Introduction to 8085 Microprocessor (μP)
Summary
TLDRThis educational video introduces the concept of word length in microprocessors, focusing on the Intel 4004, the first 4-bit microprocessor introduced in 1971. It explains how the 4-bit word length allows for processing, input, and output of 4-bit data chunks, performing 4-bit additions, and using place values for unsigned and signed representations. The video also covers the evolution to 8-bit microprocessors like the Intel 8085, highlighting their expanded range of numbers due to increased word length.
Takeaways
- 😀 The 8085 microprocessor is the focus of the session, with an introduction to its capabilities and specifications.
- 📚 The word length of a microprocessor is defined as the size of data it can handle at once, which is a key characteristic of its processing power.
- 🛠️ The Intel 4004, introduced in 1971, was a 4-bit microprocessor, meaning it could process four bits of data simultaneously, including performing 4-bit additions.
- 🔢 The place values for a 4-bit microprocessor in unsigned representation are 1 (2^0), 2 (2^1), 4 (2^2), and 8 (2^3), allowing it to represent numbers from 0 to 15.
- 💡 The Intel 4004 could only perform addition and subtraction within the range of 0 to 15 for unsigned numbers due to its 4-bit word length.
- ➕ In binary addition, a carry is generated when the sum of bits exceeds the bit capacity, which is then added to the next significant bit.
- 🚫 The Intel 4004 cannot perform operations resulting in numbers outside its 4-bit range, such as adding 8 and 9, which would result in 17.
- 🔄 The 4-bit microprocessor uses two's complement for subtraction, where the negative inverse of a number is found by inverting the bits and adding one.
- 🔢 In two's complement representation, the place values for a 4-bit microprocessor are 1, 2, 4, and -8, allowing it to represent numbers from -8 to 7.
- 🔄 The most significant bit in signed representation has a negative place value, affecting the range of numbers the microprocessor can represent.
- 📈 The evolution of microprocessors is highlighted, with the Intel 8085 being an 8-bit microprocessor capable of handling larger data sets and more complex operations than its 4-bit predecessors.
Q & A
What is the word length of a microprocessor?
-The word length of a microprocessor is the size of data it can handle at once, measured in bits.
What was the word length of the Intel 4004 microprocessor?
-The Intel 4004 microprocessor, introduced in 1971, had a word length of 4 bits.
How did the Intel 4004 perform 4-bit addition?
-The Intel 4004 performed 4-bit addition by taking two 4-bit inputs, processing them, and generating a 4-bit output, including handling carry operations.
What is the range of values that can be represented with a 4-bit unsigned representation?
-With a 4-bit unsigned representation, the range of values is from 0 to 15.
How does the Intel 4004 handle subtraction?
-The Intel 4004 handles subtraction using the two's complement technique, effectively performing addition of the minuend and the two's complement of the subtrahend.
What is the two's complement of a binary number?
-The two's complement of a binary number is found by inverting all the bits of the number and then adding one to the result.
What is the range of values that can be represented with a 4-bit sign representation?
-With a 4-bit sign representation, the range of values is from -8 to +7.
What is the significance of the most significant bit in sign representation?
-In sign representation, the most significant bit indicates the sign of the number, with 0 typically representing positive and 1 representing negative.
What is the difference between unsigned and sign representation in a microprocessor?
-Unsigned representation uses all bits to represent the magnitude of a number, while sign representation uses the most significant bit to indicate the sign of the number, with the remaining bits representing the magnitude.
Why is the range of values different for unsigned and sign representations in a microprocessor?
-The range is different because unsigned representation uses all bits for magnitude, allowing for a larger range of positive values, while sign representation dedicates one bit to indicate the sign, reducing the range of positive and negative values it can represent.
What is the word length of the Intel 8085 microprocessor?
-The Intel 8085 microprocessor has a word length of 8 bits.
What range of numbers can the Intel 8085 handle in unsigned representation?
-In unsigned representation, the Intel 8085 can handle numbers from 0 to 255 (2^8 - 1).
What range of numbers can the Intel 8085 handle in sign representation?
-In sign representation, the Intel 8085 can handle numbers from -128 to +127.
Outlines
📘 Introduction to the 8085 Microprocessor and Word Length
This paragraph introduces the 8085 microprocessor, focusing on its word length, which is the amount of data it can handle simultaneously. The Intel 44, a 4-bit microprocessor from 1971, is used as an example to explain the concept of word length. The 4-bit word length allows the Intel 44 to process, input, and output data in 4-bit chunks, enabling 4-bit addition operations. The explanation includes a step-by-step addition example using binary numbers, demonstrating how the carry bit is handled during the process. The paragraph also discusses the representation of unsigned numbers using 4 bits, which can range from 0 to 15, and how the least and maximum values are determined.
🔢 Understanding Binary Addition and the Range of 4-bit Numbers
The paragraph delves into the specifics of binary addition using the Intel 44 microprocessor as a reference. It explains the process of adding 4-bit numbers and the resulting carry bit, using a detailed example of adding two numbers, 101 and 0101, to produce a sum and carry. The explanation includes the place values for 4-bit numbers and how they are used to determine the decimal equivalent of the binary inputs and outputs. The paragraph also discusses the range of values that can be represented with 4 bits in unsigned representation, emphasizing that the Intel 44 can perform unsigned additions within this range.
🔄 Two's Complement and Binary Subtraction with the Intel 44 Microprocessor
This section explains the concept of two's complement and its application in binary subtraction using the Intel 44 microprocessor. It describes the process of finding the two's complement of a binary number to perform subtraction, using an example where the minuend is 1101 and the subtrahend is 0101. The paragraph details the step-by-step method of toggling bits to find the negative inverse of the subtrahend. It also discusses the place values in two's complement representation, which differ from unsigned representation by having the most significant bit's place value as negative. The example concludes with the addition of 9 and -5 in two's complement, resulting in a difference of 4.
🔢 Sign Representation and the Range of 4-bit Signed Numbers
The paragraph discusses sign representation in 4-bit binary numbers, explaining how the most significant bit is used to indicate the sign of the number. It details the place values for signed numbers, which include negative values for the most significant bit. The explanation includes a step-by-step calculation of decimal equivalents for various binary sequences, demonstrating how both positive and negative numbers can be represented. The range of values that can be represented with 4-bit sign representation is also discussed, showing that it extends from -8 to +7. The paragraph clarifies the difference between the ordered decimal numbers and the unordered binary bit sequences.
🚀 Evolution of Microprocessors and the 8085's 8-bit Word Length
The final paragraph summarizes the evolution of microprocessors, mentioning the introduction of the Intel 8008, 8080, and the 8-bit Intel 8085 microprocessor. It highlights the increased word length of the 8085, which allows it to handle 8 bits of data at once. The paragraph outlines the range of numbers that can be represented in both unsigned and signed representations for an 8-bit microprocessor, emphasizing the broader range of values that can be processed compared to the 4-bit Intel 44. The session concludes with a preview of the next topic, which will cover the pins of the 8085 microprocessor.
Mindmap
Keywords
💡8085 Microprocessor
💡Word Length
💡Unsigned Representation
💡Two's Complement
💡Binary Addition
💡Place Value
💡Carry
💡Minuend
💡Subtrahend
💡Sign Representation
💡Evolution of Microprocessors
Highlights
Introduction to the 8085 microprocessor and its significance in the evolution of microprocessors.
Definition of word length in microprocessors and its importance for data handling capabilities.
Historical context: The Intel 4000 family and the introduction of the 4-bit Intel 44 microprocessor in 1971.
Explanation of 4-bit data handling, including input, processing, and output capabilities of the Intel 44.
Demonstration of 4-bit addition using the Intel 44 microprocessor and its truth table.
Illustration of carry generation and management during 4-bit addition.
Calculation of place values for 4-bit numbers and their decimal equivalents.
Range of unsigned numbers representable with 4 bits, from 0 to 15.
Introduction to two's complement for subtraction in microprocessors.
Process of finding the two's complement of a binary number for subtraction.
Place values in two's complement representation, including negative values.
Example of performing subtraction using two's complement in a 4-bit microprocessor.
Range of signed numbers representable with 4 bits, from -8 to 7.
Comparison between unsigned and signed representations in 4-bit microprocessors.
Transition from 4-bit to 8-bit microprocessors, with the introduction of the Intel 8085.
Explanation of the range of numbers representable in unsigned and signed formats for the 8-bit Intel 8085 microprocessor.
Summary of the session on the word length of microprocessors and预告of the next session on the pins of the 8085 microprocessor.
Transcripts
[Music]
hello everyone and welcome to the
chapter fundamentals of 8085
microprocessor so this is the first
session of our new chapter and in this
session we will be introduced to the
8085
microprocessor so without any further
Ado let's get to
learning coming to the topic that we are
going to cover today today we will try
to understand what is the word length of
microprocessor so the word length of a
microprocessor is the size of data which
the microprocessor can handle at once if
you remember when we were learning about
the evolution of microprocessor we
learned that in 1971 as a member of the
Intel 4000 family
44 that is the first microprocessor was
introduced now if you remember the word
length of this was 4 bit now what is
meant by that
the Intel 44 microprocessor could handle
four bits of data at once that is it can
take a chunk of 4bit as input it can
process that 4bit chunk also it can
generate a 4bit output now since the
word length of Intel 44 was 4bit
therefore it can perform 4bit additions
now in the previous session we have seen
the truth table for addition use using
this logic inel
44 since it has the word length of 4bit
could perform 4bit addition so clearly
if we feed the aent as 1 01 and if we
also feed the addent 01 01 until 404
could perform the addition let's Now
quickly perform the addition for it 1 +
1 is supposed to generate the sum as Zer
and the carry as one so we are going to
have the sum as zero now what will be
done with the carry well the place value
of these two bits is 2 ra to the^ 0 and
the result which is just generated by
performing the addition between these
two is actually one Zer that is a two
bit number so the carry will be added
with the bits in the 2 ra to the power
one's place so adding one with zero we
will get one also if we again add that
one with 0o we will receive the sum as
one isn't it so no carry only the sum as
one because we are adding one with zero
and that to twice let's move on to the
next unit 0 + 1 we are supposed to get
the result as 1 right finally adding one
with zero we will again get the result
as one let's now check whether we have
perform the addition correctly or not
and for that we will be needing the
place values now if you notice the aent
addent and also the result all are of
four bits now for four bits what will be
the place values well 1 2 4 and 8 to be
precise 2 ra to the^ 0 that is 1 2 ra to
the^ 1 that is 2 2 squar that is 4 and
finally 2 cubed that is 8 now
considering these wets what is the
decimal equivalent value of the aent
well 8 + 1 that is 9 what about the
addent 4 + 1 that is 5 let's move on to
the result now 8 + 4 that is 12 + 2 will
give us 14 and yes it is correct because
9 + 5 is
14 now focus on the weights with these
wordss we can only represent unsigned
numbers so these are used for unsigned
representation now think about it if we
talk about unsigned representation we
are going to use the place values 1 2 4
and 8 to be precise 2 ra to the^ 0 2 ra
to the^ 1 2^ 2 and 2 Cube now with four
bits how many sequences can we have well
we will begin with 0 0 0 0 then 0 0 0 1
so on and at last we will have 1 1 1 1
that is from four zeros to 4 1es now if
we also consider the equivalent decimal
values 0 0 0 0 is 0 then we will have 1
2 3 4 5 and so on and the last value
that is all ones will be
15 so using four bits we can represent
the values from 0 to 15 and all of these
are positive values that is
unsigned now from this particular list
can you figure out the least number well
it is zero right so if we are talking
about the range with four bits with
unsigned representation that is using
these place values the least number that
we can have is zero now what is the max
number it is 15 and how did we get it we
got it by placing all ones underneath
these place
values now for four bits I already have
told you multiple times what are the
actual place values 2 ra to the^ 0 2 ra
to the^ 1 2^ 2 2 Cub now think about the
next place if we had that place what
would have been the place value 2 ra to
the power 4 right and what is 2 ra to
the^ 4 it is
16 and we got the maximum number in this
list as 15 that is 1 less than the next
place value so in order to define the
range from the least value which is zero
we can have the max value that is 2 to ^
4 - 1 or in other words 16 - 1 that is
15 now you could ask me why we are using
this particular notation why can't we
write 15 instead well we can there is
nothing wrong about it but it's a
generalized formula why because in case
of five bits if we just Place five in
this place we will have the maximum
number of five bits using unsigned
representation and it's also meaningful
since we are using four bits so we will
only have to place the number of bits in
here is it clear so this is the range
for four bit numbers using unsign
representation now focus on the Range it
is 0 to 2 to ^ 4 -1 and we are talking
about the unsigned
representation so if we talk about the
number line we are having the range from
0 to 15 and all these 16 numbers we are
representing using the binary sequences
starting from 0 0 0 0 0 0 0 1 till all
ones that is four ones so the Intel 4004
since it has the word length of four bit
it can perform unsigned additions of two
4bit numbers which will involve both the
aent and the addent from this particular
list Additionally the result or the
outcome will also have to be from this
list only for an instance if we add 5
which in binary is 0 1 0 1 with 10 that
is 1 1 0 the result is going to be 15 or
1111 and this can be done since 11 1 1
is a 4bit binary number and additions as
such can be performed by Intel 44 4bit
micro processor however if we add 8 with
N9 that is if we add 1 tri0 with 1
01 the result is going to be 17 which is
not in this particular range so this
microprocessor is not going to produce
that result I hope the word length for
addition is clear to you now like this
Intel 4004 can also perform
subtraction now if you remember when we
learned about the subtraction in binary
we learned about two different
approaches one the pen and paper method
and on the other hand we learned about
the machine's approach therefore when we
are talking about subtraction performed
by the 44 4bit microprocessor it is
going to take this approach under
consideration and specifically it will
perform the subtractions using tw's
complement technique
let me show you one such subtraction
where we are going to have the minu end
as one1 and the subtra this time is 0 1
01 now as I told you earlier we are not
going to perform a minus B rather we are
going to perform A+ minus B because
that's how using addition we can also
perform subtraction which is followed in
both ones and two's complement
approaches now here we have got the B
that is the subtend at as 0 1 0 1 now in
order to get the negative inverse
remember we are implementing two's
complement just like the 44 4bit
microprocessor so we will find out the b
complement that is a negative inverse of
B and for that how to do it if you
remember the technique for tw's
complement we are going to Traverse the
subtend from the least significant bit
towards the most significant bit and we
will keep retaining the data until we
encounter one once we encounter one we
will retain it as it is however the rest
of the bits are going to be toggled that
is zero will become one and one is going
to be zero so finally in case of the
most significant bit we toggled it to
one so this is the two's complement of
the subtend 01 01 now we have discussed
about the conversion process but I
didn't tell you what is happening in
here in reality that is how it is min -
5 because if we consider 01 0 1 it is 5
and negative inverse of that is supposed
to be minus 5 right now we just have
observed the place values in case of
unsigned representation for four bit now
since we are talking about these numbers
what is going to be the place values in
two's complement the place values are 1
2
4 and minus 8 that is apart from the
most significant bed all the remaining
places have the same place values only
the most significant bit has the place
value in
negative now look at
this 1 1 1 that is - 8 + 2 which will
give us - 6 + 1 that is we are adding -
6 with 1 we will get
-5 now you get it how did we get minus5
well the place value of the most
significant bear is Min -
A remember this is for four bits in case
of five bits apart from the most
significant bit the rest of the bits
will have the same place values just
like unsigned representation however the
most significant bit is going to have
the place value minus 16 because we are
talking about five bits then since we
are dealing with four bits in here for
four bits the most significant bits
place value is 8 that is 2 Cub which in
case of two's complement approach is - 8
or -2 to^ 3 or - 2
Cub now with - 5 if we add 1 0 0 1 now
here do remember this is in unsigned
representation so it is going to be 9
because here the place value is 8 only
so 8 + 1 is going to give us 9 now let's
observe if we add these two what is
going to be the result 1 + 1 will give
us the sum as zero and the carry is
going to be 1 1 + 0 is 1 then 1 + 1 will
again give us the sum as 0o and the
carry is going to be one now adding one
with all these zeros we will receive the
sum as one now what about the most
significant bits 1 + 1 will give us 1 Z
remember we are performing the
subtraction in two's complement
therefore whenever the carry is
generated we are going to discard it now
consider the result it is 0 1 0 0 what
is the place value of this one it is
four and yes from nine if we subtract 5
we are getting four however we didn't
perform subtraction this time rather we
added 9 with minus 5 and got the
difference as four now let's focus on
the negative
inverse we got this value - 5 using S
representation now in case of sign
presentation since we are talking about
four bits the place values are going to
be 1 2 4 - 8 remember whenever we are
talking about sign representation
whatever the number of bits may be the
most significant bit will always have
the place value in
negative now just for a moment forget
about these signs just think about it we
are using four bits now with four bits
how many binary sequences can we have
well starting from all zeros till all
ones there is 16 different
sequences let's now try to find out the
equivalent decimal numbers although this
time we are going to consider these
place
values now if you notice these patterns
till this particular point we haven't
placed one in the most significant bits
place so till this point the decimal
equivalence will be the same as we had
in case of unsigned numbers that is from
0 to
7 now once we have placed one in the
most significant bits place of the sign
representation considering the place
value what is going to be the decimal
equivalent well minus 8 isn't it now
what about the next sequence -8 + 1 that
is - 7 focus on the next sequence -8 + 2
that is - 6 then - 8 + 2 that is - 6 + 1
that is - 5 which we already used didn't
we now what about the next sequence -8 +
4 that is
-4 then -8 + 4 that is -4 and with -4 if
we add 1 we will get
-3 focus on the next sequence - 8 + 4
that is -4 + 2 that is
-2 now what about the last sequence 1 1
1 1 1 - 8 + 7 because 3 1's is 7 right
that is
min-1 now from all these decimal values
can you figure out the minimum and the
maximum
values well the minimum value is min - 8
right now what is the maximum value it
is 7even that is + 7 or postive 7 so
clearly with the sign representation of
four piit
we can represent from -8 till
pos7 now what is - 8 this is actually -
2 Cub however since we are considering
four bits we can also state it as - 2 to
^ 4 - 1 notice - 2 cubed 4 - 1 is 3 so
from -2 ra to^ 4 - 1 that is - 8 using
four bits sign representation we could
represent till the maximum value that is
seven and how did we get this we got
this by adding all these ones place
values now what is seven it is 1 less
than 8 so we could also write 7 as 2 ra
to the^ 4 - 1 - 1 that is 2 cubed or 8
-1 so this is the range for 4 bit sign
represent
resentation now we just also have seen
the unsigned representations range it
was from 0 to
15 and we use the sequence from all
zeros that is four zeros to 4 ones to
represent this particular
range now when we are talking about the
sign representation here the most
significant bits they represent the
negative
weightage so the range is not going to
be this anymore rather we will have - 8
- 7 - 6 - 5 -4 -3 - 2
-1 now if you notice this particular
line this is wrong from our perspective
because we deal in decimal number
systems right so if we want to represent
the range that is from - 2 to^ 4 - 1 to
2^ 4 - 1 - 1 we should re arrange this
that is - 2 to ^ 4 -1 or - 2 cubed or -
8 and 2 to ^ 4 - 1 - 1 that is 2 cubed -
1 or 8 - 1 that is
7 now notice the decimal numbers are
organized that is we are having the
least value as -8 and the maximum value
is + 7 however now we don't have the
order in the binary bit
sequences so remember Intel 44 4bit
microprocessor if it used sign
representation could deal with the numb
starting from minus 8 that is 1 0 till
the positive number + 7 that is 0le
1 and that's the buy of the W length of
4 bit so to sum it up we could say in 44
4bit mu p in unsigned representation
could use the numbers starting from 0
till 2 to ^ 4 - 1 that is from 0 to 15
16 - 1 that is 15 now coming to sign
representation it could deal with the
values from - 2 to^ 4 - 1 till 2 to ^ 4
- 1 - 1 that is from - 2 cubed or - 8
till 2 cubed - 1 or POS 7 remember this
all these exponents are four because the
word length of 44 was 4
bit now during the evolution of
microprocessors we have learned in
1972 Intel introduced the next
microprocessor that is Intel
8008 in 1974 it introduced Intel
880 and then in 1977 Intel came up with
intel 80 85 notice this is an 8bit
microprocessor in other words it has the
word length of 8 bit so if we follow
this particular format we could say
intel
8085 if we talk about the unsigned
representation can deal with the range
of numbers from 0 to to ra to the^ 8
minus one and similarly following this
for the 8bit microprocessor in
8085 we can also say for sign
representation it can deal with the
numbers within the range - 2 ra to^ 8 -1
2 2 to ^ 8 -1
-1 and we are having these ranges of
both unsigned and signed representation
due to the word length 8
bit in other words Intel
8085 can deal with 8 Bits of data at
once
also feel free to write down the ranges
in decimal for these two representations
in the comment
section so in this session we covered
the topic word length of
microprocessor all right people that
will be all for this session in the next
session we are going to learn about the
pins of 8085 microprocessor so I hope to
see you in the next one thank you all
for
watching
[Music]
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