# small-signal analysis

Discussion in 'Homework Help' started by zdzislavv, Jun 8, 2009.

1. ### zdzislavv Thread Starter Member

Jun 5, 2009
22
0
Hello!
Here there is what I want to solve: http://images38.fotosik.pl/135/35e73fda57d25875.jpg
I found some informations on Q-point stabilization: http://images43.fotosik.pl/139/09ebd31e0c12d268.jpg
And about small-signal analysis: http://images45.fotosik.pl/139/9a75b7c041376cf6.jpg
And finally I tried to draw scheme with substituting transistors to models (but some of wires are connected to both transistors - and I don't know what to do with them, e.g. Rc and Rb): http://images49.fotosik.pl/139/c2964a1dbbc28db2.jpg
I want to solve the whole exercise but the most important part for me is small-signal analysis. May you help me with drawing the proper circuit and give some advices so that I can try to finish this exercise?
Thank you very much in advance
Greetings!

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
If I am not-mistaken the correct small-signal schematics should look like this.
And good luck with the Q-point calculation, And remember that current flow through RB must be equal I=0.6V/Rb

I nie za trudne te zadanie dla ciebie?

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3. ### zdzislavv Thread Starter Member

Jun 5, 2009
22
0
Thanks very much, Jony130 .
I see that capacitors are shorted - that's right. But shouldn't Rg be connected directly to G (gate) of JFET, not through Rb? In Your schematics it looks like there is connection with no resistors between D (drain) of JFET and emitter of BJT. Shouldn't it be rather connection between D and collector through Rc? And what about diode, Rz and Re?
Greetings!

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Rg is directly connected to gate. Look at Uge voltage arrow.
For AC signal, Drain, Rc are shorted by rz=0Ω (nie mylić z "rz" ) of Zener diode, and emitter is short by Ce.

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5. ### zdzislavv Thread Starter Member

Jun 5, 2009
22
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http://images39.fotosik.pl/147/266ced692a1e4832.jpg

Thanks! I've got two questions. First one - why is this Rc placed here? In your schematic it is between C and E, ok. And there is just wire between D and E (they are the same point). In other words D-C is the same as E-C so Rc is placed in proper way between drain and collector. So I know why it is placed in this way - but what for? Wouldn't it be more natural to draw it in this way: http://images44.fotosik.pl/151/de2e3e90b40662a9.jpg?

The other question is connected with rather basic knowledge about circuit theory. Why cannot I replace diode with this voltage source as shown in my schematic (second photo)?

Greetings!

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Well maybe, but Zener diode has rz=0Ω and that means that Drain and upper side of Rc is short to the GND.

The same question you can ask about the BJT and base to emitter junction and Vbe=0.7V.
Take look on this link
http://www.edw.com.pl/pdf/k01/35_05.pdf
http://www.edw.com.pl/pdf/k01/37_02.pdf
http://www.edw.com.pl/pdf/k01/18_12.pdf