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Devices and Applications Ctec 201. Semi-Conductor Diodes Supplement Prepared by Mike Crompton. (Rev. 25 July 2003) 2 P.N. Junctions Semi-conductors have a valence of 4. The only semi-conductors of concern to us at present are carbon, silicon and germanium. Carbon we use to make resistors. Silicon and germanium are used to make basic semi-conductor devices such as diodes and transistors. Any atom that has a valence of 8 is considered stable and does not readily give up or accept other electrons. An atom with a valence of 7 will readily accept another electron to make its valence layer 8. It would therefore be an “acceptor” atom. An atom with 9 electrons in its outer layer would establish a valence layer of 8, making the extra electron a “free” electron that it would readily give up. This makes it a “donor” atom. This concept of “accepting” or “donating” electrons is used to create a P.N. Junction. If we take silicon with a valence of 4 and add to it an impurity that has a valence of 3 (Aluminum, Gallium) we would end up with molecules where each atom would “see” 7 electrons in its outer (valence) layer. The atoms would be combined in such a manner as to leave a gap or hole where an eighth electron should be. Since this hole means the absence of an electron, and since electrons are –ve, we refer to the hole as +ve hole. Not all the silicon atoms will join with the impurity, only some, the exact quantity would be determined by the amount of impurity added. This process of adding impurities is known as “Doping”. Heavily doped material has a lot of impurity, lightly doped, a little. Silicon, which is doped with a valence of 3 impurities, is called “P type silicon” (The P being for positive). P type silicon is an acceptor! By adding an impurity with a valence of 5 (Arsenic, Antimony) the molecule would form a bond with 4 silicon and 4 of the impurity allowing each atom to “see” a valence of eight. The extra electron (the fifth from the impurity) can bond with nothing, and therefore becomes a free electron. Since electrons are negative, we call silicon doped with an impurity with a valence of 5, “N type silicon” It is a donator! See Fig. 1 When a piece of P type and N Type Silicon P Type Silicon N type silicon are physically Doped with Arsenic Doped with Aluminum joined, the junction area or Antimony or Gallium becomes active as the N type (donor) material gives + + + + up free electrons, which + + + migrate to the P type + + (acceptor) material and “fill + in” the positive holes. This +ve Hole Extra Electron creates an area where the Fig. 1 molecules all have a valence of eight and are completely stable, creating, in effect, a zone of resistance. The activity continues until the zone of resistance is of such a width that the force of attraction between free electrons (-ve) and positive holes is insufficient to overcome the resistance, and all activity ceases. The zone of resistance has no free electrons and no +ve holes. 3 Since the electrons and holes are the vehicle for current flow, called the carriers, the zone of resistance is known as the “Depletion Zone” (Depleted of current carriers). See Fig. 2 P Type Silicon N Type Silicon + + + + + + + + + + Free electrons from N type migrate across the junction to “Fill in” +ve holes in P type. + Fig. 2 + + + + + + + + + + + “Depletion Zone” formed of molecules where each atom “Sees” 8 electrons in its valence layer. i.e. It is an “Insulator” or relatively high resistance. When the activity around the junction ceases, we have a “PN Junction” device. By attaching a wire to each end, which will allow us to connect the device into a circuit, we have created a silicon semi-conductor “Diode”. See Fig. 3 below. + P + N + + + Anode - Pictorial representation Cathode Diode Symbol Actual Diode Fig. 3 Current (Electron) Flow A diode is an electronic one-way street. It will allow current (electrons) to flow in one direction only, from cathode (N type) to anode (P type). In order to understand how this works, we must apply a voltage across the diode. This is often referred to as “Biasing”. “Forward Biasing” a PN Junction is when we put a positive voltage on the P type and/or a negative voltage on the N type. In the case of the diode this would be +ve on the anode and/or –ve on the cathode. “Reverse Biasing” is the opposite, -ve on the anode and/or +ve on the cathode. We will begin by reverse biasing the diode. With a –ve voltage on the anode, (P type) the +ve holes are attracted to the voltage and tend to migrate to the end of the P type material. This effectively increases 4 the width of the depletion zone. In the N type material the free electrons are attracted to the +ve voltage, and they too migrate to the end of the material, further increasing the width of the depletion. zone. This has the effect of increasing the resistance by an amount sufficient to prevent any current flow. See Fig. 4 below Depletion. Zone increases with Reverse Bias Fig. 4 + + N P + + With an increase in the reverse Reverse Bias bias there is a further increase in the Depletion Zone width. This can continue until the whole diode is depleted. Any further increase in reverse bias will force electron flow, but the electrons will not be free electrons. This current flow in the reverse direction, called ‘Avalanche Current’, will destroy the diode. The Max Reverse Bias that can be applied to any diode is one of the two most important parameters and is specified as the Peak Inverse Voltage (P.I.V.) - - + + When Forward Bias (+ to anode, - to cathode) is applied, the reverse of the above occurs. The +ve holes and –ve electrons are repelled by the same polarity bias being applied to their respective terminals. See Fig. 5. Current (Electron) Flow This causes a movement towards the Fig. 5 depletion zone having a “squeezing” + + + effect and reducing the resistance as + N P + + + holes and electrons (carriers) are + + + forced towards and into the depletion zone. The reduced resistance allows + electrons to flow through the zone. This constitutes current flow. Increasing the forward bias will cause further “Squeezing”, further reduction in resistance and corresponding increase in current. Forward bias can be increased to the point where every available electron/hole combination is being utilized. This condition is called “Saturation”. Any further increase in forward bias will cause non-free electrons to start moving, and will destroy the diode. To prevent this, every diode is given a limit of maximum forward current. This is the second of the two most important parameters. We now see that current will flow in the forward bias state but not in the reverse bias state. We can also see that the circuit voltage (forward or reverse) will cause the depletion zone to change, which in turn changes the resistance of the diode. This means the forward bias resistance of the diode is inversely proportional to the circuit voltage. It also means that the voltage across the diode will always be the same as the barrier potential. i.e. 0.7V for Silicon 0.3V for Germanium. That is, it requires a minimum voltage of 0.7V to overcome the initial resistance of the depletion zone barrier (0.3V for Germanium). Once 5 overcome, the combination of current & resistance (V=IxR) gives us a 0.7V drop. When circuit voltage increases, diode resistance decreases and current increases. With one going down (R) and the other going up (I), the voltage drop remains at 0.7V. A decrease in circuit voltage increases resistance, decreases current and the voltage drop remains at 0.7V. This is one of the most important concepts to remember in order to understand semi conductor diodes & transistors. The voltage drop across a conducting (forward biased) PN Junction is 0.7V (0.3V for Germanium). ALWAYS. The diode can therefore be regarded as an open switch when reverse biased, and a closed switch (virtual short circuit) when conducting. Remembering that when conducting, there will actually be 0.7V across a “closed switch diode” and not 0V. See Fig. 6. Remembering that this is a series circuit, total current and the resistance of the diode can easily be calculated. 0.7V D1 Vsup D1 RL 1k RL 1k + 10V 9.3V Equivalent Cct +ve Half Cycle (Forward Bias) Fig. 6 10V D1 Vsup D1 RL 1k - 10V RL 1k + Equivalent Cct -ve Half Cycle (Reverse Bias) For forward bias: ITOTAL = IRL = VRL / RL = 9.3V/1k = 0.0093A RDIODE = VDIODE / ITOTAL = 0.7V/ .0093 = 75 For reverse bias: RDIODE = VDIODE / ITOTAL = 10V/ 0A = When connected to an AC supply voltage the diode will conduct on one half cycle and cut-off during the other half cycle. On exactly which half cycle it conducts or cuts off will be determined by the orientation of the diode. Regardless of which way round the diode is connected, one of the half cycles will appear across the diode (when the diode is an “open switch” or cut off), and the other half cycle across any series components, (when the diode is a “closed switch” or conducting). The process of “removing” one half cycle is often referred to as Rectification, the diode being the Rectifier. This is the basis of most power supplies that convert AC voltages to DC voltages. For a brief explanation of this refer to Fig. 7 on the following page. An in depth explanation will be given later in the text. 0V 6 Refer to Fig. 7a. On +ve half cycles D1 is forward biased (closed switch) and therefore has no voltage across it, current flows through the circuit and the +ve half cycles appear as a voltage across R1. On the –ve half cycles, D1 is reversed biased (open switch), no current flows therefore there is no voltage (0V) across R1 and VSUP (the –ve half cycle) appears across D1. D1 Vsup Fig.7a RL 1k D1 With D1 reversed in the circuit, (See Fig. 7b) it conducts on the –ve half cycle and cuts-off on the +ve half cycle. This gives the opposite results, -ve half cycles appearing across R1 and +ve across D1. R1 Vsup Fig.7b Fig. 8 D1 Vout Vin +5V +10V Vin -10V +5.7V -10V Vout RL 1k Diodes can also be used to “clamp” a particular point in a circuit to a given voltage by “clipping” off any voltage, above or below the chosen value. See Fig. 8. Remember, that to conduct the diode anode must be more +ve than its cathode. If a +ve voltage was connected to the diode’s cathode, the anode would have to rise to +0.7V above that voltage before the diode would conduct. In the example shown the diode will not conduct until it’s anode reaches 5.7V. This means the positive 10V peak is clipped to +5.7V while the full –ve peak will remain. The reverse would be true if –5V was connected to the anode; the cathode would have to fall to –5.7V for the diode to conduct and clip the –ve half cycles. It is possible to use diodes to provide the logic “OR” & “AND” functions. The output of the logic “gates” can be logic ‘HI’ or logic ‘LO’ as desired. On the following page, both gates are depicted and in both cases the output will be a logic Hi when the gates are activated. 7 Fig. 9 shows a diode circuit that will perform a logic “OR” function with Vout as 4.3V representing a logic ‘Hi’, and 0V a logic Lo. With switches A and B both open there is no path for current to flow. Without current flow there will be 0V across R1, representing a logic Lo. If switch A or B or both are closed, one or both diodes conduct and 4.3V (5V-0.7V) appears across R1 giving a logic 1 O/P. (As long as the supply voltage is +5V the diodes are actually redundant, however, they do prevent a –ve voltage from appearing across R1) A D1 B D2 +5V R1 Vout Fig. 9 The circuit in Fig. 10 will perform a Logic “And” function with Vout as 5V representing a Logic ‘Hi’ and 0.7V a logic Lo. R1 Fig. 10 +5V D2 D1 Vout A 1 0 1 B 0 With switches A & B or, A or B in the 0 (right) position D1, D2 or both, have their cathodes connected to ground and their anodes to +5V through R1. Vout will be 0.7V as one or both diodes will conduct and Vout is across the diodes. This will represent a logic Lo. With switch A and switch B in the 1 position (left), both diodes will have +5V on their cathode (reverse bias) and will cut-off (open switch). The full 5V supply will appear across the diodes which represents a logic Hi at Vout.