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4

In a word, no. The practical hurdles between you and getting something fabricated are huge - both monetary and logistical. The first problem you are going to have is getting a company to work with you - never mind the fact that designing an integrated circuit is a whole world of its own. I would suggest instead that you look at FPGAs or CPLDs if the circuit ...


4

I don´t think there is a "best software/website", you will need to look at websites like distributors/manufactors for this. There are some very helpfull websites like: Octopart Parts.io Also, Digikey and Mouser are great places to find parts. Hope it helps!


3

This will work. When the switch is open, the pin does not float, it is pulled down via R2. A couple of caveats, however. You need to ensure that the ground of the MCU is connected to the ground of the 12V supply. The motor is an inductive load, and as such you will get voltage spikes when it is switched off. To counteract this you should clamp the input ...


3

You are mixing up the ideas of digital signals with binary signals. A digital signal is one that takes only discrete values, or, in the real world, one in which we distinguish only a discrete set of values. It might have 2 possible values, or 3, or 10, or whatever. A binary signal is a specific case of a digital signal that takes two different values. A ...


3

This is an example of a classic Open Collector bus architecture using MOSFETS to achieve the open drain equivalent. This technique was very common before the days of 3 state logic (TriState is a registered trademark incidentally) in discrete computing, where each processor register was in fact an octal D type latch. The setup you have should work fine for ...


3

Imagine a load resistor (say 10K) from output to Vdd, and make X_1 and X_2 low. Ideally, output should be 0.0V. You can figure out what it is (not 0V). Now imagine a 10K load resistor from output to GND and make X_1 low and X_2 high. Ideally, output should be Vdd. You can figure out what it is (not Vdd). simulate this circuit – Schematic ...


2

Well I haven't stumbled upon a direct current to digital converter yet, so I'll give an approach for an ideal situation: The current you want to measure is run through a shunt resistor and will create a voltage across it according to ohms law. Let's say you want to measure a current of \$1A\$ and you have a shunt resistor of \$1\Omega\$. The voltage drop ...


2

Assuming you're not doing the "theoretically pure" method of programming a FSM in VHDL, just do something simple like this: fsmCount: process(clk, reset) begin if reset = '1' then --Reset code here elsif(clk'event and clk = '1') then --on the rising edge of the clock case state is when s0 ...


2

This is an example, on how a FSM can control a counter. If the second state is entered, the counter counts up until 15, then the FSM goes to the next state. The counter is reset in every other state. Declarations: type T_STATE is (ST_IDLE, ST_SECOND, ST_FINISHED); signal State : T_STATE := ST_IDLE; signal NextState : T_STATE; signal ...


2

Briefly, although one-hot encoding uses more FFs than binary encoding, it is much simpler to decode the states, and this often allows significantly higher performance (clock rate). It also maps particularly well onto most FPGA architectures, which usually have a 4 to 6-input LUT followed by a FF. On the other hand, FPGAs also often have dedicated structures ...


2

HCMOS (74HC) is still very popular these days due to its wide voltage range and the fact that it still has through-hole packages, but LVC (74LVC) is useful for its lower bottom supply limit, 5V tolerance regardless of supply, partial power-down capability, reduced footprint packages, and reduced gate count devices (e.g. 74LVC2G00).


2

A CD4001B with both inputs connected together will act as a NOT gate. Each package contains 4 NOR gates. It will operate from a 9V battery with no other components- the + goes to pin 14 and the - to pin 7. You could also ground the unused inputs (say pins 5,6, 8, 9 , 12, 13) and leave the unused inputs 4, 10,11 open. The input (1,2) should be near 0V ...


2

Alternatively, you can employ an optoisolator. simulate this circuit – Schematic created using CircuitLab The FOD817 isolates any spikes from the motor, means you don't have to have the grounds connected, and keeps the input voltage to the chip within proper bounds. Edit: Incidentally, note that none of these circuits actually measure the ...


2

Normally I would say tie the unused inputs to pin 14 (analog ground). But you say you can't use that. Why not?


2

Update to version 1.1.34, then place a time marker where you want the protocol to start being analyzed, the in the Analyzers section click on the gear button and hit "Re-run starting at timing marker..." and it reanalyzes the data starting from the marker... Hope this helps


2

As others have mentioned the circuit you linked to is a 3-input NAND gate (7410), not an AND. This is a pretty standard circuit for a TTL logic gate, except that instead of the diode between the two totem-pole output transistors shown in Keith's answer for most 74xx gates, your circuit uses a Darlington pair (T4 and T5) to do the same function as the diode: ...


2

You can find the original source of this circuit in the SN7400 datasheet: It's a NAND gate. Wikipedia explain it thus: When all the inputs are held at high voltage, the base-emitter junctions of the multiple-emitter transistor are reverse-biased. [A] small “collector” current (approximately 10µA) is drawn by each of the inputs. This is because the ...


2

In theory the chip does not need a decoupling capacitor, what it needs is a power that is within its specifications. To achieve that you can use any means you want, including magic. But the most common way to do that is to have a decoupling capacitor (or mybe more than 1) nearby, and not too much inductance in the feed line and ground. So you can say that ...


2

It really does cost around $10,000 to set up equipment that can make a particular IC. Once that's done, it costs pennies to make more. That's the reason for the quote you got. If your logic is not time constrained, you can do the same thing using a microprocessor. If you do need it to be fast, an FPGA is the way to go. Note that with either alternative, ...


2

In very most cases (maybe 99.999%) the transistors of digital circuits are triggered by the output signal of other transistors. The remaining few cases are interfaces to the non-electrical or analog world: peripherals like keys (interfaces to digital but not electrical signals), ADCs (interfaces to electrical signals thar are not digital), oscillators, etc. ...


1

First, I am not 100% sure what your question is, is it about how the gates are arranged or about the logic.. But what I have interpreted is the following: Present state is stated as Q=0 Q'=1 from the truth table below this indicates that the SR is RESET. Now S=1 and R=0 (stated) so this puts the Latch into SET state, giving Q=1 and Q'=0. So everything ...


1

There is a really cool trick that you can use under specific conditions that allows you to build a 2-input OR gate or AND gate using only 1 diode and 1 resistor. The conditions are: 1) The circuit is low speed. 2) The load impedance is high, such as a CMOS input. simulate this circuit – Schematic created using CircuitLab This can be expanded ...


1

A digital signal is measured in volts with a logic 1 typically being 3.3 volts or 5 volts and logical zero being 0 volts. So, a digital signal can be regarded as having logical values (i.e. 0 or 1) but also can be regarded as having real values such as 0V and 3.3 volts. A combination of digital signals can be regarded as a binary number of several digits in ...


1

\$n\$ bits (either 0s and 1s) can be combined to code intergers in then range \$0..2^n-1\$. So both statements are true. That's just like in the decimal system several digits (either 0s, 1s, 2s, .. or 9s) can be combined to code integers in the range \$0..10^n-1\$. Note that it is not possible to code any interval of real numbers by a finite number of ...


1

Look in the datasheet for the flipflop, and there will be two specs that address this issue: Setup time and hold time. The setup time is how long the input data needs to be held fixed before the clock, and hold time is how long it needs to remain fixed after the clock. If either of these specs is violated, then the answer is you don't know what state the ...


1

Aside from @Oka's suggestion, you could also use a pair of common NPN 2n3904. simulate this circuit – Schematic created using CircuitLab R1 and Q1 act as a simple inverted switch, so when the SPI chip select is enabled (i.e. logic low/grounded), Q1 turns off, allowing Q2 and the LED to turn on. This is smaller/simpler than a ULN2003D and a Not ...


1

Any transistor will work. If you need 7 transistors, using transistor array is a good choice. Note that your above circuit will turn on the LED when the input in a positive logic (~5V), however, ~CS pin selects the chip with low voltage. Consider using PNP transistor switch that turn on the LED with low logic input (~0v) R2 limits the base current. Make ...


1

"But we know that for input voltage 0 , PMOS will allow current flow and for input voltage VDD or logic value 1 , NMOS will allow current flow." That is not true in general, it holds only when the NMOS source is at 0 and the PMOS source is at 1, which is not (always) the case in your circuit.



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