# What the RL as view by this circuit

simulate this circuit – Schematic created using CircuitLab

In this condition, RL=infinity so the voltage gain of the output at the node R5 and Q1 can be obtain as below

$$A_v=\frac{4800}{450}=10.67$$

Verified with simulator which is in good agreement

simulate this circuit

However, now the output is connected to a bypass capacitor of 5.3micro farad and my simulator showing a gain of only 3, understood as there is current flow into the 5.3micro farad capacitor that pull down the voltage gain and from youtube https://www.youtube.com/watch?v=-MyVscG-Pew&t=363s I understood the gain is gain is $$A_v=\frac{4800||R_{effective}}{r_e'+450}$$

Is the $$R_{effective}=R_7||R_8$$ so that theoretical value of $$A_v=3.73$$ or the calculation is actually more complicated than this?

– jonk
Commented May 27, 2021 at 6:29
• How about R7? Isn't in the analysis, the DC supply is replace by a short circuit? This is the part I not sure and need clarification Commented May 27, 2021 at 6:35
• R7 is a large value. It doesn't have the same impact. Not even close.
– jonk
Commented May 27, 2021 at 6:41
• Also, your added capacitor in the 2nd stage means a heavy load that is even worse that R7.
– jonk
Commented May 27, 2021 at 6:54
• Quick unrelated note: If it's not an exercice but something you actually want to build, you should ensure that the component values you choose do exist. (Look at the "E series of preferred numbers" on Wikipedia). Almost all the resistors and capacitors of your design don't exist at theses specific values. Commented May 27, 2021 at 7:04

The first thing to do is to work out the DC quiescent point for your 2nd stage. Assuming $$\\beta\approx 200\$$, I get $$\I_{\text{B}_2}=\frac{V_{_\text{TH}}-V_\text{BE}}{R_{_\text{TH}}-\beta\cdot R_\text{E}}\approx 5.5\:\mu\text{A}\$$ and therefore $$\I_{\text{E}_2}\approx 1.1\:\text{mA}\$$. This means $$\r_e^{'}\approx 24\:\Omega\$$. And if your frequency is $$\1\:\text{kHz}\$$, then $$\X_\text{C}\approx 5.3\:\Omega\$$. Reflected back to the base I get about $$\5.9\:\text{k}\Omega\$$. So the total load on the first stage is about $$\5.9\:\text{k}\Omega\mid\mid 26\:\text{k}\Omega\mid\mid 2.9\:\text{k}\Omega\$$ or about $$\\approx 1.81\:\text{k}\Omega\$$. This means that only about 27.4% of the 1st stage voltage gain is retained. About $$\27.4\%\cdot 10.67\approx 2.92\$$, or so.