Diodes & Power Supplies

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1 Diodes & Power Supplies Diode Basics A diode is a one way valve for electricity. It lets current flow in one direction but not the other. Diodes are used for Converting AC DC Detecting Radio Signals Doing Logic in Computers Also, transistors have diodes in them, so we will have to understand diodes to understand transistors. Suppose we graph current versus voltage for a diode.

2 An ideal diode would be one that turned on completely at 0 volts. For comparison, a resistor has the following current versus voltage curve. Power Supplies Power is transmitted as A.C. (60 Hz in the U.S.) This is because 1. Generators make A.C. A generator generates voltage by spinning a coil of wire in a magnetic field. 2. To transmit power efficiently over long distances, it must be transmitted at very high voltage. Otherwise the resistance in the wires eats up all of the power. Transformers can raise the voltage for transmission and then lower it at the destination, but transformers only work with A.C..

3 Most electronic circuits use D.C. (Radio, T.V., Computers, etc.) A power supply circuit takes A.C. from a wall socket and produces DC from it. It usually also changes the voltage. Usually the voltage is changed by a transformer before it s changed to D.C.. A typical power supply has three parts 1. Transformer 2. Rectifier 3. Filter

4 Rectifier Circuits Half Wave Rectifier Circuit The half wave rectifier circuit is very simple, it s just a diode, but it throws away half of every cycle. Full Wave Bridge Rectifier The full wave bridge rectifier requires four diodes, but does not waste half of the sine wave. This is a popular rectifier circuit. Full Wave Center Tapped Power Supply The center tapped full wave rectifier uses the full wave also, and uses only two diodes, but it requires a center tapped transformer. The output voltage is half of the transformer voltage.

5 Half Wave Rectifier with Filter A single capacitor makes a fair filter, if it s large enough. Suppose you start with the capacitor discharged, and plug the circuit in at exactly T = 0. As V S rises, V A becomes larger than V B. This turns on the diode and starts charging the capacitor. Assume an ideal diode. With the diode on, V B = V A & V L = V S. V L = V C = V S until V S reaches the first peak.

6 After the peak, V A starts to decrease. The only way V C can decrease is to discharge through R L. It can t drive current backward through the diode. If C is large, R L C >> 1/60 Second. Therefore, V S will soon fall faster than V C can follow it. When V A becomes less than V B, the diode turns off. C discharges slowly through R L. Eventually V S will start rising again, and will rise above V B. When V S becomes greater than V B, the diode turns on again. This makes V L = V S until after the peak. V C charges back up to V P. V L is almost D.C. V L = V P + ripple. (ripple < 0)

7 Suppose we plug it in at a time that s not zero. There are at least three cases to consider. V ripple is easy to calculate approximately if we assume it is small. I L = V L /R L V P /R L if V ripple << V P. Q decreases as the capacitor discharges. Q = I L t I L (1/f) =. Note that 1/f is 1/60 second. V C = Q/C, so V C = Q / C =. Notice that this is actually an upper bound on V C.

8 V ripple = Large C makes small ripple. Improvements With a full wave rectifier, it only discharges ½ as long, so V ripple is half as big. ( ) versus V ripple = for a full wave rectifier power supply. For one or two dollars, you can buy a sophisticated integrated circuit, called a voltage regulator, which will remove almost all of the remaining ripple.

9 Diode Current Versus Voltage Curves The diode current versus voltage curve is given by the following equation. I D = I 0 ( 1), where V t = volts at room temperature (300 degrees kelvin) η = 1 for germanium diodes 1 n 2 for silicon diodes I 0 depends on the size of the diode. A typical value is I 0 = Amps. A typical silicon diode at room temperature would have I D = amps (. (. ) 1) 1.5 x = 0.039, so this can be simplified to I D = amps (. 1) The best way to understand how this equation produces the diode curve is to plot some values and graph them.

10 I D = amps (. 1) Reverse Leakage Current The reverse leakage current is the diode current when the voltage is negative. One can easily see that the reverse leakage current is I 0. I D (-1 volt) = I 0 (. 1) = I 0 (. 1) = I 0 (7 x ) = - I 0.

11 The Ideality Factor, η (eta) η is called the ideality factor. The equation I D = I 0 ( 1) was derived from physics using basic principles. Certain approximations were made when deriving it. These approximations work almost perfectly for a diode made of the element germanium, but not for silicon. Germanium is considered ideal in this respect. This equation works almost perfectly for germanium diodes. It does not work well for silicon diodes. The equation can be made reasonably accurate for silicon diodes by adding an extra factor, η, to the denominator of the exponent. η is sort of a fudge factor that makes the diode equation work for elements like silicon. The Thermal Voltage, V t V t = kt/q, where K = Boltzman s constant = 1.38 x Joules/degree kelvin q = charge of electron = 1.6 x Coulombs t = temperature in degrees kelvin (above absolute zero) V t is proportional to absolute temperature. Room temperature is considered to be 300 degrees kelvin. 300 degrees kelvin = 27 degrees centigrade = 80 degrees Farenheit V t (room temperature) = V t (300 degrees kelvin) = volts. 0 degrees centigrade = 273 degrees kelvin A degree kelvin is the same size as a degree centigrade. It is easy to calculate V t at other temperatures because V t is proportional to temperature. e.g., V t (350 degrees kelvin) = (0.026 volt) = 0030 volt.

12 Solving for V D as a Function of Current I D = I 0 ( 1) I D / I 0 = 1 = I D / I = ln(i D / I 0 + 1) For our example diode, ηv t = 1.5(.026 volt) =.039 volt. Therefore, V D =.039 volt ln(10 10 I D + 1).039 volt ln(10 10 I D ). Diode Power Dissipation P = IV. If the voltage is less than the cutin voltage, or is negative, then the current is approximately zero and the power dissipation is approximately zero. Notice that the diode voltage is approximately equal to the cutin voltage any time significant current is flowing in the diode. This is an important fact, which we will use a lot. MEMORIZE THIS! The cutin voltage for a silicon diode is approximately 0.7 volts. The voltage across a silicon diode will be approximately 0.7 volts any time significant current is flowing in it. If significant current is flowing in a silicon diode, then the power being dissipated as heat by it is P = (0.7 volt)i D.

13 Diode Reverse Breakdown Voltage If you put too much reverse voltage on a diode, it will break down and reverse current will flow. Reverse breakdown does not harm a diode unless the power is too great and the diode melts. Some diodes are made to be used in the breakdown region. These are called Zener diodes, and they have a special symbol. The main difference between a Zener diode and an ordinary diode is that a Zener diode has a precisely specified breakdown voltage. With normal diodes, the manufacturer specifies a minimum breakdown voltage, but does not give the actual breakdown voltage. For instance, he might just guarantee that the breakdown voltage is at least 100 volts. On the other hand, one can buy a zener diode whose breakdown voltage is guaranteed to be withing 5% of 12 volts.

14 Zener diodes can be used to make voltage regulators. i = (7 volts 5 volts)/1k. V out = 5 volts. Reductio Ad Absurdum Proofs Recall reductio ad absurdum proofs from high school geometry class. A reductio ad absurdum proof proves a theorem is true by assuming it is false, and showing that assuming it s false leads to an impossibility. This kind of proof is very useful in diode and transistor circuits. Many times you ll see a diode or transistor circuit in which you can kind of guess what the circuit is doing, but can t be sure. One can often prove what the circuit is doing using a reductio ad absurdum proof. The following circuit is a classic example. If the diodes are ideal diodes and the inputs are less than 5 volts, then the output is the minimum of the inputs. It s a minimum finding circuit. Looking at the circuit, it appears likely that D2 is on since it has lower voltage on the left end than D1, and since both input voltages are less than 5 volts. If D2 is on, then the voltage at point P is 3 volts. This makes D1 reverse biased and off. V out = 3 volts. This seems reasonable, but it s not really a proof. The following is a proof.

15 1. Assume D2 is off. 2. If D2 is off, then D1 must be off as well because it has more reverse voltage. 3. Since both diodes are off, the current through the 10 k resistor is zero. Therefore, the voltage drop across the 10 k resistor is zero volts. 4. This implies that the voltage at P is 5 volts. 5. V P = 5 volts makes a forward voltage across D2 of 5 3 = 2 volts. This is impossible since D2 is off. Assuming D2 is off leads to an impossibility. Therefore, the assumption must be false. We now know for sure that D2 is on. This implies that the voltage at P is 3 volts. This makes V out = 3 volts and it puts a reverse voltage of one volt on D1. Do diode problem from in class practice problem set #10.

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