EGE217: Electronics 1 Lecturer: Siti Hamimah Sh. Ismail January 2009 ELECTRONICS 1 CHAPTER 2: DIODES AND APPLICATIONS Introduction In previous chapter, we have learnt how P type and N type semiconductor can be combined to form a diode. Now, we are going to have a close look at the characteristics and the application. Objective At the end of this chapter you will be able to: 1. Understand the use of rectifier diodes and filters; 2. Differentiate between half wave, full wave and bridge rectifier; and 3. Understand the applications of diodes as clippers and limitters and how it can be used a voltage multiplier. 2.1 Basic Ideas An ordinary resistor is a linear device because the graph of its current vs voltage is a straight line. A diode is a non-linear device because when the diode voltage is less than the barrier potential the diode current is small, but when the diode voltage exceeds the barrier potential, the diode current increases rapidly. The schematic symbol of a diode is as shown in Figure1 below: 13
In previous chapter, we have learnt how P type and N type semiconductor can be combined to form a diode. Now, we are going to have a close look at the characteristics and the application.
Objective
At the end of this chapter you will be able to:
1. Understand the use of rectifier diodes and filters;2. Differentiate between half wave, full wave and bridge rectifier;
and3. Understand the applications of diodes as clippers and limitters
and how it can be used a voltage multiplier.
2.1 Basic Ideas
An ordinary resistor is a linear device because the graph of its current vs voltage is a straight line. A diode is a non-linear device because when the diode voltage is less than the barrier potential the diode current is small, but when the diode voltage exceeds the barrier potential, the diode current increases rapidly.
The schematic symbol of a diode is as shown in Figure1 below:
The p side is called the anode and the n side the cathode.
2.1.1 The Reverse Region
The diode current vs diode voltage can be plotted as shown:
Figure 2
When the diode is forward biased, there is no significant current until the diode voltage is greater than the barrier potential. On the other hand, when the diode is reverse biased, there is almost no reverse current until the diode voltage reaches the breakdown voltage, Then, avalanche produces a large reverse current which will destroy the diode.
2.1.2 The Forward Region
In the forward region, the voltage at which the current starts to increase rapidly is called the knee voltage of the diode, which equals to the barrier potential. Analysis of diode circuits usually comes down to determining whether the diode voltage is more or less than the knee voltage. If it's more, the diode conducts easily, if it's less, the diode conducts poorly.
2.2 Data sheet
If the current in a diode is too large, the excessive heat can destroy the diode. For this reason, a manufacturer's data sheet specifies the maximum current a diode can safely handle without shortening its life or degrading its characteristics.
The maximum forward current is one of the maximum ratings given on a data sheet. This current may be listed as Imax, IF(max), Io, etc. For instance, a 1N456 has a maximum forward current rating of 1mA; which means that it can safely handle a continuous forward current of 135 mA.
This is the simplest approximation. In the most basic terms, a diode acts like a perfect conductor (zero resistance) when forward biased, and like a perfect insulator (infinite resistance) when reverse biased. Therefore an ideal diode acts like a switch that closes when forward biased and opens when reverse biased.
Example:
Use the ideal diode to calculate the load voltage and load current in figure below:
Solution:
Since the diode is forward biased, it is equivalent to a closed switch. Then, you can see that all of the source voltage appears across the load resistor:
VL = 10V
With ohms law, the load current is:
IL = 10V / 1K = 10mA
2.3.2 Second Approximation
This is a more accurate approximation. There will be no current exists until 0.7V appears across the diode, where at this point the diode turns on. Thereafter only 0.7V can appear across the diode, no matter what the current is.
Use the second approximation to calculate the load voltage, load current and diode power in figure below:
Solution:
Since the diode is forward biased, it is equivalent to a battery of 0.7V. This means that the load voltage equals the source voltage minus the diode drop:
VL = 10V – 0.7V = 9.3V
With ohms law the load current is:
IL = 9.3V / 1K = 9.3mA
The diode power is:
PD = (0.7V)(9.3mA) = 6.51mW
2.4 Rectifier Diodes
Most electronic devices like TVs, stereos and computers need a dc voltage to work properly. Since the power line voltage is alternating, the first thing we need to do is to convert the ac line voltage to dc voltage. The section of the electronic device that produce this dc voltage is called the power supply. Within the power supply are circuits that allow current to flow in only one direction. These circuits are called rectifiers.
The ac source produces a sinusoidal voltage. Assuming an ideal diode, the positive half cycle of source voltage will forward bias the diode, thus the switch is closed; and the positive half cycle of the source voltage will appear across the load resistor.
On the negative half cycle, the diode is reverse biased and it appears as an open switch and no voltage appears across the load resistor.
A half wave signal shown in figure below, is a pulsating dc voltage that increases to a maximum, decreases to zero, and then remains at zero during the negative half cycle. This is not the kind of dc voltage that we need for electronics equipment. To get a constant dc voltage, we need to filter the half wave signal, which will discuss in the next topic.
Figure 4
Ideal Half Wave
The peak output voltage is equals to the peak input voltage.
The dc value of a signal is the same as the average value.
Half wave:
Output Frequency
The output frequency is the same as the input frequency.
Half wave:
Second Approximations
We don't get perfect half wave voltage across the load resistor. Because of the barrier potential, the diode does not turn on until the ac source voltage is much greater than 0.7V.
2nd half wave: Example
1. Find peak load voltage and dc load voltage for both ideal and second approximation. Given that Vrms = 0.707 Vp .
Solution
From given formula,
With an ideal diode, the peak load voltage is:
The dc load voltage is:
With the second approximation, we get a peak load voltage of:
The full wave rectifier is equivalent to two half-wave rectifiers. Diode D1 conducts on the positive half cycle and diode D2 conducts on the negative half cycle. As a result, the rectified load current flows during both half cycles. The full wave rectifiers act the same as two back to back half-wave rectifiers, shown below:
Figure 6
Dc or Average Value
Since the full wave signal has twice as many positive cycles as the half wave signal, the dc or average value is twice as much.
The ac line voltage has a frequency of 60 Hz. Therefore the input period is
Because of the full wave rectification, the period of the full-wave signal is half the input period:
The frequency of the full wave signal is double the input frequency.
Full wave:
Second Approximation
Since the full wave rectifier is like two back to back half wave rectifiers, we can use the second approximation given earlier. The idea is to subtract 0.7V from the ideal peak output voltage. See example.
Example
Find input voltage to each diode. And find the output voltage ideally and using second approximation.
Solution:
The peak primary voltage is:
Because of the 10:1 step down transformer, the peak secondary voltage is:
The full wave rectifier acts like two back to back half wave rectifier. Because of the center tap, the input voltage to each half wave rectifier is only half the secondary voltage:
Ideally the output voltage is:
Using the second approximation:
2.4.3 The Bridge Rectifier
Figure below shows a bridge rectifier circuit.
Figure 7
Bridge rectifier is similar to a full-wave rectifier because it produces a full-wave output voltage. Diodes D1 and D2 conduct on the positive half cycle, and D3 and D4 conduct on the negative half cycle.
Figure below shows the equivalent circuit for the positive half cycle.
Figure 8
Note that D1 and D2 are forward biased so it produces a positive load voltage as indicated by the plus minus polarity across the load resistor.
We can see that when D2 is short circuit, the circuit acts like a half-wave rectifier.
Figure below shows the equivalent circuit for the negative half cycle.
Figure 9
This time, D3 and D4 are forward biased. This also produces a positive load voltage. We can also see that when D3 is short circuit, the circuit looks like a half-wave rectifier. So the bridge rectifier acts like two back to back half-wave rectifier. Advantage of bridge rectifier: All the secondary voltage is used
as the input to the rectifier. So, using the same transformer, we can get twice as much peak voltage and twice as much dc voltage with a bridge rectifier compared to full wave rectifier mentioned in previous section.
Summaries of the three rectifiers and their properties:
Half wave Full Wave Bridge Number of Diodes 1 2 4Rectifier InputPeak output (Ideal)Peak Output (2d)DC Output
Ripple Frequency
= peak secondary voltage
= peak output voltage
2.5 Rectifier Filters
Think!!!!
Why do you think that we need to filter the output of a rectifier?
There are a few rectifier filters that will be discussed here. They are:
The choke-Input Filter The Capacitor Input Filter
2.5.1 The Choke Input Filter
Figure below shows the circuit for choke-input filter.
Figure 10
The ac source produces a current in the inductor, capacitor and resistor. The ac current in each component depends on the inductive reactance, capacitive reactance and the resistance.
The choke (or inductor) has the primary characteristic of opposing a change in current. Because of this, a choke input filter ideally reduces the ac current in the load resistor to zero.
Requirement for a well designed choke-input filter:
1. at the input frequency is much smaller than . Thus the load resistance can be ignored. The circuit can be redrawn as shown below:
Figure 11
2. is much greater than at the input frequency. Thus the ac output voltage equals:
FILTERING THE OUTPUT OF A RECTIFIER
Figure below shows a choke-input filter between a rectifier and a load. Note that the rectifier can be either half-wave, full-wave or bridge rectifier.
Figure 12The rectifier output has two different components: a dc voltage (the average value) and an ac voltage (the fluctuating part). As shown below:
Each of these voltages acts like a separate source.
Consider the ac voltage: is much greater than .Thus, a very little ac voltage across the load resistor.
Consider the dc voltage: At 0 Hz, the inductive reactance is zero and the capacitive reactance is infinite.Thus, only the series resistance of the inductor windings remains. is much smaller than causes most of the dc component to appear across the load resistor.
Almost all of the dc component is passed on the load resistor and almost all of the ac component is blocked. In this way, we get an almost perfect dc voltage, almost constant like the voltage output of a battery.
Figure below shows the filtered output for a full wave signal. Note that there is a small ac load voltage which is called a ripple.
Figure 14
MAIN DISADVANTAGE
1. Large inductances have to be used to get enough reactance for adequate filtering.
2. Large winding resistance will create a design problem with large load currents.
3. Too much dc voltage is dropped across the choke resistance.4. Bulky/big and heavy inductors are not suitable for modern
semiconductor circuit.
2.5.2 The Capacitor-Input Filter
Think!!!
Identify the difference in the dc output voltage produced in choke-input filter and capacitor-input filter.
BASIC IDEA OF CAPACITOR CIRCUIT (without load resistor)
Refer to the figure below and see what this simple circuit (without load resistor) does during the first quarter cycle.
Figure 15
Initially, the capacitor is uncharged. During the first quarter cycle of figure below, the diode is forward biased. Since it ideally acts like a closed switch, the capacitor charges and its voltage equals the source voltage at each instant of the first quarter cycle. The charging continues until it reaches the maximum value. At this point the capacitor voltage equals .
Figure 16
After the input voltage reaches the peak, it starts to decrease. As soon as the input voltage is less than , the diode turns off, and it acts like an open circuit. During the remaining cycle, the capacitor stays fully charged and the diode remains open. This is why the output voltage in figure above is constant and equals to .
For the capacitor-input filter to be useful, we need to connect a load resistor across the capacitor as shown below:
Figure 17
The capacitor remains fully charged as long as the time constant is much greater than the period. Thus the load voltage is
approximately , and there is a small ripple in the output.
Figure 18
Between peaks, the diode is off and the capacitor discharges through the load resistor, (the capacitor supplies the load current). Since the capacitor discharges only slightly between peaks, the peak to peak ripple is very small. When the next peak arrives, the diode conducts briefly and recharges the capacitor to the peak value.
FULL WAVE FILTERING
If we connect a full-wave or bridge rectifier, to a capacitor-input filter, the peak-to-peak ripple is cut in half. See figure below:
The diodes we have mentioned earlier are used in low frequency power supplies and is called rectifier diodes. These diodes have little use because most circuits inside electronics equpment are running at much higher frequencies.
In this section, we will be using small-signal diodes. These diodes are optimized for use at high frequencies.
2.6.1 The Positive Clipper
A clipper is a circuit that removes either positive or negative part of a waveform. This kind of processing is useful for signal shaping, circuit protection and communications. Figure below shows a positive clipper:
Figure 20
Note that the circuit removes all the positive part of the input signal. The output signal has only negative half cycle.
Circuit Operation:During the positive half cycle, the diode turns on and looks like a short circuit across the output terminals. Ideally, the output voltage is zero, but to a second approximation, the diode voltage is 0.7V when conducting.
On the negative half cycle, the diode is open. Thus a negative half cycle appears across the output.
The series resistor is much smaller than the load resistor, thus, the negative output peak is shown as .
2.6.2 The Negative Clipper
Reverse the polarity of diode used in the positive clipper to get a negative clipper, as shown below:
Figure 21
The negative part of the signal is removed. Ideally the output waveform has nothing but positive half cycle. Because of the diode offset voltage (barrier potential), the clipping level is at –0.7V.
2.6.3 The Limiter or Diode Clamp
In previous section, we have seen that, clipper is useful for waveshaping. Now, we will see how the limiter or diode clamp is used to protect sensitive circuits from too much input.
See figure below:
Figure 23
The normal input to this circuit is a signal with a peak of only 15mV. Therefore the normal output is the same signal because neither diode is turned on during the cycle. The diode remains off during normal operation, and will only conduct when the signal is too large.
Thus, if the input tries to rise above 0.7V, the output is limited to 0.7V. On the other hand if the input tries to drop below –0.7 V, the output is limited to –0.7V.
1. Draw the current vs. voltage graph for a diode. Please indicate clearly the knee voltage, reverse current, breakdown region, reverse region and forward region. Explain briefly what happened in each region.
2. (a) Describe the purpose of using rectifier diodes.(a) Draw the circuit and output to each rectifier:
i) half wave rectifierii) full wave rectifieriii) bridge rectifier
3. Calculate the average value for each rectifier when the voltage ac source is 10V and 60 Hz:
a. half wave b. full wave
4. Draw the circuit and the output for choke filter and capacitor input filter if they are connected to a half wave rectifier.
5.
(a) Name the rectifier and filter used in this circuit.(b) Show the output waveform of this circuit. (c) find the dc load voltage and ripple voltage.
6. Give the name of the filter used in this figure and find the dc load voltage and ripple in this figure by assuming an ideal diode and second approximation. Show the output waveform.
7. Find input voltage to each diode. And find the output voltage ideally and using second approximation.
8. Find the name of this rectifier and find peak load voltage and dc load voltage for both ideal and second approximation. Given that Vrms = 0.707 Vp .
