3

Click here to load reader

Diode logic crkts

Embed Size (px)

Citation preview

Page 1: Diode logic crkts

Guru/LGCkts/DiodeLogic/ March 20, 2006 Logic Circuits 8

2. DIODE LOGIC CIRCUITS

The switch in a logic circuit can be implemented with an electromagnetic relay. However, a

relay is not a fast acting device. Therefore, it is not suitable for high-speed switching action. We

use 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 acts

as a short circuit (closed switch) when it is forward biased. A reverse biased diode acts as an open

circuit (open switch).

A

BC

R

Figure 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 us

model the diode by an ideal diode in series with a forward voltage drop of 0.7 V. When both the

inputs A and B are zero and the supply voltage is also zero, then the output is also zero and

neither diode is conducting. All possible outcomes for the two-diode OR logic are given below

when the low value of either input is zero and high value is 5 V. Instead of grounding the one end

of 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

[1] as long as it is at least 4.3 V and low [0] whenever it is less than or equal to 0.7 V.

Page 2: Diode logic crkts

Guru/LGCkts/DiodeLogic/ March 20, 2006 Logic Circuits 9

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 = + 5

V. The minimum value of each signal is zero and the maximum value is 5 V. Let us assume that

the input signal A is high (5 V) and B is low (0 V). The current can now flow from the 5-V supply

through R and the diode D2. Thus, the output voltage is 0.7 V assuming that the diode voltage drop

is 0.7 V. Likewise the output voltage would also be 0.7 V when A is low and B is high. The current

in the circuit will now complete its path through D1. When both inputs are set at their low values and

the two diodes are identical, the current will divide equally between the two diodes and the output

voltage will be 0.7 V. The output voltage would be 5 V only when both input signals are set at their

high values (5 V). These are the traits of an AND gate and the entire logic is tabulated below.

A

B

+ V = 5 V

C

D

R

D1

D2

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 [0] when it

is less than or equal to 0.7 V. Once again, when the output voltage is greater than or equal to 4.3

V, the output voltage is high [1].

Page 3: Diode logic crkts

Guru/LGCkts/DiodeLogic/ March 20, 2006 Logic Circuits 10

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 above

equation. This circuit employs two OR circuits; one for A+B and the other for A+C. The output of

these OR circuits is the input to the AND circuit. The output F of the AND circuit provides us the

desired logic. The truth table is given on Page-7. Let us check for one of the condition in the truth

table 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. Let

us assume that each diode has a forward voltage drop of 0.7 V when it is conducting.

+ 5 V

A

B

C

- 5 V

- 5 V

D

E

F

D1

D2

D3

D4

D5

D6

R

R

R

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 lower

than that at D, D6 is ON; D5 is OFF; and the voltage level at F is 0 V. You can select another

condition from the truth table and verify the outcome.