Question at position 1 Consider the following circuit where Ra=1k, Rb=5k, RL=32. Assume that the Op-Amp is ideal and for the BJT β=100 and VBE=0.7V Find: a) The output voltage and the collector current of the BJT assuming Vin=-0.5V Vout = Question Blank 1 of 3[input] V Ic = Question Blank 2 of 3[input] mA Hint: Notice that the current on Rb and RL must come from the emitter of the BJT. b) The maximum values that |Vin| can have while keeping the Op Amp out of saturation. Notice that Vout can not be negative as current can't flow into the emitter of the transistor. |Vin,max | = Question Blank 3 of 3[input] V

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Question restatement and options: - Problem: A circuit with Ra = 1 kΩ, Rb = 5 kΩ, RL = 32 Ω (likely a load resistor), an ideal Op-Amp, a BJT with β = 100 and VBE = 0.7 V. Given Vin = -0.5 V, find: a) Vout and Ic (BJT collector current). b) The maximum |Vin| such that the Op-Amp stays out of saturation; note Vout cannot be negative because current cannot flow into the emitter of the transistor. - Provided answer choices (as blanks to fill): The final numeric entries shown in the prompt are: a) Vout = -0.98 V b) Ic = -30.37 mA c) |Vin,max| = 0.00 V - The answer field list confirms the numeric results: ["-0.98", "-30.37", "0.00"] Option-by-option analysis: Option 1 (Vout = -0.98 V): - Reasoning path: Start by considering that Vin is applied to the non-inverting input of the ideal Op-Amp, while the inverting input is connected through Ra and the BJT emitter network such that the Op-Amp output (Vout) adjusts to satisfy the virtual short condition (V+ ≈ V−) when the circuit is active. Since Vin = -0.5 V, the Op-Amp must drive Vout to a level that makes the emitter/feedback node prior to the BJT adjust so that the inverting input equals -0.5 V. The hint in the problem statement states that the current through Rb and RL must come from the emitter of the BJT, which constrains the emitter node behavior and indicates that Vout cannot swing to strongly positive values without violating current directions, and also that the emitter node cannot sink current in a way that would requi......Login to view full explanation