How to determine ICQ and voltage across load in common emitter?

[zhi_yi]

x'cuse me, I have questions about transistor configuration,
1. if we want to for example create a common emitter with BC547, of course we need Vcc, beta value, and Icq, am I right or wrong? please correct me if i does any mistakes, okay, the question is, how can i know the value of Icq ? if the beta, we've already known it from the databook of BC547, it's said the beta value is 50, and then breakdown voltage for example 30 V, so I use 15 V, and then in the databook told me that the Icq max is for example 50mA. so, how much does the Icq should I use for determine the value of Rc, Re, R1, and R2?? (after i've known the Icq, i can determine the value of Ve = 0.1Vcc, and then i can determine the value of Re = Ve/Icq, and then Rc = 4.Re, and then vb, ib, and then R1 and R2)

2. which one should we determine at first? is it the value of Icq, and then I determine how much the gain of the circuit I want it to resulted. and then I determine the R, or i've to determine the value of the resistor for have the value of Icq?

thanks for the answer ^^

 

[Humungus]

Re: about common emitter

You are forgetting the swing of your output voltage as well as the maximum current you load will sink. So, make your Icq~1.05 times your maximum load current. Then, make your Ve~1.05 times your maximum voltage across the load. Now continue the biasing of the transistor using beta min.

 

[zhi_yi]

Re: about common emitter

thanks for the answer ^^.. how do i know the maximum voltage across the load? should i determine it myself?

and.. why it should be 1.05 times???

thanks ^^

 

[baza]

Re: about common emitter

You'll choose Icq in function of your application purpose.
Work at low signal ( the variable component should be 10% or less of the continous application ).

It depends where you'll use BC547 ( amplifier, mixer, SMPS etc etc )

 

[zhi_yi]

Re: about common emitter

i want to use it in amp, how to determine the icq? still i can't understand it >_<

 

[zhi_yi]

Re: about common emitter

please help me... please give me the answer >.<

and i have another question, in transistor configuration,
                                   ____ Vcc
               _____________|
              |
              >
              <R1
              >              C
Vin     C |        B | /
o------|(-----------|/
              |           |\
              |              *E
              >
              <R2
              >
               |

if the current from Vin is flow to that circuit, all the current will flow only to basis, but they neither flow to R1 nor R2, why? is it because there is not resistance present in basis, and there is resistances in R1 and R2, so the current will flow through to the lowest resistance path, the basis, is it?

what is the Rb here? is it R1//R2??

what makes the AC current from Vin cannot flow to the circuit if i doesn't use C?

can we use a non polar capasitor as a rectifier for make a power supply?

please help me, thanks for the answer ^^

 

[baza]

Re: about common emitter

Hmm, let's see:

Rin=(R1||R2)||hie , where hie is the impedance between base and emitor in common emitor mode.
hie=β/gm, where gm=Ic/Vth, where Vth=K*T/q.
gm=40*Ic ( mA/V ) @ 25 Celsius ( ambient temperature )

The voltage gain is about Av = Vo/Vi= -gm*Rc, where Rc is the collector resistance.

U may use non-pol. caps, but the maximum value for them is about 10uF, and they're relatively big. If u use electrolytic caps, watch at the polarity of signal in c.c.

Xc=1/ωC, where C- capacitance, see the lowest frequency u need and calculate the C value for Xc= 10 - 100 ohms ( use a value in this range ).

If you need high input impedance, try boostrap technique, or use a FET.
The d.c. current in the R1-R2 divider should be about 5 - 10* Ib, where Ib=Ic/β
Make a Kirchoff II on the R1-R2 -base-emiter eye, give a value for R1 or R2 ( according to your input impedance, and dc voltage in the base of transistor ( about 0.6-0.7 volts ) ) and find the value for R2 or R1.

Hope this is useful

Regards,

baza