Triode emitter calculation problem! Now the characteristic curve measured by experiment determines the method of finding Rb and Rc according to the curve! How to calculate

1 step? Negative feedback resistor without emitter? In this way, the static voltage of the C pole is about 0.5*Vcc? . This step needs you to understand? Think about it. Why? Only in this way can the maximum voltage amplitude be output.

Two steps? Make sure IC:? IC= 10mw/0.5VCC

Three steps? Determine RC: RC =? 0.5VCC/IC

Four steps? Find IB:Ib=IC/HFE?

Five steps? Find RB=(VCC-0.7)/Ib.

Example: suppose VCC= 10V? HFE=200

Really? IC = 0.0 1W/0.5 * 10V = 0.002 a

Rc = 0.5 *10v/0.002a = 2,500 euros.

IB = 0.002 a/200 = 0.0000 1A = 10uA

Rb = (10v-0.7)/0.00001a = 930k ohm.

The following is verified by software! Because the HFE value of triode in software is not easy to set! He is over 200. So when RB= 1230K, the static voltage of the C pole is about 0.5*Vcc? When the output voltage is close to 5Vp-p, a little distortion has been seen in the negative half cycle (oscilloscope 1V/ grid). If the static potential of the C pole is far away from 0.5*Vcc, the distortion of the output voltage will be smaller. It can be verified by experiments or mathematics.