View stunning SlideShares in full-screen with the new iOS app!Introducing SlideShare for AndroidExplore all your favorite topics in the SlideShare appGet the SlideShare app to Save for Later — even offline
View stunning SlideShares in full-screen with the new Android app!View stunning SlideShares in full-screen with the new iOS app!
2. DIODE LOGIC CIRCUITS The switch in a logic circuit can be implemented with an electromagnetic relay. However, arelay is not a fast acting device. Therefore, it is not suitable for high-speed switching action. Weuse electronic devices for high-speed switching. The simplest of all electronic devices is the diode.It operates in one of the two states depending upon the voltage drop across it. An ideal diode actsas a short circuit (closed switch) when it is forward biased. A reverse biased diode acts as an opencircuit (open switch). A C B RFigure 2.1: A two-diode OR gate 2.1 A Two-Diode OR Gate A simplest possible two-diode OR circuit is shown in Figure 2.1. In our discussion, let usmodel the diode by an ideal diode in series with a forward voltage drop of 0.7 V. When both theinputs A and B are zero and the supply voltage is also zero, then the output is also zero andneither diode is conducting. All possible outcomes for the two-diode OR logic are given belowwhen the low value of either input is zero and high value is 5 V. Instead of grounding the one endof the resistor, it could have also been connected to a negative supply as well. Voltages, V Corresponding truth table A B C A B C 0 0 0 0 0 0 0 5 4.3 0 1 1 5 0 4.3 1 0 1 5 5 4.3 1 1 1 The truth table is based upon the understanding that the output voltage is considered high as long as it is at least 4.3 V and low  whenever it is less than or equal to 0.7 V.Guru/LGCkts/DiodeLogic/ March 20, 2006 8 Logic Circuits
2.2 A Two-Diode AND Gate A two-diode AND gate is shown in Figure 2.2. In this case, let us assume that + VD = + 5V. The minimum value of each signal is zero and the maximum value is 5 V. Let us assume thatthe input signal A is high (5 V) and B is low (0 V). The current can now flow from the 5-V supplythrough R and the diode D2. Thus, the output voltage is 0.7 V assuming that the diode voltage dropis 0.7 V. Likewise the output voltage would also be 0.7 V when A is low and B is high. The currentin the circuit will now complete its path through D1. When both inputs are set at their low values andthe two diodes are identical, the current will divide equally between the two diodes and the outputvoltage will be 0.7 V. The output voltage would be 5 V only when both input signals are set at theirhigh values (5 V). These are the traits of an AND gate and the entire logic is tabulated below. + VD = 5 V R D1 A C D2 B Figure 2.2: A Diode AND Gate Voltages, V Corresponding truth table A B C A B C 0 0 0.7 0 0 0 0 5 0.7 0 1 1 5 0 0.7 1 0 1 5 5 5 1 1 1 In this case, the truth table is based upon the fact that the output voltage is zero  when itis less than or equal to 0.7 V. Once again, when the output voltage is greater than or equal to 4.3V, the output voltage is high .Guru/LGCkts/DiodeLogic/ March 20, 2006 9 Logic Circuits
2.3 Diode Logic For the Special Distributive Law Let us develop the diode circuit for the special distributive law (A + B) (A + C) = A + B C Figure 2.3 shows the diode logic circuit based upon the left-hand side of the aboveequation. This circuit employs two OR circuits; one for A+B and the other for A+C. The output ofthese OR circuits is the input to the AND circuit. The output F of the AND circuit provides us thedesired logic. The truth table is given on Page-7. Let us check for one of the condition in the truthtable to illustrate the technique. Let us assume that the minimum voltage of each signal is zero and the maximum is 5 V.Let us select the condition when A = 0, B = 5 V and C = 0. The expected output should be low. Letus assume that each diode has a forward voltage drop of 0.7 V when it is conducting. +5V D1 D5 D B R D2 R A F D3 -5V E C D6 R D4 -5V Figure 2.3: Diode logic for (A + B) (A + C) Since A = 0 and C = 0, D3 and D4 are ON and the voltage level at E is – 0.7 V. Since B =5 V, the voltage level at D is 4.3 V. Thus, D2 is ON and D1 is OFF. Since the voltage at E is lowerthan that at D, D6 is ON; D5 is OFF; and the voltage level at F is 0 V. You can select anothercondition from the truth table and verify the outcome.Guru/LGCkts/DiodeLogic/ March 20, 2006 10 Logic Circuits