Question
In double ended input, double ended output configuration of a differential amplifier, the collector resistors RC1 RC2= 51kΩ, emitter resistor RE = 47 ΚΩ and supply voltage is ± 10V. Calculate the dc collector voltage and collector current in the circuit. For the matched pair of n-p-n transistors with ẞ = 100, calculate the common mode gain of the circuit. What is the advantage of using a constant current source in the emitter circuit of the differential amplifier?
Answer :
Word Count : 317
चलो इस समस्या को चरण-दर-चरण हल करते हैं। ### दिया गया डेटा: - Collector Resistors: \( R_{C1} = R_{C2} = 51k\Omega \) - Emitter Resistor: \( R_E = 47k\Omega \) - Supply Voltage: \( V_{CC} = +10V, V_{EE} = -10V \) - Transistor Gain: \( \beta = 100 \) ### Step 1: Emitter Current Calculation \[ I_E = \frac{V_{EE}}{(2R_E + \frac{2R_C}{\beta + 1})} \] \[ I_E = \frac{10V}{(2 \times 47k + \frac{2 \times 51k}{101})} \] \[ I_E = \frac{10V}{(94k ___ _________ ________ _______ ________ _________ ____ __________ _____.
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चलो इस समस्या को चरण-दर-चरण हल करते हैं। ### दिया गया डेटा: - Collector Resistors: \( R_{C1} = R_{C2} = 51k\Omega \) - Emitter Resistor: \( R_E = 47k\Omega \) - Supply Voltage: \( V_{CC} = +10V, V_{EE} = -10V \) - Transistor Gain: \( \beta = 100 \) ### Step 1: Emitter Current Calculation \[ I_E = \frac{V_{EE}}{(2R_E + \frac{2R_C}{\beta + 1})} \] \[ I_E = \frac{10V}{(2 \times 47k + \frac{2 \times 51k}{101})} \] \[ I_E = \frac{10V}{(94k ___ _________ ________ _______ ________ _________ ____ __________ _____.
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