Why does emitter degeneration create negative feedback?

[likephysics]

I'm trying to understand emitter degeneration.
Instead of the entire input applied across just the VBE junction, it is now applied across the VBE junction plus an emitter resistor RE.
So when collector current increases, some additional voltage is dropped across RE, so that voltage across VBE does not increase by a lot.
Let's say Ic is 1mA, RE is 1k. Voltage across RE is 1v, VBE is 0.650v
Now Ic increases by say 10% (due to beta change or whatever), Ic=1.1mA, Voltage across RE is 1.1v, what's VBE? How do I calculate VBE.
I don't understand why it's called negative feedback.
Anything that reduces collector current is called negative feedback?

 

[technician]

Your figures provide some indication of what will happen.
Your 1.1volt increase due to an increase in collector current suggests that Vbe would decrease to 0.55V.
This tendency of Vbe to decrease will tend to make the collector current DECREASE.
i.e the emitter resistor tends to prevent changes in collector current.
This is what negative feedback means and it a way of stabilizing the operating point (bias) of the transistor

 

[yungman]

I'm trying to understand emitter degeneration.
Instead of the entire input applied across just the VBE junction, it is now applied across the VBE junction plus an emitter resistor RE.
So when collector current increases, some additional voltage is dropped across RE, so that voltage across VBE does not increase by a lot.
Let's say Ic is 1mA, RE is 1k. Voltage across RE is 1v, VBE is 0.650v
Now Ic increases by say 10% (due to beta change or whatever), Ic=1.1mA, Voltage across RE is 1.1v, what's VBE? How do I calculate VBE.
I don't understand why it's called negative feedback.
Anything that reduces collector current is called negative feedback?

First, it is Ie that you need to look at, not Ic. Ic is only the result of Ie.

Vbe vary logarithmilly with current, that is: [itex] ΔV_{BE}= V_T\;ln(\frac {I_{E1}}{I_{E2}})\;[/itex] where VT is about 25mV at 25 degree C. As you can see, going from 1mA to 2mA only increase the Vbe by about 25mV!

You can see if you increase emitter current by only 10%, the change [itex] ΔV_{BE}\;[/itex] is very small. So just consider it unchanged compare to the voltage across the emitter resistor.

 

[jsgruszynski]

The easiest way to see it is realize you have a voltage divider between Vbe and Re. Increasing current increases the drop across Re so you have reduced Vbe. Reduced Vbe, from the diode equation, means you've reduces the minority carrier injection into the base region which decreases Ic which decreases Ie (Ie=Ic+Ib). Ergo: negative feedback loop.

 

[es1]

Let's say Ic is 1mA, RE is 1k. Voltage across RE is 1v, VBE is 0.650v
Now Ic increases by say 10% (due to beta change or whatever), Ic=1.1mA, Voltage across RE is 1.1v, what's VBE? How do I calculate VBE.

For all these calculations you have to assume Vb is fixed.
So using the data from the first point: Vb = V(RE)+VBE=1V+0.65V=1.65V.

For the case where Ic increases to 1.1mA we use the same equation:

Vb = V(RE) + VBE -> 1.65 = 1KΩ*1.1mA + VBE -> VBE =0.55V

Notice VBE, the control signal, decreased for an increasing output signal, Ic. This is negative feedback. Consequently, the decreasing VBE will reduce Ic correcting for the disturbance.