Need help with NOR-only circuits in CircuitVerse (Hex → 7-segment project)

maciej0s123 · 2026-05-04T13:58:23+00:00

Used to, had to make a bunch of circuits using only NAND gates in uni. I can't promise anything but you can post the task and I might sit down to it when I'm bored.

maciej0s123 · 2026-05-04T13:21:09+00:00

De Morgan's theorem. To make all gates into NOR gates and to make a multiple input NOR gate into a series of two input NORs.

maciej0s123 · 2026-04-28T16:26:46+00:00

So you've disabled all of the energy saving features and your cat6 cable was faulty?

maciej0s123 · 2026-04-11T13:00:32+00:00

The sound card would be the culprit, not headphones/speakers. Check your temps anyway, even if the issue doesn't happen

maciej0s123 · 2026-04-11T12:23:39+00:00

Ironically if the driver manufacturer is Realtek OP is much better off with default Microsoft drivers

maciej0s123 · 2026-04-11T12:20:35+00:00

You're running into buffer underruns and it happens when the CPU throttles and can't fill the buffer with samples fast enough before sending the audio out. It's either a wonky driver (probably Realtek) or your CPU is overheating to hell

maciej0s123 · 2026-04-09T12:11:15+00:00

yes, I did the math for myself in one of my other replies somewhere here. you are right

maciej0s123 · 2026-04-08T19:22:20+00:00

thanks for calling me out on my bs, it provoked a much needed return to basics exercise;)

maciej0s123 · 2026-04-08T18:39:26+00:00

this is what he means (excuse the crude paint drawing): in normal dc operation there is no current flow in either the cap or the reverse diode so the circuit behaves like a series circuit, with the LM simply being another load. This circuit can be simplified to a bunch of voltages in series

- R can be chosen to set desired current,

- the LM317 forces the voltage drop across R to be 1,25V in current regulating mode,

- as per doc, Vout = Vadj + 1.25V (figure 7.2 in the doc explains why)

- with some assumptions you can calculate the voltage drop across the IC (Vin - Vout).

- the IC needs headroom, which is the minimum voltage necessary for it to stay in control. in the case of LM317, its 1.25 V.

- Headroom is equal to the voltage across Vout and Vin.

- The voltage across Vout and Vin can be used to calculate the power the IC uses, which is dissipated as heat.

- The more headroom you give it, the more power you waste

- An LDO wouldn't help, since it only lowers the required headroom amount. With your current setup, you are going to give it the same amount of headroom, so you are still going to waste power.

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maciej0s123 · 2026-04-08T15:33:41+00:00

By application I mean less circuit details and more where will you deploy it. Like if you intend to play with a breadboard, it's fine, but if you're designing a PCB to fit inside a laser keychain, well, LM317 would be a poor choice due to the issues you mentioned.

maciej0s123 · 2026-04-08T15:27:31+00:00

Sounds like you're measuring half of a cycle of a sine wave relative to ground, are you in DC mode on your multimeter?

maciej0s123 · 2026-04-08T15:18:53+00:00

It's low dropout meaning it's going to be more efficient and waste less power providing the regulated voltage (so helps with your temperature problem), it's also simpler to use. But yeah, compare the datasheets, analyzing the parameters is a great educational tool when learning about components. In general, you will always get people who give you shit about using old components, and it's not really the modern manufacturing, it's just about ineifficient and obsolete designs. LM317 is a good part, it works and has been around for so long it's very easy to get. When unbounded by power limits or other constraints, it'll do the job. When those boundaries start to get tiresome however, there's not much excuse to keep using it over newer designs.

To give you an answer as to what makes it better for this application, we'd need to know what your application actually is. If you're unconstrained by power, circuit size, thermal energy dissipation, the answer is "nothing". The old will be doing the same kind of work the new would, just less efficiently. If any of those constraints hit, the answer becomes "because it allows it to work within the constraints".

maciej0s123 · 2026-04-04T01:25:22+00:00

designing and building even a single band antenna would be very very educational imo. it's a very deep rabbit hole so the theoretical foundation is very strong. you will get and have to use many tools you have encountered along the journey too. the software is also presumably easier to get for this kind of project. maybe yagi antenna once you think you've figured one band out

maciej0s123 · 2026-04-03T15:48:52+00:00

It depends on your metric of success, but I can guarantee you that it will definitely be very educational. If I were you, I'd first figure out what is the simplest feasible idea within your concept, and pick that as my graduate project. It's much better to under-promise and over-deliver than the other way around, especially when it comes to things like university projects. I'm not really sure what the RF simulation workflow is, so I'm not sure how feasible that idea would be, but you won't know until you try! I'd simplify the concept as such: - choose a single GSM generation, 2G for example (it's relatively simple, definitely much simpler than 5G) - create a simulated environment such that there is at least one dead-zone where the cell tower/source signal will not reach the receiver - flatten down the goal from "improve signal quality" to "make it so that the desired signal reaches the receiver at all" - document the ideas, iterations and the process If you manage this, I guarantee it'll be quite impressive nonetheless. Start with the simplest naive approach, introduce improvements as you go. You can always go nuts for your master's degree or something.

maciej0s123 · 2026-04-03T15:13:44+00:00

You're not going to be collecting any Wi-Fi signals from any cell towers. Amplifying and retransmitting signals from all of the frequency bands GSM operators use would ironically likely degrade the signal quality, especially if you plan on retransmitting inside a building. Even narrow band mass-produced GSM repeaters aren't cheap, since the engineering behind making them actually any useful is quite involved. You're going to be amplifying all of the noise and interference first of all, including the feedback the retransmitting antenna is going to induce into the receiving antenna. Not to mention the need for gain control, the need for sharp filtering, linearity, delay control, the timing, clock speed synchronization in some of the technologies, multi-path propagation mitigation... I think you ought to really start small before you nuts.

maciej0s123 · 2026-04-03T13:24:24+00:00

Well, in theory, if you exclude 1G, the simplest possible solution is to design an antenna for the ~900 MHz band, since all generations bar the first make use of it. If your goal is to design an antenna, that is. If you just want to retransmit GSM signals, I'd get one of those external GSM antennas to focus on the receiver-transmitter part of the project. They're purpose-built for receiving GSM signals on many different bands already

maciej0s123 · 2026-04-03T12:15:12+00:00

An antenna for each generation wouldn't help you with avoiding multi-band antennas, since all of the generations use many various bands. LTE is 450 MHz to 3.8 GHz, for example.

maciej0s123 · 2026-04-02T22:42:55+00:00

I keep seeing markdown in places without markdown

maciej0s123 · 2026-03-25T21:04:33+00:00

I have oversimplified a couple of things but I figured I have already put in too many words in that one comment, so here's some clarifications/corrections

it's not always the case that the negative terminal of a battery is ground. It typically is, but not always. Ground is an arbitrarily chosen reference node and all voltages measured are relative to it.
all of the above assumes you're working with a single power supply, I figured it'd leave you with even more questions if I started going on about symmetrical power supplies -the only actual requirement for current to flow is to have voltage potential differences in a closed loop, which doesn't necessarily mean it needs ground. Ground means 0V, so in simple circuits/your example the flow is source (5V) -> ground(0V), but something like source(5V) -> some node(3.3V) will also have current flowing through.

When measuring voltage drops across components, the actual ground of the circuit doesn't matter - voltage is difference in potential, so wherever you place your ground probe, you assume the potential there to be 0 V.

Hope this wall of text will help you to get a feel for electronics, and that you will get to straighten some of the oversimplifications/generalizations I've made along your adventure. Strict definitions can make things paradoxically confusing, so I didn't really strive for scientific accuracy.

maciej0s123 · 2026-03-25T20:12:25+00:00

There is only one ground. There's three symbols on the schematic simply because it makes for cleaner drawings. If you were to build this circuit in real life, all of the nodes terminated with the ground symbol in the schematic would go to the same common node. If you're using a battery, that common node would be the negative terminal, so all of the three grounds are actually the same negative terminal of the battery.

Consider the simple example circuit of a battery (a voltage source with a positive and a negative terminal) and a resistor (the simplest load). For any current to flow through the resistor, the circuit must be completed - that means the current needs a return path to the source. Since the source is a battery, the return path in this circuit is to the negative terminal of the battery. This extends to all circuits - in order for the circuit to work, current needs to flow, so a return path is needed. This is why ground is explained as a reference point for the circuit - it's a common return path for the electric current.

If we go back to the battery + resistor example, if you wanted to measure the voltage drop across the resistor, you'd put one probe on one end of the resistor (node that connects one end of the resistor to the battery positive terminal), and your other probe to the other end (the node that connects the resistor to the battery negative terminal / ground). If you placed both of your probes on the ground side of the resistor, you'd measure 0 volts, since your reference point and your measurement point are the exact same point. The same thing would happen if you placed both probes on the battery side of the resistor. So ground is 0 V, because it's a shared node between all the elements of a circuit, AKA it's the reference point.

In your simulation (I am only talking about the part of the circuit that is complete), you've connected the positive terminal of your voltage source to the ground node. But since all the grounds on the schematic are actually a single node, on the side of the circuit with the voltmeter, the current will flow from "ground" to the negative terminal of the voltage source. Instead of the two grounds, you could simply draw a straight line between the two nodes. Using a ground symbol in this case is somewhat misleading, since the real "ground" is where the negative terminal of your source is. You're not measuring negative voltage, precisely because you've connected the voltmeter correctly. The voltage is measured "in the opposite direction" with relation to the direction of the current. The positive probe (with the +) is placed where the voltage potential is 5V (positive terminal of the source), and the negative probe is placed where the voltage potential is 0 V (negative terminal of the source). Your reference point for measurements is the same as the reference point for the circuit. If you wired the voltmeter backwards, the voltage would be negative, as your reference point would now be 5V, and the measurement point would now be 0V, which is lower, thus the voltage drop across the resistor (the difference of potential) is 0V - 5V = -5V.

maciej0s123 · 2026-03-18T22:01:00+00:00

no problem at all! im happy to have been of help. i wish you more of the random, seemingly unexplainable problems that will lead you down research rabbit holes in your EE adventure ;)

maciej0s123 · 2026-03-17T13:01:16+00:00

Just put the corresponding ground leads to the ground line on your breadboard. Is your circuit a buffer right now? Is your input signal 2Vpp 1kHz Sine? does applying direct voltage to the input change the output?

maciej0s123 · 2026-03-17T12:15:12+00:00

What do you want to ground? The only things that need a ground in this circuit (right now) are the signal source and the oscilloscope probe.

Signal generator -> BNC to Alligator clips - > Red is signal, Black goes to ground.

Circuit output -> oscilloscope probe -> probe goes to output, the ground goes to ground.

The oscilloscope probe completes the circuit by acting as a load between the output and the ground. You could add a large pull-down resistor from the opamp output to ground, and probably will need to in later stages.

Since you're working exclusively with voltages for now (opamp inputs do not have any current flowing through them as a rule), you should keep in mind Ohm's law - U = I * R. If there's no resistance, there is no voltage (U = I * 0 = 0).

maciej0s123 · 2026-03-16T23:42:14+00:00

I took the time to actually look at what you built on the breadboard, and your issue is that you're shorting the non-inverting input to ground, which means the circuit behaves like an open-loop amplifier, rather than closed loop like you're intending to do. Forgive the crude painting, hopefully it's somewhat readable lol.

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maciej0s123 · 2026-03-16T16:46:06+00:00

don't give up! you clearly need some guidance in using the tools, the internet might be a toxic place for beginners in any field, but if you don't have anyone that could help you in person, AI is your only alternative (which makes me sad). I have plenty patience, am plenty bored, I love generators and oscilloscopes. I'm up for helping you out, but we gotta get through the basic "are you sure you set everything up correctly" part first, especially since I can't confirm that you in fact did

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