OK,
So Vc=1v and Ve=1v, Vbb =1.7v
Assuming a transistor with a Vbe On of +0.7v.
With those values and the negative feedback action of the 1K in the Emitter, the transistor will be just out of saturation.
Do you follow that OK.?
You should now be able to prove that mathematically.
E
hi,
The middle of Vcut off and Vsat is Vsupply minus Vsat which is 12v -1v = 11v/2 = 5.5V
So you have to calculate the value of Collector current that will give a voltage drop across the 11k of 5.5V.
Assume the same current thru the 1k Emitter resistor, from that current you can calculate the Ve and adding a transistor Vbe 0.7v to Ve will give Vbb voltage required.
The Gain of the Common Emitter is Collector Resistor divided by the Emitter resistor, 11k/1k =11
hi hup,
This is an LTSpice simulation of the circuit.
Note: I have used a value of 0.6v for Vb not 0.7v, this is more accurate, the Vc swings close to 12v and 1v, ie: +/-5.5v
E
First, find the saturation current. Solve 12 - 11000*I = 1000*I which gives I = 1 ma for saturation current. One half of 1 ma is where the transistor should be for max dynamic range. Ve = 1000*0.5 ma = 0.5 volts. Add 0.7 gives 1.2 volts.