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So I'm currently doing an assignment where I'm meant to design my own common emitter. I'm supposed to choose the Q-point of my collector current and choose my own signal amplification (Av). During this design, I'm assuming (I_C=I_E, if I_B << I_C) and (I_R1=I_R2, if I_B << I_R1)

I'm using the hybrid-pi model.

I got the following value:

  • Vcc = 22V
  • Voltage over R_C: 11V
  • Collector resistance: 3.7k ohm
  • I_CQ (Q-point of collector): 3mA
  • Small signal amp: Av = 6 times.
  • Emitter resistance: R_E = -(R_C/|Av|) = 615 ohm
  • Voltage across emitter resistance V_Re: 1.85V
  • Voltage at base V_B = Vbe + V_Re = 0.7 + 1.85 = 2.55V
  • I_R1 = I_R2 = I_C/10 = 3mA/10 = 0.3mA
  • V_R2 = 2.55V (Same as V_B)
  • R_2 = 8.5k ohm
  • V_R1 = 19.45V
  • R_1 = 65k ohm
  • My voltage (V_CE) across the transistor is 9.15 V

When I'm going over to small-signal analysis, to calculate R_in and R_out. I need to know the base current I_b or current amplification beta/hfe, however, I know neither with makes me get stuck.

  • r_pi = beta/gm

Circuit diagram in LTSpice enter image description here

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  • \$\begingroup\$ Can't we calculate Ic and Ie, and from that get Ib? \$\endgroup\$ – aMike Oct 5 '20 at 1:21
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    \$\begingroup\$ When you assumed \$I_C=I_E\$ you already made an assumption about \$\beta\$. \$\endgroup\$ – The Photon Oct 5 '20 at 2:05
  • \$\begingroup\$ The Photon, you're correct. I apparently did make an assumption without acknowledging it. In the assignment, I'm supposed to choose my own current amplification Beta. \$\endgroup\$ – gripen Oct 5 '20 at 10:14
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That`s quite normal.

You are choosing a suitable DC operational point (collector current Ic, resistors Rc and Re). The voltage gain depends on the transconductance gm (slope of the voltage-control function Ic=f(Vbe) at the DC operating point). The so-called "current gain" has no direct influence on the voltage gain.

For calculating the resistive base divider values you must make a rough estimation for the DC current gain B (in case you do not have any value). As a first step you may assume an infinite B value (Ie=Ic). In many cases, this is sufficient if the divider current is much larger than the (unknown) base curent. Very often, the error caused by such a simplification is even smaller than the error caused by parts tolerances and by the rough estimation Vbe=0.7 V.

And yes - it is correct that the input resistance at the base is part of the overall input resistance. However, if you have no information about the actual base current you have no other choice than to make a good estimate.

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  • \$\begingroup\$ Hi LvW, thank you for your response. It was good reading! I called my professor this morning regarding it, the message I got from him was that I was supposed to assume or choose a current amplification Beta, which solves the entire problem for me! \$\endgroup\$ – gripen Oct 5 '20 at 10:12

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