# Ch2 Diode SEE2063

Document Sample

```					Chapter 2 DIODE Part 1 Resistance Levels

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 1

I-V Characteristics : Silicon vs. Germanium
Peak Inverse Voltage (PIV) is the maximum reverse-bias Potential that can be applied before entering the Zener g region. Si: PIV ~ 1000V Ge: PIV ~ 400V Si: temperature up to 200 C. Ge: temperature below 100 C. Si: VT ~ 0.7V Ge: G VT ~ 0 3V 0.3V

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 1

1

Resistance Levels
i) DC or Static Resistance
The application of DC voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not g change with time. The dc resistance levels at the knee and below will be greater than the resistance levels obtained for the vertical rise section of the characteristics. The lower the current through a diode the higher the dc resistance level.

Resistance, RD at the operating point

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 2

Example : DC or Static Resistance
Q: Determine the dc resistance levels for the diode at (a) ID= 2 mA (b) ID = 20mA (c) VD = -10V 10V Solution

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 3

2

ii) AC or Dynamic Resistance
If a sinusoidal voltage(signal) is applied,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. With no applied varying signal the point of operation signal, would be the Q-point (quiescent) determined by the applied dc levels. Quiescent: “still” or “unvarying” AC Resistance, rd

The lower the Q-point of operation (smaller current or lower voltage) the higher the AC resistance.

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 4

Example : AC or Dynamic Resistance

Solution

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 5

3

iii) Average AC Resistance
If the input signal is sufficiently large to produce a broad swing, the resistance associated with the device for this region is called the average ac resistance. The average ac resistance is the resistance determined by a straight li d d t i db t i ht line drawn b t between the two intersections established by the maximum and minimum values of input voltage.

maximum

minimum

The lower the level of currents used to determine the average resistance the higher the resistance level.

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 6

Summary of Resistance Levels

SEE 2063

Chapter 2 DIODE, Part 1 Resistance Level 7

4

Chapter 2 DIODE Part 2 Diode Equivalent Circuits

Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 1

An equivalent circuit is a combination of elements properly chosen to best represent the Actual terminal characteristics of a device, system, or such in a particular operating region.

i) Piecewise-Linear Equivalent Circuit
One technique for obtaining an equivalent circuit for a diode is to approximate the characteristics of the device by the straight-line segments. For a sloping section,the average ac resistance is the resistance level appearing in the equivalent circuit of the actual device.

The ideal diode is included to establish that there is only one direction of conduction through a device and device, a reverse-bias condition will result in the open-circuit state for the device. Since a diode does not reach the conduction state until VD reaches 0.7V with the forward bias, a battery VT opposing the conduction direction must appear in the equivalent circuit.

Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 2

5

ii) Simplified Equivalent Circuit
For most applications, the resistance rav is sufficiently small to be ignored in comparison to the other elements of the network.

It states that a forward-bias silicon di d i an electronic system under d t t th t f d bi ili diode in l t i t d dc conditions has a drop of 0.7V across it in the conduction state at any level of diode.

Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 3

iii) Ideal Equivalent Circuit
Let us take it a step further and establish that a 0.7V level can often be ignored in comparison to the applied voltage level.

Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 4

6

Summary of Diode Equivalent Circuits

Chapter 2 DIODE, Part 2 Diode Equivalent Circuits 5

Chapter 2 DIODE Part 3 Diode Notation and Testing

Chapter 2 DIODE, Part 3 Diode Notation and Testing

1

7

The notation frequently used for semiconductor diodes.

Various type of junction diodes.

Chapter 2 DIODE, Part 3 Diode Notation and Testing

2

“ON” state :
The meter has an internal constant current source (~2mA) that will define the voltage level of “~0.67V”.

Digital Tester

“OL” indication:Open

Chapter 2 DIODE, Part 3 Diode Notation and Testing

3

8

Curve Tracer

Chapter 2 DIODE, Part 3 Diode Notation and Testing

4

Chapter 2 DIODE Part 4 Load Line Analysis

Chapter 2 DIODE, Part 4 Load Line Analysis

1

9

The applied load will have an impact on the point or region of a device. A line can be drawn on the I-V characteristics of the device that represents the applied load.
Kirchoff’s voltage law

Chapter 2 DIODE, Part 4 Load Line Analysis

2

Example 1: Determine (a) VDQ and IDQ (b) VR

Chapter 2 DIODE, Part 4 Load Line Analysis

3

10

Example 2: Change R to 2kOhm. Determine (a) VDQ and IDQ (b) VR

Chapter 2 DIODE, Part 4 Load Line Analysis

4

Example 3: Repeat example 1. Determine (a) VDQ and IDQ using approximate equivalent model

Chapter 2 DIODE, Part 4 Load Line Analysis

5

11

Example 4: Repeat example 2. Determine (a) VDQ and IDQ using approximate equivalent model

Chapter 2 DIODE, Part 4 Load Line Analysis

6

Example 5: Repeat example 1. Determine (a) VDQ and IDQ using ideal diode model

Chapter 2 DIODE, Part 4 Load Line Analysis

7

12

Approximate and Ideal Diode Models

rav was not employed because rav is typically much Less than the other series elements of the network.

rav was not employed because rav is typically much Less than the other series elements of the network.

Chapter 2 DIODE, Part 4 Load Line Analysis

8

Chapter 2 DIODE Part 4 Series Diodes Configurations with DC Inputs
Chapter 2 DIODE, Series Diode Configurations 1

13

Series Diode Configuration with DC Inputs
(i) ‘ON’ state

Series diode configuration (1) Determining the state of the diode (2) Substituting the equivalent model for the “on” diode

Chapter 2 DIODE, Series Diode Configurations

2

(ii) ‘OFF’ state

Series diode configuration (1) Determining the state of the diode

(2) Substituting the equivalent model for the “off” diode

Chapter 2 DIODE, Series Diode Configurations

3

14

Example 6: Determine VD, VR and ID. Solution
The applied voltage establishes a current in the clockwise direction to match the arrow of the symbol and th di d i i th “ON” f th b l d the diode is in the state.
Series diode configuration

Chapter 2 DIODE, Series Diode Configurations

4

Example 7: Repeat example 6 with the diode reversed. Determine VD, VR and ID.
Solution The direction of current is opposite to the arrow in the diode symbol. Diode is in “OFF” state g p resulting to “open” circuit.

Series diode configuration

1. An open circuit can have any voltage across its terminals, but the current is always 0 A. 2. A short circuit has a 0V drop across its terminals, but the current is limited only by the Surrounding network.

Source notation

Chapter 2 DIODE, Series Diode Configurations

5

15

Example 8: Determine VD, VR and ID.

ON

Series diode configuration

OFF

Operating point

The level of the applied voltage is insufficient to turn the silicon diode “ON”. The point of operation on the characteristics shown in the figure establishing the open circuit equivalent.

Chapter 2 DIODE, Series Diode Configurations

6

Example 9: Determine VD, and ID. Solution:
The resulting current has the same direction as the arrowheads of the symbols of both diodes. Here, E 12V > (0.7V + 0.3V)= 1V H E=12V (0 7V 0 3V) 1V.
Series diode configuration Determining the state of the diode and substituting the equivalent model for the diode

Chapter 2 DIODE, Series Diode Configurations

7

16

Example 10: Determine VD, Vo and ID.

The combination of a Short circuit in series With an open circuit always Results in an open circuit And ID =0A.

Series diode configuration

Determining the state of the diode and substituting the equivalent model for the diode

Chapter 2 DIODE, Series Diode Configurations

8

Example 11: Determine I, V1 V2 and Vo .

Series diode configuration

Determining the state of the diode and substituting the equivalent model for the diode

Chapter 2 DIODE, Series Diode Configurations

9

17

Chapter 2 DIODE Part 5 Parallel and Series Diodes Configurations C fi ti with DC Inputs
Chapter 2 DIODE, Parallel and Series Diodes Conf. 1

Example 12: Determine Vo , I1 , ID1 and ID2 .

Parallel diode configuration Determining the state of the diode and substituting the equivalent model for the diode

Assuming diodes of similar characteristics,

Chapter 2 DIODE, Parallel and Series Diodes Conf.

2

18

Example 13: Determine I .

Parallel diode configuration

Determining the state of the diode and substituting the equivalent model for the diode

Chapter 2 DIODE, Parallel and Series Diodes Conf.

3

Example 14: Determine Vo .

The applied voltage will turn both diodes “on”. However, if both were “on”, the 0.7V drop across the silicon diode would not match 0.3V across the Ge diode as required by the fact that the voltage across parallel elements must be the same.

Parallel diode configuration

Determining the state of the diode and substituting the equivalent model for the diode

Chapter 2 DIODE, Parallel and Series Diodes Conf.

4

19

Example 15: Determine I1,I2 and ID2 .

Series and Parallel diode configuration Determining the state of the diode and substituting the equivalent model for the diode

KVL

Chapter 2 DIODE, Parallel and Series Diodes Conf.

5

Chapter 2 DIODE Part 6 AND / OR Gates

Chapter 2 DIODE, AND / OR Gates

1

20

Example 16: Determine V0 .

Redrawn OR Gates

Positive logic OR Gates First, assume that D1 is “on” and D2 is “off”. Vo = E - VD = 10V – 0.7V = 9.3V (level 1) With 9.3V at the cathode (-) side of D2 and 0V at the anode (+) side, D2 is in the “off” state. Our assumption is correct.

Chapter 2 DIODE, AND / OR Gates

2

Example 17: Determine V0 .

Positive logic AND Gates

Redrawn OR Gates

With 10V at the cathode side of D1, it is assumed that D1 is in the “off” state even though There is a 10V source connected to the anode of D1 through the resistor. D2 is assumed to be in the “on” state due to the low voltage at the cathode side and the on Availability of the 10V source through the 1kOhm resistor. The voltage at Vo is 0.7V due to the forward-biased diode D2. With 0.7V at the anode of D1 and 10V at the cathode, D1 is definitely in the “off” state.

Chapter 2 DIODE, AND / OR Gates

3

21

Ideal Diode

Chapter 2 Ideal Diode

1

22

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 799 posted: 12/10/2009 language: English pages: 22