Basic Electronics Concepts
Diodes
DIODE – PART 3 Lecture Delivered @ HITEC University By Atif M. Khokhar
Table of Contents Ideal diode Model to Practical Diode Model Q-Point Resistance Level Types of Diodes
Lecture Delivered @ HITEC University By Atif M. Khokhar
DIODE MODEL
Lecture Delivered @ HITEC University By Atif M. Khokhar
Diode Circuit Models The Ideal Diode Model
The diode is designed to allow current to flow in only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal diode approximation is acceptable.
Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias) RS = 50 a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so: ID ID = VA/RS = 5 V / 50 = 100 mA + VA 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. Lecture Delivered @ HITEC University By Atif M. Khokhar
Lecture Delivered @ HITEC University By Atif M. Khokhar
Diode Circuit Models The Ideal Diode with This model is more accurate than the simple ideal diode model because it includes the Barrier Potential approximate barrier potential voltage. voltage. Remember the barrier potential voltage is the + V voltage at which appreciable current starts to flow. 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 ID VA
+ _
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
V
+
ID = VA - V = 4.7 V = 94 mA RS 50
Lecture Delivered @ HITEC University By Atif M. Khokhar
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 transconductance curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant. For lowlow-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
Linear Portion of transconductance curve
RF
ID
RF = VD ID
VD VD
Lecture Delivered @ HITEC University By Atif M. Khokhar
Diode Circuit Models The Ideal Diode with Barrier Potential and Linear Forward Resistance
Example: Assume the diode is a lowlow -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:
ID VA
+ _
0 = VA – IDRS - V - IDRF V
+
ID = VA - V = 5 – 0.3 = 85 85..5 mA RS + RF 50 + 5 RF
Lecture Delivered @ HITEC University By Atif M. Khokhar
Diode Circuit Models Values of ID for the Three Different Diode Circuit Models
ID
Ideal Diode Model
Ideal Diode Model with Barrier Potential Voltage
Ideal Diode Model with Barrier Potential and Linear Forward Resistance
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. Lecture Delivered @ HITEC University By Atif M. Khokhar
The Q Point The operating point or Q point of the diode is the quiescent or nonosignal 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. RS = 1000 ID VA = 6V
First the load line is found by substituting in different values of V 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 – V
+ _
V
RS
+
Using V 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.
Lecture Delivered @ HITEC University By Atif M. Khokhar
The Q Point ID (mA) 12
10
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 mA..
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 0 .7
Lecture Delivered @ HITEC University By Atif M. Khokhar
1 .0
1 .2
1 .4
RESISTANCE LEVELS • As the operating point of a diode moves from one region to another the resistance of the diode will also change due to the nonlinear shape of the characteristic curve • The type of applied voltage or signal will define the resistance level of interest • Three different types of applied voltage – DC or Static Resistance – AC or Dynamic Resistance – Average AC Resistance
Lecture Delivered @ HITEC University By Atif M. Khokhar
DC or Static Resistance • The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time • The resistance of the diode at the operating point can be found simply by finding the corresponding levels of VD and ID • The lower current through a diode the higher the dc resistance level
Lecture Delivered @ HITEC University By Atif M. Khokhar
EXAMPLE 1.1 • Determine the dc resistance levels for the diode of Fig. 1.31 at • (a) ID = 2 mA • (b) ID = 20 mA • (c) VD = -10 V
Lecture Delivered @ HITEC University By Atif M. Khokhar
Figure 1.31
Lecture Delivered @ HITEC University By Atif M. Khokhar
AC or Dynamic Resistance • The varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage as shown in Fig. 1.1
Figure 1.1 Lecture Delivered @ HITEC University By Atif M. Khokhar
• A straight line drawn tangent to the curve through the Q-point as shown in Fig. 1.2 will define a particular change in voltage and current that can be used to determine the ac or dynamic resistance for this region of the diode characteristics • In equation form,
• In general, therefore, the lower the Q-point of operation (smaller current or lower voltage) the higher the ac resistance. Lecture Delivered @ HITEC University By Atif M. Khokhar
Figure 1.2 Determining the ac resistance at a Q-point.
EXAMPLE 1.2 For the characteristics of Fig. 1.3: (a) Determine the ac resistance at ID = 2 mA. (b) Determine the ac resistance at ID = 25 mA. (c) Compare the results of parts (a) and (b) to the dc resistances at each current level.
Lecture Delivered @ HITEC University By Atif M. Khokhar
Figure 1.3
Lecture Delivered @ HITEC University By Atif M. Khokhar
Lecture Delivered @ HITEC University By Atif M. Khokhar
Lecture Delivered @ HITEC University By Atif M. Khokhar
Average AC Resistance • If the input signal is sufficiently large to produce a broad swing such as indicated in Fig. 1.4, the resistance associated with the device for this region is called the aver-age ac resistance • The average ac resistance is, by definition, the resistance deter-mined by a straight line drawn between the two intersections established by the maximum and minimum values of input voltage
Lecture Delivered @ HITEC University By Atif M. Khokhar
Figure 1.4 determining the average ac resistance between indicated limit
Lecture Delivered @ HITEC University By Atif M. Khokhar
Lecture Delivered @ HITEC University By Atif M. Khokhar
Summary Table
Lecture Delivered @ HITEC University By Atif M. Khokhar