When υ s increases in a positive direction, it increases the transistor base-emitter voltage (V BE). This can be understood by considering the effect of a positive-going input signal. It is seen that there is a 180° phase shift between the input and output waveforms. Please, pay attention to the fact that we strictly should discriminate between (a) ohmic resistors (capital letter R) and (b) small-signal differential resistances (small letter r).The current and voltage waveforms for the CE circuit in Fig. Start with an external test signal voltage at the emitter and find the current into the emitter node (as a first step forget the ohmic resistor Re which is in parallel to the resistance into the emitter node). The calculation of r,out is rather straightforward using ohms law and Kirchhoff`s rules. Transconductance ge=d(Ie)/d(Vbe) or gm=d(Ic)/d(Vbe)=hfe/hie Input resistance at the base node: hie (or h11 or rbe) Ĭurrent ratio ic/ib: hfe (or h21 or beta) For corresponding calculations you need, therefore, small-signal characteristic parameters for the BJT (which are depending on the selected DC bias point). Input and output resistors (as well as the gain factor of an amplifier) are small-signal parameters. (Counter example: Zsource=1k and beta=100. Note that both expressions for Zin and Zout as given in AoE are rough approximations only! AoE is a book - more or less - for practical design purposes, but NOT a good book for learning how a BJT and BJT amplifiers work. sorry, can't get MathJax to display inline with paragraph using single $ sign.So I am either missing or misunderstanding something. But simulations are ideal with fixed values and formulas and I should be getting an exact 454.000uA instead of 454.095uA. Where is this or why did I get this extra 0.0905uA in the calculation? I understand practically there are miscellaneous factors affecting the beta and voltage drop across Vbe, Vbc and Vce, and we are subject to approximation, availability of components, and trials and errors to choose a resistance or output a current to get the job done. If I just get the voltage drop after the emitter by V_Re = V_cc - V_Rb - V_be = 10V - 8.99V - 0.555905 = 454.095mV, and hence I_e = 0.454095V/1kohm = 454.095uA from my calculation which is 0.095uA more than the simulated value of 454.000uA.What I did wrong with my output impedance measurement?.The current does not match that in the simulation, 454mA. With Rb and Rc in place, according to AoE formula, Z_source = 2Mohm/(100+1) + 10kohm + 1kohm = 30,801.980198 ohm (the resistance looking into emitter in the perspective of the load, or R_e). Output Impedance: if removing Re and Rc, Zout = 2Mohm/(100+1). The calculated values matches those displayed in the simulation. With a voltage drop of 555.905mV in V_be inside the BJT, the Thevenin voltage into the base is 10V - 555.905V = 9.444095V. According to formula 2.3, the input resistance of the emitter follower looking into the base would be ((100+1)*1k)+2M = 2,101,000ohm. Input Impedance: take the circuit simulation in the image as an example, where R_b (2Mohm), R_c (10kohm), and R_e (1kohm) each denote a resistor on the base, collector and emitter respectively, and where BJT has a constant of 100 beta or Hfe by default. In AoE's chapter on BJT (specifically section 2.2.3B Input and Output Impedances of Emitter followers), it derives the input and output impedances to be:
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