I'm reading a book on feedback amplifiers at the moment and it makes comprehensive use of a simplified hybrid-pi model. A parameter that keeps popping up is hie and I've always considered this to be equivalent to $$\left ( \beta + 1 \right )* re$$ in the re transistor model and for good reason too: previous stackexchange question and here (page 12).

However, what's confusing is the size of the "typical" resistances given to hie (if you're lucky enough to find it!): 1k to 10K. Really? Lets look at an example, using that classic hFE of 100 and an Ie of 50mA. $$101 * \frac{26mV}{50mA}\approx 52\Omega$$

I'm clearly missing something, because those figures don't compare at all. In this thread a correspondent makes the same point (that the actual input resistances are much smaller). However, there's a suggestion that maybe hie is a small signal AC parameter that's distinct to the hybrid-pi model and I'm left wondering if the corresponding re model equivalent equations are correct?

Can somebody set me straight on this? Is the comparable equation justified and why are those typical values so out of whack?

  • 1
    \$\begingroup\$ Yes, Hie is equal to (hfe+1)*re. But for a large Ie current, you need to include a built-in resistances into the BJT's. The base spreading resistance and emitter resistance. \$\endgroup\$
    – G36
    Commented Apr 17, 2018 at 17:31
  • \$\begingroup\$ @G36 any chance you could expand on this in an answer? It's hard to find answers in my usual sources. \$\endgroup\$
    – Buck8pe
    Commented Apr 17, 2018 at 17:42
  • \$\begingroup\$ Gah! I see my mistake now! The units are in mV/mA, I should have scaled up by 1000. The annoying thing is I knew this, but had forgotten. \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 7:54

3 Answers 3


\$Hybrid~ Model\$

more complex

\$h_{ie}=\dfrac{\delta V_{BE}}{\delta I_B} @Q_{point}\$ This is the small signal input impedance. This is a product of the dynamic current gain and many variables. \$h_{ie}\$ is normally in kΩ range for low linear currents. The transistor has both a DC current gain of \$\beta =H_{fe}\$ and a smaller AC current gain of \$h_{fe}\$ .

Simpler \$r_e~Model\$

-simpler model for DC to mid-frequency use with only 2 variables \$h_{fe}\$ and temperature.

Vt = γ = kT/q as the thermal voltage with temp. T [°K] and k=Boltzman's constant and charge, q
Vt = 26mV @ 30°C , 25mV @ 20°C

\$r_e = V_t/I_e~~~~\$ [Ω= mV / mA ] while \$r_e\$ is typically in the 10~70 ohm range.

since \$I_e=I_c+I_b ~,~~ ~ I_c=βI_b~,~~ I_e/I_b=(β+1)\$



Xi'an University of Electronic Science and Technology

Other stuff

Now if we were to add an emitter resistor Re=100Ω;

\$r_e=50,~~ Z_{in(f)} = β * [r_e+Ze(f)]=100 * (50+100)=15kΩ\$

But if Vce is saturated as a switch, β drops to 10-20 and Zin(f) drops to 10% of the above value.

When Vbe saturates this become more linear and depends on the PN junction size inverse to Pd of the diode junction. But we know that Ic doubles for every 17mV rise in Vbe if at constant temp.

But consider when Vbe is saturated and consider just diodes with no current gain.

I generalize this /$r_e~/$ as ESR~1/Pd so a 20mW junction is 50 Ohms, a 65mW LED as 16 Ohms, a 1W LED as 0.5 to 1 Ohm at 85'C but due to tolerances this is generally +/-25% typ and +/-50% worst case over temp.

Thermal design affects this as does temp rise, so generally SMD parts for ESR=k/Pd(max) k is drops from 1 to 0.5 and the very best thermal bulk power diodes are k=0.25 .

You won't find this in datasheets or textbooks, but it is my observation of doing above over thousands of devices.

The same can be done for the Rce value when a transistor is used as a switch.

Vce /Ice is saturated Vce(sat) @ I spec, then you can expect an Rce value similar to a power diode Rce = x/Pd but depends on design of chip and thermal capacity. so k has a wider tolerance and an offset that depends on doping differences of BE-CB, but x = 0.25 to 1 is a good fit from SMD power transistor to small signal plastic transistors.

  • \$\begingroup\$ "re is typically in the 10~70 ohm range" are you sure about this? For smaller emitter currents (see OP), I get < 1 Ohm. I made sure to divide mV by mA. Have I made a mistake somewhere? \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 6:44
  • \$\begingroup\$ Sorry Tony, rookie error - I forgot to scale 1/gm up by 1000. I'll give you the tick since it was you're estimates of re that gave me the hint. The important thing is that the universe is, once again, set to right (or is it). \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 7:58

To be more precise the \$h_{ie}\$ is not strictly equivalent to \$(\beta+1)r_e\$.

\$h_{ie}\$ also known as \$h_{11}\$ comes directly from a two-port network theory. Where they treat the transistor as a black box. http://ux.brookdalecc.edu/fac/engtech/andy/engi242/bjt_models.pdf

On the other hand, the \$r_e\$ is coming directly from the Shockley equations.

But despite this in hand calculation, we are using this approximation

\$h_{ie} \approx r_\pi = (\beta+1)r_e\$ in small signal analysis (when BJT working in linear/active region).

But BJT also has some "real" resistances "built into" the BJT's terminals.

The base spreading resistance (\$r_{bb}\$)

https://4donline.ihs.com/images/VipMasterIC/IC/ONSM/ONSMS04099/ONSMS04099-1.pdf?hkey=EF798316E3902B6ED9A73243A3159BB0 (figure 5)

And "bulk" emitter resistance around \$0.5\Omega\$ for small-signal BJT's.

  • \$\begingroup\$ OK, but the question I'm really asking is why are the quoted typical hie values so large compared with the values you get for ordinary small signal transistors with ordinary Ie quiescent currents? Can your additional resistances account for 100's of Ohms of difference? \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 6:50
  • \$\begingroup\$ Actually, if you look at Tony's answer he's calculating re as > 10 (which when multiplied by beta would get you there). But, I get < 1 Ohm when I use milli units (i.e., mV/mA see question). Have I made an error here? \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 6:55
  • \$\begingroup\$ Gah! I see my mistake now! The units are in mV/mA, I should have scaled up by 1000. The annoying thing is I knew this, but had forgotten. \$\endgroup\$
    – Buck8pe
    Commented Apr 18, 2018 at 7:54

Here is my comment to the "re-model":

There are two important BJT characteristics:

(1) Ic=f(Vbe) . This is Shockleys fundamental exponential relation. For gain calculation, the SLOPE of this function is important - and this is the transconductance gm=d(Ic)/d(Vbe) .

(2) Ib=f(Vbe) . This is the dynamic input characteristic of the BJT. The SLOPE of this curve gives the inverse of the dynamic input resistance hie=d(Vbe)/d(Ib).

(3) The product of both expressions is gm*hie=d(Ic)/d(Ib)=h21=beta.

(4) Hence: hie=beta/gm. This equation relates the most important BJT parameters (a) dynamic input resistance, (b) current gain and (c) transconductance.

(5) As we can see, there is no resistive element called "re" at all.

In some books the inverse transconductance is called re=1/gm - however, this is misleading because re=1/gm is NOT a resistive element (although it has the unit V/A=ohms). It is the inverse transconductance - nothing else and it must not called "intrinsic emitter resistance" (this error can be found in some books or papers).

Therefore, we should use the transconductance gm only (and NOT re) - as the most important and gain determining quantity.

Comment (explanation): Dividing the voltage between two nodes by the current through these two nodes gives the resistance between the two nodes (Ohms law). However, this definition does NOT apply to the quantity re=1/gm=d(Vbe)/d(Ic), because Vbe is the voltage between B and E - and Ic is the current through a third node (C).

  • \$\begingroup\$ But no one here claims that "re" is a true resistance. Also "re" can also be derived from the Shockley's equation. And can also be viewed as a slope. And T-model is a valid small signal model that will give you exactly the same final result as hybrid-pi model. \$\endgroup\$
    – G36
    Commented Apr 18, 2018 at 17:33
  • \$\begingroup\$ There are some books and papers in which the invers transconductance 1/gm is called "intrinsic emitter resistance" - and the corresponding T-model uses for re=1/gm the classical resistor symbol . For my opinion, this is misleading and does not help at all to understand the working principle of the BJT - in contrary!! \$\endgroup\$
    – LvW
    Commented Apr 18, 2018 at 18:43
  • \$\begingroup\$ Regarding the T-model: Looking at the model, I see a resiistance re=1/gm between the node B and E. But the B-E resistance is NOT identical to 1/gm, as you know. Accordinf to my experience with students, this can create problems .... \$\endgroup\$
    – LvW
    Commented Apr 18, 2018 at 19:18

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