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How are diodes used in logic circuits?
Diode logic gates use diodes to perform OR and AND logic functions as shown in the circuit diagram. Connection of the LED at the output is optional which simply displays the logical state of the output, i.e. the logic state of output is 0 or 1, if LED is off or on, respectively.
What are the characteristics of diode transistor logic?
Diode-Transistor Logic, or DTL, refers to the technology for designing and fabricating digital circuits wherein logic gates employ both diodes and transistors. DTL offers better noise margins and greater fan-outs than RTL, but suffers from low speed, especially in comparison to TTL.
What is advantage of diode transistor logic?
As every emitter is nothing but a diode, this logic eliminates the use of all diodes. This is the major advantage. As transistor becomes ON and OFF much rapidly than a diode, switching time will be faster. TTL, or Transistor-transistor logic replaced resistor-transistor logic, and used much less power.
How are diodes and transistors in a logic circuit?
It consists of two diodes and a transistor. The two diodes D A, D B and the resistor R 1 form the input side of the logic circuit. The common emitter configuration of transistor Q 1 and resistor R 2 forms the output side. How does a 2-input DTL NAND gate operate?
What are the diodes in a NAND gate?
The following figure shows the circuit for the 2-input DTL NAND gate. It consists of two diodes and a transistor. The two diodes D A, D B and the resistor R 1 form the input side of the logic circuit. The common emitter configuration of transistor Q 1 and resistor R 2 forms the output side.
Why is the transistor in cut off mode?
So the current due to the supply voltage +V CC = 5 V will go to the ground through R 1 and the two diodes D A and D B. The supply voltage gets dropped in the resistor R 1 and it will not be sufficient to turn ON the transistor. So the transistor will be in cut off mode.
Which is diodes conduct in the forward direction?
When both the inputs A and B are LOW, the diodes D A and D B become forward biased and so both diodes will conduct in the forward direction. So the current due to the supply voltage +V CC = 5 V will go to the ground through R 1 and the two diodes D A and D B.