E84 Home Work 10

1. The circuit shown in the figure contains a voltage source $V=10V$, two resistors $R_1=300\Omega$ and $R_2=200\Omega$, and a silicon diode. Find the voltage $V_D$ across and the current $I_D$ through the diode. Solve this problem in two different methods: (a) assume voltage $V_D=0.7$ (as the diode is always forward biased), and (b) use the graphic approach to find the intersection of the load line and the diode equation:
   
   $$I_D=I_0 ( e^{V_D/\eta V_T}-1 )$$
   
   
   Sketch the plot of the two curves and estimate the solution $(I_D,V_D)$ at their intersection. (Note that you can assume $I_0=10^{-10}$ and $V_T=0.026V$ at room temperature 300K.)

   

   Solution: As the diode is forward biased, the voltage across it is 0.7V, and the current through $R_2$ is $I_2=0.7V/200\Omega=3.5mA$, the current through $R_1$ is $I_1=(10-0.7)/300=31 mA$, and $I_D=I_1-I_2=31-3.5=27.5 mA$.

   Also, using Thevenin's theorem, we have $R_T=R_1\vert\vert R_2=300\vert\vert 200=120$, $V_T=10\times 200/(300+200)=4$, and then the linear equation
   

   $$I_D=\frac{V_T-V_D}{R_T}$$
   
   
   Plotting both equations, we get

   

2. The circuit shown in the figure contains a voltage source $V=10V$, three resistors $R_1=300\Omega$, $R_2=200\Omega$, and $R_3=100\Omega$, and two silicon diode. Find the voltage across the two parallel branches.

   

   Solution: As the diode is forward biased, the voltage across it is 0.7V. Assume the voltage in question is $x$, then we have the following equation:
   

   $$\frac{10-x}{300}=\frac{x-0.7}{200}+\frac{x-0.7}{100}$$
   
   
   which can be solved for $x$ to get the voltage to be $2.39V$. Alternatively, the voltage can also be found as
   
   $$\frac{R_2\vert\vert R_3}{R_1+R_2\vert\vert R_3}(10-0.7)+0.7=2.39$$
   
   

3. The circuit shown in the figure is a converter (adaptor) based on a full-wave rectifier, which gets an AC voltage input of 115V 60 Hz, and produces a 12V DC output. The voltage variation or ripple of the output should not exceed 5% when the load current is no more than 2 mA. Design the converter in terms of the turn ratio of the transformer and the value of the capacitor.

   

   Solution: This is a full-wave rectifier circuit, which turns the negative half cycle of the input to positive, so that the period of the output is $T=1/120=0.0083S$. The charge of the capacitor reduced during this period is $Q=IT=2A\times 0.083 S=0.0167 C$. If the voltage across the capacitor is dropped by less than $V_C=0.05\times 12=0.6V$, its capacitance has to be greater than $C=Q/V_C=0.0167/0.6=0.0278 F$.

   The peak voltage of input is $115\times \sqrt{2}=162.6$, the peak voltage of the output is $12V$. If the 1.4 V voltage drop across the two diodes is considered small enough when compared with the required 12 V, then the turn ratio of the transformer is $162.6/12=13.55$. Otherwise, the the turn ratio is $162.6/(12+1.4)=12.13$.

4. The input voltage to the circuit in the following figure is $v(t)=10 \sin (\omega t)$. The two DC voltages are both 5V. Sketch the waveform of the output voltage $V_{out}$.

   

   Solution: When $v(t)$ is in the range from -5.7V to 5.7V, neither of the two diode branches is conducting and no current is drawn from the voltage source. As the result, the output is identical to the input. When $v(t)>5.7$, the diode branch on the right is conducting and it will hold the output at 5.7V. When $v(t)<5.7$, the diode branch on the left is conducting and it will hold the output at -5.7V.

   

5. Assume each of the input voltages $V_1$ and $V_2$ takes one of two values, either 0V or 5V. Find the output voltage $V_{out}$ in the following combinations of two input. As always, all voltages are measured with respective to ground, although it is not explicitly shown.

   Solution: If at least one of the inputs is 0V, the corresponding diode is conducting and the output voltage is held to 0.7V. Only when both inputs are 5V, both diodes are cut off and draw no current from the voltage source. Since there is no voltage drop across the resistor, the output voltage is the same as the voltage source of 5V.

   This circuit can be used to implement logic AND, i.e. only when both input $V_1$ AND input $V_2$ are high (for true or logical 1), will the output be high (true or logical 1). Otherwise the output is low (false or logical 0).

   $V_1$	0V	0V	5V	5V
   $V_1$	0V	5V	0V	5V
   $V_{out}$	0.7V	0.7V	0.7V	5V