A common circuit using a parallel KC circuit is the bypass capacitor across the

Question

A common circuit using a parallel KC circuit is the bypass capacitor across the emitter resistor in a transistor amplifier, as illustrated in Figure 26-4. The frequency at which the resistance RE is equal to the capacitive reactance XC is called the cutoff frequency. Compute the cutoff frequency for the circuit by setting RE =Xc and solving for f. How do the branch currents compare at the cutoff frequency? Explain what happens above this frequency to the current in the parallel KC circuit. If the capacitance in Figure 26-4 is increased, what happens to the cutoff frequency?

Explanation / Answer

a)Re = 2.7k= Xc = 1/2fC= 1/2f 10e-6. Shuffle the equation around to solve for f where the impedances are equal gives f = 1/(2**2.7e3*10e-6) = 5.89462752 Hz.

b) Since the impedances are equal at the cutoff frequency (we know this because we set them equal to find the cutoff frequency), we know that the currents through each will be equal.

c)As the frequency increases, more current will flow through the capacitor than through the resistor. THis can easily be seen when you look at the equasion for Xc; as f increases the value of Xc decreases creating a lower impedance to ground thereby shunting most of the current.

5)  fcutoff = 1/(2**R*C), as from the equation above, so as C increases, the cutoff frequency will then decrease.

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