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This is a paragraph from "Engineering circuit analysis" by W H Hayt. I really don't understand the statement "it is difficult to open circuit the open terminals...." What does he mean in the paragraph? I also don't understand if we are able to open circuit for measurement of Z parameters why is it difficult here?

  • \$\begingroup\$ Maybe you read it wrong because it says "open circuit the output terminals". Does this solve your issue? \$\endgroup\$
    – sarthak
    Jul 8, 2019 at 15:41
  • \$\begingroup\$ @sarthak I really don't understand that as we are able to do it in calculation of Z parameters \$\endgroup\$ Jul 8, 2019 at 15:55
  • \$\begingroup\$ As it is mentioned in the answer, at low frequencies the Z-parameter measurements are possible. But every node has some capacitance associated with it and at higher frequencies the capacitor starts to look more and more like a short circuit. In very simplistic terms, you might know the capacitor blocks DC current and allows AC current to pass. This makes the measurement of Z-parameters difficult at high frequencies. \$\endgroup\$
    – sarthak
    Jul 8, 2019 at 16:17
  • \$\begingroup\$ it actually says to open-circuit the output terminals ... open-circuit is used as a verb, so the statement means to make the output terminals into an open circuit or to disconnect the output terminals \$\endgroup\$
    – jsotola
    Jul 9, 2019 at 5:14

3 Answers 3


Suppose you want to characterize the transistor at 1GHz, yet that node has 10pF or 20pF parasitic capacitance what with 1) scope probe 2) biasing circuit 3) mechanical structure to hold the device

At 1GHz, 1pF is -j159 ohms.

At 1Ghz, 20pF is -j 8 ohms.

Thus the realworld prevents "open circuit" tests at high frequency.


If a transistor is a current sink you expect infinite impedance output, but due to the Early Effect shunt resistance and output capacitance and layout stray impedances including probing method, this can be >1M and <1pF depending on size , ratings and frequency applied.

Using a short circuit current with Rc=0 is the easiest way uA to Amps but then Vce= Vcc may not be your operating condition or temperature rise due to Ic*Vce *Rja = temperature rise above 25’C, which affects the results.

Leakage current, increases as well as hFE(dc), Hfe(ac) and GBW (MHz) with temperature as Vbe reduces.

This can be measured with a micro amp to Amp range of collector currents compared with input currents in either Vdc or Vac at some frequency on a DC bias.

Then z21 is the ratio of currents for v21/i21=z21. But v2 the collector is fixed Vcc and v1 is your incremental Vbe bias voltage and same for i1 is the incremental bias current.

You know Ic is a function of Vbe, but for the AC model you are only interested in the incremental or small signal impedance gain (transfer function) at some Q operating point of Ic(dc) with a swept frequency for the s or z(f) domain transfer function over the range of interest which may include near DC.


Suppose you have a transistor having a \$h_{oe} \$ equivalent to a 10k resistor. You wish to make a measurement of \$h_{oe} \$ at a DC collector current of 100uA. To prevent loading \$h_{oe}\$ in your measurement test jig, you must supply a DC collector bias voltage through a much larger resistor...say 100 times larger....like 1Meg ohm. In this manner, the bias resistor affects little the \$h_{oe} \$ you're trying to measure.

Your voltage-measuring device like an oscilloscope might increase loading via its 10M internal resistance.


simulate this circuit – Schematic created using CircuitLab

Biasing that transistor into its linear range around 100uA would require a DC voltage fed through that 1M collector biasing resistor in excess of 100V. That's awkward, and risks destroying the transistor from over-voltage. Trying to make a measurement of a common-base (where \$h_{ob}\$ is even higher) is even more awkward, if a larger bias resistor must be used.

As other's have mentioned, the AC test frequency to make the measurement must be very low, else capacitances influence the measurement.


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