# Tag Info

40

74xx05 ICs have open collector outputs. Therefore, when the input is high, the output is pulled down, but when the input is low, the output is in high impedance. To get an high level output, you should connect a pull-up resistor between the output and VCC (5V). Or, use the push-pull (totem pole) output version, i.e. 7404. simulate this circuit – ...

26

Don't think for one minute that just because you have an FPGA that learning about 74xx is obsolete. For designing with FPGA you must be able to 'see' the logic working in your head at a discrete gate level (you will learn this skill from discrete logic chips 74xx, cmos 40xx ). Programming an FPGA is NOT like writing a computer program, it looks like it is, ...

26

If this is for an industrial project you are probably better off using an opto-isolator to do your conversion. It provides several benefits. Current loop on input side is impervious to common mode noise. Isolated grounding gets rid of loops and different ground level issues. Galvanic isolation provides protection to both sides. And of course, the level ...

19

Two kinds of discrete logic I see still used a lot: Buffers. If you need 60 mA to drive a long bus line, or you have an incoming signal from off your board that you don't want to give a chance to fry your FPGA, you still need a discrete buffer device. Buffers are also used as level-shifters between 5 V legacy interfaces and low-voltage FPGA I/O's. Little ...

18

Two overlapping circles represents a current source. In this case, it is being used to sink a certain minimum amount of current through the output transistor, in order to keep its dynamic impedance low and improve the overall frequency response. Current sources (and sinks) are commonly used in IC design, because they're actually easier to implement than ...

16

Given the goals of the class, I think the TTL approach is fine, and I say this as an "FPGA guy". FPGAs are a sea of logic and you can do all sorts of fun stuff with them, but there's only so much that's humanly possible to do in a semester. Looking at your syllabus, your class is a mix of the logic design and "machine structures" courses I took in ...

16

In short, the answer is yes. If you look through the DATASHEET, you can see that it is fine to use ceramic capacitors. If you go to page 10, and look at the 'Typical Applications', you will find some notes at the bottom. In case that wasn't clear enough, it is also stated in 10.2.2, that ceramic capacitors can be used. In fact, you will probably find ...

16

A custom chip for 10k units is unlikely to be cheap, as the masks will almost certainly cost more than $1k, which means your chip cannot possibly be made for$0.10. The reason the one you've found costs \$0.10 is because they sell in a much greater volume (100k-millions), so the cost of masks and setup for a production run is spread over many more units. ...

12

The first comprehensive logic series was the TTL series 74xx. This used BJTs (Bipolar Junction Transistors). Later there came variants like the often used 74LSxx, where the "LS" stands for Low-power Schottky TTL. As the name implies these used less power than the rather power-hungry TTL, and were faster too. At the same time the CMOS 4000 series was ...

12

No, the circuit structures to produce gates in TTL and CMOS are very different. It's actually a very complex topic, because at this level, you can't just treat transistors (BJTs or FETs) as simple "switches". It becomes an analog circuit design problem in which many issues need to be considered: how static and dynamic currents flow, where charges are stored ...

12

It's the symbol of a current source. There are different symbols used to represent a current source:

12

The 74HC14 isn't TTL, it's CMOS, and what matters is that the input pins be kept from floating. The reason is because the impedance of the input pins is so high when they're floating that random charges accumulating there can/will cause unpredictable internal behavior in the chip, including oscillation and high power dissipation. The outputs don't need to ...

12

You have many options. If you need to connect very few optocouplers, you can connect them directly to the GPIO of your microcontroller (through a resistor), provided that: You do not exceed the GPIO output current. You do not exceed the total port current. You do not exceed the total gnd/vdd current. If you need to connect more optocouplers, you can try ...

11

When no device is pulling down the line, the "left side" (with lower voltage) is in high state by pull-up resistor. The voltage between the gate and the source is below the threshold voltage and MOSFET isn't conducting. So the "right side" (with higher voltage) is pulled up by pull-up resistor too. When the "left side" pulls down the line to a low state, ...

11

A simple BJT like MMBT3904 or any switching BJT will do the job. You can get a reel of 100 for two bucks.

10

It's a Bad Habit, just like thoughtlessly using a 1k$\Omega$ base resistor for a switching transistor, or as thoughtlessly using a 100nF decoupling capacitor. Millions of engineers get away with it. Even without much calculation we can see that 1k$\Omega$ is a bad choice. From a +5V supply it won't allow more than 5mA, and that's even without taking ...

10

That open collector output ... when off, it will have some leakage (possibly microamps) depending on temperature. That can supply enough base current to turn Q2 on, at least partially. The simulation might not model that leakage accurately. R1 pulls Q2 base to 5V against that leakage, ensuring Q2 is fully off.

9

The 2.4 V is the minimum for a high level TTL output. A TTL input needs at least 2.0 V, to give you a 400 mV noise margin. But that's for TTL devices, or TTL compatible, like HCT-CMOS. An HCMOS device will typically need 0.7 Vdd as minimum for a high level. For a 5 V supply that's 3.5 V, so you probably won't make it. You can use an HCT single gate IC as ...

9

Do you know that the horizontal strips of a breadboard are all connected together (except sometimes in the middle)? To me this looks like you have shorted the input to the output of the 7805. (Or maybe the output to te ground, my view is blocked by the capacitor.) ==================================== OK, next step. You are using a plain old 7400, no LS or ...

9

The input stage of a TTL device acts as a reverse-biased diode. As such, any high impedance at the input stage, whether due to expansion of the depletion zone of the B-E junction or due to an entirely disconnected pin, will appear as a high input.

9

The differential line drivers are not designed for driving LEDs. These buffer chips drive (or receive) a differential signal on two wires. The voltage swing may be 1.3 volts to 1.7 volts. Not enough to turn an LED on or off. The TTL buffers are ideal for this application, but rather than connect to the high side of the LED as drawn in your schematic they ...

8

Where to start... Logic Gates used in VLSI chips are not constructed from Bipolar Junction Transistors (BJT) but rather use Field Effect Transistors (FET) (Metal Oxide Semiconductor (MOS) to be exact). The topologies used in this realm are commonly composed in Complementary P-type pull-up and N-type pull-down networks. Basically I think these types are ...

8

You don't say in so many words, but the "idle high" suggests you mean a UART. UARTs a point-to-point connected to line-transceivers, like the ubiquitous but dated MAX232 (there are far better solutions nowadays). The line between microcontroller and transceiver will also be short; if there's distance to be bridged it will be between transceivers. The ...

8

The SN74HC266N is CMOS, not TTL, and has open-drain outputs. Try putting a pull-up resistor (1k, say) on the output. You should then get the expected outputs, according to the truth table in the data sheet.

8

"Open Drain" output devices operate like a switch. They don't source current of their own. They are usually used to either drive higher current devices, or to connect to a bus along with lots of other devices. To use as a standard output device you have to supply the output with a voltage. This is in the form of a pull-up resistor (a resistor connected ...

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.

8

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

8

It sounds like you want an octal buffer with tri-state output. For your 5 V application, a device such as a 74HCT245 will do it.

8

However looking through the datasheets I found that MAX9010 outputs TTL levels, while 74VHC86 accepts CMOS levels (0.7 * Vcc). That's a good spot and I agree with you - maybe you should inform Maxim on their dodgy circuit. Shame on them. Should I pay special attention to this issue - what are the conditions when circuit may fail producing proper ...

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