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For the circuit shown,assume=100=hfe -10v 3k vo 7k wwwm 3v 500 (a) find if the silicon transistor is in cutoff saturation or in the active region (b) find vo (c) find the minimum value for emitter resistor r, for which the transistor operates in the active region?
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For the circuit shown,assume=100=hfe -10v 3k vo 7k wwwm 3v 500 (a) fin...
Answer:
(a) Transistor Region:
To find the region of operation of the transistor, we need to check the base-emitter voltage (Vbe) and collector-emitter voltage (Vce).
Vbe = Vin - 0.7V = -10 - (-0.7) = -9.3V
Vce = Vo ≈ 0.7V (assuming the transistor is in active region)
Since Vbe is negative, the transistor is in the cutoff region.
(b) Output Voltage (Vo):
When the transistor is in cutoff, the output voltage (Vo) is equal to the supply voltage (Vcc). Hence, Vo = Vcc = 3V.
(c) Minimum Emitter Resistor (Rmin):
To operate the transistor in the active region, we need to ensure that the base current (Ib) is sufficient to drive the transistor into saturation. This can be achieved by selecting an appropriate value for the emitter resistor (Re).
The minimum value of Re can be calculated using the formula:
Rmin = (Vbe)/(Ibmin)
where Ibmin is the minimum base current required to ensure that the transistor is in saturation.
Assuming a saturation voltage (Vsat) of 0.2V and a collector current (Ic) of 1mA, we can calculate Ibmin as follows:
Ic = hfe * Ib
Ib = Ic/hfe = 1mA/100 = 10uA
Ibmin = Ib * (Vsat/Vbe) = 10uA * (0.2/0.7) = 2.86uA
Substituting the values in the formula for Rmin, we get:
Rmin = (-9.3V)/(2.86uA) ≈ 3.24MΩ
Therefore, the minimum value of the emitter resistor (Re) for which the transistor operates in the active region is 3.24MΩ.
Summary:
- The transistor is in cutoff region.
- The output voltage (Vo) is 3V.
- The minimum value of the emitter resistor (Re) for which the transistor operates in the active region is 3.24MΩ.
(a) Transistor Region:
To find the region of operation of the transistor, we need to check the base-emitter voltage (Vbe) and collector-emitter voltage (Vce).
Vbe = Vin - 0.7V = -10 - (-0.7) = -9.3V
Vce = Vo ≈ 0.7V (assuming the transistor is in active region)
Since Vbe is negative, the transistor is in the cutoff region.
(b) Output Voltage (Vo):
When the transistor is in cutoff, the output voltage (Vo) is equal to the supply voltage (Vcc). Hence, Vo = Vcc = 3V.
(c) Minimum Emitter Resistor (Rmin):
To operate the transistor in the active region, we need to ensure that the base current (Ib) is sufficient to drive the transistor into saturation. This can be achieved by selecting an appropriate value for the emitter resistor (Re).
The minimum value of Re can be calculated using the formula:
Rmin = (Vbe)/(Ibmin)
where Ibmin is the minimum base current required to ensure that the transistor is in saturation.
Assuming a saturation voltage (Vsat) of 0.2V and a collector current (Ic) of 1mA, we can calculate Ibmin as follows:
Ic = hfe * Ib
Ib = Ic/hfe = 1mA/100 = 10uA
Ibmin = Ib * (Vsat/Vbe) = 10uA * (0.2/0.7) = 2.86uA
Substituting the values in the formula for Rmin, we get:
Rmin = (-9.3V)/(2.86uA) ≈ 3.24MΩ
Therefore, the minimum value of the emitter resistor (Re) for which the transistor operates in the active region is 3.24MΩ.
Summary:
- The transistor is in cutoff region.
- The output voltage (Vo) is 3V.
- The minimum value of the emitter resistor (Re) for which the transistor operates in the active region is 3.24MΩ.
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For the circuit shown,assume=100=hfe -10v 3k vo 7k wwwm 3v 500 (a) find if the silicon transistor is in cutoff saturation or in the active region (b) find vo (c) find the minimum value for emitter resistor r, for which the transistor operates in the active region?
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