0 like 0 dislike
 Lecture -3: Diode Operating Point Example - 1: When a silicon diode is conducting at a temperature of 25°C, a 0.7 V drop exists across its terminals. What is the voltage, VON, across the diode at 100°C? Solution: The temperature relationship is described by  VON (TNew) – VON(Troom) = KT (TNew – Troom) or,           VON (TNew ) = VON (Troom) + KT (Tnew – Troom) Given     VON (Troom) = 0,7 V, Troom= 25° C, TNew= 100° C Therefore, VON (TNew ) = 0.7 + (-2 x 10-3 ) (100-75)  = 0.55 V Example - 2: Find the output current for the circuit shown in fig.1(a).  Fig.1- Circuit for Example 2 Solutions: Since the problem contains only a dc source, we use the diode equivalent circuit, as shown in fig. 1(b). Once we determine the state of the ideal diode in this model (i.e., either open circuit or short circuit), the problem becomes one of simple dc circuit analysis. It is reasonable to assume that the diode is forward biased. This is true since the only external source is 10 V, which clearly exceeds the turn-on voltage of the diode, even taking the voltage division into account. The equivalent circuit then becomes that of fig. 1(b).with the diode replaced by a short circuit. The Thevenin's equivalent of the circuit between A and B is given by fig. 1(c). The output voltage is given by or, If VON= 0.7V, and Rf= 0.2 W , then Vo = 3.66V

IIT ROORKEE

0 like 0 dislike
Lecture - 3: Diode Operating Point

Example - 3

The circuit of fig. 2, has a source voltage of Vs = 1.1 + 0.1 sin 1000t. Find the current, iD. Assume that

nVT = 40 mV

VON = 0.7 V

 Solution: We use KVL for dc equation to yield Vs= VON+ ID RL  Fig.2

This sets the dc operating point of the diode. We need to determine the dynamic resistance so we can establish the resistance of the forward-biased junction for the ac signal. Assuming that the contact resistance is negligible Rf= rD Now we can replace the forward-biased diode with a 10 W resistor. Again using KVL, we have,

vs= Rf id + RL id The diode current is given by

I = 4 + 0.91 sin 1000 t mA

Since iD is always positive, the diode is always forward-biased, and the solution is complete.

IIT ROORKEE

0 like 0 dislike
Lecture - 3: Diode Operating Point

Small Signal Operation of Real diode:

 Consider the diode circuit shown in fig. 3. V = VD + Id RLVD = V- IdRL This equation involves two unknowns and cannot be solved. The straight line represented  by the above equation is known as the load line. The load line passes through two points,                          I = 0, VD = V and                  VD= 0, I = V / RL. Fig. 3

The slope of this line is equal to 1/ RL. The other equation in terms of these two variables VD & Id, is given by the static characteristic. The point of intersection of straight line and diode characteristic gives the operating point as shown in fig. 4. Fig. 4 Fig. 5

Let us consider a circuit shown in fig. 5 having dc voltage and sinusoidal ac voltage. Say V = 1V, RL=10 ohm.

The resulting input voltage is the sum of dc voltage and sinusoidal ac voltage. Therefore, as the diode voltage varies, diode current also varies, sinusoidally. The intersection of load line and diode characteristic for different input voltages gives the output voltage as shown in fig. 6. Fig. 6

In certain applications only ac equivalent circuit is required. Since only ac response of the circuit is considered DC Source is not shown in the equivalent circuit of fig. 7. The resistance rf represents the dynamic resistance or ac resistance of the diode. It is obtained by taking the ratio of Δ VD/ Δ ID at operating point.

Dynamic Resistance Δ rD = Δ VD / Δ ID Fig. 7

IIT ROORKEE