Speaking for NPN Bipolar transistor:

In many practical examples parallel resistance of base-emitter junction (aka hie) and finite collector-emitter parallel resistance resistance (aka hoe) are usually neglected. Why? Because their values shouldn't significantly change input and output resistance of the transistor circuit.

In my opinion their values are actually very important - why? Because "hie" and "hoe" are usually small enough to cause drastic changes in input and output resistances of transistor circuit.

Differential resistance of base-emitter junction is defined as rbe = β/gm, where gm = Ic/Ut (where rbe is from few kΩ to few tens of kΩ for Ic values of few mA and β values 100 - 500). Example shown below: enter image description here

It can be seen that parallel differential resistance of base-emitter junction has significant effect on input resistance small-signal calculus (e.g. voltage divider).

  • So why does it get neglected so many times if practical values of hie shows important meaning of it in calculus of input impedance?
  • Same thing for differential collector-emitter resistance which is defined as ro = Vce+Va/Ic, where Va has values from few tens of kΩ and on - which can also have significant effect on output resistances.
  • \$\begingroup\$ But nowadays almost no one uses a single stage BJT amplifiers. In 90% of a circuits designs now, the BJT's are used as a saturation switch. Hence is this application we do not care about hie,hoe. As for the amplifiers I always include hie (hie = r_pi = (beta +1)*re = (beta+1)*26mV/Ie). Sometimes you can ignore hie you RE is large or BJT is used inside a feedback loop. And hoe = h22 = 1/ro = VA/Ic and we usually ignore it in hand calculations. But we have it in mind when we selecting Rc resistor. Also, can you show us the example when hie was neglected? \$\endgroup\$
    – G36
    Jul 26, 2017 at 15:12
  • \$\begingroup\$ Where hoe = Vearly/Ic. \$\endgroup\$ Jul 26, 2017 at 16:20
  • \$\begingroup\$ @analogsystemsrf But the hoe is the admittance is Siemens. hoe = 1/ro = 1/rce = Ic/Va. But I also made this slip of the pen. \$\endgroup\$
    – G36
    Jul 26, 2017 at 16:43
  • \$\begingroup\$ Yes. I also slipped up. For higher Ic, the tilted output curves have HIGHER tilt. \$\endgroup\$ Jul 26, 2017 at 17:04

4 Answers 4


Congratulations. You are starting to confront and understand the business of modelling.

A model is used to simplify the horrible reality, and make it easier to handle, when that simplification can be justified. We use as simple a model as we can get away with, no simpler, and no more complicated.

H parameters are a model for a transistor, valid at a certain set of bias conditions. Sometimes, let's say when the collector load is << hoe, we can neglect hoe in calculating the gain. We can't neglect it if trying to build a good current source. When there is an emitter degeneration resistor, or we are using the transistor in common collector, then the hie gets bootstrapped by the transistor, and can almost always be ignored. It can often be ignored compared to 50 ohms too.

A good engineer will know when it's OK to say a transistor has a Vbe of 0.7v and an infinite hfe, when to use finite hfe, when full h parameters are needed, or Ebers-Moll if those aren't adequate, or Gummel-Poon when that isn't adequate.

Whenever it's asserted that this or that parameter can be neglected, check the context to make sure that we're agreed on what level of model we should be using.

  • \$\begingroup\$ You wrote: "When there is an emitter resistor, or we are using the transistor in common collector as an emitter follower, then the emitter resistor or hie gets multiplied by the current gain etc." -> Are you sure about that? Because I'm studying about transistor many different ways and I have never seen this multiplication of hie that you were talking about. \$\endgroup\$
    – lucenzo97
    Jul 26, 2017 at 15:51
  • \$\begingroup\$ @Keno I didn't express that well. I've edited the answer. Thx for pointing it out. \$\endgroup\$
    – Neil_UK
    Jul 26, 2017 at 16:30

It can be seen that parallel differential resistance of base-emitter junction has significant effect on input resistance small-signal calculus (e.g. voltage divider)

If you choose to design simple BJT amplifiers that have no emitter resistor then you are right - the base-emitter impedance varies quite a lot and can cause biasing problems. However, if you design amplifiers this way you cannot conveniently control the voltage gain, the temperature drift is bad, and you get high distortion and other hFE related issues.

Most sensible folk will put in an emitter resistor and this allows voltage gain to be controlled, vastly magnifies input impedance and reduces distortion. Plus hFE variations with temperature barely have any performance changes.

Same thing for differential collector-emitter resistance which is defined as ro = Vce+Va/Ic, where Va has values from few tens of kΩ and on - which can also have significant effect on output resistances.

Yes, we have to be aware of this and from this knowledge, we choose not to design BJT amplifiers that have too high a collector resistor. Another knock-on benefit of not choosing too high a collector resistor is reduced miller capacitor problems and therefore flatter frequency responses.

So we design common emitter amps that include an emitter resistor so that the input impedance is high, and we make sure that the collector resistor is not so high that its output impedance is not troubled when connected to the next BJT stage (now having a fairly high input impedance due to Re).


hie, hoe etc. are parameters in a linearized transistor model. They are used for ac component calculations when ac components of currents and voltages are much smaller (=few percents of) than the idle state DC values. Linearization makes possible to utilize linear equations when estimating the gain, frequency response, noise, stability etc... Those equations are not simple especially when the capacitances and time delays are taken into the account, but they were the way to make plausible calculations before the era of computers.

In circuits where currents and voltages change not only few percents, but tens or even 100% the linearized model is useless. It's useful only for small signal circuits. Even in there it's not a must because computers allow easy simulations without consciously carrying sorrow about hie, hre, hfe and hoe.


To design a circuit, even on silicon, requires understanding each individual transistor (bipolar, fet) and how its operating conditions (DC and small signal) contribute to the overall circuit.

The simulator cannot understand the circuit. Only the human brain can.

In opamps or other signal chain building blocks, where nanoVolts of inputs become volts of output ( whether precision opamp instrumentation amplifiers, or a radio receiver for -130dBm signals ---- 200 nanoVolt peakpeak ), the crucial transistors operate as linear elements whether amplifying or merely providing biasing.

If the designer is not aware of small signal numbers (because the modelers may have made errors in data collection or in curve-fitting), then surprises happen and circuits do not work.

The designer must own, that is, perform sanity-checks, on all the tools.

  • \$\begingroup\$ Problem that causes that certain parameter can't be calculated is also the lack of data from devices data sheet (e.g. Early Voltage) - which can be common for most designers which need these values, whereas those are not given my manufacturer. Because of that errors occur. \$\endgroup\$
    – lucenzo97
    Jul 26, 2017 at 18:48
  • \$\begingroup\$ If Early Voltage is important, then use a stack of CE and CB devices. \$\endgroup\$ Jul 27, 2017 at 17:23
  • \$\begingroup\$ What is stack ? \$\endgroup\$
    – lucenzo97
    Jul 27, 2017 at 19:55

Your Answer

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

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