ave that you designed in Assignment 1. From superposition of the AC signals vs and v. on top of your DC bias points Vs and Vr. i.e. Va Va (DC)+vo (AC) VE = VE (DC) + v. (AC) Va (DC) + vo (AC) ■ VE (DC) + Avr vo (AC) Ensure that both va and vr with their AC swings will stay between GND and Vcczx, i.e. 0 < VB < Voca 0<VE < Voca Check your conclusions on the swing limits by filling in this table: va (max)= va (min) = VE (max)= Ve (min) = OPA344 output G -22 R-120 Earbud Load resistance Fig. 4. NPN Common-collector or Emitter Follower driver for 322 earpiece The choice of PNP or NPN will depend on the Second-last Digit of your Matriculation Number. If it is odd, use PNP as in Fig. 3. If it is even, use NPN as in Fig. 4. For example: U2212334A Design the PNP (2N3906) emitter follower in Fig. 3 U2212345B Design the NPN (2N3904) emitter follower in Fig. 4 Note that for the convenience of working with positive DC voltages, DC calculations from 3.1-3.3 must again reference the most negative supply in the circuit as GND (OV), and hence the total dual-supply potential is referred to as Vccxx = 2xVcc V where Vcc 2.xx V and xx the last two digits of your Matriculation Number. Note in Fig. 3 and Fig. 4 we represented the OPA344 output as a Thévenin-equivalent source with voltage source vo and RTH =-02 which reflects an ideal opamp output. 3.1) We must design a DC operating point to satisfy the requirement of AvT > 0.95. a) Write out the equation for Avr. b) What is the minimum value for gmRL' such that Avr > 0.95? c) Observe from R = Re//Rt that Re' will attain a maximum value no matter how much Re increases. Prove this maximum value of RL = RL. d) Assuming the maximum value of RL', find the minimum value of gm required to achieve the minimum value for gmRL' in (b). e) Hence derive the minimum value of Ic. In practice, R.' will be <RL due to finite Re hence: gm must be larger than the value found in (d), and Ic must be larger than the value found in (e). Note that the value of Ic is set by your design choice of Ve and Re. For PNP: le le (Vecz - Ve)/Re For NPN: Ic le Ve/Re Which are in turn set by Ve and Veel = 0.75V For PNP: Ve=Vs+ Ves For NPN: VE V-VSE Hence you can maximize gm by adjusting the value of Va (using the Rr and Rz voltage divider), and by reducing Re. However, observe that reducing Re also reduces RL. f) Hence find a value of Va and then select Re from the Resistor Choices which is low enough to produce high g, but not so low as to kill the value of R', and thereby meet the minimum gRL found in (b). (Note Rr and Ra will be chosen later, in 3.2) Ensure that your final design choice of g and Re is compatible with physical values for both Va and Ve that must lie between your Vcca supply rail and GND Summarize your design choices below, assuming [Va] = 0.75V. Va (V)= VE (V)- Ic (mA) = gm (1/2)= BJT PNP/NPN RE (0) R₁(0)= g-R= g) Calculate the maximum AC swing on vo given mic 2uApk. Use the same values of Ruc and Ave that you designed in Assignment 1. From superposition of the AC signals vs and v. on top of your DC bias points Vs and Vr. i.e. Va Va (DC)+vo (AC) VE = VE (DC) + v. (AC) Va (DC) + vo (AC) ■ VE (DC) + Avr vo (AC) Ensure that both va and vr with their AC swings will stay between GND and Vcczx, i.e. 0 < VB < Voca 0<VE < Voca Check your conclusions on the swing limits by filling in this table: va (max)= va (min) = VE (max)= Ve (min) = OPA344 output G -22 R-120 Earbud Load resistance Fig. 4. NPN Common-collector or Emitter Follower driver for 322 earpiece

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