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7

In an unsigned binary representation, only positive numbers can be represented, and the weight of each bit including the most significant bit is a power of two. So with a word size of 8 (byte), 00000000 => 0 01111111 => 127 10000000 => 128 11111111 => 255 Two's-complement, which is used for signed binary notation, encodes both positive and ...


6

You are connecting to the output of the '02, not the input:


6

In 2's complement the MSb is defined as -(2n) where n is the bit position of the MSb. Due to how numbers work, all bits other than the MSb add up to (2n)-1. Adding those together results in -1.


5

It's been a long time since I've done this the long way. The Truth Table The first step for learning what these circuits do is to create a truth table. Perhaps you already know how to do this, but didn't know it was the first step, we'll go over it in any case. I'll work out the first example circuit. Make columns for each of your inputs and outputs. ...


4

Cutting corners in this area is well summed up by a metaphor based on a long ago Bell Helmets ad - "If you have a $10 head then buy a $10 helmet" Specifications that allow body diode conduction ALWAYS relate to worst case specifications - ie will not DIE but are not guaranteed to WORK. Allowing body diodes to conduct during NORMAL operation violates ALL ...


4

It's not necessarily better, in fact, it depends on which timing is harder to respect. Quoting Wikipedia, "Positive clock skews are good for fixing setup violations, but can cause hold violations. Negative clock skew can guard against a hold violation, but can cause a setup violation." The best practice is to design circuits that work well with zero skew. ...


4

As you said, 32bit CPUs can handle numbers larger than 32 bit, so why shouldn't a 8bit CPU be able to do this? If you add two numbers, you start with the last bit of both. If one of them is 1, the result is 1, if both are 1, the result is 0, and you have to carry a 1 to the calculation of the second bit. For the second bit, you have the two bits of the ...


4

It's certainly possible to work with very large numbers even on 8 bit or 4 bit computers. It's not very efficient, but it's possible. The way this is done is by operating on the numbers in pieces, with the support of specific processor instructions. A common 8 bit microcontroller is the Atmel AVR series. To add 8-bit numbers, it uses an instruction ...


3

Because all 1s is -1 in twos-complement encoding. Think about it. With a N-bit number, adding 2N doesn't change the value because the change is in the first bit past the number. For example, consider adding 16 to a 4-bit number. In binary that is: XXXX + 10000 ------- 1XXXX which is still just XXXX because by definition of a 4 bit number you ...


3

There is this this nice small (and incomplete) set of rules about digital circuits, about the little balls to be more precise: little balls can travel around over wires (not always at T sections) little balls can travel across logic gates little balls neutralize each others when they collide The second needs a little expansion. If you have a little ball ...


3

Both are SR latches. The SR NOR latch will have the following truth table: ---------- S R Q ---------- 0 0 no change 0 1 0 1 0 1 1 1 not allowed ---------- SR NAND latch is an inverted version of SR NOR latch. The truth table of which is: ---------- S R Q ---------- 0 0 not allowed 0 1 1 1 0 0 1 1 no change ----------


3

+Vcc simply means the positive supply voltage. The inputs work because when you connect A OR B to 0V they pull the output Q down to a LOW voltage. (typically 0.7V for silicon diodes). The current that can flow through them is limited by the size of the resistor, R. What the book didn't mention was that if the inputs are not connected to anything the ...


3

You could use a set of NOT gates and AND gates to get the output you desire like so: simulate this circuit – Schematic created using CircuitLab This would make the inputs in decreasing priority, but never more than one would be high.


3

Doublers are possible using digital circuits, the following diagram is one such example. The trouble with it is that it relies on propagation delays in delay chains in order to generate the doubled frequency. This means that you don't get a guaranteed duty cycle, it will vary depending on how long your chain is. The clock will probably also be fairly ...


3

First of all, your title and question do not match. I assume your question is 'how to design a 32k x 4 memory using two 16k x 4 RAM chips?'. This can be done as shown below. 14 address lines access the 16k nibbles from each ram chip. The 15th address line is given as chip select. When A14 is low, the lower bank is selected and when it is high, the upper ...


2

I do not follow how you get C dv/dt = 3mA. But here is a more accepted way of calculating the power requirement. The energy to charge 80pF to 5V is $$ E_c = \frac{1}{2}CV^2 = 1 \cdot 10^{-9} J $$ Power is: $$ P_c = E_c \cdot F(SwitchingFrequency) $$ Therefore the power requirement is proportional to frequency. Examples: F = 10MHz, Pc = 10mW, I_average = ...


2

It is possible to construct a simple SR flip flop using NOR or NAND gates. There isn't much difference in the output. The only minor difference occurs because of the properties of a NOR or a NAND gate. Consider a SR flip flop using NAND gates:- The truth table can be given as:- Now, consider SR flip flop using NOR gates:- The truth table can be ...


2

The Early latch: The truth table can be derived from the circuit diagram: ---------------------- Clock Data Out(n+1) ---------------------- 0 0 Out(n) 0 1 Out(n) 1 0 0 1 1 1 ---------------------- One useful feature of Earle's latch is that it has a constant delay of 2 gates for any input. And this makes ...


2

You could interface the MAX233 to the MCU using an Open Drain or Open Collector arrangement. The MAX233 activates a transistor (NPN BJT or N-channel MOSFET) which in turn pulls the MCLR pin low. simulate this circuit – Schematic created using CircuitLab This is also one of the recommended ways of activating a reset line from multiple sources - ...


2

The Verilog standard guarantees that all events within a begin..end block scheduled for the same simulation time will be processed in the order they are declared. Other than in this specific scenario every other event may be taken out of the queue in any order. With this said, any behavior can be expected because both events are scheduled for the same ...


2

Do \$2^N\$ first. For \$2^N\$ you need a 4 bit to 1 of 16 decoder. You can make one using 2 74LS138 3 bit to 1 of 8 decoders. \$2^{2^N}\$ can be done by just re-labeling the outputs!


2

Here is a 4-bit ripple carry adder (taken from here): The number of gates for each bit in each full adder is 5. The number of gates just for the carry logic is 3. So the total number of gates for 32 bits would be 5 * 32 = 160. There are also plenty of circuits on the web that use 6 gates instead of 5, e.g. this one, which would be 6 * 32 = 192 gates, ...


2

Like so: (A+B+C) (A+B+C') (A+B'+C') = ((A+B+C) (A+B+C')) ((A+B+C') (A+B'+C')) = (A+B+(CC')) (A+C'+(BB')) = (A+B+0) (A+C'+0) = (A+B) (A+C')


2

Boolean operators AND, OR, and NOT are the basic building blocks: any Boolean expression can be expressed with these operations. But that's not the only possible orthogonal set of Boolean operators. NAND by itself is sufficient to express all Boolean logic expressions, because it's possible to build each of the basic AND, OR, and NOT equivalents out of ...


2

Yes, they are the same. In a Moore machine, the outputs are associated with the states, while in a Mealy machine, they are associated with the edges. This means that a Mealy machine can often have fewer states than the corresponding Moore machine. In this specific example, your states s2 and s3 could be combined, since they have the same set of output edges ...


2

Assuming the decoder is active, the Truth table is as follows : XZ Dec_Out HA1 HA2 a b cin F G Sum | C Sum | C 00 1000 1 0 0 0 1 1 0 0 1 01 0100 1 0 0 0 1 1 0 0 1 10 0010 0 0 1 0 0 1 1 0 1 11 0001 0 0 1 0 0 1 1 0 1 The truth ...


2

See below. When the input is at 0V, Q1 is on, the output voltage is pulled up to +2.3V. When the input is at 5V, the output is at -2.3V. Note the logic reversal. You can use dual transistors etc.- they're not critical.


2

You mean "even" parity rather than "level" - the left bit is the parity bit and the remaining 4 bits form the number 5 (0101). If the number of bits in the remaining 4 bits is "even" then the left bit is unset. Does this make sense now? Note that I'm trying hard not to feed this answer on a plate to the questioner!


2

When the inputs \$\small{\text{A}}\$, \$\mathsf{\small \overline{\text{A}}}\$, \$\small{\text{B}}\$, and \$\mathsf{\small \overline{\text{B}}}\$ are latched in, they create four minterms \$\mathsf{\small \overline{\text{A}}}\$ ⦁ \$\mathsf{\small \overline{\text{B}}}\$, \$\small{\text{A ⦁ B}}\$, \$\mathsf{\small \overline{\text{A}}}\$, and \$\mathsf{\small ...


2

No, there's no single IC that does exactly that (that I'm aware of), mainly because that isn't a commonly-required function. It's generally much more useful to provide a binary representation of the highest-priority input at the output of a priority encoder. But you could take such a priority encoder chip, and connect its outputs to a decoder chip (another ...



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