(Bipolar Junction Transistor) The Current Gain of Common Base Amplifier

[Terrycho]

Homework Statement

I think there is a typo in my textbook (Basic electronics for scientists and engineers)

Relevant Equations

Bipolar Junction Transistor

Everything is okay besides the current gain of this circuit, which is "g" there

The formula of current gain is I_out/I_in

If I call the current flows to R_e I_2, according to KCL, the following formula must be satisfied.

I_in + (β+1)I_b = I_2

Therefore, to know I_in, we should use the following

I_in = I_2 - (β+1)I_b

After a long calculation, I got g= - Re / (r_be + Re(β+1)) * Rc/(R_L+ Rc), which is the opposite sign of the value above in my textbook.

I cannot find the errata of the book, so I decided to ask here!

Thanks in advance.

 

[NascentOxygen]

If you examine the circuit with its basic biasing shown, you can see at a glance whether a low-frequency circuit is going to be inverting or non-inverting. Looking at the circuit attached, consider the moment when emitter voltage is slightly increased (such as by the positive peak of an AC component superimposed on the DC bias). This increase in emitter voltage reduces the base-emitter voltage, reducing the collector current. The reduced collector current causes less drop across the collector resistor, R_C in your image, in turn causing a rise in collector voltage. This increased collector voltage drives more current through the load, designated R_L in your images (and connected between collector and base). So the positive peak of input current coincides with the positive peak of current through the load: a non-inverting current amplifier.

Had we done things differently, and instead designated the load to be the collector resistor itself, as in my attachment, then because the positive peak of input current coincides with a drop in current through the collector resistor, we would say this altered arrangement to be an inverting amplifier of AC current.