0
\$\begingroup\$

diagram of 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?

\$\endgroup\$
  • \$\begingroup\$ 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\$. \$\endgroup\$ – glen_geek Oct 26 '19 at 14:15
  • \$\begingroup\$ Ok thanks. I didnt short Vo \$\endgroup\$ – Manav Shetty Oct 27 '19 at 12:44
0
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot! Now I understood \$\endgroup\$ – Manav Shetty Oct 27 '19 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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