Problem #2 - Common-Emitter Amplifier Using the rule of thirds, design a single-supply Common Emitter Amplifier using a 12 V supply for Vcc, with a collector current Ic = 0.5 mA. Use a current through R₁ equal to 25% of Ic. Vcc R₁ R₂ (a) Specify all the values for RC, R£, R₁, R₂ (b) Specify the voltages VB, VE Rc RE Recall the rule of thirds states that 1/3 of Vcc flows across Rc, 1/3 across R₂, and assume a very high ß such that Iç = IĘ, i = 0 (c)Assuming the lowest value for ß is 70, calculate the actual value the collector current will have with the circuit design in parts (a) and (b).

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Problem #2 - Common-Emitter Amplifier
Using the rule of thirds, design a single-supply Common Emitter Amplifier using a 12 V supply
for Vcc, with a collector current Ic = 0.5 mA. Use a current through R₁ equal to 25% of Ic.
Vcc
R₁
R2
(a) Specify all the values for Rc, RẼ, R₁, R₂
(b) Specify the voltages VB, VĒ
Rc
RE
Recall the rule of thirds states that 1/3 of Vcc flows across Rc, 1/3 across R₂, and assume a
very high ß such that Iç = IĘ, ig = 0
(c)Assuming the lowest value for ß is 70, calculate the actual value the collector current will
have with the circuit design in parts (a) and (b).
Transcribed Image Text:Problem #2 - Common-Emitter Amplifier Using the rule of thirds, design a single-supply Common Emitter Amplifier using a 12 V supply for Vcc, with a collector current Ic = 0.5 mA. Use a current through R₁ equal to 25% of Ic. Vcc R₁ R2 (a) Specify all the values for Rc, RẼ, R₁, R₂ (b) Specify the voltages VB, VĒ Rc RE Recall the rule of thirds states that 1/3 of Vcc flows across Rc, 1/3 across R₂, and assume a very high ß such that Iç = IĘ, ig = 0 (c)Assuming the lowest value for ß is 70, calculate the actual value the collector current will have with the circuit design in parts (a) and (b).
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