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Dynamics with AC input

Transistor circuit (common-emitter configuration) is commonly used for voltage amplification, as shown in the example below.

transistoramplifier.gif

Here we assume $V_{CC}=15V$ and $R_C=1.5K\;\Omega$, and $\beta=40$. Given the input voltage $V_{BE}$, the base current $I_B$ can be found by the input characteristic plot as shown below:

transistorinput.gif

From the plot above we see that the voltage between B and E is a superposition of DC component $V_{BE}=0.7$ and an AC component $v_{be}=0.02\;\cos\;\omega t\;V$:

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

The resistance, the reciprocal of the slope of the curve around $V_{BE}=0.7V$, is approximately $r_{be}=40 mV/0.1\;mA=400 \;\Omega$. Correspondingly the base current is also a superposition of a DC component $I_B=0.1\;mA$ and an AC component with an amplitude $20\;mV/400\Omega=0.05\;mA$:

\begin{displaymath}I_B+i_b=(0.1+0.05\;\cos\;\omega t)\;mA \end{displaymath}

(Why can't we get $I_B=0.7V/400\Omega=1.75\;mA$?)

The output characteristic plot is shown below:

transistorCEplots1a.gif

Note that the $\beta$ value is $\beta=\triangle I_C/\triangle I_B=2\;mA/0.05\;mA=40$. The load line is the 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 found either algebraically or graphically:

Note:

Switch Circuit

From the current-voltage plot of the output characteristics, we see that the operation of a transistor can be in one of the three possible regions:

Severe distortion in output $v_c$ will be caused if a transistor amplification circuit is working near either the cutoff or the saturation region, as can be seen in the following sections.

OperatingRegins.gif

Example

CEswitch.gif

Assume $V_{CC}=15V$, $R_C=1.5\;K\Omega$, $\beta=80$. Given the input voltage $V_1=V_{BE}=0.2V,\;0.7V$ or $0.8V$, find the output voltage $V_2=V_{CE}$.

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 2014-04-15