Simple transistor feedback

Thread Starter


Joined Feb 8, 2005
studing a little of transistor biasing history and the uses of negative feedback i find a topic i dont understand,

i attached a simple circuit with a feedback resistor inserted into the emiter side of the transistor,

i will type exactly what the book "says":

"Historically, the first attempt at stabilizing the Q point was emitter-feedback bias, Notice in the figure that an emitter resistor has been added to the circuit. the basic idea is this: if Ic increases, Ve increases causing Vb to increase.,(until here i'm fine).
More Vb means less voltage across Rb" . this results in less Ib.<-- here starts my problems:

Vb is the voltage across Rb minus the 0.7 voltage drop across the emitter diode, i'm ok?, so if originally the collector current is 20 ma and the emmiter current is +/- 20 ma, and the voltage across the Er is 3 v, this means that the Vb is about 3.7 v that is the same across the Br .i'm ok until here?.

now suppose a rise in temperature makes the Ic rise to 30ma so the Ie is about 30ma and now the Ve is 5 v, and Vb is 5.7 volts that equals Rb voltage, checking ohms law "more voltage across the resistor = more current" so why in the book the autor states: "More Vb means less voltage across Rb"

i think my confution is with the Rb voltage and the Vb.

thanks for any help and please excuse my poor english


Joined Apr 26, 2005
Using your schematic .....

Vb is the voltage measured between the base and ground.
Ve is the voltage measured between the emitter and ground.
Vbe you established as 0.7 and is the voltage betwen the base and the emitter.

As Ic increases you increase Ie.
As Ie increases, the voltage drop across Re increases and Ve increases.
Ve + .7 = Vb so Vb increases.
As Vb increases, the voltage drop across Rb becomes less and Ib decreases.
If Ic = Ib * Beta, and Ib decreases, so does Ic.

Does that help?


Joined Jan 23, 2006

As you correctly stated, the purpose of the emitter resistor is stability. Think about that transistor circuit amplifying a sinewave. If the increase in temperature cause the transistor to saturate then the sine wave will be distorted. In order to accurately reproduce the sine wave at the collector, the transistor must not saturate.

Lets consider the rising edge of the sine wave and assume that the transistor goes into saturation before the instantaneous voltage at the base reached the peak value of the sine wave. A saturated transistor can not react to increases in base current. The common emitter circuit inverts the signal so, at the collector, the bottom part of the sine wave would be flattened. If the sine wave is an audio signal, this would result in poor quality sound.

Your analysis is correct. If the temperature causes the current Ic to increase, the increase in voltage accross Re will decrease the base current thereby bringing the collector current back down and keeping the transistor out of saturation.