# Can someone how to find this voltage gain neglecting r0

I'm trying to figure out why my professor starts off the way he does when trying to find voltage gain. How he starts off is shown in the last picture. Any explanation would be appreciated.

• Assuming by "last picture" you mean the last four equations, which of them do you already understand? (Also, just as a note, you probably should have simply modified your earlier question to improve it rather than ask something quite similar with these additions.)
– jonk
Commented Feb 14, 2020 at 0:01
• Hey, thanks for replying. I don't understand why he starts by setting Vbe equal to Vs(rpi/rpi+Rs). I know it's voltage division but I'd assume you'd start with Vout being equal to an equation with Vs so you can divide the Vs and get Vout/Vs which gets you the gain. Don't understand why he starts with Vbe and then is able to go to Vo in the second equation. Commented Feb 14, 2020 at 0:19
• Eq1. 1st he derives Vbe based on divider action of Rs and Rpi. | Eq2. He then states a standard general equation for Vout | Eq3. He substitutes Vbe from Eq1 into Eq2. | Eq4 He then rearranges Eq3 to give Vout/Vin = .... . Commented Feb 14, 2020 at 0:31
• @Johanis Well, it's probably just easier to start at the left. Especially since the voltage across $r_\pi$ affects something on the right side. Eventually, you'll get the output worked out and when you put the full equation together you can re-arrange it for what you need. But to start, you really do want to know what the voltage across $r_\pi$ is and the first equation is fine for telling you that much. Of course, you could start with the second equation and then write the first one down. Do both of them make sense to you?
– jonk
Commented Feb 14, 2020 at 0:44
• @jonk thanks for clearing it up. So essentially it best to start with Vbe due to the fact that it allows us to include rpi in the equation and we want to include rpi because it effects the right side? Commented Feb 14, 2020 at 0:59