Crystal diode and transistor amplifier circuit 7

If Δ R represents the alternating resistance of the emitter junction and R represents the load resistance, diode we can easily calculate the voltage amplification factor K. If Vo represents the output voltage generated by the collector current Δ Ic on the load R and Vin represents the input voltage, then

K=Vo/Vin=(△IcR)/(△Ie△R)≒R/△R

Due to the forward voltage applied to the emitter junction, a slight change in voltage can cause a significant change in the current flowing through the junction.diode Therefore, the emitter junction resistance △ R is very small, usually only a few tens of ohms. We know that the forward resistance of the P-N junction is very small, while the reverse resistance of the P-N junction is very large, so the resistance of the collector junction is very high, up to several hundred kiloohms. Therefore, the load resistance R can also be used very large (impedance matching). diode R is generally several thousand ohms to tens of thousands of ohms, so R/△ R is very large. Therefore, the output signal voltage Vc taken out from the load is much larger than the input signal voltage Vin, which is amplified many times. This is the principle of amplifying signals with a transistor.