Electronics: CE Transistor with Collector to Base Feedback

This article assumes a basic knowledge of transistor operation. It focuses on the circuit shown in Figure One. Note that the base current flows through the collector resistor R1 and then through the resistor R3.

First lets write the equation for the base current.

Assuming that Ic/Ie = 1

where
Ic is the collector current
Ie is the emitter current

equation one

Vcc = (Ic + Ib)* R1 + Ib*R3 + Vbe + (Ib + Ic)*R2

and

Vc = Vcc - (Ic + Ib)*R1

Where

Ic = collector current

Ib = base current

Vc = collector voltage

Vbe = voltage drop from base to emitter

To find the base current we substitute Ic = HFE * Ib into equation one

The base current is the current through R3.

Ib = (Vc - Vb)/R3

where

Vb = base voltage

The voltage across the resistor R4 is the voltage at the base of the transistor Q1. The resistors R1, R3 and R4 form a voltage divider.

The voltage across the resistor R4 is

Vr4 = Vb = (R4 * Vcc)/(R1+R3+R4)

Going from Vcc though R1, Vce and R2 to ground yields another equation.

Again assuming Ic/Ie = 1

equation two

Vcc = (Ic + Ib)*R1 + Vcb + Vbe + (Ic + Ib)*R2

where

Vcb is the voltage from collector to base

Now that we've established the equations, lets use thevenin's theorem to generate Thevenin's bias circuit.

We short out Vcc and calculate the resistance across R4:

Let R13 = R1 + R3

Then the Thevenin resistance seen across R4 is equal to

Rth = R13*R4/(R13+ R4)

The Thevenin voltage is the voltage across R4.

equation three

Vth = Vb = (R4 * Vcc)/(R1+R3+R4)

Hence, our new equivalent Thevenin circuit is shown in figure two.

The base current Ib can be calculated by dividing the Thevenin voltage by the Thevenin resistance.

Ib = Vth/Rth

The collector current Ic is equal the product of HFE and the base current.

Ic = HFE * Ib

where

HFE is the current gain

The collector current and the base current represent the quiescent point. However, the characteristics of this circuit are different from other popular common emitter circuits because of the dependence of the base current on the collector current.

Lets take a look at this via equations one and three.

Vcc = (Ic + Ib)*R1 + Vcb + Vbe + (Ic + Ib)*R2

Vb = (R4 * Vcc)/(R1+R3+R4)

Now look at figure three. Figure three shows an AC source. Assume that the AC voltage source is providing a ramp voltage to the base of the transistor Q1. Let's assume the base voltage increases by a fraction of a volt. This increase results in an increase in base current. The increase in base current results in an increase in collector current via the equation.

Ic = HFE * Ib

The increase in collector current and base current results in a larger voltage drop across R1 and hence a lower collector voltage via the equation

Vc = Vcc - (Ic + Ib)*R1

The lower collector voltage results in a lower base current via the equation

Ib = (Vc - Vb)/R3

Hence the amplification of the signal is decreased by the feedback.

This could be demonstrated in another way. Assume the AC voltage source is delivering a sine wave to the base of the transistor Q1. The common emitter circuit inverts the sine wave. This means that the amplified sine wave at the collector of Q1 is 180 degrees out of phase with the sine wave at the base. This sine wave at the collector of Q1 is fed back to the base of Q1 via R3. Hence the two sine waves add algebraically resulting in a lower amplitude sine wave at the base of Q1. This is commonly known as negative feedback.

References:

I have a Bachelor of Science In Electrical Engineering and worked as an Electronics Technician.

Electronic Principles: Third Edition
ISBN 0-07-039912-3