You are designing a common emitter amplifier using an NPN transistor. The collector current (IC) is 2mA, and the base current (IB) is 20μA. Calculate the transistor's β (current gain) and the collector-emitter voltage drop (VCE) if the supply voltage is 5V
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To calculate the transistor's β or current gain, we can use the formula:
β = IC / IB
Given that IC = 2mA and IB = 20μA, we need to convert the base current to amperes:
IB = 20μA = 20 * 10^-6 A
Now we can substitute the values into the equation:
β = IC / IB
β = 2mA / (20 * 10^-6 A)
β ≈ 100
Therefore, the current gain (β) of the transistor is approximately 100.
To calculate the collector-emitter voltage drop (VCE), we need to use the supply voltage and Ohm's law. In a common emitter configuration, the collector-emitter voltage drop can be approximated to approximately half of the supply voltage when the transistor is in active mode.
VCE ≈ (Vsupply - VBE) / 2
Given that the supply voltage (Vsupply) is 5V, we need to determine the base-emitter voltage (VBE), which typically ranges from 0.6V to 0.7V for an NPN transistor. Let's assume VBE = 0.7V.
Now we can substitute the values into the equation:
VCE ≈ (5V - 0.7V) / 2
VCE ≈ 2.65V
Therefore, the collector-emitter voltage drop (VCE) is approximately 2.65V.
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