1. Diode Applications

2. Shokley Equation where VT is ~25.8mV at room temp T= 300K or 27C
•The forward bias current is closely approximated by
where VT is ~25.8mV at room temp T= 300K or 27C
k = Boltzman’s constant = 1.38 x joules/kelvin
T = absolute temperature
q = electron charge = x 10-19coulombs
η = constant dependent on structure, between 1 and 2 (if not given assume n = 1)
IS= reverse saturation current
–Notice there is a strong dependence on temperature – 2

3. We can approximate the diode equation for i>> IS
In reverse bias (when v << 0 by at least VT), then
In breakdown, reverse current increases rapidly… can be approximated by a vertical line

4. Dynamic resistance

5. Load Line Analysis
The load of a circuit determines the point or the region of operation of a diode (or device)
The method: A line is drawn on the characteristic of the device. The intersection point gives the point of operation (called Q-point)

6. Load line
•Look at the simple diode circuit below. We can write two equations:

7. Diode in DC Series Circuit: Forward BiasVD
KVL: E – VD – VR = 0
E = VD + IDR
ID = IR= E/R → VD = 0V

8. Diode in DC Series Circuit:
Reverse Bias
E = VD + IDR →ID =0A
• VD = E
• VR = IDR = 0V VD 8

9. Example1
For the series diode configuration below, employing the diode characteristics of figure below, determine VDQ, IDQ and VR.
In general, approximate model of diode is used in applications because of non- ideal real life conditions (tolerance, temperature effect, etc) never
allow an ideal case to be applied 9 9

10. Solution Step1: Find the maximum ID. VD = 0V→ ID = IR= E/R= 10mA
Step 2: Find the maximum VD. ID =0A → E = VD + IDR=10V
Step 3: Plot the load line
Step 4 : Find the intersection between the load line and the characteristic curve. This is the Q-point
From curve : VDQ=0.78 V and IDQ=9.25mA
Step 5: Checking : VR=IRR=IDQR=9.25
Or VR =E-VD =9.22V 10 10

11. Example
Repeat the analysis of previous example using the practical diode model
VDQ=0.7 V and IDQ=9.25mA 11

12. Example
Repeat the analysis of previous example using the ideal diode model
VDQ=0 V and IDQ=10 mA

13. Repeat the analysis of Example 1 with R = 2 k
VD = 0V→ ID = IR= E/R= 5mA
ID =0A → E = VD + IDR=10V
VDQ=0.7 V
IDQ=4.6 mA 13

14. Example
Repeat the analysis of previous example using the practical diode model
VDQ=0.7 V
IDQ=4.6 mA 14

15. PWLDM (Piecewise Linear Diode Model)

16. SERIES DIODE CONFIGURATIONS WITH DC INPUTS
For the series diode configuration below, determine VD, VR and ID, given that E=8V and R=2.3 k VD
Solution: KVL
VR = E – VD = = 7.3V
ID = IR = VR/R = 7.3/2.2k
= 3.32m A Si 8V
2.2 k 16

17. Example Determine ID, VD2 and Vo for the circuit.
Remember, the combination of
short circuit in series with an
open circuit always results in
an open circuit and ID=0A.

18. Example
Determine I, V1, V2 and Vo

19. Example : Determine Vo and ID for the series cct below:
Vo=E-VD1-VD2= =11V
ID=IR=11/5.6=1.96 mA 19

20. Parallel Configurations
Solve this circuit like any Parallel circuit, knowing VD = 0.7V (or up to 0.7V) in forward bias and as an open in reverse bias.
VD1 = VD2 = Vo =0 .7V
VR = 9.3V
Diodes in parallel are used to limit current:
IR = E – VD = 10V -0 .7V = 28mA
R k
ID1 = ID2 = 28mA/2 = 14mA

21. Example
Determine the resistance R for the network when
I=200mA. Si Si

22. Example
Determine the currents I1, I2, and ID2 for the network

23. Diode as logic gate
OR Gate

24. AND Gate