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Derive algebraic expression for a Thevenin\'s equivalent resistance Rjh for this

ID: 1929395 • Letter: D

Question


Derive algebraic expression for a Thevenin's equivalent resistance Rjh for this amplifier circuit as seen at terminals a'-a from the emitter terminal E to the GND, including the resistor Re and all the other resistors in the circuit. Use a test source technique to complete your assignment. You may use a voltage-test- source or a current-test-source. You should try using a couple of different methods to solve the circuit (i.e. node voltages and loop currents). Show all your work. Arrange your final algebraic expressions in a meaningful way. Next calculate the value of the Thevenin's equivalent resistance RTH for this circuit. Given : vs(t) = 0.1 sin(377t)v,R=500 ohm, r pi = 2.6k ohm, RE = 100 ohm, RL = 1k ohm, and beta = 100.

Explanation / Answer

the Rin = R + (1+)Re

now when we calculate thevinin resistance between two points, the current sources are broken and the voltage sources are shorted.

so, when the current source is broken, and voltage is shorted, the circuit becomes

Rth = Rin || Re || Rs ( the voltage source Vs is shorted.)

hence the

Rth = Rin || Re || Rs = 12.7 K || 100 || 500

= 83.333

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