Question 8.2: The Current of Diode Find the current, I, flowing through th...
The Current of Diode
Find the current, I, flowing through the diode of the circuit shown in Figure 8.9 when:
a. R = 1 kΩ
b. R = 10 kΩ
This problem can be solved using either of two methods.
Method 1: Because the voltage available to directly drive the diode is unknown (i.e., V_1-V_2 of Figure 8.7), an assumption must be made of the status of the diode (ON or OFF). This assumption is used to solve the problem, and then the validity of the assumption can be checked by observing the direction of the diode current. For example, when assuming the diode is ON, the assumption will be valid if the current through the diode flows from the anode to the cathode, because that situation is consistent with an ON diode.
a. R = 1 kΩ
Assume the diode is ON. Therefore, the voltage across it is:
V= V_\gamma=0.7 VAccordingly, the voltage across the 350-Ω resistor is 0.7 V and the current through it corresponds to:
I_2=\frac{0.7}{350}=2 mAThe current through R corresponds to:
I_1=\frac{12-0.7}{1}=11.3 mAApplying KCL, the current through the diode corresponds to:
I=I_1-I_2=11.3-2=9.3 mAThis current flows through the diode from anode to the cathode, which confirms the validity of the assumption.
b. R = 10 kΩ
Again, assume the diode is ON. Therefore, the voltage across it is:
V=V_\gamma=0.7 VAccordingly, the voltage across the 350-Ω resistor is 0.7 V and the current through it corresponds to:
I_2=\frac{0.7}{350}=2 mAThe current through R corresponds to:
I_1=\frac{12-0.7}{10}=1.13 mAApplying KCL, the current through the diode corresponds to:
I=I_1-I_2=1.13-2=-0.87 mAThe negative sign means that the current flows in an opposite direction to that shown in Figure 8.9 (i.e., from cathode to anode). The diode passes current only in one direction (i.e., from anode to cathode). Thus, the assumption of the diode being ON is not valid.
Now, consider another assumption: the diode is OFF. In this case: I = 0. Therefore, the currents I_1 and I_2 are equal and are given by:
I_1=I_2=\frac{12}{10+0.35}=1.1594 mAThe voltage across the diode corresponds to:
V = 1.1594 × 0.35 = 0.41 V
It is notable that the voltage across the diode is less than V_\gamma. This confirms that the diode is OFF.
Method 2: For simplicity, find the Thévenin equivalent circuit between the diode terminals.
a. R = 1 kΩ
The Thévenin voltage corresponds to:
V_{th}=12 \times \frac{0.35}{1+0.35}=3.11 VIn addition, the Thévenin resistance corresponds to:
R_{th}=\frac{1 \times 0.35}{1+0.35}=0.2593 k \OmegaThe Thévenin equivalent circuit is shown in Figure 8.10. Because 3.11 V is greater than V_\gamma, the diode is ON and the current through it corresponds to:
I=\frac{3.11-0.7}{259.3}=9.3 mAb. R = 10 kΩ
The Thévenin voltage corresponds to:
V_{th}=12 \frac{0.35}{10+0.35}=0.41 VIn addition, the Thévenin resistance is:
R_{th}=\frac{1 \times 0.35}{10+0.35}=0.3382 k \OmegaThe Thévenin equivalent circuit is shown in Figure 8.11.
Because 0.41 V is less than V_\gamma = 0.7 V, the diode is OFF and no current flows through it. Thus, I = 0 and V = 0.41 V.
