# Measuring input and output resistance of the hybrid model

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
• Ok thanks. I didnt short Vo – Manav Shetty Oct 27 '19 at 12:44

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