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find voltage of node 1... please elaborate solution is expected |
March 26, 2021 |
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The simulator gives you the voltage just by mousing over the |
March 26, 2021 |
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yes. i have to solve this question with pen and paper for a test but i failed to solve that . so i needed help so posted the circuit. |
March 26, 2021 |
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Assuming R2 and R3 are a stiff voltage divider (to be true, the current at the base of Q1 should be small in comparison with the one through any of these two resistors), then Vr2 = Vr3 = 6 volt. Voltage loop Vr3, Vbe_q1 and Vr4: 6 = 0.7 + Vr4 (assuming Q1 is active but not blocking and its Vbe = 0.7 volt), leads to Vr4 = 5.3 volt and Ir4 =2650 uA. Since the current at the base of Q1 will then be not more than Ir4 / beta (and would be less if the current from Q2 is not zero), the stiff voltage divider assumption is considered validated. The voltage loop through Vr1 and Veb_q2 gives Vr1 =0.7 volt (assuming Q2 is active non blocking). Ir1 = 70 uA. 70 uA = Iq1_emitter ( taking Icollector = I emitter ) The sum of currents at the intersection of Q1 (its emitter), R4 and Re: Ire = ir4 - iq1_emitter = 2650 uA - 70 uA = 2580 uA Vre = ire * R2 = 2.58 volt Vnode1 = Vr4 + Vre = 5.3 volt + 2.58 volt = 7.88 volt The simulator is more precise about the Vbase_emitter values that we assumed at 0.7 volt, neither used the assumption of a stiff voltage divider. |
by vanderghast March 27, 2021 |
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