0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I am doing a small signal analysis of the following common-emitter BJT stage and I am trying to calculate the voltage across r_π (v_π). I am applying Thevenin's theorem.

Do I keep r_π in the Thevenin circuit? Is the Thevenin resistance Rthev = Ri//Rb1//Rb2//r_π OR Rthev = Ri//Rb1//Rb2 + r_π ?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You need to think about what part of the circuit you want to replace with a Thevenin equivalent. If you are trying to use a Thevenin equivalent to find the voltage across r_π then you probably do not want r_π as part of the Thevenin equivalent. \$\endgroup\$
    – Elliot Alderson
    Dec 5, 2021 at 16:08
  • \$\begingroup\$ It entirely depends on what you are trying to achieve. Sometimes you'll keep it; sometimes you won't. Given that you haven't said what you are aiming to do then it's impossible to say. I think, on this occasion, if you are truly trying to calculate the voltage across r_π then you should include r_π in your analysis. Whether you use Thevenin or not, the presence pf r_π does affect the voltage across it. \$\endgroup\$
    – Andy aka
    Dec 5, 2021 at 16:27

2 Answers 2

Reset to default
1
\$\begingroup\$

If you break the circuit as shown, then \$\text r_{\pi}\$ is not included in the calculation of the Thévenin resistance.

enter image description here

So you can replace everything left of the green 'X' with a voltage source of V1*(K/Ri) in series with a resistor of value K, where K = Ri||Rb1||Rb2 and treat it like a simple voltage divider. But keep in mind that the voltage applied to the divider is not V1, but V1*(K/Ri).

P.S. If you include \$\text r_{\pi}\$ you can get the answer in a single step, but then you are not invoking M. Thévenin.

\$\endgroup\$
1
\$\begingroup\$

Do I keep r_π in the Thevenin circuit? Is the Thevenin resistance Rthev = Ri//Rb1//Rb2//r_π OR Rthev = Ri//Rb1//Rb2 + r_π ?

Answer: You put them all in parallel: \$R_{thev} = R_i//R_{b1}//R_{b2}//r_π\$ While \$r_π=h_{IE}\$

Input impedance across Vbe is defined as hie,
Where;

hfe : It is called the current gain aka \$\beta \$
hie : It is the internal resistance, hie=hfe/gm aka \$r_\pi\$
hre : It is called reverse voltage gain
hoe : It is the output resistance at emitter.

Let's compare a small signal diode incremental load ΔVf/If=Rd to a base-emitter diode in a small transistor.

enter image description here

\$h_{ie} \approx (hfe+1)*\frac{26mV}{Ie}=\dfrac{26}{Ib [mA]}\$ @ 25 'C

hie for input impedance of a common emitter with Re=0

2N3904

For some odd reason Plotting Hie , hfe for 2N3904 , has better matching for \$hie=30.5 hfe/Ic\$ otherwise 17% error. (k=30.5 vs k=26 temperature coefficient) , let hfe = hfe+1 for hfe >100

enter image description here

For the long answer, read Ian Getreau's book 1979 from Tektronix. But there is a simple approach for small non-saturated signals.

For an accurate answer compute hie from the datasheet and estimate tolerance.

\$\endgroup\$
10
  • 1
    \$\begingroup\$ May I give a short comment to the above answer? In your contribution you have defined r-pi as r_pi=1/gm. This definition seems to be not in accordance with r_pi as used in the the original question. For my opinion, the inverse transconductance 1/gm is called "intrinsic emitter resistance r_e" (although I do not like this nomenclature at all). As a consequence, the quantity you call Rbe (capital letter?) is identical to the commonly used dynamic small-signal base-emitter resistance r_pi=r_be=h11=hie. \$\endgroup\$
    – LvW
    Dec 5, 2021 at 20:02
  • \$\begingroup\$ @LvW, I appreciate your assistance to edit my answer and correct my formula and syntax to show the Rin for Vbe when Ic = 1mA to illustrate the loading effects for small signal and simplify the problem rather than an academic one that includes Early Effects, forward and reverse current gain and ignores bulk resistance. rb and re or using i.stack.imgur.com/h72AX.png (+1) \$\endgroup\$
    – Tony Stewart EE75
    Dec 5, 2021 at 20:59
  • \$\begingroup\$ Tony Stewart, I must admit that I really do not fully understand your reply. Please, do not hesitate to state if I am wrong or not. In your contribution I read Rbe=(beta+1)*r_pi. However, in the original question as well as in Spehro Pefhany`s answer it is clear that r_pi is the differential resistance of the B-E junction [dVbe/dIb]. So what is the quantity you call Rbe in the first line of your contribution? I think it is also in the interest of the questioner that this discrepancy is cleared up. \$\endgroup\$
    – LvW
    Dec 6, 2021 at 8:27
  • \$\begingroup\$ Perhaps the most simple and direct way to avoid misunderstandings is to use words for defining the quantities. To me, the commonly used symbol for the slope of the input voltage-current characteristic (identical to the small-signal differential B-E resistance) is r_pi=h11=hie=d(Vbe)/Ib. Simple question: Right or wrong? \$\endgroup\$
    – LvW
    Dec 6, 2021 at 8:38
  • \$\begingroup\$ Right, but describe in terms of Ic/Ib ratio with bulk resistance in junction for different transistors and thermal effects of offset drift from Pd. Since gm is not plotted in datasheets it is less directly useful yet the dominant parameter. \$\endgroup\$
    – Tony Stewart EE75
    Dec 6, 2021 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.