For the common emitter amplifiers why is the voltage at emitter always less than the voltage at base??

 

Here is a link into our Ebook - http://www.allaboutcircuits.com/vol_3/chpt_4/5.html

 

The emitter-base junction is basically a diode. Current passes through a diode in only one direction. The voltage drop is due to the diode's resistance and is about .6v for silicon and .2 volts for germanium. The resistance is given by the thermal voltage divided by the current. At room temp this is about 26mV and at a current of 1mA we get a diode resistance of about 26 ohm.

 

The emitter-base junction is basically a diode. Current passes through a diode in only one direction. The voltage drop is due to the diode's resistance and is about .6v for silicon and .2 volts for germanium. The resistance is given by the thermal voltage divided by the current. At room temp this is about 26mV and at a current of 1mA we get a diode resistance of about 26 ohm.

The resistance you are describing is the small-signal or dynamic resistance. It does not explain the ≈0.6V Vbe. The forward drop can be calculated from the diode equation as a function of current, if you know the saturation current and the emission coefficient.

 

Thanks Ron. I stand corrected. But Electricshock, the diode resistance with the base current passing through it is what reduces the voltage.

 

Thank you got it