# How to derive transistor input impedance?

Given the following circuit,

When Ro is not infinite, from the book, ZB is given as,

However, when it's infinite, the input impedance is easily obtained as the parallel combination of RB and β(re+RE). Given that it's not infinite and I need to derive the input circuit to find the input impedance as well as ZB, how would I draw it and derive the formula to gain intuition? Similarly, the output impedance how would the circuit look?

• See the example of finding Zout = Vx/Ix using "test voltage techniques" electronics.stackexchange.com/questions/265046/…
– G36
Commented Apr 9, 2023 at 18:54
• Closely related question asked by the OP yesterday. Commented Apr 9, 2023 at 19:24
• May I ask you a question? What do you want to achieve? What is the point of finding Zb that includes the ro in the first place (and doing all this math)? When we never use this equation in the real world when we design the amplifier circuit.
– G36
Commented Apr 9, 2023 at 19:34
• @G36 I want to learn to derive the formula to gain intuition because the book has exercises with ro being not infinite. When ro is infinite Zb is β(re+RE), making the Zi = Rb || Zb or Rb || β(re+RE). I'm trying to find Zb when ro is not infinite and how that would change the circuit's Zi. With that how the input circuit would look like and gain another perspective. Commented Apr 9, 2023 at 22:07

To be able to find $$\Z_b\$$ that includes the translator output resistance $$\r_o\$$. We need to find the voltage at the $$\V_e\$$ node in the first place. Because I'm going to use the test voltage method to find $$\Z_b\$$

$$Z_b = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{\frac{V_{test} - V_e}{r_{\pi}}}$$

The circuit looks like this:

simulate this circuit – Schematic created using CircuitLab

I will use brutal force -> nodal analysis. And the math software will do the math part for me.

We have two nodes. Thus, the first equation:

$$\frac{V_e}{R_E} +\frac{V_e + V_{test}}{r_{\pi}} +\frac{V_e - V_o}{r_o} - \beta \times\frac{V_{test} - V_e}{r_{\pi}} = 0$$

And the second node equation:

$$\frac{V_o}{R_C} + \frac{V_o - V_e}{r_o} + \beta \times\frac{V_{test} - V_e}{r_{\pi}} = 0$$

And after we solve this

$$V_e = V_{test} *\frac{R_E (R_C +\left(\beta +1) r_o \right)}{r_o r_{\pi} + R_C (R_E + r_{\pi}) + R_E ( r_{\pi} +(\beta +1) r_o)}$$

$$V_o = V_{test} \times\frac{R_C\: R_E - R_C\: r_o \beta}{r_o\: r_{\pi} + R_C(R_E + r_{\pi}) + R_E ((\beta +1)r_o + r_{\pi})}$$

We plug the $$\V_e\$$ into $$\Z_b\$$ and solve it.

$$Z_b = \frac{(R_C + R_E + r_o)r_{\pi} + R_C R_E + R_E \:r_o + R_E\: r_o \:\beta}{R_C + R_E + r_o}\tag{Equation 1}$$

And now we can simplify it further but we have to do it manually because the software cannot give as a answer in a "human wanted form".

$$= \frac{(R_C + R_E + r_o)r_{\pi} + R_C R_E + R_E \:r_o + R_E\: r_o \:\beta}{R_C + R_E + r_o} = r_{\pi} + \frac{R_C R_E + R_E \:r_o + R_E\: r_o \:\beta}{R_C + R_E + r_o}$$

$$= r_{\pi} + \frac{\frac{R_C R_E}{r_o} + R_E + R_E\:\beta}{\frac{R_C + R_E}{r_o} + 1}$$ $$= r_{\pi} + \frac{\frac{R_C R_E}{r_o} + R_E\:(\beta + 1)}{\frac{R_C + R_E}{r_o} + 1}$$ $$= r_{\pi} + \frac{R_E \left(\frac{R_C}{r_o}+(\beta + 1)\right)}{\frac{R_C + R_E}{r_o} + 1}$$

And this is the end. Also, notice that this method does not give us any circuit intuition.

To gain some intuition we could try to apply the Miller theorem (a very strong simplification).

And view the $$\r_ o\$$ as a resistor with the smaller resistance $$\\frac{r_ o}{A_V}\$$ connect in parallel with the $$\R_E\$$ resistor.

simulate this circuit

And now we can clearly see why $$\Z_b\$$ decreases in value when we include $$\r_o\$$.

Equation 1 can be formed into: $$Z_{b}=r_{\pi}+\frac{R_{E}\left(R_{C}+r_{o}\right)}{\left(R_{C}+R_{E}+r_{o}\right)}+\frac{R_{E}r_{o}\beta}{\left(R_{C}+R_{E}+r_{o}\right)}$$ which reveals a circuit intuition of three resistances in series: $$Z_{b}=r_\pi+R_E//(r_o+R_C)+R_\beta$$ where $$R_\beta=\frac{R_{E}r_{o}\beta}{\left(R_{C}+R_{E}+r_{o}\right)}=R_\beta'\beta$$

The first two terms are easily found by removing the dependent current source ($$\\beta = 0\$$) in the OP's original diagram.

The resistance modulated by $$\\beta\$$ is almost $$\R_E//r_o\$$ but with the denominator increased by $$\R_C\$$, Not quite so intuitive but acceptable.

$$\R_\beta'\$$ can be further analyzed but may not add to intuition but is interesting none the less: $$R_{\beta}^{'}=\frac{R_{E}r_{o}}{R_{C}}//R_{E}//r_{o}$$

• G36, Do you mind if I add a circuit intuition to your answer? You have done the work that I would have done in my own answer. If you don't agree with the intuition you can roll back my edits Commented Apr 10, 2023 at 0:08
• @RussellH I see no problem with that. You can add it.
– G36
Commented Apr 10, 2023 at 7:26