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23

The basic difference is a gating or clocking mechanism. For example, let us talk about SR latch and SR flip-flops. An SR Latch will look like this In this circuit when you Set S as active the output Q would be high and Q' will be low. This is irrespective of anything else. (This is an active low circuit so active here means low, but for an active high ...


19

A flip flop is built from two back to back latches with opposite polarity clocks, which form a master slave topology. The type of latch is irrelevant (JK, SR, D, T) to this constraint, but it is important that the transparency is controlled by some pin (call it clock or enable or whatever you like). SR latches throw everyone for a loop because the most ...


19

Here's what wikipedia says: - According to P. L. Lindley, a JPL engineer, the flip-flop types discussed below (RS, D, T, JK) were first discussed in a 1954 UCLA course on computer design by Montgomery Phister, and then appeared in his book Logical Design of Digital Computers. Lindley was at the time working at Hughes Aircraft under Dr. Eldred ...


19

To "gate the clock" means put a logic gate in the clock line to switch it on or off. simulate this circuit – Schematic created using CircuitLab The diagrams above show and AND and OR used to gate the clock. One forces the clock low the other high. To prevent clock pulses which are 'too short' either high or low ("runt pulses"), we must make sure ...


18

One reason we clock flip flops so that there isn't any chaos when the outputs of flip flops are fed through some logic functions and back to their own inputs. If a flip-flop's output is used to calculate its input, it behooves us to have orderly behavior: to prevent the flip-flop's state from changing until the output (and hence the input) is stable. This ...


17

If the flip-flop's setup time is 20 ns, it means that data has to be stable atleast 20ns before the capturing clock-edge. Similarly hold time is the amount of time, data has to remain stable after a clock edge has appeared. So together they define a "setup-hold-window", in which data has to remain stable. If the data changes/toggles within this window, the ...


15

In all likelihood it isn't really a "set/reset" signal, so much as a signal that can be configured to be either set or reset depending on how the device is configured. Note that in figure 3.2 (copied below for clarity), it says "flip-flop with optional [...] set or reset controls". Similarly it will also be the configuration bits that disable the signal ...


14

A edge-triggered latch (flipflop) ideally samples the data line instantaneously on one of the edges of the clock. However, nothing is truly instantaneous, so the data must be valid for some finite amount of time around the clock edge. The time it must be fixed before the clock edge is called the setup time, and the time it must be fixed after the clock ...


14

FFs on FPGAs don't have explicit "Q-bar" outputs, because inverters are basically available "for free" as a result of how logic is implemented in LUTs (lookup tables). You can just add the inverter, and it will be incorporated into every LUT that it feeds. In any case, a ripple counter like the one you have shown is a poor choice for FPGA implementation. ...


13

You could use the circuit below with 3 D flip-flops and one 3-input AND gate. You would also need to use the reset input of the flip-flops to bring the output back to zero (not indicated in the schematic). simulate this circuit – Schematic created using CircuitLab


12

Of course there's an external power supply. Take a look at a NOR gate chip (say, 74HC02) and you'll see it there clear as day: Logic symbols purely show the logic flow of a circuit, not the power flow. Many schematic capture programs will take the power connections as read and automatically connect them to the right nets for you. Internally a typical CMOS ...


11

Normal flip-flops are volatile, in that they don't remember their state when power is lost. Usually CPU registers are made from such flip-flops, which is why you need to save stuff you want to remember accross a powerdown to special memory for that purpose. Since non-volatile is a important attribute, this will be clearly mentioned in any datasheet for a ...


11

Assume ideal logic gates (no propagation delay) like this (image from wikipedia): We know that the output of NOR gate is 1 if and only if both inputs are 0; and 0 otherwise. When S = 1, Q = 1 and therefore \$\bar{Q} = 0\$; when R = 1, Q = 0 and \$\bar{Q} = 1\$. But if you set both R and S to 1 we have that Q = 0 and \$\bar{Q} = 0\$ at the same time. This ...


11

The circuit you cited is a ripple counter, not a synchronous counter. It actually has eleven states, 0000 through 1010, but as soon as the last state is reached, the NAND gate immediately (asychronously) resets the flip-flops to the 0000 state. In a synchronous counter, all of the flip-flops would share a common clock, and you'd control the sequence of ...


11

I have been thinking about this definition a lot today. As others pointed out, the exact meanings will vary. On top of that, you will probably see more people get this wrong, even on this site, than right. I don't care what wikipedia says! But in general: A flip flop will change it's output state at most once per clock cycle. A latch will change its ...


11

Metastability cannot be 'cured', but if you wait long enough, the likelihood of it occurring can be made arbitrarily small. Once you've got it down to once in the age of the universe, it's probably unlikely to cause you trouble. It's like balancing a pencil on its point. It's likely to fall over, and the longer you wait, the less likely it is to remain ...


10

Fan-out is the number of logical inputs an output can drive. (Both Kaz's answer as well as the Wikipedia entry about it are wrong.) Even for HCMOS fan-out is expressed as the number of LS-TTL inputs it can drive. The input current of an HCMOS gate is negligible. Fan-in is a rarely used term simply meaning the number of inputs. 1 input has a fan-in of 1. ...


10

They start out undefined, that is they could be set to either. When you switch power on, assuming a real latch with no input signals, both gates will want to output high. However due to no two gates being exactly the same (and other real world effects), one will "win" the race to bring it's output high first, and set the others output to low. The same ...


10

Because the electronic device reminded engineers of both this item: and this one: In the first case, the devices have two states: flipped (in contact with heel) and flopped (not in contact with heel). In normal use they alternate continuously between those two states at about 1Hz. In the latter case, you can change the object's state from open (unlatched) ...


10

You're forgetting a couple of important facts: A flip-flop isn't a single atomic gate, but made up of multiple gates. It takes time for a signal to pass through a gate (or propagate). There is no such thing as a pure square wave. Take this diagram of a transparent latch: Assume each gate requires one "time unit" to propagate the signal. The D signal ...


10

Given it is an assignment, I'm not going to work through the logic. However in all such questions you can boil it down to a circuit with two steps: You have a binary 0 to n-1 counter, where n is the number of required states (13 in your case). This can be a bog standard synchronous binary counter that starts at 0000 and then when it reaches n is reset back ...


10

We can see from the datasheet, that the CD4020 has the following block diagram: Notice the naming of the outputs, you have Q1, and Q4-Q14. Note also that it is a 14-stage counter, which means the counter internally has 14 outputs. From your data you can see that Q1 (the LSB) is toggling on every negative edge pulse as you would expect. Q1 is the first bit ...


9

Alan Turing used the letter q to denote states in what came to be known as Turing machines. Presumably the q stood for quanta, emphasizing a state's discrete rather than continuous nature. This happened in the 30s when quantum theory was permeating the scientific æther.


9

In general, "flip-flop" and "gated latch" are synonyms. However, the term "flip-flop" is often used when referring to "edge-triggered flip-flop" (just because it is shorter). When people speak about "D flip-flop" the usually refer to "D edge-triggered flip-flop", however when you hear "JK flip-flop" it can refer to both "JK edge-triggered flip-flop" and "JK ...


9

You're asking why there is a double negative. Typically in a circuit diagram, the elements will be real, purchaseable or manufactureable things. Many logic chips have active-low signals, such as CL(ea)R, so they come with that "bubble" baked in, indicating their inverted logic. So, when a designer needs a flip-flop which has that active-low input, they ...


9

The "bubble" is part of the symbol and means the input is inverted, notice how the label inside the flip-flop symbol has no "bar" or overline, hence inside the symbol the signal is "normal", SET or CLR. The "bar" or overline is part of the label given to the signal in the circuit diagram, sometimes referred to as net-name when working with a circuit editor....


9

Flip-flops are single bit devices with two stable states. The outputs are typically Q and \$\mathsf{\small \overline{\text{Q}}} \$. There are several kinds, here are probably the most common: SR flops have two inputs, S (Set) and R (Reset). As they name implies, asserting either of these either sets or resets the flip-flop. After the input is de-...


9

There is nothing in the specifications, nor in the internal schematics, that defines what the outputs of a JK flip-flop should be at power-on. The only way of getting a known state is to use the CLR input of the chip, which resets the whole chip into a known state. The normal way of doing this is to hold CLR low for a short period while the chip powers up, ...


9

If it is an output, simply leave it unconnected. If you try to connect it to a power rail directly, it will just short out the supply when the flip flop output is driving to the opposite rail. If you connect via a resistor, it will just waste power - there will be a voltage drop across the resistor when the output is opposite from the resistor's bias supply ...


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