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BJTrf

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This page exists to hold shortcuts, understanding, formula, and notes on BiPolar Junction Transistors. Its necessity for me became apparent during a youtube tutorial by W2AEW on the devices. I understood things each step of the way but when done was unable to track all of the knowledge. The notes here are on the first video listed. Despite the title of this page he is running his tests at 1Khz.

• Common Emitter or Emmitter Follwoer Amplifier Input and Output Impedances
• Back to Basics the Emitter Follower
• Tips & Tricks on Common Emitter Amplifiers Common Emitter Amplifier Design Tips Current Sources
• Back to Basics Transistor Current Sources: Video Notes

Parts List for Input and Output Impedance video.

1. Q1 2N222 or 2N3904 will work equally well.
2. CIn 4.7pF
3. Bias Voltage Divider ~1.6-1.7 volts
4. R1 (12V to Base) 47kΩ
5. R2 8.2KΩ
6. RC 6.8KΩ
7. RE 1KΩ
8. COut 4.7uF
9. Powersupply bypass cap 100uF
   • beta he assumes to be 100
   • configuration is often used as a buffer
   • bias current is 1.7V -.7V drop in the transistor = 1V across the RE of 1K giving one milliamp of current
   • he sets up with a 1Khz signal generator several hundred millivolts (700) through a resistor substitution box just to check his calculations later. It is initially set to 0Ω
   • gain is going to essentially be RC/RE in this case gain is~ 6.8 measured 6.7
   • emmitter follower is inverting.

• Output impedance
  • "The collector looks like a very high impedance,, high enough that we can ignore it".
  • Consequently the output impedance is the load impedance, RC.
  • Load on the Collector appears as in parallel with RC ,, this will reduce the gain RC/RE
  • With a 10K load attached gain drops to ~3.9
  • 6.8KΩ in|| 10KΩ= 1/ 1/6.8K + 1/10K = 4K Ω
  • RC/RE = 4KΩ/1KΩ= 4. measured at 3.94
  • So the input impedance of a cascaded amplifier is going to affect the gain of this amplifier.
• Input impedance
  • As far as the signal is concerned,,,, the AC portion flowing to the base,,,, the filter capacitor, the bypass cap, effectively makes the 12V rail shorted to the ground. Therefore R1 and R2 are effectively in parallel as far as any input signal is concerned. These resistances ~7KΩ
  • R1 and R2 are also in parallel with the input impedance of the base through RE. This appears as ß*RE or 100 * 1K = 100KΩ . 7K||100KΩ=6.5KΩ
  • Because the frequency is low he can use the resistor matching trick to check the input impedance.
• Bybassing the emitter Resistor
  • you need to calculate the small signal emmitter resistance re (little re) (intrinsic to the transistor) this is essentially the thermal voltage over Ic/ = 26 ohms. ß* re =2600. Gain becomes 260, input impedance becomes 7KΩ||2.6K= 1.9KΩ

1. https://www.qsl.net/w2aew/youtube/W2AEW_video_index.pdf W2AEW Video Index