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The small-signal equivalent circuit shown below has Rc = 2 KΩ . The transistor parameters are gm = 56 mA/V and β= 100. The time-varying output voltage is given by vo = 1.2 sin(ωt) V. Find the small-signal amplitude of base-emitter voltage vbe in mV.?
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The small-signal equivalent circuit shown below has Rc = 2 KΩ . The tr...
Given information:
- Rc = 2 KΩ
- gm = 56 mA/V
- β = 100
- vo = 1.2 sin(ωt) V

Objective:
To find the small-signal amplitude of base-emitter voltage vbe in mV.

Explanation:
To find the small-signal amplitude of vbe, we need to analyze the given circuit and apply small-signal analysis techniques.

Step 1: Identify the circuit elements and their values:
- Rc = 2 KΩ (collector resistor)
- gm = 56 mA/V (transconductance)
- β = 100 (current gain of the transistor)
- vo = 1.2 sin(ωt) V (output voltage)

Step 2: Derive the small-signal equivalent circuit:
To simplify the analysis, we can consider the transistor as a linear device and neglect the non-linear effects. This allows us to derive the small-signal equivalent circuit.

The small-signal equivalent circuit for the given circuit can be derived as follows:
- Replace the transistor with its small-signal model, which consists of a current source (gm*vbe) in parallel with a resistor (re).
- Replace the time-varying output voltage (vo) with a small-signal AC voltage source (vout) with an amplitude of Vout.

Step 3: Apply small-signal analysis:
Once we have the small-signal equivalent circuit, we can apply small-signal analysis techniques to find the small-signal amplitude of vbe.

In the small-signal equivalent circuit, the base-emitter voltage (vbe) is the input voltage to the transistor. The small-signal current through the transistor can be given by:
ib = gm*vbe

Using Ohm's law, we can find the small-signal voltage across the collector resistor (Rc):
vout = -ib * Rc

Since vout is the output voltage, we can equate it to the small-signal amplitude of vo, which is Vout:
Vout = -gm*vbe * Rc

Step 4: Solve for vbe:
Now we can solve the equation for vbe, the small-signal amplitude of the base-emitter voltage.

Vout = -gm*vbe * Rc
Vout = -56 mA/V * vbe * 2 KΩ

Simplifying the equation, we get:
vbe = -Vout / (gm * Rc)

Substituting the given value of Vout = 1.2 V and solving the equation, we get:
vbe = -1.2 V / (56 mA/V * 2 KΩ)
vbe ≈ -10.71 mV

Conclusion:
The small-signal amplitude of the base-emitter voltage vbe is approximately -10.71 mV.
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The small-signal equivalent circuit shown below has Rc = 2 KΩ . The transistor parameters are gm = 56 mA/V and β= 100. The time-varying output voltage is given by vo = 1.2 sin(ωt) V. Find the small-signal amplitude of base-emitter voltage vbe in mV.?
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