Small-signal Model for BJT

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In a Grounded-emitter Amplifier Circuit below

When we ignore the existence of AC signal, we have

{IBQ=VCCVBEQRbICQ=βIBQ\left\{\begin{array}{l}I_{BQ}=\frac{V_{CC}-V_{BEQ}}{R_b}\\I_{CQ}=\beta I_{BQ}\end{array}\right.

so we can have the static working point Q as


based on the output curve, at the working point Q, we can have a small-signal model ONLY FOR AC as

If you find it hard to remember, just remember that base and emitter serves as a resistor and will always be linked with a rber_{be}, and collecter and emitter serves as a current-controlled source and will be always linked with a power source. As rce>>0r_{ce}>>0 and sometimes even near \infty, we can totally ignore it.

Based on the working Q point and the signal source, we can easily have all the other component aside from rber_{be}, and rber_{be} is accually the slope of the output curve(VbeiBV_{be}-i_{B}), with what we know about BJTs, we have

iC=βibi_C=\beta i_{b}

so that with KCL we have


let's take a look at the equivalent resistence in a BJT

as the dosage concentration is a lot higher at collecter than emmiter, rbe>>rcr_{b'e}>>r_c. Also, when we don't take heat loss into account, rer_e can be ignored.

As a result, the rber_{be} for ibi_b shold be written as


we know rbbr_{bb'} for it is a constant for every BJT, so we only need to get rber_{b'e}, which is the slope for UBEIeU_{B'E}-I_e,As we know what the form of the current on a PN node is, we can have


based on the fact that rbbr_{bb'} is small compared ro rber_{b'e}, we can simplify as


derivate on both sides, we have


so that we can have


we can have


and totally simplify it into a small-signal model.

We can also know the parameters for perfermance, which the input resistence is Ri=Rb//rbeR_i=R_b//r_{be},the output resistence is Ro=RCR_o=R_C, and the amplification is

AV=uoui=βiBRC//RLiBrbe=βRC//RLrbeA_V=\frac{u_o}{u_i}=\frac{\beta i_BR_C//R_L}{i_Br_{be}}=\frac{\beta R_C//R_L}{r_{be}}