BJT HELP

Common emitter crash course: First we have to setup a DC quiescent state. Set Re value to determine Ice. You'll also want to pick a value for the voltage at the emitter. You'll see 1V quite often, but it's by no means any kind of rule. If we know what Ice we want, if we know the beta the BJT (100 for many common, small signal transistors), we know that Ibe will be 100x less than that. As a general rule of thumb, put a factor of ten more than that through the voltage divider to be sure there is enough current to source. Our can say that we 10x less than Ice through the voltage divider. We'll call to of the voltage divider R1, bottom R2, and set Ibe via R1. There will also be a voltage drop of about 700mV across the base-emitter junction. To keep voltage at the emitter at the desired voltage, set the voltage between R1 and R2 to .7V above that. R1 Is already fixed, so the the voltage between the two via R2. Finally, set the collector voltage to half of V+ (7.5V for your example) via Rc. This will allow for an AC voltage swing that's (almost) rail-to-rail, or 0-15V. Now, you use caps to couple the input and output. Determine your input impedance (can look up how, I've written enough) to help with determining the input cap you'll need. And finally Av = Rc/Re. In your example, the gain is roughly 4mV/V, or for every 1V that you input, it outputs just 4mV.

thnx that was very helpful .. but I set betta to 122 DC offset to 5V etc

I want to know how to calculate the value of Re and Rc the values I used are useless

A beta of 122 isn't unreasonable. You just change your calculations a little. Is there a specific reason you went with that value? Also, is there a specific reason you have a 5V DC offset on the input? It doesn't really matter though, once the signal passes the the decoupling cap, the offset will be removed.

yeah that I know but its like a homework