1. Diode Circuit Analysis
Electronics 1 
Lecture 5
Diode Circuit Analysis
Ahsan Khawaja
Lecturer
Room 102
Department of Electrical Engineering

2. Small-signal model
Small-signal modeling is a common analysis technique in electrical engineering which is used to approximate the behavior of nonlinear devices with linear equations.
This linearization is formed about the DC bias point of the device (that is, the voltage/current levels present when no signal is applied), and can be accurate for small excursions about this point.

3. 27-2: The PN Junction Diode
Diodes Have Polarity
(They must be
installed correctly.)

4. The Ideal Diode
The ideal diode: (a) diode circuit symbol; (b) i-v characteristic; (c) equivalent circuit in the reverse direction; (d) equivalent circuit in the forward direction.

5. Diode Equation
where IS = reverse saturation current (A) vD = voltage applied to diode (V) q = electronic charge (1.60 x C) k = Boltzmann’s constant (1.38 x J/K) T = absolute temperature n = nonideality factor (dimensionless) VT = kT/q = thermal voltage (V) (25 mV at room temp.)
IS is typically between and 10-9 A, and is strongly temperature dependent due to its dependence on ni2. The nonideality factor is typically close to 1, but approaches 2 for devices with high current densities. It is generally taken as 1.

6. Diode Voltage and Current Calculations (Example)
Problem: Find diode voltage for diode with given specifications
Given data: IS=0.1 fA ID= 300 mA
Assumptions: Room-temperature dc operation with VT=0.025 V
Analysis:
With IS=0.1 fA
With IS=10 fA
With IS=0.1 fA , ID= 1 mA 6

7. Diode Current for Reverse, Zero, and Forward Bias
Reverse bias:
Zero bias:
Forward bias: 7

8. Diode Approximations
Three different approximations can be used when analyzing diode circuits.
The one used depends on the desired accuracy of your circuit calculations.
These approximations are referred to as
The first approximation
The second approximation
The third approximation

9. Diode Approximations
The first approximation treats a forward-biased diode like a closed switch with a voltage drop of zero volts, as shown in Fig.

10. Diode Approximations
The second approximation treats a forward-biased diode like an ideal diode in series with a battery, as shown in Fig.

11. Diode Approximations
The third approximation of a diode includes the bulk resistance, rB.
The bulk resistance, rB is the resistance of the p and n materials.
The third approximation of a forward-biased diode is shown in Fig.
Fig
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

12. Diode Circuit Analysis: Basics
Loop equation for given circuit is:
This is also called the load line for the diode. Solution to this equation can be found by:
Simplified analysis with ideal diode model.
Simplified analysis using constant voltage drop model.
Graphical analysis using load-line method.
Analysis with diode’s mathematical model.
Objective of diode circuit analysis is to find quiescent operating point for diode, consisting of dc current and voltage that define diode’s i-v characteristic. 12

13. Analysis using Ideal Model for Diode
If diode is forward-biased, voltage across diode is zero. If diode is reverse-biased, current through diode is zero.
vD =0 for iD >0 and vD =0 for vD < 0
Thus diode is assumed to be either on or off. Analysis is conducted in following steps:
Select diode model.
Identify anode and cathode of diode and label vD and iD.
Guess diode’s region of operation from circuit.
Analyze circuit using diode model appropriate for assumed operation region.
Check results to check consistency with assumptions. 13

14. Diode Circuit Models The Ideal Diode Model ID + _ RS = 50 
a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so:
ID = VA/RS = 5 V / 50  = 100 mA
b) With VA < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = 0. ID + VA _

15. The Ideal Diode with Barrier Potential
Diode Circuit Models
The Ideal Diode with Barrier Potential
This model is more accurate than the simple ideal diode model because it includes the approximate barrier potential voltage. Remember the barrier potential voltage is the voltage at which appreciable current starts to flow. + V
Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts (forward bias).
RS = 50 
With VA > 0 the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find ID:
0 = VA – IDRS - V
ID = VA - V = 4.7 V = 94 mA
RS  ID + VA _ + V

16. Diode Circuit Models
The Ideal Diode with Barrier Potential and Linear Forward Resistance
This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant.
For low-power germanium and silicon diodes the RF value is usually in the 2 to 5 ohms range, while higher power diodes have a RF value closer to 1 ohm. ID + V RF
Linear Portion of curve
RF = VD ID
ID VD
VD

17. The Ideal Diode with Barrier Potential and Linear Forward Resistance
Diode Circuit Models
The Ideal Diode with Barrier Potential and Linear Forward Resistance
Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts.
RS = 50 
Once again, write a KVL equation for the circuit:
0 = VA – IDRS - V - IDRF
ID = VA - V = 5 – 0.3 = 85.5 mA
RS + RF ID + VA _ + V RF

18. Diode Circuit Models ID
Values of ID for the Three Different Diode Circuit Models
Ideal Diode Model
Ideal Diode Model with Barrier Potential Voltage
Ideal Diode Model with Barrier Potential and Linear Forward Resistance ID
100 mA
94 mA
85.5 mA
These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.

19. Analysis using Ideal Model for Diode: Example
Since source is forcing positive current through diode assume diode is on.
Q-point is(1 mA, 0V)
Since source is forcing current backward through diode assume diode is off. Hence ID =0 . Loop equation is:
Q-point is (0, -10 V)

20. Two-Diode Circuit Analysis
Analysis: Ideal diode model is chosen. Since 15V source is forcing positive current through D1 and D2 and -10V source is forcing positive current through D2, assume both diodes are on.
Since voltage at node D is zero due to short circuit of ideal diode D1,
Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)
But, ID1 <0 is not allowed by diode, so ?.

21. 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)

22. Load line corresponding curve plus load line.
•Look at the simple diode circuit below and its
corresponding curve plus load line.

23. The Q Point
The operating point or Q point of the diode is the quiescent or no-signal condition. The Q point is obtained graphically and is really only needed when the applied voltage is very close to the diode’s barrier potential voltage. The example below that is continued on the next slide, shows how the Q point is determined using the transconductance curve and the load line.
First the load line is found by substituting in different values of VD into the equation for ID using the ideal diode with barrier potential model for the diode. With RS at 1000 ohms the value of RF wouldn’t have much impact on the results.
ID = VA – VD RS
Using VD values of 0 volts and 1.4 volts we obtain ID values of 6 mA and 4.6 mA respectively. Next we will draw the line connecting these two points on the graph with the transconductance curve. This line is the load line.
RS = 1000  ID + VA
= 6V _ + VD

24. The Q Point ID (mA) VD (Volts)
The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.
ID (mA) 12 10 8
Q Point: The intersection of the load line and the transconductance curve. 6
5.3
4.6 4 2
VD (Volts)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0.7

25. Example
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 25

26. 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 26

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

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

29. Repeat the analysis of the previous example 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

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

31. Example Remember, the combination of short circuit in series with an
Determine ID, VD2 and Vo for the circuit.
20V
Remember, the combination of
short circuit in series with an
open circuit always results in
an open circuit and ID=0A.

32. Example
Determine I, V1, V2 and Vo

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

34. 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
IR = E – VD = 10V -0 .7V = 28mA
R k
ID1 = ID2 = 28mA/2 = 14mA

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

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

37. Summary: Diode Approximations