E84 Home Work 11

1. A BJT transistor with $\beta=24$ is set up as a common-base configuration as shown in the figure below.
   • If it is known that $I_{CB0}=10 nA$ and $I_E=2 mA$, find $I_C$ and $I_B$.
   • If $V_{CB}=15V$ and $V_{EB}=0.7$, estimate $I_E$, $I_B$ and $I_C$ based on the plots below. Re-estimate these currents if $V_{EB}$ is increased to $0.8V$.

     Note: in the figure above, as the assumed polarities of both $I_E$ and $V_{EB}$ are the opposite to those assumed in plot (b) below, the negative signs for both $I_E$ and $V_{EB}$ in the plot should be dropped.

   

   Solution: $\alpha=\beta/(1+\beta)=24/25=0.96$, $I_C=I_E \alpha + I_{CB0}=2 mA \times 0.96+0.01=1.921 mA$, $I_B=I_C/\beta=1.92/24=0.08$, or $I_B=I_E-I_C=2-1.92=0.08 mA$.

   If $V_{EB}=0.7 V$ and $V_{CB}=15V$, $I_E=5 mA$ can be estimated from figure (b), and correspondingly, $I_C=4.8 mA$ and $I_B=0.2 mA$. When $V_{EB}=0.8 V$, $I_E=20 mA$, $I_C=19.2 mA$, and $I_B=0.8 mA$.

2. In the figure below, the transistor with $\alpha=0.99$ and $I_{CB0}=10^{-11} A$ is set up as a common-emitter circuit. The base-emitter pn-junction is forward biased with $I_B=20 \mu A$, $V_{CC}=10V$, and $R_C=2k\Omega$. Find $I_{CE0}$, $I_C$, $I_E$, $V_{CE}$, and $V_{CB}$. Is the collector-base pn-junction forward or reverse biased? (Assume the voltage across a forward biased pn-junction is $0.7 V$.)

   

   Solution:
   

   $$\beta=\frac{\alpha}{1-\alpha}=99$$
   
   
   
   
   $$I_{CE0}=(1+\beta) I_{CB0}=(1+99) 10^{-11}=10^{-9}$$
   
   
   
   
   $$I_C=\beta I_B+I_{CE0}=99 \times 2 \times 10^{-5} + 10^{-9}=1.98 mA$$
   
   
   
   
   $$I_E=I_C+I_B=1.98+0.02=2 mA$$
   
   
   
   
   $$V_{CE}=V_{CC}-R_C I_C=10-2 \times 10^3 \times 1.98 \times 10^{-3} \approx 10-4=6 V$$
   
   
   Since the base-emitter pn-junction is forward biased with $V_{BE}=0.7V$, we have $V_{CB}=V_{CE}-V_{BE}=6-0.7=5.3 V$, the collector-base pn-junction is reverse biased.

3. In the same CE transistor circuit above, now assume $V_{CC}=20V$, $R_C=1\;K\Omega$, and the transistor's output characteristics is shown below. Also assume $I_B=0.2\;mA$. Find $V_{CE}$, $I_C$ and the $\beta$ and $\alpha$ values of the transistor.

   

   Solution: The load line goes through the points $(I_C=0mA, V_{CE}=20V)$ and $(I_C=V_{CC}/R_C=20 mA, V_{CE}=0V)$. Its intersection with the curve $I_B=0.2 mA$ is approximately at $(I_C=10mA, V_{CE}=10V)$, where we can also find $\beta=I_c/I_B=10/0.2=50$ and $\alpha=\beta/(\beta+1)=0.98$.

4. The figure (A) below shows a common-emitter transistor applification circuit (silicon) with$\beta=100$, $V_{cc}=20V$, and $R_C=2\;K\Omega$.
   • The base current is $i_b(t)=(50+30 \cos \omega t)\;\mu A$. Find $i_c(t)$ and $v_c(t)$ as functions of time.
   • Repeat the above for $i_b(t)=(90+30 \cos \omega t)\;\mu A$.
   • Repeat the above for $i_b(t)=(10+30 \cos \omega t)\;\mu A$. However, note that the actual base current cannot be negative as it can only flow from base to emitter but not the other direction, i.e., all negative parts of $i_b(t)$ are zero.
   For each of the three cases above, sketch the output (collector) characteristics ($i_c$ vs $v_c$) as shown in class show the wave forms of $i_b(t)$, $i_c(t)$ and $v_c(t)$, following the example in the lecture notes:

   Note: As the convention in the schematics of transistor circuits, the bottom horizontal line is treated as the ground, and all voltages, such as $V_b$, $V_c$ and $V_e$ are measured with respect to the ground as the reference point.

   Hint: The relationship $I_C=\beta I_B$ is only valid in the linear region in the middle range of the load line. However, in the cut-off region (close to the horizontal axis) and the saturation region (close to the vertical axis), the above relationship no longer holds and the actual output current $I_c$ and $V_{ce}$ can only be found graphically in the output characteristic plot.

   

   Solution:

   • 
     
     $$i_c(t)=\beta i_b(t)=(5+3\cos \omega t)\;mA$$
     
     
     
     
     $$v_c(t)=V_{cc}-R_c i_c(t)=20-2000 (5+3\cos \omega t)\times 10^{-3} =10-6\cos \omega t$$
     
     

   • 
     
     $$i_c(t)=\beta i_b(t)=(1+3\cos \omega t)\;mA$$
     
     
     
     
     $$v_c(t)=V_{cc}-R_c i_c(t)=20-2000 (9+3\cos \omega t)\times 10^{-3} =2-6\cos \omega t$$
     
     
     Clipping happens during the negative half-cycle due to saturation.
   • 
     
     $$i_c(t)=\beta i_b(t)=(10+3\cos \omega t)\;mA$$
     
     
     
     
     $$v_c(t)=V_{cc}-R_c i_c(t)=20-2000 (1+3\cos \omega t)\times 10^{-3} =18-6\cos \omega t$$
     
     
     Clipping happens during the positive half-cycle due to cut-off.

   

5. In the circuit shown below, the two base resistors $R_B=4.3K\Omega$, the collector resistor is $R_C=1K\Omega$. Assume the two transistors have the same $\beta=100$ value and they each receive an input ($V_1$ and $V_2$) at either $0.2V$ or $5V$. Find the output voltage $V_{out}$ for the following combinations of inputs. (Hint: $5V$ input to a transistor will drive it to saturation.)

   $V_1$	$V_2$	$V_{out}$
   $0.2$	$0.2$
   $5.0$	$0.2$
   $0.2$	$5.0$
   $5.0$	$5.0$

   

   Solution: When $V_1=0.2V$, $T_1$ is cut-off. When $V_1=5V$, $I_{b1}=(5-0.7)/4.3=1\;mA$, $T_1$ is saturated with $V_{CE}=0.2V$ and $I_C=(5-0.2)/1=4.8\;mA$ (instead of $I_C=\beta I_b=100\;mA$). The same is true for $T_2$. From the table below we see that the circuit is a NOR (not OR) gate (high voltage for True and low voltage for False).

   $V_1$	$V_2$	$V_{out}$
   $0.2$	$0.2$	$5.0V$
   $5.0$	$0.2$	$0.2V$
   $0.2$	$5.0$	$0.2V$
   $5.0$	$5.0$	$0.2V$