Diode Diode circuit Type of diodes Diodes 1

Diode (Diode circuit)

Type of diodes Diodes 1 – Power diode 2 – Zener diode 3 – Signal diode 4 – Light Emitting diode ID + VD Symbol

Diode I-V Characteristic Silicon knee voltage = 0. 7 V, Germanium knee voltage = 0. 3 V Note: ØIn actual case, current in reverse bias is not zero ØCurrent starts to flow at high voltage in reverse biased region

Modeling diode Shockley Equation Where Is is reverse saturation current VD is the forward biased voltage n is the ideality factor (usually equals to 1) VT is thermal voltage (usually 26 m. V) -- When VD is negative (reversed biased), ID ≈ -IS -- When VD is positive (forward biased), ID ≈

Diode Dynamic and Static Resistance • Diode Static resistance – Forward biased resistance of diode due to DC voltage source • Diode Dynamic resistance – Forward biased resistance of diode due to AC voltage source

Diode model Models Forward bias 1. Ideal – replace diode with switch 2. Practical – replace diode with voltage source, Si (0. 7 V), Ge (0. 3 V) 3. Complete – replace diode with voltage source, VT and internal resistance, rd Reverse bias Replace the diode with open circuit Note: *Open circuit; no current *short circuit: has current

Ideal Diode model

practical Diode model

Complete Diode model

Diode model Current Actual characteristics Ideal model practical model complete Open circuit Knee voltage, VT @VK Voltage

Diode circuit analysis Series configuration 1. Determine the state of the diode whether it is in ON or OFF state. • The applied Voltage is matched with the arrow of the diode symbol or not. • The VD > Vknee for the diode to be operated. 2. Substitute the equivalent circuit for ON or OFF diode. 3. Perform KVL to find node voltages 4. Perform Ohm’s Law to find current

Diode circuit analysis Series configuration Forward bias diode: ØUsing practical model diode: ØVK for Si = 0. 7 V and Ge = 0. 3 V ØIf voltage source (E) > knee voltage(Vk), diode is assume replaced by a battery 0. 7 V for Si or 0. 3 V for Ge. ( E = Vs ) Øhence, using • Kirchoff's voltage law (KVL): E = VR + VD • ID=IR , Ohm’s law : VR = IRR = IDR

Diode circuit analysis Series configuration Reverse bias diode: - VD + Current flow is approximately 0 A (ID=0 A) Using Kirchoff's voltage law (KVL): E = -VD + VR E = -VD + ID R where ID= 0 A

Diode circuit analysis Example 1 Series configuration Find VD, VR and IR in the circuit below. Using KVL, 3 V = 0. 7 V + VR VR = 2. 3 V VD = 0. 7 V Using Ohm’s Law,

Diode circuit analysis Example 2 Find VD, VR and IR in the circuit below. Since it is an open circuit, IR = 0 Using Ohm’s Law, VR = IRR = 0 Using KVL, VD = 3 V Series configuration

Diode circuit analysis Series configuration Example 3 Find VR and IR in the circuit below.

Diode circuit analysis Series configuration Example 3 cont. . Using KVL, 12 V = 0. 7 V + 0. 3 V + VR VR = 11 V Using Ohm’s Law,

Diode circuit analysis Series configuration Example 4 Find V 1, V 2 and Vo in the circuit.

Diode circuit analysis Series configuration Example 4 cont. . Using KVL, 10 V = V 1 + 0. 7 V + V 2 – 5 V 1 + V 2 = 10 – 0. 7 + 5 = 14. 3 V Using Ohm’s Law to get current Therefore, V 1 = IR 1 = 2. 07 m. A x 4. 7 kΩ = 9. 73 V V 2 = IR 2 = 2. 07 m. A x 2. 2 kΩ = 4. 55 V V 2 = Vo + 5 = 4. 55 V Vo = 4. 55 – 5 = -0. 45 V

Diode circuit analysis Example 5 Series configuration -- Find Vo 1 and Vo 2 in the figure below Solution: Vo 1 = -9 V Vo 2 = - 6. 6 V

Diode circuit analysis Parallel configuration

Diode circuit analysis Parallel configuration Example 1 Calculate the value of current when the diode is ideal.