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10

Can An Operational Amplifier Circuit Be Made Entirely Out Of Diode Nand And Nor Gates? This apparently simple-enough question is somewhat ambiguous and can be answered several ways. Spehro has assumed that you convert the input to a digital value and perform digital arithmetic on it. So he says the answer is yes. ScottMcP takes your question at face ...


10

Let \$\text{SAND}(A,B) = \overline{A}B\$ NOT gate \$\overline{A} = \overline{A}.1 = \text{SAND}(A,1)\$ Connect B to '1' and feed input to A. AND gate \$AB = \overline{(\overline{A})}B= \overline{(\text{SAND}(A,1))}B = \text{SAND}(\text{SAND}(A,1),B)\$ Invert the 1st input (using NOT gate implemented above) before feeding to SAND. OR gate ...


10

Both answers are correct. Let: $$ f_1(A,B,C) = AB+AC+\overline{C}\\ f_2(A,B,C) = A+\overline{C} $$ Let's build the thruth table: $$\begin{array}{|c|c|c|c|c|} \hline A & B & C & f_1 & f_2 \\ \hline 0 & 0 & 0 & 1 & 1\\ \hline 0 & 0 & 1 & 0 & 0\\ \hline 0 & 1 & 0 & 1 & 1\\ \hline 0 & 1 & ...


8

For practical purposes of getting it working on your desk, you probably want an FPGA. In order to make sure it works, you should simulate it first with a program like Modelsim. This enables you to iron out the bugs before buying any hardware.


5

Other answers have focused on why you might be approaching this the wrong way. Although I agree with those answers, what you're asking for does exist, so I'll go ahead and give you a straight answer. You'll likely find that this approach is more expensive than alternatives though. What you want is a 2 GHz voltage-controlled oscillator (VCO) with 3.3-V ...


5

A possible alternative is to use one of the newer FPGAs with high speed transceivers (5 to 10Gb/s). These are intended for fast Ethernet, SATA and other high speed serial interfaces. They are relatively cheap, common, faster than the aforementioned ECL device, and internally deserialise (presenting a serial stream of bits as a parallel word). I understand ...


5

The point where you arrived is almost right: $$ C+\overline{A}B\overline{C}=C(1+\overline{A}B)+\overline{A}B\overline{C}=\\= C+\overline{A}BC+\overline{A}B\overline{C}=C+\overline{A}B(C+\overline{C})=\\= C+\overline{A}B $$ That should be the same answer the K-map and whatever reduction software should give you.


5

you have not connected the body/bulk connection properly in your schematic. PMOS to VSS. But that circuit leaves the output node floating for the A=B=1 condition. Scan through the various solutions here, there is a robust TG version of XOR and XNOR, it uses 4 transistors though. If you look at all the possible states, you have 2 transistors each in 2 ...


4

The heart of the operational amplifier is the difference amplifier, where a current is divided between a pair of highly matched input transistors. This is what allows high gain combined with low offsets. Logic gates simply don't have the geometry to do this. So the answer is no. That said, logic inverters (and by extension almost any inverting gate such ...


4

The most positive input of the differential pair must be within that voltage range. In the case of this chip, it varies from Vee+2.0V to Vcc, so for a 3.3V Vcc (PECL) it will be 2V ~ 3.3V. (the diagram below is for Vcc = 5.0V rather than 3.3V) I don't see Vpp specified in the datasheet for that chip, but it's usually 150-800-1200mV ...


4

Just as a starting point, you may want to consider looking at "time-of-flight" range finder circuits using LASER. Considering RF and light travel at the ~same speeds, I would suspect that the time counting sections would be a good comparative match. A quick google search for "laser range finder circuits" shows a schematic on Parallax.com as the first ...


4

You should try loading the output with something like a 10K resistor, first to ground then to +12V and see what happens. One way or the other you will likely see a reliable logic change, the opposite way you'll see no change in voltage. Sounds like this might be an industrial standard output that can only source or sink current. If you don't put an ...


3

Vbb - reference voltage for ECL used in single-ended input mode or for some types of termination. Usually you bypass to Vcc and tie the differential input that's not being used to Vbb, or tie inputs through back-to-back Schottky diodes to Vbb. Some devices have a Vbb output or it can be generated with an ECL gate. Vtt - termination voltage- usually a ...


3

Texas Instruments, et al. have almost unbelievably extensive stables of digital logic functions available which can be assembled into pretty much anything you want. Go here and here , then check out HC (HCMOS) to get started.


3

Although the 4017 is not an equivalent to the 4026 you can use a 4017 if you build your own 7-segment encoder, e.g. as diode matrix:


3

Sure, if you have digital numbers for the input voltages an op amp merely does the calculation Vout = Av(Vin+ - Vin-). Easily done with gates (and either NAND or NOR gates would be sufficient). NAND and NOR gates require more than just diodes, they require transistors, of course. OR and AND gates are not sufficient. If you do the calculation ...


3

Although you have already selected Spehro's answer, I thought you ought to learn a little about using LS series ICs. LSTTL, like regular TTL, has an input which is essentially an NPN transistor with the emitter exposed. It also has an output which, while it can drive both high and low, is much stronger pulling low than driving high. This makes sense, since ...


3

The concept looks sound. The main thing I see is that your 5.6K resistor is way too high to guarantee a low level for a 'LS04. The 100 ohm resistors you used elsewhere are okay (but they'll get warm if the switch is held). Suggest you use ~150 ohms for the pull down resistors and maybe 4.7K for the LED resistors. It would be better to use a spare ...


3

One of the simplest op-amps looks like this: simulate this circuit – Schematic created using CircuitLab And you can build it using NAND gates: simulate this circuit simulate this circuit So the answer is YES. Keep in mind that this will be crap, and will operate only over avery narrow range. The primary tool for operation that ...


3

Your question makes no sense since 910 can't be expressed by a 8 bit 2s complement number. Such a number can only take on values from -128 to +127.


3

A schematic would be helpful as there is no way to know what "the component on the right" is; however here is what I can tell you: From the datasheet for the 74AHC574, you can see that the 74AHC574 is "OCTAL EDGE-TRIGGERED D-TYPE FLIP-FLOPS WITH 3-STATE OUTPUTS". The tri-state (3-state) outputs are probably the reason for the pull-up resistor. The pull-up ...


3

Take a look at the data sheet: - When the output is driving low (trying to drive to 0V), with an 8 mA load, the specification tells you that it is guaranteed to drive as low as 0.36 volts at ambient temperature and as low as 0.55 volts at full temperature of the device. If it's running from a 5 volt supply and the resistor is 1 kohm, the most current ...


3

Try searching yourself at a distributor such as Digikey to get some idea of availability. I'll take you through a step-by-step if I was looking for, say, a NOR gate with at least 4 inputs. I'll be primarily interested in 4000 series CMOS or 74HC CMOS families (usually it would not be for the same application). First, get on the site and search for NOR ...


2

Welcome to the forum. To begin with, I hope you know what you're doing. Messing around at GHz speeds is not for the faint of heart, or those with shallow pockets. The most obvious way to get a clock is to start with a 1.4 GHz sine wave oscillator, and use the circuit here to convert it to ECLinPS. Once that is done, you will need to look at other chips ...


2

An operational amplifier must have gain. There is no way to get gain out of diodes.


2

In a word; no. The 4017 is a decade counter with decoded one-hot outputs, while the 4026 is a decade counter with decoded 7-segment outputs. Here are the data sheets for the 4017 and the 4026


2

Ohms law can be shown to imply higher resistances produce more power when the supply to them is constant current - clearly as resistance increases, the voltage across the resistor increases thus power also increases because power = voltage * current. However, on your circuit you have a constant voltage supply so current reduces as resistance increases and ...


2

Vtt is used in some ECL devices. It is typically 2 V below Vcc and is internally connected to the differential inputs of a receiver through a pair of resistors, ensuring the inputs are correctly terminated to a transmission line.


2

This (untested*) circuit should do exactly what you specify. Adding the optional inverter and diode should cause it to toggle to and fro between batteries once the current one is charged. Operation: Push button starts charging at battery 1. IC1 + IC2 + Rl form a latch. Pushbutton High sets latch output to high. Chg1 low sets latch output to low. Rl ...


2

While the above answers are correct, in the late 70's through the early 90's sometimes simple CPU's were constructed from TTL logic. While you may or may not want to get into such a project, it is interesting concept. Because the CPU is spread out and hand wired you can probe around and see it in its entirety rather than just as a small black box. Here's ...



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