0
\$\begingroup\$

Given current gain (hfe = 100), internal resistance (hie = 2000), reverse voltage gain and output resistance (hre = hoe = 0), compute input impedance for the following circuit. enter image description here

The problem is that it is not a classic scheme and I don't know how to compute the input impedance (i.e. can apply formulas directly). My guess is that there will be some capacitor shortage in the circuit but I don't know how and where. I'm looking for ways to transform this circuit into a simpler one in order to apply the formulas.

\$\endgroup\$
  • \$\begingroup\$ Voting as "too broad" as OP is unfamiliar with small signal analysis which is required to solve this. \$\endgroup\$ – Bimpelrekkie Jan 3 at 16:18
  • \$\begingroup\$ I'm looking for ways to transform this circuit into a simpler one in order to apply the formulas. Uhm yeah, that is exactly what small signal analysis is all about. \$\endgroup\$ – Bimpelrekkie Jan 3 at 16:19
2
\$\begingroup\$

To determine the input impedance of this circuit, it is necessary to replace the bipolar transistor by its hybrid-pi model. This is what is shown below where I have left \$C_3\$ out and considered an unloaded circuit:

enter image description here

Now the question is "how to determine the input impedance of this circuit?". Well, you can apply the brute-force method or the fast analytical circuits techniques known as FACTs. The FACTs will lead you to the answer by determining the time constants involving each of the energy-storing elements of this circuits, \$C_1\$ and \$C_2\$. You have two energy-storing elements, this is a second-order circuit.

The first step is to determine the resistance in dc, when all the caps are open. Clearly, when \$C_1\$ is open-circuited, the input impedance is infinite. However, for the sake of later rearranging the results, I place a finite-value resistance \$R_{inf}\$ across the current source. As such, the dc resistance is \$R_0=R_{inf}\$.

Now, to determine the time constants, I reduce the stimulus to zero. The stimulus here for this impedance determination is the current source \$I_T\$. Turning it off means open-circuiting it. This is shown in the below drawing and you have to determine the resistances seen from the caps connections in this mode:

enter image description here

Clearly, the first time constant \$\tau_1\$ is zero because of \$R_{inf}\$ is in the path. However, we keep \$R_{inf}\$ in the expression for further factorization. The second time constant \$\tau_2\$ involving \$C_2\$ requires a few equations. Finally, short \$C_1\$ and determine the resistance "seen" from \$C_2\$'s connections to form \$\tau_{12}\$. The denominator is equal to \$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}\$.

For the zero, you have to null the response. The response is the voltage \$V_T\$ generated across the current source. When it is nulled, the current source can be replaced by a short circuit. The new circuit from which time constants must be determined is now:

enter image description here

Doing this exercise also requires a few equations to obtain the time constants you want. I have gathered all the results and the transfer function in the below Mathcad sheet. Needless to say that simplifications and factorization can be undertaken especially to arrange the numerator with an inverted zero and keep a simple \$1 + s\tau\$ type of denominator. The goal is to unveil a leading term giving the resistance value obtained in the plateau. A little more work but it's too late here to proceed : )

enter image description here

The results look okay with the SPICE simulation:

enter image description here

\$\endgroup\$
0
\$\begingroup\$

Try this simplified schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Can you do it now?

\$\endgroup\$
  • 1
    \$\begingroup\$ to be honest, I don't quite understand what happened \$\endgroup\$ – Sargen Revlon Jan 3 at 16:08
  • \$\begingroup\$ Simplify assume all capacitor act as a short circuit. VCC as a GND and replace the transistor with his Hybrid-PI model \$\endgroup\$ – G36 Jan 3 at 16:12
  • \$\begingroup\$ I don't quite understand what happened Then you need to educate yourself on the subject of small signal models and small signal equivalent circuits. This isn't something that can be explained in a few words in an answer. You MUST have had the change to follow a class or course or read a textbook about this. \$\endgroup\$ – Bimpelrekkie Jan 3 at 16:17
  • \$\begingroup\$ I understand the need for simplification and shorting \$C_1\$ and \$C_3\$ is okay with me. However, I would certainly not short \$C_2\$ as it must be part of the result unless you want to compute the gain for \$s\$ approaching infinity? \$\endgroup\$ – Verbal Kint Jan 3 at 16:20
  • 1
    \$\begingroup\$ Oui, exactly, this is part of the circuit the student has to analyze. Besides, thinking back of it, if this is really \$Z_{in}(s)\$ that is needed, I am afraid \$C_1\$ has to be brought back in the circuit which now becomes 2nd order. I can have a look with the FACTs if needs be. \$\endgroup\$ – Verbal Kint Jan 3 at 16:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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