Maximizing gain in a bipolar transistor amplifier: capacitor vs. resistor

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SUMMARY

The discussion focuses on maximizing gain in a bipolar transistor amplifier, specifically using a 2N3904 transistor. The output voltage (Vout) should theoretically be around 7.5V when biased correctly at the Q-point with a quiescent collector current of approximately 1mA. The presence of a 15µF capacitor is crucial for increasing output voltage and gain, as removing it reduces gain. To further enhance gain, cascading stages or employing a current source instead of a collector resistor is recommended.

PREREQUISITES
  • Understanding of bipolar transistor operation and biasing techniques
  • Familiarity with the 2N3904 transistor pinout and characteristics
  • Knowledge of amplifier gain calculations, specifically -Rc/Re
  • Basic concepts of capacitive coupling in amplifier circuits
NEXT STEPS
  • Research the impact of emitter degeneration on amplifier stability and gain
  • Learn about cascading amplifier stages for increased gain
  • Explore the use of current sources in transistor amplifier design
  • Investigate the frequency response of capacitive coupling in amplifiers
USEFUL FOR

Electronics engineers, hobbyists designing audio amplifiers, and students studying transistor amplifier configurations will benefit from this discussion.

math&science
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I have the amplifier in the attached file and I measured Vout without any input and got a really big number. Should it be really big or really small? I mean theoretically, what should it be equal to because I'm not sure that my answer makes any sense. Thanks!
 

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I'll admit I didn't bother doing any calcualtions so take this as a best hunch: With no input on the base you should have ~7.5V out. The transistor should probably be biased somewhere in the middle of the Q-point thus the CE current should induce about a 7.5V drop across the Collector resistor. This is a rough assessment.
 
To bias the output at half of the 15V supply, you want about 1mA of quiescent collector current (given the collector resistor that you show). That puts the emitter at about 1V, and the base at about 1.6V. The input resistor divider should be sized to provide that bias voltage, accounting for the base current of about 1mA/beta. I didn't check the base resistor divider to see if it accomplishes this -- check the values, and then probe the E,B,C points again to see if you can figure out what might be wrong. When you say a "big value" for Vc, I assume you mean up by the 15V rail? If so, figure out why the transistor is off. And make sure that you have the pinout right for the 2N3904.
 
Oh thank you! I got around 8 V but then someone told me it should be a really small number and so I got confused. Thank you for clarifying.
One more question, if you have an input but took out the 15microF capacitor, is it possible to increase the gain in any way? I know that if you take out the capacitor, you get a lower gain because that capacitor increases the output voltage.
 
Last edited:
math&science said:
One more question, if you have an input but took out the 15microF capacitor, is it possible to increase the gain in any way? I know that if you take out the capacitor, you get a lower gain because that capacitor increases the output voltage.
The gain of that configuration is about -Rc/Re, where Re is the sum of the intrinsic emitter resistance re and the external emitter resistor. You can ground the emitter to maximize the gain, but then your bias point is not as stable over temperature and transistor component variations. Using the capacitor in parallel with the emitter resistor gives you an effective grounding of the emitter at frequencies above 1/2piRC, which gives you max gain there. If you need more gain, you will need to cascade (not cascode...) a couple stages in series, or use a current source instead of the collector resistor.
 

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