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To ensure a stable gain, an common emitter amplifier would be designed so the resistance Rc is much smaller than the load resistance, RL. This is done so any variation in R₁ would have a minimal impact on the gain of the amplifier. With this in mine, the equation can be further reduced to: Vout Gain = FGEN gm Rc (1+9m) RE As you will see in lab, we are using a 100KQ resistor for RÅ and thus can be neglected in our calculations. If you are analyzing or designing an amplifier and R₁ is within an order of magnitude (x10) of Rc, it should be included in the calculations. Once you have the gain equation, you would then solve for gm using the following formula and the DC bias conditions: gm = where VÃ is the thermal voltage that is equal to 25mV at room temperature. VI Once we have this information, we can solve for the small signal gain. The gain for a common-emitter amplifier with an emitter resistor (feedback), RE, is: Gain = gm Rc 1+ gm RE If the emitter resistor is bypassed using a bypass capacitor (at the pass frequencies, the bypass capacitor equivalent impedence would be much smaller than RE and can be replaced with a short RE=0), the gain would be equal to: Gain = -gm Rc Common Emitter Inverting Amplifier with Feedback For the circuit in Figure 16, determine the DC bias conditions (solve for IC, VCE, and the voltage at the Collector). VBE=0.7V, ẞ=150, R₁=22ks2, R2=4.3ks, Rc=1.2ks, R=2202, R₁=100ks. Assume Forward Active Region. Hint: For DC bias analysis, treat all caps as open circuits. For AC small signal analysis, treat all caps as shorts. +15. FGEN GND R₁ Rc R₂- RE Probe O-Scope Q₁ R₁ Vout + Probe GND Figure 16: Common-Emitter Inverting Amplifier For the circuit in Figure 16, use small signal analysis (see Background Information) and calculate the gain for the inverting common-emitter amplifier (you can use the appropriate formulas given in the Background Information). If a 300mV 10kHz signal is input into the circuit at FGEN, determine the output at Rr. Assume the capacitors are chosen to be negligible (shorts) at 10kHz.

To ensure a stable gain, an common emitter amplifier would be designed so the resistance Rc is much smaller than the load resistance, RL. This is done so any variation in R₁ would have a minimal impact on the gain of the amplifier. With this in mine, the equation can be further reduced to: Vout Gain = FGEN gm Rc (1+9m) RE As you will see in lab, we are using a 100KQ resistor for RÅ and thus can be neglected in our calculations. If you are analyzing or designing an amplifier and R₁ is within an order of magnitude (x10) of Rc, it should be included in the calculations. Once you have the gain equation, you would then solve for gm using the following formula and the DC bias conditions: gm = where VÃ is the thermal voltage that is equal to 25mV at room temperature. VI Once we have this information, we can solve for the small signal gain. The gain for a common-emitter amplifier with an emitter resistor (feedback), RE, is: Gain = gm Rc 1+ gm RE If the emitter resistor is bypassed using a bypass capacitor (at the pass frequencies, the bypass capacitor equivalent impedence would be much smaller than RE and can be replaced with a short RE=0), the gain would be equal to: Gain = -gm Rc Common Emitter Inverting Amplifier with Feedback For the circuit in Figure 16, determine the DC bias conditions (solve for IC, VCE, and the voltage at the Collector). VBE=0.7V, ẞ=150, R₁=22ks2, R2=4.3ks, Rc=1.2ks, R=2202, R₁=100ks. Assume Forward Active Region. Hint: For DC bias analysis, treat all caps as open circuits. For AC small signal analysis, treat all caps as shorts. +15. FGEN GND R₁ Rc R₂- RE Probe O-Scope Q₁ R₁ Vout + Probe GND Figure 16: Common-Emitter Inverting Amplifier For the circuit in Figure 16, use small signal analysis (see Background Information) and calculate the gain for the inverting common-emitter amplifier (you can use the appropriate formulas given in the Background Information). If a 300mV 10kHz signal is input into the circuit at FGEN, determine the output at Rr. Assume the capacitors are chosen to be negligible (shorts) at 10kHz.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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To ensure a stable gain, an common emitter amplifier would be designed so the resistance Rc is much smaller than the load resistance, RL. This is done so any variation in R₁ would have a minimal impact on the gain of the amplifier. With this in mine, the equation can be further reduced to: Vout Gain = FGEN gm Rc (1+9m) RE As you will see in lab, we are using a 100KQ resistor for RÅ and thus can be neglected in our calculations. If you are analyzing or designing an amplifier and R₁ is within an order of magnitude (x10) of Rc, it should be included in the calculations. Once you have the gain equation, you would then solve for gm using the following formula and the DC bias conditions: gm = where VÃ is the thermal voltage that is equal to 25mV at room temperature. VI Once we have this information, we can solve for the small signal gain. The gain for a common-emitter amplifier with an emitter resistor (feedback), RE, is: Gain = gm Rc 1+ gm RE If the emitter resistor is bypassed using a bypass capacitor (at the pass frequencies, the bypass capacitor equivalent impedence would be much smaller than RE and can be replaced with a short RE=0), the gain would be equal to: Gain = -gm Rc
Transcribed Image Text:To ensure a stable gain, an common emitter amplifier would be designed so the resistance Rc is much smaller than the load resistance, RL. This is done so any variation in R₁ would have a minimal impact on the gain of the amplifier. With this in mine, the equation can be further reduced to: Vout Gain = FGEN gm Rc (1+9m) RE As you will see in lab, we are using a 100KQ resistor for RÅ and thus can be neglected in our calculations. If you are analyzing or designing an amplifier and R₁ is within an order of magnitude (x10) of Rc, it should be included in the calculations. Once you have the gain equation, you would then solve for gm using the following formula and the DC bias conditions: gm = where VÃ is the thermal voltage that is equal to 25mV at room temperature. VI Once we have this information, we can solve for the small signal gain. The gain for a common-emitter amplifier with an emitter resistor (feedback), RE, is: Gain = gm Rc 1+ gm RE If the emitter resistor is bypassed using a bypass capacitor (at the pass frequencies, the bypass capacitor equivalent impedence would be much smaller than RE and can be replaced with a short RE=0), the gain would be equal to: Gain = -gm Rc
Common Emitter Inverting Amplifier with Feedback For the circuit in Figure 16, determine the DC bias conditions (solve for IC, VCE, and the voltage at the Collector). VBE=0.7V, ẞ=150, R₁=22ks2, R2=4.3ks, Rc=1.2ks, R=2202, R₁=100ks. Assume Forward Active Region. Hint: For DC bias analysis, treat all caps as open circuits. For AC small signal analysis, treat all caps as shorts. +15. FGEN GND R₁ Rc R₂- RE Probe O-Scope Q₁ R₁ Vout + Probe GND Figure 16: Common-Emitter Inverting Amplifier For the circuit in Figure 16, use small signal analysis (see Background Information) and calculate the gain for the inverting common-emitter amplifier (you can use the appropriate formulas given in the Background Information). If a 300mV 10kHz signal is input into the circuit at FGEN, determine the output at Rr. Assume the capacitors are chosen to be negligible (shorts) at 10kHz.
Transcribed Image Text:Common Emitter Inverting Amplifier with Feedback For the circuit in Figure 16, determine the DC bias conditions (solve for IC, VCE, and the voltage at the Collector). VBE=0.7V, ẞ=150, R₁=22ks2, R2=4.3ks, Rc=1.2ks, R=2202, R₁=100ks. Assume Forward Active Region. Hint: For DC bias analysis, treat all caps as open circuits. For AC small signal analysis, treat all caps as shorts. +15. FGEN GND R₁ Rc R₂- RE Probe O-Scope Q₁ R₁ Vout + Probe GND Figure 16: Common-Emitter Inverting Amplifier For the circuit in Figure 16, use small signal analysis (see Background Information) and calculate the gain for the inverting common-emitter amplifier (you can use the appropriate formulas given in the Background Information). If a 300mV 10kHz signal is input into the circuit at FGEN, determine the output at Rr. Assume the capacitors are chosen to be negligible (shorts) at 10kHz.
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