AC Signal Amplification

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 $\beta=40$.

The current and voltage on both input and output sides can be obtained either algebraically or graphically as shown below.

The input voltage and current

• Input voltage:

  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 $V_{BE}$ and a small AC input $v_{be}(t)=0.02\;\cos(\omega t)\,V$:

  $$V_{BE}+v_{be}(t)=0.7+0.02\;\cos(\omega t)$$ (22)

• Input current:

  Due to the small dynamic range ($40\,mV$) 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 $V_{BE}=0.7$:

  $$r_{be}=\frac{\triangle v_{be}}{\triangle i_b}=\frac{40\,mV}{0.1\;mA}=400 \;\Omega$$ (23)

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

  $$I_B+i_b(t) \approx 0.1+\frac{\triangle v_{be}}{r_{be}} =0.1+\frac{20\;\cos(\omega t)}{400} =(0.1+0.05\;\cos(\omega t))\;mA$$ (24)

  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 in the following way?

  $$I_B+i_b(t) =\frac{V_{BE}+v_{be}(t)}{r_{be}} =\frac{0.7+0.02\;\cos(\omega t)}{400} =(1.75+0.05\;\cos(\omega t))\;mA$$ (25)

The output voltage and current

• The DC operating point (Q-point):

  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:

  $$\left\{\begin{array}{l} I_C=0\\ V_{CE}=V_{CC}=15V\end{array}\right. \;\;\; \;\;\; \left\{\begin{array}{l} V_{CE}=0\\ I_C=V_{CC}/R_C=15V/1.5\,k\Omega=10\, mA \end{array}\right.$$ (26)

• The output current, with $\beta=40$:

  $$I_C+i_c(t)=\beta (I_B+i_b(t))=40\times (0.1+0.05\, cos(\omega t)) =(4 + 2\;\cos\;\omega t) \; mA$$ (27)

• The output voltage:

  $$V_C+v_c(t)=V_{CC}-R_C(I_C+i_c(t))=15-1.5 (4 + 2\;\cos(\omega t ) =(9-3\;\cos(\omega t)) \; V$$ (28)

Comparing the AC sinusoidal component $0.02\,\cos(\omega t)\,V$ of the input and the AC component $-3\,\cos(\omega t)\,V$ of the output, we see that the CE transistor ircuit is a voltage amplifier by which the input is amplified by $-3/0.02=-150$ times. Also, the negative sign indicates the output voltage ( $-\cos\;\omega t$) is $180^\circ$ out of phase compared to that of the input signal ( $\cos\;\omega t$), i.e., the circuit is a reverse amplifier.

Waveform distortion

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 either the positive or negative peaks of the sinusoidal component to exceed the linear range of the output characteristic plot, as illustrated below:

We see that severe distortion in output $v_c$ will be caused if a transistor 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 cutoff region. Specifically,

• $I_B$ needs to be large enough to avoid distortion due to the nonlinearity of the input characteristics (close to the cutoff region);
• $I_B$ and thereby $I_C=\beta I_B$ need to be small enough to avoid distortion due to the nonlinearity of the output characteristics (close to the saturation region).

Example

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 $V_2=V_{CE}=V_C$.

• $V_{BE}=0.2V < 0.7V$, the forward bias of BE PN-junction is insufficient for it to conduct current, we have $I_B=0$, $I_C=\beta I_B=0$, $V_2 =V_{CC}-I_C R_C=V_{CC}=15V$. The transistor is cutoff (the switch is open or open-circuit).
• $V_{BE}=0.7V$, the BE PN-junction is forward biased, we can find $I_B$ from the input characteristics, here assumed to be $0.1\;mA$, and get $I_C=\beta I_B=5\;mA$ and $V_C=V_{CC}-I_C R_C=15-5\times 10^{-3} \times 1.5\times 10^{3} =7.5\; V$. The transistor is in linear region.
• $V_{BE}=0.8V>0.7V$, the BE junction is forward biased, we can find $I_B$ from the input characteristics, here assumed to be $0.4\;mA$. If the linear relationship $I_C=\beta I_B$ were to hold, we would get $I_C=\beta I_B=20\;mA$ and $V_C=V_{CC}-I_C R_C=15-20\times 10^{-3} \times 1.5\times 10^{3}=-15 V$. This result is obviously wrong, indicating that the transistor is actually in the saturation region (the switch is closed or short-circuit), i.e., the linear relation $I_C=\beta I_B$ does not hold. In fact, it is impossible for the transistor to draw $I_C=20\;mA$ from the voltage source, as the maximum current is $I_C=V_{CC}/R_C=15\;V/1.5\; k\Omega=10\;mA$ when $V_{CE}=0$. In this case, the actual $V_{CE}$ can be approximated on the output characteristics to be about $0.2V$, the intersection of load line and the curve corresponding to $I_B=0.4\;mA$), and $I_C=(15-0.2)/1.5\approx 10\;mA$.

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 cutoff, linear, and saturation region, respectively. $I_C=\beta I_B$ is only valid when the transistor is in the linear region.