Time Delay in Microprocessor

Time Delay in Microprocessor – In real-time applications, such as traffic light control, digital clock, process control, and serial communication, it is important to keep track of time. In some applications, a time delay is required between the execution of two instructions.

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Microprocessor Time Delay

There are two techniques to design the time delay in a microprocessor.

Hardware technique
Software technique

Hardware Technique

For the hardware time delay technique, additional hardware like 8253/8254 (Programmable Interval Timer) is required. The hardware technique has the advantages like it is very accurate. It can be used for real-time applications and frees the microprocessor to perform other tasks at the same time. The fundamental limitation of the hardware technique is the cost.

Software Technique

In the software time delay technique, no additional hardware is required, so it is cheaper than the hardware technique. But it has some limitations, like it is not accurate, makes the microprocessor busy for the entire delay period, etc. The tedious calculation is required to decide the count. The following methods are used to provide a time delay using a software technique.

Using the NOP instruction
Using counter
Using a nested loop

Time Delay using NOP Instruction

Meaning of instruction NOP is NO operation. This instruction does nothing but simply makes the microprocessor busy for 4T states. So by executing a NOP instruction in between two instructions, a delay of 4T states may be provided.

1T-state = \frac{1}{Operating frequency}

Let’s take an example question to understand the time delay using the NOP instruction

Question – How many times does the instruction NOP have to be executed to provide a time delay of 1m sec? between two instructions. The crystal frequency is 4MHz.

Solution – Given that

Crystal frequency = 4MHz

Hence, the operating frequency (f) = \frac{1}{2} (Crystal frequency)

= \frac{4}{2} MHz

= 2MHz

Time required to execute NOP instruction = 4T

= 4\left(\frac{1}{f}\right)

= \frac{4}{2 \times 10^6} sec.

= 2μ sec.

Number of times the instruction NOP has to be executed to provide a time delay of 1m sec. between two instructions is (1m sec)/(2μ sec) = 500 times. From the above example solution, it is clear that the provided delay using the NOP instruction is not a good solution. This is because the programmer has to store a large number of NOP instructions in memory.

Time Delay using Counter

A counter can be used to provide the time delay. According to this technique, a counter is initialized with the desired value and decremented by one. This looping operation continues till the count value reaches zero. Since the execution times of the instructions used in a counting are known, the initial value of the counter can be determined. The counter can be designed by using a register (8-bit counter) or a register pair (16-bit counter). Figure 1 shows the flow chart for time delay using a counter.

Using One Register (8-bit counter)

Lable	Instruction Opcode	Instruction Operand	Comments	Machine Cycles	No. of T-states
	MM	C, n	Load the count of register C.	OF + MR	4T + 3T = 7T
LOOP	DCR	C	Decrement counter	OF	4T
	JNZ	LOOP	If the count is not zero, go to the location loop	OF + MR + MR	10/7T

Using one Register (8-bit counter)

Here, OF = Opcode Fetch and MR = Memory Read

The instruction JNZ is a three-byte instruction. The number of T-states for the JNZ instruction is shown as 10/7 in the program. It means that the microprocessor 8085 requires 10T-states to execute the instruction when the condition is true. But when the condition is false, the loop is terminated, requiring only 7T states. Time delay in the 8-bit counter using one register is given as below:

T_{Delay} = 7T + (n-1)(4T + 10T) + (4T + 7T)

= 7T + 14nT – 14T + 11T

= 7T + 14nT – 3T

T_{Delay} = 4T + 14nT

Let’s take an example question to understand the time delay using a one-register (8-bit Counter)

Question – Find the maximum time delay that can be provided using one register. The operating frequency of the microprocessor is 2MHz.

Solution – For the maximum time delay, the count value must be maximum. The maximum count value can be FFH or 25510 using one register. Therefore, the maximum time delay is given as follows:

T_{Delay} = 4T + 14nT

= 4 \times \frac{1}{2 \times 10^6} + \frac{14 \times 255}{2 \times 10^6} sec.

= 1.785m sec.

Using Register Pair (16-bit Counter)

The maximum time delay (1.785 sec.), which can be introduced between two instructions using one register, is not sufficient for some applications. In those applications, we need to initialize the counter more than FFH or 255. Hence, a counter using a register pair instead of one register is used.

Label	Instruction Opcode	Instruction Operand	Comments	Machine Cycles	No. of T-states
	LXI	D, N	Load counter with 16-bit number N	OF + MR + MR	4T+3T+3T = 10T
LOOP	DCX	D	Decrement counter	OF	6T
	MOV	A, E	Copy the contents of register E to register A	OF	4T
	ORA	D	Logically OR the contents of register A and register D to check the counter	OF	4T
	JNZ	LOOP	Repeat until the DE register pair reaches zero.	OF + MR + MR	10/7T

Using a register pair (16-bit counter)

Here, OF = Opcode Fetch and MR = Memory Read.

In the above program, the DCX instruction is used to decrement the contents of the DE register pair. This instruction has 6T states in its opcode fetch machine cycle and does not affect any flag. When the contents of the DE register pair reach zero, the zero flag remains unchanged. Therefore, another technique is used to affect the zero flag as shown in programming by performing an ORing operation between the lower and higher pair, as shown below:

T_{Delay} = 10T + (N-1)(6T + 4T + 4T + 10T) + (6T + 4T + 4T + 7T)

= 10T + 24NT – 24T + 21T

T_{Delay} = 7T + 24NT

Let’s take an example question to understand the time delay using the Register Pair (16-bit Counter).

Question – Write a program to generate a delay of 0.4 sec. If the crystal frequency is 5MHz.

Solution – In the 8085 microprocessor, the operating frequency is half of the crystal frequency, hence

Operating \space frequency (f) = \frac{5 \times 10^6}{2} Hz = 2.5MHz

Time \space for \space one \space T-state = \frac{1}{2.5 \times 10^6} sec = 0.4μ sec

Now, \space we \space know \space T_{Delay} = 7T + 24NT \space for \space 16-bit \space counter

\Rightarrow N = \frac{T_{Delay} – 7T}{24T} = \frac{0.4 – 7 \times 0.4 \times 10^{-6}}{24 \times 0.4 \times 10^{-6} }=41666_{10} \space or \space A2C2H

The value of count is 41666₁₀ or A2C2H. The delay loop using a 16-bit counter will be as follows:

Label	Instruction Opcode	Instruction Operand	Comment
	LXI	D, A2C2H	Load the counter with a 16-bit number A2C2H.
LOOP	DCX	D	Decrement counter.
	MOV	A, E	Copy the contents of register E to register A.
	ORA	D	Logically OR the contents of register A and register D to check the counter.
	JNZ	LOOP	Repeat until the DE register pair reaches zero.

Using nested Loop

This technique is also known as time delay using a loop within a loop. In this, there is more than one loop. The outer loop sets the count to repeat the delay provided by the inner loop. A flowchart for time delay using a nested loop is shown in Figure 2.

Label	Instruction Opcode	Instruction Operand	Comments	Machine Cycles	No. of T-states
	MM	B, N_out	Load the outer counter	OF + MR	4T + 3T = 7T
	MM	C, N_in	Load the inner counter	OF + MR	4T + 3T = 7T
LOOP2	DCR	C	Decrement the inner counter	OF	4T
LOOP1	JNZ	LOOP1	If not zero goto LOOP1	OF + MR + MR	10T / 7T
	DCR	B	Decrement the outer counter	OF	4T
	JNZ	LOOP2	If not zero goto LOOP2	OF + MR +MR	10T/7T

Let’s take an example program to understand time delay using a nested loop.

Question – Write a program to count continuously in hexadecimal from FFH to 00H in a system with a 0.5μ sec. clock period. Use register C to set up a 1m sec. Delay between each count and display the numbers at one of the output ports. Also, show the delay calculation.

Solution – The program has two parts; first is to design a given delay of 1m sec. between two counts, and the second is to set up a continuous down counter from FFH to 00H.

Delay Calculation: The required delay is 1m sec. which is less than the maximum delay provided by using one register (1.785m sec.). Hence, the delay counter can be designed with one register.

T_{Delay} = 4T + 14nT

\Rightarrow 1 \times 10^{-3} = (4 + 14n) \times 0.5 \times 10^{-6}

\Rightarrow n = 142.6_{10} ≃ 143_{10}

= 8FH

Let the accumulator be used as a counter to display, and register C is used as a counter for the delay loop. The following algorithm is used to design the counter.

Algorithm:

Step 1: Initialize the counter (accumulator) with FFH to display.
Step 2: Display the value of the accumulator.
Step 3: Provide a desired delay.
Step 4: Decrement the counter and go to step 2.

Flowchart for Hexadecimal Counter with Delay of 1m sec — **Figure 3:** Flowchart for Hexadecimal Counter with a Delay of 1m sec

The assembly language program is given below for this problem.

Memory Address	Hex Code	Label	Instruction Opcode	Instruction Operand	Comments
XX00H XX01H	3E FF		MM	A, FFH	Initialize the display counter
XX02H XX03H	D3 Port #	RPT	OUT	Port #	Display the contents of the accumulator at port #
XX04H XX05H	0E 8F		MM	C, 8FH	Initialize the delay counter
XX06H	0D	LOOP	DCR	C	Decrement the delay counter
XX07H XX08H XX09H	C2 06 XX		JNZ	LOOP	If not zero, goto memory location LOOP
XX0AH	3D		DCR	A	Decrement the display counter
XX0BH XX0CH XX0DH	C3 02 XX		JMP	RPT	Jump to location RPT

Assembly Language Program

Microprocessor Subject Tutorials & Notes | Beginners Guide

Time Delay in Microprocessor

Microprocessor Time Delay

Hardware Technique

Software Technique

Time Delay using NOP Instruction

Time Delay using Counter

Using nested Loop

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