VOLTAGE DIVIDER BIAS CIRCUIT

could anyone help with voltage divider bias.

need to find IC and VCE, if the following info is given:

VCC = 24V
BETA = 200
VBE = 0.7V
R1 = 68K
R2 = 33K
RC = 2.2K
RE = 1K

How do you mean we're supposed to help you without a chematic ?

well you can find the analysis in any good text book.

Since B*RE< 10*R2 we need to follow exact analysis.

Apply thevinin theorem at the input side which will make a voltage source of [R2/(R1+R2)]*Vcc and Rth = R1||R2.
apply Vth-vbe = Ib*Rth + Ie*RE
where u can put Ie= (1+b) Ib

solve for Ib, hence u will get Ic, Ie

vcc-vce= Ic*Rc solve them

The base current has a negligible effect on the voltage divider, we can mentally open the connection between the voltage divider and the base. We then get an voltage divider output of Vbb = (33k/(68k+33k))*24v so the base supply voltage equals Vbb = 7.84 volts. We then take the supply voltage minus the voltage base-emitter (.7 volts) to get the emitter voltage VE = 7.14 volts. To get the emitter current we divide 7.14 volts by 1k ohms to get 7.14mA. We know that emitter current is approximately equal to the collector current so this answers your first question. We then take the emitter/collector current times the collector resistor to get 15.7 volts. To get the voltage of the collector emitter we then subtract 15.7 volts from 7.14 to get 8.57 volts.