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The assignment is to design a common-base amplifier from a supplied schematic, and quite frankly amplifiers are a significant weakness of mine.

Given:

  • Small-signal voltage gain A_V≥50 V⁄V
  • Input resistance R_in=75Ω
  • Peak output voltage swing without distortion ≥2V
  • β_F=90
  • V_A=800V

Schematic

Based on my hand calculations I've got:

  • g_m=1/75 S
  • I_C=346.666 uA
  • r_o=2.308 MΩ
  • I_B=3.852 uA
  • I_E=350.5 uA

But I still have no idea what to pick for R1 and R2. I want to see what the underlying math is.

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Good job so far. Going off your calculations, you already have the emitter current. All of this current has to flow through R1 (there's no other DC path) so R1=[V(emitter)-VEE]\I_E. Assuming a 0.7V base-emitter voltage (depends on doping as well as current), you'd have -0.7V at the emitter; 9.3V/351uA=26.5kOhm.

To bias the collector for max output swing, you'd want it at about VCC/2, so it can swing +-VCC/2 without clipping. That gives you 5V, and since you know your collector current, R2=(VCC-5V)/IC =5V/347uA = 14.4kOhm.

As a quick check, A=gm*(R2//R3)=(1/75)S*12.6kOhm=178, which is well over 50.

Hope this helps!

Edit: I missed the series resistance with the input. This is 75 Ohms, and the input resistance to the emitter is also 1/gm= 75 Ohms, meaning you get a 0.5x voltage divider from your source to the emitter voltage. This cuts your total gain in half, so you'd get a gain of 89 or so instead of 178.

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  • \$\begingroup\$ R2 / R3 is NOT 12.6kO. What is going on with that line? \$\endgroup\$
    – StarSword
    Dec 14, 2023 at 22:01
  • \$\begingroup\$ Also according to equations for common-base amplifiers (electronics-tutorials.ws/amplifier/common-base-amplifier.html), A_V = alpha(R_C/R_E), so the ratio from collector to emitter theoretically ought to be about 51:1. \$\endgroup\$
    – StarSword
    Dec 14, 2023 at 22:30
  • \$\begingroup\$ @StarSword thanks for bringing the input resistance to my attention; the voltage gain from emitter to collector is actually RC/r'e in the source you described, which is the number I got. RC/RE is true for a different set of assumptions \$\endgroup\$ Dec 15, 2023 at 5:54

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