(A) Vcc = 12V QbreakP 11.3V Dbreak QbreakN Dbreak + 15V OPAMP, 1012 Vout Vop Vin 0.01 sin (2007) + QbreakP + QbreakN -15

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answerhappygod
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(A) Vcc = 12V QbreakP 11.3V Dbreak QbreakN Dbreak + 15V OPAMP, 1012 Vout Vop Vin 0.01 sin (2007) + QbreakP + QbreakN -15

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A Vcc 12v Qbreakp 11 3v Dbreak Qbreakn Dbreak 15v Opamp 1012 Vout Vop Vin 0 01 Sin 2007 Qbreakp Qbreakn 15 1
A Vcc 12v Qbreakp 11 3v Dbreak Qbreakn Dbreak 15v Opamp 1012 Vout Vop Vin 0 01 Sin 2007 Qbreakp Qbreakn 15 1 (77.16 KiB) Viewed 24 times
A Vcc 12v Qbreakp 11 3v Dbreak Qbreakn Dbreak 15v Opamp 1012 Vout Vop Vin 0 01 Sin 2007 Qbreakp Qbreakn 15 2
A Vcc 12v Qbreakp 11 3v Dbreak Qbreakn Dbreak 15v Opamp 1012 Vout Vop Vin 0 01 Sin 2007 Qbreakp Qbreakn 15 2 (53.89 KiB) Viewed 24 times
(A) Vcc = 12V QbreakP 11.3V Dbreak QbreakN Dbreak + 15V OPAMP, 1012 Vout Vop Vin 0.01 sin (2007) + QbreakP + QbreakN -15V VEE = -12V = 200k12 1k12 1. The Hi-Fi circuit is shown in the figure below, please calculate Vout .model Qbreakn NPN Is=le-14 .model Qbreakp PNP Is=le-14 .model Dbreak D Is=le-14 Cjo=1pF Rs=1 2. Please change the positive and negative of Vin to observe whether Vout has push pull phenomenon
(B) Vcc = 12V QbreakP 11.3V QbreakN(Q1) Dbreak a Dbreak + 15V OPAMP RL 1012 Vout Vin Vop 0.01 sin(2007) QbreakP + QbreakN -15V Vee = -12V 200k.12 *Qs=Qbreaks 1kΩ 3. Please choose an appropriate resistor r, so that when R, is accidentally short-circuited, it can drop the Ic of Q1 to protect it, and try to explain how it works.
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