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Next: DC Biasing Up: ch4 Previous: Bipolar Junction Transistor (BJT)

Dynamics with AC input

The common-emitter transistor circuit is commonly used for voltage amplification, as shown in the example below. Here we assume $V_{cc}=15V$ and $R_C=1.5K\;\Omega$, and $\beta=40$.

transistoramplifier.gif

The input voltage and current

Assuming the transistor is properly biased so that $V_{be}=0.7\,V$, we get the input voltage $v_{in}(t)=v_{be}(t)$ as the superposition of DC component $V_{be}\,V$ and a small AC inpupt $v_{be}=0.02\;\cos\omega t\,V$:

\begin{displaymath}
v_{be}(t)=V_{be}+v_{be}=(0.7+0.02\;\cos\;\omega t) \;V
\end{displaymath}

and find the base current $I_b$ by the input characteristic plot:

transistorinput.gif

Due to the small dynamic range of the input voltage, the non-linear (approximately exponential) input characteristic can be linearized locally as a resistance, the reciprocal of the slope of the input characteristic curve around $V_{be}=0.7V$:

\begin{displaymath}
r_{be}=\frac{\triangle v_{be}}{\triangle i_b}=20\,mV/0.05\;mA=400 \;\Omega
\end{displaymath}

The base current $i_b(t)$ can be approximated to be

\begin{displaymath}
i_b(t)\approx \left(0.1+\frac{\triangle v_{be}}{r_{be}}\righ...
...ga t}{400}\right)\;mA
=(0.1+0.05\;\cos\;\omega t)\;mA =I_b+i_b
\end{displaymath}

which is also a superposition of a DC component $I_b=0.1\;mA$ and an AC component with an amplitude $0.05\;mA$. (Why can't we get the base current as $i_b(t)=v_{be}(t)/r_{be}$?)

The output voltage and current

The output characteristic plot is shown below:

transistorCEplots1a.gif

Here the $\beta$ value is $\beta=\triangle I_c/\triangle I_b=2\;mA/0.05\;mA=40$. The load line is the plot of equation $V_{ce}=V_{cc}-I_c R_C$, a straight line that goes through the two points:

\begin{displaymath}
(I_c=0,\;V_{ce}=V_{cc}=15V) \;\;\;\mbox{and}\;\;\;
(V_{ce}=0,\;I_c=V_{cc}/R_C=15V/1.5K\Omega=10 mA)
\end{displaymath}

The collector current $I_c$ and the output voltage $V_{ce}$ can be obtained either algebraically or graphically:

Note:

The waveform of the output $v_c(t)$ may be distorted if the DC component of the input voltage $V_{be}$ (and thereby, the base current $I_b$) is either too low or too high, causing the positive or negative peaks of the sinusoidal component to exceed the linear range of the output characteristic plot, as illustrated below:

InputOutputPlots4.png

We see that severe distortion in output $v_c$ will be caused if a transistor amplification circuit is working near either the cutoff or the saturation region. It is therefore desirable to properly set the DC operating point around the middle of the linear range along the load line, to avoid to be too close to either the saturation or cut-off region.

OperatingRegins.gif

Example

CEswitch.gif

Assume $V_{cc}=15V$, $R_C=1.5\;K\Omega$, $\beta=50$. Given the input voltage $V_1=V_{be}=0.2V,\;0.7V$ or $0.8V$, find the corresponding output voltage $V_2=V_{ce}=V_C$.

Conclusion: a change in input from 0.2V to 0.8V switches the output current from 0 to about 10 mA, and the output voltage from 15V to 0.2V, and the transistor is in cut-off, linear, and saturation region, respectively. $I_c=\beta I_b$ is only valid when the transistor is in the linear region.


next up previous
Next: DC Biasing Up: ch4 Previous: Bipolar Junction Transistor (BJT)
Ruye Wang 2014-08-02