the question can be found here: http://www.mailfreeonline.com/uploader/04AEED12.jpg
i'm having issues with parts (a) and (d). the rest i think i've done correctly, but it would be helpful if someone could verify my answers.

here are my attempts:

(a) my guess is that the capacitors are there to filter out low-frequency components, and that the 2 resistors are there for biasing purposes. why it would be better to have the second as well as the first, i have no idea.

(b) i got R1 = 43 kO, R2 = 57 kO and R_E = 4.95 kO, but i don't understand why the current flowing through the 2 resistors has to be at least 1 order of magnitude bigger than I_b.

(c) i got my gain to be 1

(d) it has a very high input impedence and could be used as a buffer, right?


again, it's mainly parts (a) and (d) that i need help with. the rest just needs a little verification. thanks in advance

 

a) The capacitors are there to cutoff any DC current and to set up your bandwidth (cutoff frequencies). You are correct with the resistors being used for biasing. They are there to compensate for various β values. With this circuit biased correctly, the β can vary by a large degree, and will still have the same relative functionality. This is useful because transistors aren't manufactured identically. If you pick a transistor with a β that can vary by ±70, you're screwed without biasing, because the uncertainty will most likely offset your entire amp stage.

 

a) The capacitors are there to cutoff any DC current and to set up your bandwidth (cutoff frequencies). You are correct with the resistors being used for biasing. They are there to compensate for various β values. With this circuit biased correctly, the β can vary by a large degree. This is useful because transistors aren't manufactured identically. If you pick a transistor with a β that can vary by ±70, you're screwed without biasing, because the uncertainty will most likely offset your entire amp stage.

great answer. thank you.

 

another quick question -- do you know what the resistor at the emitter end of the transistor (R_E) is for?

 

Re limits current, and helps stabilize biasing with a bit of negative feedback.

 

Re limits current, and helps stabilize biasing with a bit of negative feedback.

i see. thanks.