# Transfer Function of two Common Emitters

I'm sorry about the pictures and used circuitlab.

I am dealing with a circuit that has two common emitter transistors. Until now I was only dealing with a circuit that that is common emitter.

This is an example So when the following question comes : Draw the AC- circuit and find the transfer function of the function

I would draw the ac-circuit and then find the transfer function which is this I am faced with the following circuit so I am a bit lost on what would the final transfer function be • Can’t you find a transfer function from U1 to the base of T2. Then find another transfer function from the base of T2 to your output U2. Then, multiply them together for the overall transfer function. – relayman357 Feb 5 at 0:36
• The first circuit's voltage gain is more like $\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{V_T}\cdot\frac{R_\text{L}}{R_\text{C}+R_\text{L}\cdot\left(1+\frac{V_\text{CC}-V_{\text{C}_\text{Q}}}{V_\text{A}}\right)}$. – jonk Feb 5 at 5:09
• Do you need to include the two series capacitors? It is a second-order system then with a double zero at the origin. – Verbal Kint Feb 5 at 12:18
• I am not sure if he really wants the "TRANSFER FUNCTION". Most probably (have a look on his formulas) he is interested in the midband gain only. – LvW Feb 5 at 15:08

To determine this transfer function, many methods exist. I will use the fast analytical techniques or FACTs described in the book I wrote. The principle consists of individually determining the time constants associated with each energy-storing element. You do that to express the denominator which, is the present case, is defined as: $$\D(s)=1+sb_1+s^2b_2\$$. The numerator, where the zeroes lie, is determined using a generalized expression as follows: $$\N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{21}\tau_2\tau_{21}\$$

$$\H_0\$$ is straightforward: open all the caps and the output is 0 V then $$\H_0=0\$$.

To determine the time constants, you go through two small sketches in which you determine the resistance driving the capacitor. You do that using inspection, without writing a line of algebra. For $$\\tau_1\$$ and $$\\tau_2\$$, you examine the below sketch: Once you have the time constants on hand, you write $$\b_1=\tau_1+\tau_2\$$.

For $$\\tau_{21}\$$, capacitor $$\C_2\$$ is replaced by a short circuit and you determine the resistance driving $$\C_1\$$ in this mode: Then you form $$\b_2=\tau_2\tau_{21}\$$. This is it, you have the denominator which is of second order and considering a low-$$\Q\$$ approximation, you can factor it as $$\D(s)\approx \frac{1}{(1+\frac{1}{\omega_{p1}})(1+\frac{1}{\omega_{p2}})}\$$ with $$\\omega_{p1}=\frac{1}{b_1}\$$ and $$\\omega_{p2}=\frac{b_1}{b_2}\$$.

To determine the zeroes, you will alternatively set the capacitors in their high-frequency state (a short) while one remains in its dc state (open) and calculate the gain in this mode. If you do that, you find $$\H_1=H_2=0\$$. So the term in $$\s\$$ in the numerator is null. Then, to determine the second-order zeroes, put all the caps in their high-frequency state (a short) and determine the gain $$\H^{12}\$$ of the below circuit: In this case, $$\H(s)=s^2H^{12}\tau_2\tau_{21}\$$ reusing the time constants already determined. This is it, the transfer function is obtained and given in the below Mathcad sheet rearranged in a way to show the gain in the plateau region: remember, all of these techniques are for design-oriented analyses or D-OA, a term forged by the late Dr. Middlebrook.  The final transfer function involves two inverted poles which allow you to factor the plateau gain as the leading term since this is what is wanted. The terms could obviously be simplified if we ignore the biasing resistors but I wanted to show the complete expression for the sake of the exercise. As you could see, all was done by inspection, without a single line of algebra and this is the power of the FACTs.

• Glad to see you here. While I expect the OP to not grab up all this, I on the other hand will benefit from a new (for me) perspective to consider and learn from. Thanks! – jonk Feb 5 at 14:28
• Glad to read from you too : ) I agree that this is quite complicated considering the simple arrangement. But things get quickly nasty with a second-order system. If you neglect the biasing resistors, it should lead to a simpler result. – Verbal Kint Feb 5 at 15:14
• @Verbal Kint, thank you so much for the super detailed explanation. I was hoping for something more simple. I have added a picture of the way I have solved this part. I was sorry it is just a picture of my notes but I can help if something is not understandable .. Here is the link to picture PICTURE – be1995 Feb 5 at 15:24
• The mid-band gain you have derived is correct, I checked it against my Mathcad file when $R_{b2}$ goes infinite. It is difficult to do simpler than what I proposed as long as you consider the two capacitors introducing two zeroes and two poles. – Verbal Kint Feb 5 at 17:49

Your U1 has zero output impedance, so your maths of the 1-transistor circuit can ignore the input resistance at the base of the gain stage.

For the 2-transistor circuit, the input resistance of T2||rb2 must be included in the small signal equations.

• if I may ask what would the final answer be ? – be1995 Feb 5 at 11:02