bandpass filter design using a NPN BJT

bandpass filter design using a NPN BJT - Practice: Design a...

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Practice: Design a bandpass filter using a NPN BJT. Design Process: 0 QbreakN Q1 I VCE 0Vdc 0 IB 0Adc 0 V_VCE 0V 1V 2V 3V 4V 5V 6V 7V 8V 9V 10V 11V 12V IC(Q1) 0A 0.5mA 1.0mA 1.5mA 2.0mA 2.5mA 3.0mA 3.5mA Set: V CC = 10V V CC = 0.5V CC = 5V I BQ = 10uA Thus, β = I CQ / I BQ = 100 1 / 7
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Draw and get: I CQ = 1.0mA Thus, I CQ = V CC / R C = 1.0mA R C = 5kΩ I 0 VCC 10Vdc 0 VBE 0Vdc 0 Qbreakn Q1 V_VBE 0V 0.2V 0.4V 0.6V 0.8V 1.0V 1.2V 1.4V 1.6V IB(Q1) 0A 0.5mA 1.0mA 1.5mA 2.0mA (776.540m ,10.000u) Thus, V BEQ = 0.77654V R2 0 R1 RC 5k 0 0 Qbreakn Q1 VCC 10Vdc From, V BQ = ( R 2 / R 1 + R 2 ) V CC 0.77654V = ( R 2 / R 1 + R 2 ) (10V) 2 / 7
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R 1 = 11.8776 R 2 Choose, R 2 = 1kΩ Thus, R 1 = 11.8776kΩ Tweak the R 1 and R 2 values until getting V CE = 5V: VCC 10Vdc 5.002V RC 5k 0 10.00V R1 11.72k 0V Qbreakn Q1 0 0 R2 1k 777.0mV 0 0 RC 5k CL Vout C1 VCC 10Vdc Qbreakn Q1 R2 1k RL 1k Vin 0 0 R1 11.72k 0 3 / 7
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High pass filter ( 29 ( 29 = + + + = = + = -
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bandpass filter design using a NPN BJT - Practice: Design a...

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