
#109
Nov2812, 03:38 PM

P: 133

thx for answer 



#110
Dec512, 03:26 PM

P: 133





#111
Dec512, 03:36 PM

P: 391





#112
Dec512, 04:01 PM

P: 133

 this your example i just simulate (do you see the foto the value) in quote 69 Vb = Ve + Vbe = Ic*(Re1+Re2)+ Vbe = 5mA* (180 + 14) + 0.65V = 1.62V If we assume Hfe = 150 I (I use transistor 2N3903 I don’t know other there are a lot of types) Ib = Ic/hfe = 5mA/150 = 34μA R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R2 = Vb/(10*Ib) = 4.7KΩ Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω 



#113
Dec512, 04:20 PM

P: 391

But your circuit is different than mine. Also you can read from simulation DC collector current.
And your circuit has a voltage gain equal to: Av = Vout/Vin = 44.33mV/2mV = 22.165[V/V] My example look like this: And has a voltage gain Av = 45.8V/V But we can easily change the voltage gain by changing the Re1 resistance. 



#114
Dec612, 06:24 AM

P: 133

can you explain how you can change the voltage gain by changing Re1  Joney do you remember i ask you to make one amplifier with voltage gain of 50, only we Re1 Without Re2, and you show me this example, look at quote 66, IF YOU HAVE TIME. Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω, and i think you did not show me complete example, becuse i am searchin here but i cant find, do you remember now Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω for the rest value i think of maybe i misunderstand , are the same like this circuit the last one.... if not ? PLEASE CAN YOU MAKE ONE EXAMPLE ONLY WITH Re1 PLEASE THNX 



#115
Dec612, 11:12 AM

P: 3,844

I did not read the whole thread. Remember when I left off, I showed you how to calculate the gain? It's the impedance seen at the collector divided be the impedance seen by the emitter? Adding C2 don't change this, the total emitter resistance is
r'e+ Re1+(Re2//Xc2) where [itex] X_{C2}=\frac 1 {j\omega C}[/itex] You change the Re2, you change the impedance on the emitter side and change the gain. You need to get the solution manual of Malvino and work through the problems one by one. I thought I left you in good hands already. There comes a point of time you just work on the problems one by one and struggle through it. You are spending too much time writing posts here instead of working through the problem in the book. These questions are in the book. 



#116
Dec612, 11:38 AM

P: 133





#117
Dec612, 06:22 PM

P: 3,844

Make sure you get the correct edition, if you manage to download a version you don't have, go on Amazon and find a used text book of that version. They are very cheap used. I went on Amazon to look for one for you just now and can't find one cheap at the moment. In fact I just ordered a copy of Malvino a few minutes ago just to keep it in my library collection because it's that good. I only paid $US 8.00 including shipping!!! But you can go on Amazon later and see whether they have a copy cheap. 



#118
Dec712, 06:02 AM

P: 133

i cant find, but is safe amazone site? can you order it with facture? of only with card 



#119
Dec712, 06:20 AM

P: 133





#120
Dec712, 11:05 AM

P: 3,844





#121
Dec712, 11:29 AM

P: 133





#122
Dec712, 11:42 AM

P: 3,844

Regarding the circuit. This circuit is a voltage divider bias using R1 and R2 to set up the bias voltage of 10(4.7K/26.7K)=1.76V for Q1 . The emitter of Q1 is about 1.06V. Read this in Malvino. R5 together with R4 is to set up the DC of about 5mA current through Q1 and re' is about 25/5=5ohm. But without C2, gain of the stage is Rc/(Re1+Re2+re')=5. That's very low to be useful.
C2 is to provide a low impedance path to bypass Re2 at higher frequency. With the C2, the gain of the stage is Rc/(re'+Re1+(Re2//Xc)). The impedance of C2 is [itex] X_C=\frac 1 {j2\pi f C}[/itex]. But this is complicated for you. So you can use approximation. 1) At very low frequency, Xc is very high, so you can ignore it. So the gain is Rc/(re'+Re`1+Re2). 2) At frequency where [itex] Re2=\frac 1 {2\pi f C}[/itex], the total resistance of Re2// C2 decrease and the gain of the stage start to rise as show in the graph. It is not important to know the exact frequency as the final gain is usually the important one. 3) As frequency goes higher, Xc is getting lower and lower. Re1<<Re2, you can simplify by ignore Re2. At frequency where [itex] \frac 1 { 2\pi fC}=Re2[/itex], the effect of C2 start to level out. At even higher frequency, C2 can be approximated to be a short circuit( 0 ohm). So at much higher frequency, the gain of the stage is Rc/(re'+Re1) only, as C2 is a short circuit and Re2 is shorted out. 4) the transition frequency where [itex] \frac 1 { 2\pi f C}=R_{e1}[/itex] is usually called [itex]f_c[/itex] where [itex] \frac 1 { 2\pi f_c C}=Re1[/itex] or [itex]f_c=\frac 1 {2 \pi R_{e1} C_2}[/itex]. You can see the point of fc where the graph of the gain start bending horizontal to level out. Hope this help. This is an approximation. For more accurate calculation, you really have to use complex number, but believe me, it's good enough. I use this all these years for my own design at work. 



#123
Dec712, 11:45 AM

P: 3,844





#124
Dec712, 12:33 PM

P: 133





#125
Dec712, 12:41 PM

P: 133





#126
Dec712, 03:19 PM

P: 3,844

For example, if C1 is 100uF, at 1KHz, Xc=1/(2\pi f C)=1.59ohm. 1.59 ohm is so much lower than 180 of Re2. When parallel with Re2, Re2 disappeared. So you only have Re1 and r'e left. 


Register to reply 
Related Discussions  
Transistor amplifier  Electrical Engineering  2  
Transistor AMPLifier HelP !!!!  Electrical Engineering  12  
Transistor AMPLifier HelP !!!!  Atomic, Solid State, Comp. Physics  9  
Transistor CE Amplifier  Engineering, Comp Sci, & Technology Homework  2  
Transistor Amplifier  Introductory Physics Homework  13 