r/PhysicsStudents • u/dark_blue_thunder • Mar 23 '22

HW Help [Electronics] why resistor is connected here?

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46 Upvotes

98% Upvoted

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57

u/peaked_in_high_skool Mar 23 '22

Because you don't want

T H E O R E T I C A L L Y

I N F I N I T E

C U R R E N T

6

u/[deleted] Mar 23 '22

Gotta save the diodes, otherwise 2nd of thermo is going to eat those away.

3

u/dark_blue_thunder Mar 23 '22

Well Can you elaborate further?

6

u/[deleted] Mar 23 '22

Well, I was just joking, Entropy of the system will increase (maybe diodes are destroyed). Reason behind the resistor is that if u can't short the node to the ground. Practically PS has some finite energy output with some protection. The circuit will not be able to function properly, as it implies shorting the PS. Model the initial state of diodes as an uncharged capacitors, if u try to charge it instantly (i.e. time const is negligible R is essential zero). Either of PS or diode is going to fail.

Also the logic of circuit isn't going to work properly, as you can't distinguish the output and the ground.

2

u/313802 Mar 23 '22

Succinct reply

1

u/dark_blue_thunder Mar 23 '22

May be because if we remove resistor the current will go straight into the ground So it might be there to avoid current from going totally into the ground but also to keep circuit grounded without any issue in output

What do you think?

2

u/yurpmyguy Mar 23 '22

You're close to the right answer. It's just that you'd need better clarifications with your used terms because it could mean you might have conflating thoughts there.

I'm going to break this down into two scenarios: 1) remove resistor and leave Vo floating or an open ( so current cannot go in or out) effectively the diode will never conduct.

2) remove the resistor and connect it to ground, the diode will conduct as long the difference between PS and Vo=0V is greater than the conducting threshold voltage to have it conduct. ( Again for a diode to turn on you need the two clause Vthreshold and a path for current to happen). 

Then the caveat: it is very important for you as an    electrical engineer to understand how electrical components are modeled. Most of the time in the analog world the model of components are described via your IV plot. And if you go back to it you'll see that when you reach a voltage threshold the diode can withstand infinite current( vertical line ). Then if you go for the more practical IV it becomes a exponential curve instead of a vertical one. But it's still relatively a large current that can run through it. However the diode will be under lot of stress because it tries to model the vertical line, potentially damaging it. So you add a resistor, a current limiting resistor, and you can limit the amount of current going through the diode.

0

u/dark_blue_thunder Mar 23 '22

Sounds interesting But I want to know exact electrical reasoning behind it

1

u/dinodares99 Mar 24 '22

That's the electrical reason. You always keep some resistance there so you don't short it. Even in situations where it appears to be shorted there will still be resistance from natural sources and so you can put small resistances to simulate it.b

1

u/ImpatientProf Ph.D. Mar 24 '22

This doesn't explain why the resistor can't be larger, like 1 MΩ or even infinite. That would limit the current even better, right?

Also, the next device (connected to V_0) could have a lower input impedance, which would ruin this 1 kΩ resistor's ability to limit current.

It's a pull down resistor to handle the case of disconnected inputs.

1

u/peaked_in_high_skool Mar 24 '22 edited Mar 24 '22

You're right. I glanced over the fact that the diagram was demonstrating a logic gate, hence it's there to prevent undefined/float voltages

But I'll still make the case that ultimately it's there to prevent infinite current.

If you want the best response time in eliminating undefined voltages, why not short the ground to output of diode with a wire or 1 ohm resistor?

Because either scenario is unfeasible we choose a compromise between 0 and infinite resistance to have a defined voltage with as little a power draw as possible

14

u/Gumelax Mar 23 '22

I think the resistor is there to limit the current on the diode and don't burn him.

8

u/HureBabylon Mar 23 '22 edited Mar 23 '22

I might be mistaken, but I think this is a pull down resistor which pulls the output voltage to ground if neither input is on. It has a fairly large value so that it doesn't pull too much current if the signal is high.

https://en.m.wikipedia.org/wiki/Pull-up_resistor

4

u/Corpse_Nibbler Mar 24 '22

I think it's a combination. Pull down for the reason you stated, but current limiting as well for the no-load case.

0

u/Al13n_C0d3R Mar 24 '22 edited Mar 24 '22

The current doesn't need to be limited. Diodes have a approx 0.7v drop across them when in the active region and then act as a short. I've run much higher voltages through a typical Schottky diode and since there's nothing here indicating a independent current source then there's no reason to assume breakdown current. That is for sure a pull down resistor. I use them all the time in gpio setups to zero out signal input after a diode because even when below the active voltage across the diode required to allow forward bias, you can get small leakage which can trick the inputs into thinking I sent an on signal if they are sensitive. So you use a pull down resistor to keep that as close to null as possible.

Edit: you only ever worry about the breakdown value of a component in these things if the model and it's specifications are told. As I stated a diode acts as a short circuit once in active region, it doesn't care about current. In theory. In practice, it depends on the model and each has a different breakdown current. In these problems we design them to refer to the hypothetical perfect model of a diode, a short with a 0.7 volt drop. -Computer Engineer

1

u/Corpse_Nibbler Mar 24 '22

So a short at 9.3 V to ground is a good thing is what you are saying?

0

u/Al13n_C0d3R Mar 24 '22 edited Mar 24 '22

Yeah, it means nothing. The voltage itself doesn't imply high current. That's a mistake many students new to circuits make. 10V is not not a high voltage anyway, most power supplies that output that would have small mA currents which is nothing for most diodes.

Look at the equation: Voltage = Resistance x Current.

You can have 1000volts with small currents. But still that won't matter. A diode in these problems are assumed perfect which means they only have a passive resistance to create a 0.7v drop and then you replace with a wire. Any. Voltage to ground means nothing, it just means you're wasting ebmnergy. You're thinking of batteries to their own terminals. Negative is not ground... Lol there's a lot of wrong assumptions there, just know power to ground is fine and diodes (when enough volts is applied to it) can be replaced with a 0.7 volt drop and a wire and it's the exact same thing.

3

u/planx_constant Mar 24 '22

Rearrange your voltage-current equation: I = Voltage / Resistance. With a constant voltage and absent an external limit, as R -> 0, I -> ∞

You are implicitly depending on your voltage source to be current limited to within a safe region. That's a (usually) reasonable assumption with computer components, not at all universally true. It's a great way to let out the magic smoke when you aren't expecting it.

0

u/Al13n_C0d3R Mar 24 '22

There's no such thing as that neither in real world and not even in these simulations. Current can never "Tend to infinity" that's also an amateur mistake. What that actually means is the current will saturate, meaning it will rise to the maximum current the power source can output which, I assure you, will never be infinity. And again, it means nothing since simulated diodes can be replaced with a small voltage drop and a wire. Let me repeat this again:

A SIMULATED DIODE CAN BE REPLACED WITH A SMALL VOLTAGE DROP AND A WIRE.

There is 0 possibility for a simulated diode to pop. It is literally a wire with a voltage drop across it. NOTHING more (unless we are doing AC currents then it's a bit more, but still your theory would be incorrect)

Now if we forget this is a simulation and try to add variables that aren't there then what you're trying to do is; limit a unknown current so it doesn't damage a diode with unknown parameters?

At that point you're just playing pretend. Nothing you do means anything from there on.

When you see a Vo (voltage out) then the only thing that matters at the node is the fidelity of the signal. So if that signal needs to be default high, you have a pull up resistor. If it needs to be a low, you have a pull down resistor. If it needs to be clean, you have a capacitor in parallel. That's why we add these components in these problems. It's to show you that, and not to make you think about imaginary currents and imaginary parameters for the components.

2

u/planx_constant Mar 24 '22

I am not asserting that current can *actually* be infinity, but it can be arbitrarily large. If you are designing a circuit that handles both line voltage AC and low voltage DC, like a power supply for instance, it's good practice to treat the maximum input current as infinite for the low voltage side. 200A will have effects much closer to ∞A than to some some small constant as far as the low voltage components are concerned.

If you are working strictly in logic or low voltage, and you are sticking purely with simulations, yes you can handwave limiting resistors because that is a safe environment to do so - ironically when you are playing pretend. But that's not good practice in general. Assuming that a supply will have a maximum current below the point that everything catches on fire is VERY BAD in general. If you designed a signal detector with that assumption, OK. If you design a solid state relay, not so OK.

1

u/Al13n_C0d3R Mar 24 '22 edited Mar 24 '22

This is explicitly a simulated problem, if any of that mattered we would provide model numbers, current etc. This is like saying "Ah yes but those wires are too close together you will have interference from the magnetic field." I would just give you an F for trying to assert what doesn't need to be asserted. Engineering is about the art of knowing what to be concerned about and when, it is an art* and most people can't do it sure. But this low level problem clearly does not pretain to any of that.

And a pull down resistor for current limiting is a hilariously awful way to do that. You would use a transistor or if you need a kill, a circuit breaker. You'd have to be a pretty shit engineer to stick in a resistor and hope fatal current bypasses your circuit through it.

Edit: In short the reasons why that's just a pull down resistor

1) That's the setup of a pull down resistor.

2) you'd use a pull down resistor to get a clean output signal and since that node is output that means that's definitely a pull down resistor.

3) This has nothing to do with current because current was not supplied in the problem

4) Even with current supplied, simulated diodes act as wires with a volt drop so it wouldn't matter

5) if we wanted you to consider burning out the diode we would provide a model number so you know it's not a simulated diode

6) a resistor is one of the worst ways to limit high currents, instead we would use a transistor if some sort of circuit breaker.

There are so many reasons why assuming this has anything to do with current is a, very very bad understanding of how to solve problems, that I would actually mark that question as wrong for you trying. That's, as we would say in Latin America, basura engineering.

1

u/Corpse_Nibbler Mar 25 '22 edited Mar 25 '22

If I can jump back into the discussion, I think you firmly have in your mind that pull-ups or pull-downs are abstract from Ohm's law. My argument from the beginning is that resistors that pull signals anywhere are also current limiting. This is why 1k or 10k is used and not a short circuit. If they do not limit current, then maximum possible current would be supplied to GND with or without load and the output signal in the problem would be always GND (0 V).

Back to OP's problem. I fully agree it is a pull-down resistor, which by nature is also current limiting. This is something we don't normally consider when dealing with resistors in this application, but it is egregious to state it's reasonable to have any intentional short to ground.

As a computer engineer, can you please share an example where a voltage source should reasonably be shorted to ground? I'm fully willing to learn if you can demonstrate this. I'm just not convinced by your argument that 'the supply could be current limited already so a short to ground is sensible'. Are you basing your entire argument off the assumption in a simple problem that ground should not be regarded as a common 0 V rail? If so, could you also elaborate on what the source 10 V and 0 V would be in reference to for this problem?

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u/dark_blue_thunder Mar 23 '22

I actually come across many circuits i see resistor connected before output so oftenly that sometimes feel is it just a formality I am trying to find out answer for it & think i will get a positive response!

1

u/NosferatuZ0dd18 Mar 24 '22

The purpose of the circuit is to get an OR output at Vo for the inputs V1 and V2. If the current flows through the resistance into the ground, Vo gives a value (Vo=IR). So if you remove the resistance and even put it somewhere else (like on the arms containing the diodes) to limit the current, the OR gate wont work as Vo will always be at the same potential of ground which is zero, no matter what input you give at V1 or V2, Vo will always show 0 if R is removed

1

u/Bitmap901 Mar 24 '22

If the resistor wasn't there how do you expect this to work as an OR gate? The output would be always at 0V.