Diodes
A diode is a device that conducts the current in just one direction: the direction of the arrow in the diode symbol, which looks like this:
The most important characteristics of a diode are: maximum forward current, forward voltage, maximum power dissipation and reverse voltage.
The forward current is the current flow in the direction of the arrow of the diode symbol. This current causes a voltage across the diode: the forward voltage drop. The power dissipation of a diode is the forward current multiplied by the forward voltage drop.
The reverse voltage is the voltage across a diode when it is reverse biased.
If you want to know how a diode works internally, you'll have to take a peek inside.
An AC Voltage Rectifier
Since diodes conduct current in only one direction, they can be used as an AC Voltage rectifier. Take a look at the picture below.
A triangular AC voltage is connected to the input terminals of the rectifier. The output voltage will be measured across resistor R1. When the upper input terminal is positive, there will be a current flow through the diode and the resistor. This current causes a voltage across R1. Assume the peak voltage is (plus and
minus) 9V, and the forward voltage of the diode is 0.7V. The peak current will then be (9V - 0.7V) / 1k = 8.3mA. The maximum power dissipation of the diode will be 9V • 8.3mA = 74.7mW. When the voltage at the upper input terminal becomes negative, the diode is reverse biased blocking the current flow. Since the diode has a very large resistance, all the voltage will be across the diode. This should not exceed the maximum reverse voltage.
So if you want to perform this experiment, you'll need a diode with the following requirements: the maximum forward current must be 8.3mA or higher; the maximum power dissipation must be 74.7mW or higher; and the maximum reverse voltage must be 9V or higher. Any small signal diode will meet these requirements. The resistor can be a regular 0.25W resistor since the maximum power dissipation is (8.3mA) 2 • 1k = 69mW. The circuit above is called a half wave rectifier, since the ouput contains only the positive half of the input. The circuit below shows a full wave rectifier.
This circuit works as follows. When the input signal is positive, the currents flows from the upper terminal, via diode D1, resistor R1, and diode D3 to the lower terminal. When the input signal is negative, the currents flows from the lower terminal, via diode D2, resistor R1, and diode D4 to the upper terminal. Notice that the current always flows through two diodes: either D1 and D3, or D2 and D4. This means that the output voltage will always be about 1.4 volts (two 'forward voltage drops') less than the input voltage. The circuit D1...D4 is called a bridge rectifier. When you look at a bridge rectifier, you'll probably see something imprinted like 'B80C5000/3300'. The number after the 'B' indicates the maximum (reverse) voltage, in this case 80V. The number after the 'C' indicates the maximum peak/continuous (forward) current in mA. In this case the maximum peak current is 5A and the maximum continuous current is 3.3A.
Smaller bridge rectifiers only indicate the maximum voltage and current, e.g. 'B40C800'.
LEDs
The abbreviation LED stands for Light Emitting Diode. LEDs consume less power than light bulbs, and have a much longer life time: about 100000 hours. A regular LED needs a current flow of 10...20mA, and has a forward voltage drop of 1.5 to 2 volts, depending on the color.
With the circuit below, you can test and experiment with LEDs.
Question: What will be a good value for R1? Assume that the voltage across LED D1 is 2 volts, and we want a current flow of 15mA.
Electronics Course - Diodes
Answer: The voltage across R1 will be 9V - 2V = 7V. The current flow through R1 will also be 15mA. So R1 should have a value of 7V / 15mA = 467ohms. From the E12 series, 470ohms is a good choice.
Zener Diodes
A zener diode in conducting state acts like a normal diode. It's the reverse voltage that distinguishes a zener diode from a regular diode. Take a look at the picture below.
In this picture you see a reverse connected zener diode. The 'value' of a zener diode is given in volts; this is the reverse voltage. But a zener diode doesn't blow when the voltage tends to get higher. A zener diode stabilizes the voltage at the reverse voltage. So the voltage across the zener diode in the picture above will always be 4.7 volts, even when the battery voltage increases. Again, we need to calculate the value of R1. Unfortunately, it's difficult to say what's the ideal current flow through a zener diode. (Yep, altough the diode is reverse biased there is a current flow!) In most cases 5mA is fine. Since the voltage across R1 will be 9V - 4.7V = 4.3V, a good value of R1 is 4.3V/5mA = 860ohms.
A zener diode manufacturer publishes the maximum power dissipation of a zener diode. 0.4 or 0.5W is a very common value for a small zener diode. Using this characteristic, we can calculate the minimum value of R1: Assume we use a 0.4W zener in the design above. Since the voltage across the zener is 4.7V, the maximum current flow is 0.4W/4.7V = 85mA. The voltage across R1 will be 9V - 4.7V = 4.3V. So the minimum value for R1 is 4.3V/85mA = 51ohms. So a good value of R1 ranges from 51 to 860ohms. 820ohms is a good choice. Note however that the calculations above only count for a zener without a load. Take a look at the picture below.
In this design zener D1 has a 50ohms load (R2). Again, we'll calculate a proper value for R1. Since the voltage across the load R2 is always 4.7V, the current flow through R2 will always be 4.7V/50 = 94mA. The current flow through D1 should be between 5 and 85mA. So the current flow through R1 ranges from 99 to 179mA. The voltage across R1 is always 4.3V, so the resistance should be between 24 and 43ohms.
Electronics Course - Diodes
39ohms may be a good choice. In that case, the power dissipation is 4.32 / 39 = 0.47W. So you'd better take a 1W resistor! But what should we do if the 50ohms load can be detached, e.g. because it's an external load? With the load connected, the *maximum* value of R1 is 43ohms, but without the load the *minimum* value is 51ohms!
The answer is simple: use a higher wattage zener diode, e.g. 1.3W. In that case, the maximum current flow through D1 is 1.3W/4.7V = 276mA. This means, without the load connected, a minimum value of R1 of 4.3V/276mA = 16ohms. Now we have an overlapping range of values for R1 from which you may choose one. Again, a 39ohms / 1W resistor is a good choice.
Testing diodes using a multimeter
Most digital multimeters look like this:
1 = Display
2 = Function switch
3 = Transistor socket (optional)
4...6 = Test lead jacks
If you want to test a diode, set the function switch to "diode test".
Next, connect the test leads. Mulimeters usually come with two test leads: a black one and a red one.
- Connect the black test lead to the COM jack and the red lead to the V/ jack.
- Connect the other ends of the test leads across the diode.
- Connect the black wire to the cathode and red wire to the anode. The display should now read about 0.6V (600mV). If you swap the test leads, the display will indicate an overflow. Note that in-circuit testing may lead to wrong results, since other components may be parallel-connected to the diode. Also make sure that the equipment-under-test has been turned off!
Notes:
Power Supply Unit
Power Failure (Kerosakan Disebabkan Kilat)
- Check Fius
- Check Capasitor Seramik
- Check Diod
- IC/Chip – Power IC
Contoh kerosakan diod
Probe +/- pada diod +/- , multimeter naik, probe -/+ pada diod -/+ multimeter naik
Contoh Diod Baik
Probe +/- pada diod +/- multimeter naik , probe -/+ pada diod -/+ multimeter tidak naik.
Fungsi diod ialah menukar arus dari AC ke DC
The most important characteristics of a diode are: maximum forward current, forward voltage, maximum power dissipation and reverse voltage.
The forward current is the current flow in the direction of the arrow of the diode symbol. This current causes a voltage across the diode: the forward voltage drop. The power dissipation of a diode is the forward current multiplied by the forward voltage drop.
The reverse voltage is the voltage across a diode when it is reverse biased.
If you want to know how a diode works internally, you'll have to take a peek inside.
An AC Voltage Rectifier
Since diodes conduct current in only one direction, they can be used as an AC Voltage rectifier. Take a look at the picture below.
A triangular AC voltage is connected to the input terminals of the rectifier. The output voltage will be measured across resistor R1. When the upper input terminal is positive, there will be a current flow through the diode and the resistor. This current causes a voltage across R1. Assume the peak voltage is (plus and
minus) 9V, and the forward voltage of the diode is 0.7V. The peak current will then be (9V - 0.7V) / 1k = 8.3mA. The maximum power dissipation of the diode will be 9V • 8.3mA = 74.7mW. When the voltage at the upper input terminal becomes negative, the diode is reverse biased blocking the current flow. Since the diode has a very large resistance, all the voltage will be across the diode. This should not exceed the maximum reverse voltage.
So if you want to perform this experiment, you'll need a diode with the following requirements: the maximum forward current must be 8.3mA or higher; the maximum power dissipation must be 74.7mW or higher; and the maximum reverse voltage must be 9V or higher. Any small signal diode will meet these requirements. The resistor can be a regular 0.25W resistor since the maximum power dissipation is (8.3mA) 2 • 1k = 69mW. The circuit above is called a half wave rectifier, since the ouput contains only the positive half of the input. The circuit below shows a full wave rectifier.
This circuit works as follows. When the input signal is positive, the currents flows from the upper terminal, via diode D1, resistor R1, and diode D3 to the lower terminal. When the input signal is negative, the currents flows from the lower terminal, via diode D2, resistor R1, and diode D4 to the upper terminal. Notice that the current always flows through two diodes: either D1 and D3, or D2 and D4. This means that the output voltage will always be about 1.4 volts (two 'forward voltage drops') less than the input voltage. The circuit D1...D4 is called a bridge rectifier. When you look at a bridge rectifier, you'll probably see something imprinted like 'B80C5000/3300'. The number after the 'B' indicates the maximum (reverse) voltage, in this case 80V. The number after the 'C' indicates the maximum peak/continuous (forward) current in mA. In this case the maximum peak current is 5A and the maximum continuous current is 3.3A.
Smaller bridge rectifiers only indicate the maximum voltage and current, e.g. 'B40C800'.
LEDs
The abbreviation LED stands for Light Emitting Diode. LEDs consume less power than light bulbs, and have a much longer life time: about 100000 hours. A regular LED needs a current flow of 10...20mA, and has a forward voltage drop of 1.5 to 2 volts, depending on the color.
With the circuit below, you can test and experiment with LEDs.
Question: What will be a good value for R1? Assume that the voltage across LED D1 is 2 volts, and we want a current flow of 15mA.
Electronics Course - Diodes
Answer: The voltage across R1 will be 9V - 2V = 7V. The current flow through R1 will also be 15mA. So R1 should have a value of 7V / 15mA = 467ohms. From the E12 series, 470ohms is a good choice.
Zener Diodes
A zener diode in conducting state acts like a normal diode. It's the reverse voltage that distinguishes a zener diode from a regular diode. Take a look at the picture below.
In this picture you see a reverse connected zener diode. The 'value' of a zener diode is given in volts; this is the reverse voltage. But a zener diode doesn't blow when the voltage tends to get higher. A zener diode stabilizes the voltage at the reverse voltage. So the voltage across the zener diode in the picture above will always be 4.7 volts, even when the battery voltage increases. Again, we need to calculate the value of R1. Unfortunately, it's difficult to say what's the ideal current flow through a zener diode. (Yep, altough the diode is reverse biased there is a current flow!) In most cases 5mA is fine. Since the voltage across R1 will be 9V - 4.7V = 4.3V, a good value of R1 is 4.3V/5mA = 860ohms.
A zener diode manufacturer publishes the maximum power dissipation of a zener diode. 0.4 or 0.5W is a very common value for a small zener diode. Using this characteristic, we can calculate the minimum value of R1: Assume we use a 0.4W zener in the design above. Since the voltage across the zener is 4.7V, the maximum current flow is 0.4W/4.7V = 85mA. The voltage across R1 will be 9V - 4.7V = 4.3V. So the minimum value for R1 is 4.3V/85mA = 51ohms. So a good value of R1 ranges from 51 to 860ohms. 820ohms is a good choice. Note however that the calculations above only count for a zener without a load. Take a look at the picture below.
In this design zener D1 has a 50ohms load (R2). Again, we'll calculate a proper value for R1. Since the voltage across the load R2 is always 4.7V, the current flow through R2 will always be 4.7V/50 = 94mA. The current flow through D1 should be between 5 and 85mA. So the current flow through R1 ranges from 99 to 179mA. The voltage across R1 is always 4.3V, so the resistance should be between 24 and 43ohms.
Electronics Course - Diodes
39ohms may be a good choice. In that case, the power dissipation is 4.32 / 39 = 0.47W. So you'd better take a 1W resistor! But what should we do if the 50ohms load can be detached, e.g. because it's an external load? With the load connected, the *maximum* value of R1 is 43ohms, but without the load the *minimum* value is 51ohms!
The answer is simple: use a higher wattage zener diode, e.g. 1.3W. In that case, the maximum current flow through D1 is 1.3W/4.7V = 276mA. This means, without the load connected, a minimum value of R1 of 4.3V/276mA = 16ohms. Now we have an overlapping range of values for R1 from which you may choose one. Again, a 39ohms / 1W resistor is a good choice.
Testing diodes using a multimeter
Most digital multimeters look like this:
1 = Display
2 = Function switch
3 = Transistor socket (optional)
4...6 = Test lead jacks
If you want to test a diode, set the function switch to "diode test".
Next, connect the test leads. Mulimeters usually come with two test leads: a black one and a red one.
- Connect the black test lead to the COM jack and the red lead to the V/ jack.
- Connect the other ends of the test leads across the diode.
- Connect the black wire to the cathode and red wire to the anode. The display should now read about 0.6V (600mV). If you swap the test leads, the display will indicate an overflow. Note that in-circuit testing may lead to wrong results, since other components may be parallel-connected to the diode. Also make sure that the equipment-under-test has been turned off!
Notes:
Power Supply Unit
Power Failure (Kerosakan Disebabkan Kilat)
- Check Fius
- Check Capasitor Seramik
- Check Diod
- IC/Chip – Power IC
Contoh kerosakan diod
Probe +/- pada diod +/- , multimeter naik, probe -/+ pada diod -/+ multimeter naik
Contoh Diod Baik
Probe +/- pada diod +/- multimeter naik , probe -/+ pada diod -/+ multimeter tidak naik.
Fungsi diod ialah menukar arus dari AC ke DC
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