When two asynchronous active low inputs PRESET and CLEAR are applied to a J-K flip flop the output will be
Undefined
When working with digital logic circuits, especially sequential elements like flip-flops, it's crucial to understand the role of both synchronous and asynchronous inputs. J-K flip-flops, like many other types of flip-flops, are often equipped with asynchronous control inputs, typically labelled PRESET (or SET) and CLEAR (or RESET).
Asynchronous inputs are special control signals that directly affect the output of a flip-flop, bypassing the clock and the main data inputs (J and K in this case). They are called "asynchronous" because their effect is immediate and does not depend on the clock signal. They are usually active low, meaning they perform their function when the input is at a logic 0.
The question describes a scenario where both asynchronous active-low inputs, PRESET and CLEAR, are applied simultaneously to a J-K flip-flop. This means both PRESET = 0 and CLEAR = 0 at the same time.
Since the Q output of a flip-flop cannot simultaneously be both a logic 1 and a logic 0, applying both PRESET and CLEAR at the same time creates a conflicting condition. This situation leads to an undefined or indeterminate output state. The exact output might vary depending on the internal gate delays and manufacturing variations of the specific flip-flop. This is often referred to as a "race condition" because the two inputs are "racing" to control the output in opposite directions.
It's important to avoid this state in practical digital circuit design as it leads to unpredictable behavior and can cause system instability.
| PRESET | CLEAR | Q Output | $\bar{Q}$ Output | Condition |
|---|---|---|---|---|
| 1 | 1 | No change | No change | Normal operation (clock, J, K control) |
| 0 | 1 | 1 | 0 | Set (PRESET) |
| 1 | 0 | 0 | 1 | Clear (CLEAR) |
| 0 | 0 | Undefined | Undefined | Invalid/Prohibited condition |
Therefore, when both active low PRESET and CLEAR are applied to a J-K flip-flop, the output will be undefined.
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The basic building block of sequential logic circuit is-
A single flip-flop is a modulo __________ counter.
For a JK Flip‐flop
A. When J = 0, K = 1, Q n+1 = 0
B. When J = 1, K = 1, Q n+1 = 1
C. When J = 1, K = 1, Q n+1 =\(\rm \overline{Q_n}\)
D. When J = 1, K = 0, Q n+1 = 1
E. When J = 1, K = 0, Q n+1 = 0
Choose the correct answer from the options given below: