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DIGITAL FUNDAMENTALS Microprocessors don't do anything until they are programmed. Our final chapter in the series takes a look at programming basics. LESSON 10: PROGRAMMING MICROCOMPUTERS By Louis E. Frenzel MICROCOMPUTERS BY THEMSELVES DON'T do very much. They are simply a collection of electronic circuits and other hardware waiting for directions which tell them what to do. Those directions come from the programs that cause the microcomputer circuits to perform in a specific way. Once given a program, the microcomputer will accomplish some useful work. All the programs that computers use are referred to as software. In this final lesson, we discuss the basic process of creating the software that will make the microcomputer do something worthwhile. More specifically, we will show you how to write some simple programs using the microcomputer's instruction set. · Microcomputers have, for the most part, replaced large complex collections of smaller SSI and MSI integrated circuits. By programming a microprocessor, you can cause it to perform all of those basic operations which are usually carried out by individual gates and flipflops wired in many different configurations. Microprocessors can be programmed to perform arithmetic operations, logical operations such as AND, OR and NOT, counting, shifting, comparing, and many other standard digital functions. We will show you how to accomplish many of those basic operations in this lesson. Introduction to Programming Programming is the process of converting a problem to be solved into a form that the microcomputer can interpret. During that process, a program is created. Recall that a program is a sequential list of microcomputer instructions that tell how to accomplish some specific result. Programming is basically a two-step process. The 66 SILICON CHIP first step is problem solving where you define the work to be done and develop a solution. The second step is called coding. That is the process of converting your problem into a specific computer language. In this case, that language is the specific instruction set of the microcomputer. You will hear it called machine language. Following the work we started in Lesson 9, we will use the instruction set of the type 6800/6502 microprocessors. Problem Solving The first step in programming is to clearly define the problem. That is where you ask yourself what it is that you are trying to do. The best way to begin the process is to write down a general statement of what you are trying to accomplish. Anything else you can do to explain or illustrate the problem will also be helpful to you in coming up with a solution. For example, draw yourself a picture using blocks that might show the flow of the problem from one step to the next. If necessary, make charts and tables or lists that specifically state what needs to be done. Keep in mind a picture of the basic computing process, which involves input data of some kind that is then processed to create some useful new output data. Defining the available inputs and the desired outputs will help you understand exactly what processing must take place. Next, analyse all of the material you have collected. That will result in your coming up with an algorithm. An algorithm is a detailed step-by-step procedure for solving the problem. It could be a basic algebra statement or formula to compute. Otherwise, it might be even more like a recipe with individual steps to be accomplished one after another. Fig.1: a flowchart is useful for describing the problem to be solved. This one is for determining the larger of two numbers, A and B. START ACC A CONDITION CODE REGISTER (CCR) ACCUMULATORS{' • INSTRUCTION REGISTER (OP CODE) ACC B ,...L===== NEGATIVE ZERO ---CARRY PROGRAM COUNTER STACK POINTER NO (A>B) INDEX REGISTER Fig.2: this is the basic register organisation in the 6800 microprocessor. Note that there are two accumulators. STORE B STORE A But to refresh your memory, let's take a brief look at the architecture of the device and the formats of the instructions. STOP One way to illustrate the problem you are trying to solve and test is to draw yourself a flow chart. That is a graphical description of the problem using standard symbols. Fig.1 illustrates the solving of a typical problem. The oval symbols represent start and stop conditions. The rectangular boxes state specific processing operations to be carried out on your data. The diamond-shaped figure represents a decision, which usually asks a question with a "yes" or "no" answer. Work your way through the problem in Fig.1 to be sure that you understand how it works. Coding Once your problem is well-defined and your algorithm is available, you can begin writing the program. That is the process of creating a sequential list of the microprocessor's instructions. When you finish coding the program, you should check it over carefully for errors. Programming is a very time-consuming and tedious process and it is extremely easy to make a mistake. You may goof in thinking through your algorithm and there may be some faulty logic that you will not discover until you test the program. Or you may simply make a coding error, leaving out an instruction or entering an incorrect op code or address. In any case, attempt to get out all the bugs. Next, you will load the program into the microcomputer and test it. If everything is satisfactory, the computer will run the program and will give the desired results. If not, you will need to do some debugging. Debugging is typically standard procedure, particularly for large, complex programs. Short programs can usually be written and run the first time without error. But big programs, using hundreds or even thousands of instructions, generally require hours of debugging to make them work properly. · As stated earlier, we will use the architecture and instructions of the 6800 and 6502 microprocessors. Architecture Fig.2 shows the general layout of the 6800 microprocessor. Note that it has two 8-bit accumulators labelled ACC A and ACC B. An 8-bit instruction register holds the op code, and an 8-bit condition code register (CCR) indicates CPU status. Also included is a program counter, a stack pointer and an index register. Those are 16-bit registers. We won't show the block diagram of the 6502 since it is basically similar in its organisation. The main difference is that the 6502 has only a single accumulator (A), but it does have two index registers labelled X and Y. The instruction sets are also similar and we will use instructions that are common to both. In operation, the microprocessor fetches instructions which are in the memory, puts them into the instruction register, then decodes and interprets them. The logic in the microprocessor then causes the circuitry to carry out the designated operations. Most of those operations take place on data that is in the accumulator. Data may be brought in from the memory and stored in the accumulator, then used in some arithmetic or logic operation with the result stored in the accumulator. That result can then be transferred to a storage location in memory. Once an instruction is executed, the next instruction in sequence in memory is then fetched and executed. The process continues until the end of the program is reached. Condition Codes As the program instructions are executed, various conditions are monitored and the results are stored in the flipflops of the condition-code register (CCR). Those flipflops are also known as status flags. Some of the more common conditions monitored include carry (C), zero (Z), and negative (N). For example, the carry bit is set to binary 1 in the condition-code register if the result of an addition in the accumulator causes a carry to be generated out of the most-significant-bit (MSB) position. SEPTEMBER 1988 67 OP CODE ADDRESS OR DATA OP CODE 1/2 OF ADDRESS Attit~1s { ~--------f 1/2 OF ADDRESS Fig.3: a 2-byte instruction puts the complete address or data in one memory location. A 3-byte instruction puts only half the address in a memory location. BINARY HEXADECIMAL 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 0 1 2 3 4 5 6 7 8 9 A B C D E F Fig.4: binary to hexadecimal conversion. The binary number 10001011 is the equivalent of hex 8B. The zero bit in the condition-code register is set if the result of some operation in the accumulator is zero. Should some arithmetic, logic, or other operation result in the accumulator containing all zeros, the zero flipflop is set to binary 1. In a similar way, the negative or N flag of the condition-code register will be set to binary 1 if the result of an arithmetic operation produces a negative result. A negative result is indicated by the most significant bit of the accumulator [bit 7) being a binary 1. The condition-code bits can then be monitored so that decisions can be made by special jump and branch instructions. We will illustrate these in the programs to come. Instruction Format An instruction for the microprocessor consists of one, two or three bytes stored in memory. Some simple instructions contain only an 8-bit op code that designates an operation to be performed. Some instructions require two bytes, as shown in Fig.3a. The first byte is the op code that tells the microprocessor what is to be done. The second byte usually contains an address that designates where the data to be operated upon is stored. Alternatively, the second byte may also contain the data itself to be used in the operation. A three-byte instruction [shown in Fig.3b) contains an op code in the first byte; the remaining two bytes contain the upper and lower halves of a 16-bit address, which usually designates the location where the data to be operated on is stored. The instructions, of course, are nothing more than binary numbers. However, programmers don't work with binary. It is too difficult to remember all those ls and 0s. Most programmers who write programs in 68 SILICON CHIP machine language typically use hexadecimal [hex) notation. This is the process of replacing each 4-bit binary group of numbers with its equivalent hexadecimal symbol. · The hexadecimal symbols include the digits 0 through 9 and the letters A through F. Fig.4 shows the designations. To express a given binary number in hex, simply replace each 4-bit group with its equivalent hex digit. For example, the binary number 10001011 in hexadecimal is 8B. This happens to be the op code for the ADD instruction in the 6800. All of the instruction op codes have a specific hexadecimal designation. In addition, the addresses are also expressed in hex notation. To keep things simple in this lesson, we are going to use our own special notation. We will use a simple three or four-letter mnemonic designation for each instruction and express memory locations and data in terms of decimal numbers. We will also limit ourselves to the use of two-byte instructions. Simple Example The simple program shown below is designed to add two numbers together and store their sum. Let's step through the program one instruction at a time so that you can see how to read it. LOCATION MEMORY INSTRUCTION ADDRESS OR DATA 0 LDAA 1 7 2 ADD 8 STAA 3 4 5 6 1 8 9 9 HALT 25 19 44 (SUM) The first instruction in location 0, LDAA, loads accumulator A with the number stored in memory location 7. That is designated by the second byte of the instruction in location 1. Looking down the program, you can see that in memory location 7 is the number 25. That first instruction brings the number 25 into accumulator A. The next instruction in location 2 is ADD. This instruction tells the computer to add the number stored at the memory location designated by the second byte of the instruction [in location 3) to the number contained in accumulator A. As you can see, the number 19 is stored in location 8. The sum, 44, will be stored in the accumulator. The third instruction in location 4, STAA, says to store the accumulator in location 9. That is, the sum in the accumulator will now be stored in the memory location designated by the second byte of the instruction in memory location 5. The address stored there is 9. Therefore, the sum will be stored in that location. The program ends at that point. Now that you have the format in mind, let's take a look at a number of sample programs that show how to do useful functions. In each example we will show a programming listing and explain what is happening. We will introduce new instructions in each sample, but those will be explained as we go along. Arithmetic Operations MEMORY LOCATION c-------~ INSTRUCTION ADDRESS OR DATA 0 LDAA Let's code the program 1 18 described by the flow chart 2 SUBA 3 19 in Fig.1. The purpose of the 4 BMI program is to determine 12 5 6 LDAA which of two numbers, A and 7 18 B, is larger. Our algorithm is 8 STAA 9 2D to subtract B from A, then to 10 JMP 11 test to see if the difference is 22 12 LOAA positive or negative. If the 13 19 14 STAA difference is positive, then A 15 20 is larger than B. If the dif16 JMP 17 22 ference is negative, B is 18 A= 14 larger than A. The program 19 B = 17 20 LARGER= 17 concludes by storing the 21 larger of the two numbers in 22 HALT texta designated memory location. The program for solving the problem is illustrated at the top of this page. The first instruction, LDAA, loads the first number A stored in memory location 18 into accumulator A. The second instruction subtracts number B stored in location 19 from number A. The difference between the two numbers is stored back in the accumulator. If A is 14 and Bis 17, the result in the accumulator is -3. The next instruction is a branch-if-minus (BMI) instruction. This is a decision-making instruction that may change the sequence of program execution if a certain condition occurs. In this case, the instruction tests the negative flipflop in the condition code register. Recall that various flipflops in the condition code register are set or reset depending upon the outcome of the execution of the various instructions. If the subtraction results in a positive result in the accumulator, then the negative flipflop in the condition code register is set to binary 0. If the subtraction operation results in a negative value, then the negative flipflop is set to binary 1. The branch if minus instruction looks at the negative flipflop to determine what action to take. If the subtraction operation results in a positive value, no branch occurs. The next instruction in sequence after the BMI instruction is executed as would normally be the case. Therefore, A is the larger of the two numbers. The LDAA instruction in location 6 causes that number to be loaded into the accumulator. The next instruction, STAA, stores that number in the designated memory location. Next, the JMP instruction causes an unconditional jump to occur. The program counter is loaded with the address part of the JMP instruction in location 11. Therefore, the CPU resumes executing at location 22. A HALT at that location ends the program. (Note: in this article, a HALT instruction is used to stop a program. In truth, there is no such instruction in the 6800 or 6502 . It is used here to simplify the discussion. In practice, other more complex techniques beyond the scope of this article are used). r MAIN PROGRAM I I I I INSTRUCTIONS JUMP TO SUBROUTINE - ' • SUBROUTINE I I ~ I I I ~ -- -- - RETURN I l. r J I I I JUMP TO SUBROUTINE _ _ _ _ _ _ _ _ _ ..JI Fig.5: a small program which will be used many times by the main program can be stored as a subroutine, to which the main program jumps (and returns) when it's needed. If the result of the subtraction is a negative number, as it is in this example, then the BMI instruction detects it and causes the program to branch to a memory address designated in the second byte of the BMI instruction in location 6. In other words, the usual next instruction in sequence is not executed. The BMI instruction causes the program counter to be loaded with the address designated in the BMI instruction, in this case 12. This indicates that the larger of the two numbers is B. Therefore, the instruction at 12, an LDAA, causes the B to be loaded into the accumulator. The STAA instruction stores the result in the memory location designated for the larger number. The JMP instruction then causes the program to jump to the designated memory location, in this case 22. At that point, the program ends. Subroutines Most simple 8-bit microprocessors such as the 6800 and 6502 can only perform simple arithmetic operations such as add and subtract. But many times the problem to be solved requires more complex math, such as multiplication, division, square root or some other operation. The microprocessors can perform these operations, but they must be programmed to do so. A special program is written to perform one of these math operations. Whenever the main program needs to perform, say, a multiplication, it will branch to or call this special multiplication program, perform the desired operation, and then return to the main program sequence. This is illustrated in Fig.5 . Such programs, which may be referenced several times by the main program, are called subroutines. A subroutine is simply a specialised segment of computer code that does some specific task. Rather than write that sequence of code each time it is needed, the code is written once. Then special inSEPTEMBER 1988 69 START (PARTIAL PRODUCT) chosen here is simple and is easy to understand. Since the microprocessor can add and subtract, naturally the subroutine must use these instructions. One way to perform multiplication is by repeated addition. For example, if you wish to multiply 5 by 3, you can accomplish it by simply adding 5 three times. This is what our program will do. Add to Multiply LOOP N STOP Fig.6: as this flowchart shows, multiplication is done by repeated addition. structions are used by the main program to call the subroutine, use it, then return to the main program. A jump to subroutine instruction [JSR) calls up the subroutine. This instruction's address tells where the subroutine is located. When a JSR is executed, the contents of the program counter are stored in the stack. That allows the CPU to pick up where it left off when the subroutine has been executed. At the end of the subroutine is a return-fromsubroutine instruction (RTS). It retrieves the address in the stack and loads it into the program counter. The main program then resumes executing. The program we want to illustrate here is a subroutine for performing multiplication. It will illustrate the concept of a subroutine and how higherlevel math functions are programmed using the microprocessor instruction set. There are a number of ways to produce multiplication in a microprocessor. The algorithm we have 70 SILICON CHIP Recall that in multiplication, there are usually two quantities involved. One number is called the multiplicand while the other is called the multiplier. The answer or solution to the multiplication results in a new value called the product. In the microprocessor we will set aside three memory locations to store those three values. The flow chart outlining the multiplication is shown in Fig.6. Basically, the process is to add the multiplicand for the number of times specified by the multiplier. In the program, we designate the location where the final product will be stored. During the multiplication process, the partial product obtained each time the multiplicand is added is stored in that location. Initially, the partial product is zero. Then, each time we pass through the program, the multiplicand is added to it and then restored. Each time the multiplicand is added, the multiplier is loaded into the accumulator and one is subtracted from it. That tells us that the multiplicand has been added one time. We then reduce the multiplier by 1 and restore it. We then check to see if the multiplier is zero. Naturally, if it is zero, then the multiplicand has been added the correct number of times. If not, then the program goes back and repeats the same sequence. Note in the flow chart the use of a decision block designated Is multiplier zero? If the answer is yes, then the correct product is available for use. If the multiplier is not zero, the answer is no and the procedure begins again. That forms a program loop. A loop in programming terms simply means that a particular sequence of instructions is repeated a number of times as required by the algorithm. A loop is implemented with a decision making branch or jump instruction. The code for the problem is given below. Let's walk through it step-by-step to be sure you understand the process. The first instruction, CLR, clears the memory location designated by the second byte to be cleared to zero. As you can see, that clears memory location 20 where the partial product is to be stored. The next instruction, LDAA, causes the partial product to be loaded into the accumulator. Of course, it is zero at this time. The next instruction, ADD, causes the multiplicand in location 18 to be added to the accumulator. The result in this case is simply the multiplicand itself. The next instruction, STAA, causes the contents of the accumulator to be stored back into location 20, the partial product. Next, the LDAA instruction in location 9 causes the multiplier in location 19 to be loaded into the accumulator. The DECA instruction then decrements it. MEMORY LOCATION 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 INSTRUCTION ADDRESS OR DATA CLR 20 LDAA 20 ADO 18 STAA 20 LOAA 19 OECA STAA 19 BNE 3 RTS MULTIPUCAND MULTIPLIER PRODUCT (PARTIAL PRODUCT) That instruction simply causes 1 to be subtracted from the contents of the accumulator. To decrement means to subtract one. The multiplier has now been reduced by 1, indicating that the multiplicand has been added once. We restore the reduced multiplier in location 19. Next, a BNE (branch if not equal to zero) instruction is executed. That instruction looks at the decremented multiplier in the accumulator and attempts to determine if its value is equal to zero. If it is not equal to zero, then it causes the normal program sequence to be abandoned. A branch occurs to the memory location designated in the second byte of the BNE instruction. In this case, the branch occurs back to memory location 3. This causes the instructions stored between locations 3 and 13 to be re-executed. The process, therefore, is repeated. The multiplicand is added again to the partial product and the result is restored. The multiplier is loaded and decremented by 1 and restored. Again, the BNE checks for any zero result. If none occurs, then a backwards loop is again performed. At some point, however, the multiplier will be reduced to zero. At that point the BNE instruction detects the zero so no branch or loop occurs. The next instruction in sequence at location 16 is executed. That is a return from subroutine (RTS) instruction. In a subroutine, the final instruction is called a return from subroutine. The instruction simply retrieves the content of the program counter which was previously stored in the stack by the jump to subroutine instruction that caused the subroutine to be called. At that point, the main program resumes. cumulator while the other is stored in a memory location that will be referenced by the AND instruction. The AND instruction is designated ANDA which means to AND the contents of accumulator A with the contents of the memory location designated in the instruction. The corresponding bits in the two 8-bit words will be ANDed together with the logical result being stored back in that same bit position in the accumulator (see Fig.7). The individual bits being operated on are independent of one another and do not affect adjacent bits. Because of this, you can perform various individual bit manipulations within a word if desired. OR operations can also be performed with the ORA instruction. The EXCLUSIVE OR function is performed with the EORA instruction. As for performing the NOT function, a special instruction called complement [COM) is used. Here a word is loaded into the accumulator and the COM instruction is executed. The result is that each bit in the word is complemented; that is, ls are changed to Os and Os changed to ls. With those basic logic functions, virtually any Boolean operation can be implemented with short subroutines. An example of a useful logical operation is to use the AND instruction to perform masking. Masking is the process of blocking out or eliminating a bit or groups of bits from a binary word. When the bits are eliminated, they are effectively set to or replaced by binary Os. To mask bits from a given word, that word is first stored in the accumulator. Then, a special mask word in some memory location is ANDed with it. To eliminate the desired bits from the word, the corresponding bits in the mask are set to binary 0. Those bits to be retained are ANDed with binary ls in the mask word. Here's an example: A - BINARY WORD 11001100 B - MASK WORD 00111100 C - RESULT OF ANDING 00001100 The truth table for the basic AND function is given below as a refresher. 8 C 0 1 0 1 0 0 0 1 The simple program below illustrates the use of the AND instruction for masking. The purpose of the program is to convert ASCII characters into standard BCD numbers. Logical Operations As we indicated, a microprocessor can also perform logical operations such as AND, OR and NOT. It can even perform more complex operations such as EX-. CLUSIVE OR. The 6800 and 6502 microprocessors, for example, have instructions for those operations. The instructions operate on all of the bits in the accumulator rather than one bit at a time. For example, to perform an AND operation between two binary numbers, one of them is stored in the ac- A 0 0 1 1 MEMORY LOCATION INSTRUCTION ADDRESS OR DATA 0 1 2 3 4 LDAA 5 6 7 8 9 10 8 ANDA 9 STAA 10 HALT 00111001 (ASCn CODE FOR 9) 00001111 (MASK WORD) 00001001 (BCD RESULT} ASCII Code Recall that ASCII is the 8-bit code widely used for communications between computers and peripheral SEPTEMBER 1988 71 ACCUMULATOR __ MEMORY _LOC&_J:ION Fig.7: the AND instruction is designated ANDA which means to AND the contents of accumulator A with the contents of the memory location designated in the instruction. devices or in data communications. A special 8-bit code is given to upper and lower case letters of the alphabet, numbers, punctuation marks and special symbols. The ASCII codes for the numbers 0 to 9 turn out to be the standard 4-bit BCD codes for those numbers plus a special 4-bit code preceding it, in this case 0011. For example, the BCD number for 5 is 0101. The ASCII version, therefore, is 00110101. In order to use ASCII numbers in arithmetic operations, they must first be converted into their BCD equivalents. A simple way of doing that is to strip off the first four bits, leaving the 4-bit BCD code. That can Mailbag - SILICON CHIP Implementing Counters A microprocessor can also be used to perform counting functions. For example, by using the interrupt feature of a microprocessor, external pulse trains can be counted. The pulses to be counted are simply applied to the interrupt input of the microprocessor. continued from page 3 This was because: (a) the additional switches installed in the caravan were only single pole instead of double pole; and (b) the inlet socket on the caravan had been incorrectly wired at the factory, even though it was clearly marked "A" and "N" on the appropriate terminals. Prominently displayed in the caravan was a sign to the effect that all electrical wiring had been inspected and passed. It had been used like this for many years. On the table was an old toaster with hinged sides, so it was surprising that no-ori.e had· been zapped. After this initial wiring fault was corrected, the single-pole switches were replaced with doublepole types. The third example was reported in another magazine many years ago. As I recall, a young child was killed because the electrician who wired the house had incorrectly installed a combination GPO with a coax TV socket. The insulating shroud over the back of the coax socket had been omitted and, as a result, the outer braid of the coax cable had come into contact with the Active terminal. This caused the metal roof and downpipes of the house to become live at 240V AC. The downpipes finished about 15cm above ground 72 be done with a masking code of 00001111. The first four zeros will be ANDed with the 0011 portion of the ASCII code we wish to eliminate. ANDing those digits with zero will cause the logical result to be 0000. The four bit BCD portion of the code we wish to retain will be ANDed with the 1111 portion of the mask word. ANDing bits with binary 1 's will simply reproduce those bits. In the program above, the ASCII number has been previously stored in memory location 8. That number is now brought into the accumulator with the first instruction, LDAA. Next, the ANDA instruction is executed. The mask word is stored in location 9. That number is ANDed with the contents of the accumulator. The result appears in the accumulator. In this case, the ASCII number 9 is converted to its BCD equivalent, 00001001. That number is then stored in location 10 by the STAA instruction. Then the program halts. level and when the child touched the downpipe, a fatal shock was received. This house wiring had also been inspected and passed. In conclusion, I should say that noone is perfect but as you pointed out, there is room for improvement in the electrical trade, especially with regard to council electricians and inspectors. M. C. Laybutt Ainslie, ACT Good mix of projects Congratulations on a very good magazine. It is well ahead of all the other electronics magazines in the marketplace. Electronics and other journalists have a privileged position as far as keeping up with new products and technology. Notice I did not say that it is an easy job. They cannot really be effective without keeping abreast. Manufacturers and suppliers make it easier by supplying literature, demos, samples etc in the hope of getting press releases published. Compare this with the engineer who chose to go into industry, say into designing power supplies. Now let's reduce his field of vision further by saying switching power supplies. Only those suppliers that have corn- ponents that go into switching supplies send him data - and then only half of those because he is, unlike the magazines, unknown to many. His principal job is to design new power supplies but a large proportion would consist of other items, such as investigating why a particular design fails for customer A but not for customer B. Or marketing may say they can get an order for 100 units if you can increase the output voltage by a couple of volts or increase the output current by lO0mA. Such tasks are often made worse because he was not the original designer. And of course there's the paperwork like writing engineering change instructions to manufacturing. This engineer doesn't have a chance of knowing about memory mapping of microprocessors or 100 other things. Changing the subject, magazine articles are not sufficient to act as an instruction manual. While I and many other people can build some circuits from a schematic, there is a large potential kitset builder market out there who need step by step instruction. In this day of wordprocessing and cut and paste desktop publishing, there is no reason decent instruction manuals could not be reasonably priced. D. Hire Annandale, NSW Then, each time the interrupt occurs, the microprocessor will complete the current instruction in process for the main program, then jump to a subroutine for the count operation. The simple program below is the subroutine that performs the counting operation. MEMORY LOCATION INSTRUCTION ADDRESS OR DATA 100 101 102 103 104 105 106 LDAA INCA STAA COUNT VALUE RTS Memory location 105 is set aside to store the count value. It is previously cleared to zero in the main program. When the interrupt occurs, that word will be loaded into the accumulator by the LDAA instruction. Next, an INCA instruction is executed. That causes the number in accumulator A to be incremented. Incrementing with this instruction causes 1 to be added to the accumulator value. The next instruction, STAA, causes the number to be restored in memory location 105. Finally, an RTS instruction causes a return from the subroutine. The program counter value which was stored away in the stack when the interrupt occurred is retrieved by the RTS instruction and loaded back into the program counter. Therefore, the main program resumes where it left off. Counting routines can also be used to provide for timing, sequencing and delay operations which are common in digital circuits. For example, the simple subroutine shown below generates a specific time delay when triggered by an interrupt. At the end of the delay time, the program outputs a binary 1 in the least significant bit position of the data bus. MEMORY LOCATION INSTRUCTION ADDRESS OR DATA 52 . LDAA 64 OECA STAA 64 BNE 52 LDAA 65 STAA 100 RTS COUNT VALUE 00000001 OUTPUT PORT 53 54 55 56 57 58 59 60 61 62 63 64 65 100 START Memory location 53 is assigned a count value. Next, that count number which is proportional to the length of the delay is loaded into the accumulator by the LDAA instruction. For short delays, a low value number will be used. For long delays, a high value will be used. The decrement instruction (DECA) subtracts one from it. It is then restored by the STAA. Next, the BNE or branch if not equal to zero instruction is executed. That cheoks to see if the count value has been decremented to zero. If it hasn't, the program loops back to location 52 for additional decrementing. The process repeats as the counter is decremented once each pass through the loop. When the count value reaches zero, the BNE instruction NO STOP Fig.8: this is the flow chart for the time delay program shown on this page (the one that starts at memory address 52). Starting at the top, we find that the program loads a count value, which is then decremented by 1 and eventually tested against zero. If the decrement doesn't result in zero, it is repeated until it does equal zero. When the count equals zero, a binary 1 is loaded in the accumulator and then output to port 100. detects it. It then executes the LDAA instruction in location 59. That loads the number 00000001 into the accumulator. Then, the STAA instruction outputs this value (00000001) to the address designated (100). In this program, the length of the delay is determined by the time it takes for the various instructions to be executed. Most of the time involved is that time required to load, decrement and store the count value. Since most microprocessors are controlled by a crystal clock oscillator, the clock cycle time is precisely known. With that information then, the time it takes to execute various instructions can also be determined. By knowing the amount of time it takes for each instruction to be executed, the delay count .value can be determined to create a delay of a specific duration. This program ends with the RTS, which was triggered by an interrupt. ~ SEPTEMBER 1988 73