Measuring input and output resistance of the hybrid model

Question

I understand that the input and output resistance is calculated using the Thevenins theorem. Thus the output resistance = Rc itself as the current source is open. But what about input resistance? Shouldn't it be (hi + Re)||Rb since the current source is open? What am I doing wrong here?

Why would you open-circuit the collector?...doing so invalidates the whole amplifier. What about \$V_o\$? It should be shorted....then all of \$h_{fe}I_b\$ current flows through \$R_E\$. — glen_geek, Oct 26 '19 at 14:15

score 0 · Accepted Answer · answered Oct 26 '19 at 14:19

No, you have a dependent source here (CCCS). So, you cannot turn it OFF, as we usually do with the "normal" (independent) current sources.

And \$R_{OUT} = R_C\$ is only true because when we are trying to find the output resistance we are setting/forcing the \$V_{IN}\$ to \$0V\$, hence no base current will flow due to input signal. And because your small-signal model do not contain \$h_{22} = h_{OE} = 1/ro\$. The dependent current source will be OFF in this case.

But if we add \$h_{22} = h_{OE}\$ the siutiation will be defrent.

See the example here:

BJT common-base output resistance derivation

Calculation of output impedance of CE emitter bias configuration( unbypassed) with r_0

I/O Resistance of common source MOSFET with source degeneration

Thanks a lot! Now I understood – Manav Shetty Oct 27 '19 at 12:43 — Manav Shetty, Oct 27 '19 at 12:43

Measuring input and output resistance of the hybrid model

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