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:

$$\mathbf{Z}_{in} = (\beta+1)\mathbf{Z}_{load}\ \ \ \ \ \ \ (2.3)$$ $$\mathbf{Z}_{out} = {{Z}_{source} \over \beta+1}\ \ \ \ \ \ \ \ \ \ \ \ \ (2.4)$$

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. 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. 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. Hence a I_b is 9.444095V / 2,101,000ohm = 0.000004495A or 4.495uA, and voltage drop of R_b = 4.495uA * 2Mohm = 8.99V. The calculated values matches those displayed in the simulation. OK.

Output Impedance: if removing Re and Rc, Zout = 2Mohm/(100+1). 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). Voltage at the emitter would be 10V - 8.990V - 0.555905V = 0.454095V or 454.095mV. I_e should then be 0.454095V / 30,801.980198 ohm = 0.000014742A or 14.742uA. The current does not match that in the simulation, 454mA.

Here the questions:

  1. What I did wrong with my output impedance measurement?
  2. 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. 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. But simulations are ideal with fixed values and formulas and I should be getting an exact 454.000uA instead of 454.095uA. So I am either missing or misunderstanding something.

BJT Sample Circuit

  • sorry, can't get MathJax to display inline with paragraph using single $ sign
  • \$\begingroup\$ use backslash as in \$. I don't know the reason but escaping $ is required \$\endgroup\$ – beccaboo Apr 6 at 10:43
  • \$\begingroup\$ that's for future. your post is good as it is:) \$\endgroup\$ – beccaboo Apr 6 at 10:47
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    \$\begingroup\$ KMC....what are you interested in? DC bias point (quiescent current Ic) or output resistance? \$\endgroup\$ – LvW Apr 6 at 13:51
  • \$\begingroup\$ @LvW Output resistance. I have not gotten into quiescent current nor do I know what it is yet in my reading of AoE. I was trying to verify my understanding of the equations in the book through circuit simulation program \$\endgroup\$ – KMC Apr 6 at 13:59
  • \$\begingroup\$ KMC- that is not a good approach. In fact, it is the other way round: Circuit simulation is a good tool to show and to proove if the calculations and the dimensioning is OK. But is not a good tool for learning and to understand equations. And - as I have mentioned in my answer - AoE is NOT a good book for learning how circuits work. It is - more or less - a handbook for somewhat experienced people. \$\endgroup\$ – LvW Apr 6 at 14:30

1.What I did wrong with my output impedance measurement?

Because the Zin and Zout are AC parameters, not a DC one.

Output impedance seen by the load in this emitter follower

  1. 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. Where is this or why did I get this extra 0.0905uA in the calculation?

To solve the DC bias point you can use this method:

$$V_{CC} = I_BR_b+V_{BE}+I_ER_E$$

We also know that:

$$I_E = I_B+I_C = I_B + \beta I_C = I_B(\beta +1)$$


$$I_B = \frac{I_E}{\beta +1}$$

Therefore: $$V_{CC} = \frac{I_E}{\beta +1} R_B+V_{BE}+I_ER_E$$

$$I_E = \frac{V_{CC} - V_{BE}}{ \frac{R_B}{\beta +1} + R_E} = \frac{10V - 555.905mV}{\frac{2M\Omega}{100 +1} + 1k\Omega} = 454\mu A $$

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  • \$\begingroup\$ Thanks. I now understand the steps getting the current to question (2). But I still don't see why my original conceptual or intuitive approach of just deducting the voltage drop along the path from Vcc to Re, and measuring current by dividing the remaining voltage across Re resistance does not get to the exact same value (very close to tens of uA but not exact though it should). Both your given equations and my approach are correct to me conceptually ... \$\endgroup\$ – KMC Apr 6 at 13:42
  • \$\begingroup\$ @KMC I suspect a roundup error. Or in the simulation, there is some "hidden" resistance inside BJT model what is very common. \$\endgroup\$ – G36 Apr 6 at 14:50

Well, we have the following circuit:


simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

  • $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
  • Transistor gain \$\beta\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$

Using KCL, we can write:

$$ \begin{cases} \text{I}_\text{x}=\text{I}_1+\text{I}_3\\ \\ \text{I}_3=\text{I}_\text{T}+\text{I}_4\\ \\ \text{I}_\text{x}=\text{I}_2+\text{I}_4\\ \\ \text{I}_2=\text{I}_\text{T}+\text{I}_1\\ \\ \beta=\frac{\text{I}_1}{\text{I}_\text{T}} \end{cases}\tag3 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{x}-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\ \\ \text{V}_2-\text{V}_3=\alpha \end{cases}\tag4 $$

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Those formulas with Zin and Zout are valid only for weak AC voltages which ride on the DC bias voltage. You can calculate with them only how much weak AC signals get amplified or attenuated and how strong AC currents they cause as summed to idle state DC current.

ADD: There's comments which state that the formulas are inaccurate. That's true, the formulas assume transistor works for signal AC component like an ideal by current controlled current source.

The idle state DC current and voltage (=Operating point, bias point) must be calculated with DC model where BE junction is considered to be a diode which has few hundred millivolts voltage drop, for simplicity that voltage drop is often assumed to be 600...700mV although it's actually determined in a complex way by the nonlinear diode law.

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  • \$\begingroup\$ It should be noted that both formulas are very rough approximations only - or better: They are wrong. \$\endgroup\$ – LvW Apr 6 at 17:15

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.

(Counter example: Zsource=1k and beta=100. Zout=1000/101=10 ohms is totally wrong).

Input and output resistors (as well as the gain factor of an amplifier) are small-signal parameters. For corresponding calculations you need, therefore, small-signal characteristic parameters for the BJT (which are depending on the selected DC bias point). These parameters are

  • Input resistance at the base node: hie (or h11 or rbe);

  • Current ratio ic/ib: hfe (or h21 or beta);

  • transconductance ge=d(Ie)/d(Vbe) or gm=d(Ic)/d(Vbe)=hfe/hie

The calculation of r,out is rather straightforward using ohms law and Kirchhoff`s rules. 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).

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).

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