Author Topic: Do you have to tie grounds together to get an arduinos IO to works with other...  (Read 2778 times)

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Offline BeaminTopic starter

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...other chips or devices?

Say you have an Arduino powered off USB and you have some logic chips or simple devices that have the ability to get logic level signals on a bread board running off batteries. Maybe you want the arduno to program the memory of a another chip, can't think of the name but DIP package comes to mind, very simple learning electronics level chips. This chip has a pin that goes high and low for 1000ns to write data to it and another for data. Can you just hook one of the Arduino digital out pins to the chips input pin then connect a second digital out to the data pin? Or does everything need a reference for what ground is, so connect the negative of the batteries to a gnd pin on the arduino?
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Offline dmills

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In general yes, there has to be a common path for the return current if this is to work.

You do sometimes find that it seems to sort of work without an obvious return path, but that is because leakage and stray capacitance is getting the job done for you, it should never be relied upon.

Currents flow in loops (always), make the loop the current flows in predictable and things may work well, make the loop something that happens by chance and things work less then well (or not at all).

Regards, Dan.
 

Online rstofer

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There are no one-wire circuits.  To actually be a circuit, the path must go 'around'.  The word circuit was well chosen:

https://en.wiktionary.org/wiki/circuit#Etymology
 
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Online Zero999

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Yes, a common ground is needed. As mentioned above it might work without an obvious return path, but it could be unreliable and even cause damage, if the two circuits float at high voltages, relative to one another.

In some applications it might be necessary to maintain different grounds, which is why opto-couplers, pulse transformers and digital isolators are used. One of my favourites is the Si8642, which provides two channels in each direction, across an isolation barrier.
https://www.silabs.com/documents/public/data-sheets/si864x-datasheet.pdf
 

Offline IDEngineer

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There are no one-wire circuits.
WHAT?!? You mean Dallas Semi was lying to us all this time?!?  :palm:
 

Online rstofer

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There are no one-wire circuits.
WHAT?!? You mean Dallas Semi was lying to us all this time?!?  :palm:

Maybe...  There is one DATA wire and one GROUND wire.

Quote
One distinctive feature of the bus is the possibility of using only two wires: data and ground. To accomplish this, 1-Wire devices include an 800 pF capacitor to store charge and power the device during periods when the data line is active.

https://en.wikipedia.org/wiki/1-Wire

True, sending power and bidirectional data over a single data line is quite clever but it still requires a ground return.
 

Online Mechatrommer

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will mutual inductance or errr... wire "less" count? they are 0-wire ;D
Nature: Evolution and the Illusion of Randomness (Stephen L. Talbott): Its now indisputable that... organisms “expertise” contextualizes its genome, and its nonsense to say that these powers are under the control of the genome being contextualized - Barbara McClintock
 

Offline PointyOintment

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You can also, at high enough frequencies, complete the circuit using capacitance. This is how Tesla coils work, for instance: It looks like there's only one wire going to the topload, but there is sufficient capacitance between the topload and ground (the literal ground it's standing on) to pass enough current at the operating frequency.
I refuse to use AD's LTspice or any other "free" software whose license agreement prohibits benchmarking it (which implies it's really bad) or publicly disclosing the existence of the agreement. Fortunately, I haven't agreed to that one, and those terms are public already.
 
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Offline BeaminTopic starter

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There are no one-wire circuits.
WHAT?!? You mean Dallas Semi was lying to us all this time?!?  :palm:

Are you talking about that thing on some computer mother boards that looks like a battery but is actually a security device? I always thougt that wasn't really one pin but the case was grounded. How can they possibly do that; the output pin can go above and below 0 effectively making 1 and 0 over one wire: 1t would be +5v for 1 and 0 would be -5 volts. But how would that get power?, a cap inside that recharges when you first turn it on?

So what if we don't connect the grounds; instead tie a pin on the second chip that's held logic level low will you still need to connect up the ground pins?
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Offline Nusa

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If there isn't a circuit, no current will flow, and no signal will travel. Doesn't matter what you do at the input or output pins if there's only one wire connected and no ground path to complete the circuit.

This is practically the first thing you learn in any electricity tutorial/book.
 
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Offline BeaminTopic starter

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If there isn't a circuit, no current will flow, and no signal will travel. Doesn't matter what you do at the input or output pins if there's only one wire connected and no ground path to complete the circuit.

This is practically the first thing you learn in any electricity tutorial/book.

Yes I know that but my question was how to do it. You often see in designs to build things with one output but I'm guessing they are omitting the ground because its common sense.

For instance your cable TV runs off one conductor to a station miles away. Unless that works off RF where they are sending RF down a cable instead of an antenna but that would be a signal with no return path.
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Offline Kjelt

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The only exception I know off are balanced signals, still two wires but there is within a reasonable distance no ground wire needed.
 

Offline dmills

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Cable telly rides on coax, which last time I checked has a centre conductor AND a surrounding (typically) foil shield, hence two wires.

That it is also wideband RF is irrelevant.

Regards, Dan.
 

Offline GeorgeOfTheJungle

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You can signal with a current, then don't need a common ground, but still need two wires. When the signal is a voltage, remember that a voltage is a electric potential difference, so you need to have some common ground to know what that difference is.
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Offline BeaminTopic starter

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Cable telly rides on coax, which last time I checked has a centre conductor AND a surrounding (typically) foil shield, hence two wires.

That it is also wideband RF is irrelevant.

Regards, Dan.

Think about it if you send down RF you don't have to worry about the ground you are just using copper instead of air and the ground doesn't matter its just a shield. I'm almost positive they do that because it would be easier to upgrade from the old system.
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Offline Nusa

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Cable telly rides on coax, which last time I checked has a centre conductor AND a surrounding (typically) foil shield, hence two wires.

That it is also wideband RF is irrelevant.

Regards, Dan.

Think about it if you send down RF you don't have to worry about the ground you are just using copper instead of air and the ground doesn't matter its just a shield. I'm almost positive they do that because it would be easier to upgrade from the old system.

I propose a test since you are almost positive.

Take your TV coax, pick a spot anywhere you like, strip off 1 inch of the shield without breaking the core (can be done with a knife if you don't have the right tool), and see if it still works.

Assuming you get the result I expect (not working), connect the two shields with a wire jammed in each side and see if it suddenly works again.

Now replace the cable you ruined. Or if near the end, cut it off and put a new connector on it.

[A less destructive version of this would be to undo your cable connector from the modem, stick a piece of wire in the modem middle and connect it to the center wire of the wire connector. With a micrograbber, for instance. See if it works. Now connect the outside shields, perhaps with alligator clips. Test again.]
« Last Edit: October 21, 2018, 05:52:40 am by Nusa »
 
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Offline GeoffreyF

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You can also, at high enough frequencies, complete the circuit using capacitance. This is how Tesla coils work, for instance: It looks like there's only one wire going to the topload, but there is sufficient capacitance between the topload and ground (the literal ground it's standing on) to pass enough current at the operating frequency.

Love to see your Arduino connected directly to a Tesla coil. Pictures would be really awesome to see.
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Offline Buriedcode

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If there isn't a circuit, no current will flow, and no signal will travel. Doesn't matter what you do at the input or output pins if there's only one wire connected and no ground path to complete the circuit.

This is practically the first thing you learn in any electricity tutorial/book.

Yes I know that but my question was how to do it. You often see in designs to build things with one output but I'm guessing they are omitting the ground because its common sense.

For instance your cable TV runs off one conductor to a station miles away. Unless that works off RF where they are sending RF down a cable instead of an antenna but that would be a signal with no return path.


Can you provide examples of what systems/connectors you believe to have "one output" ? I know many connectors look like they just have one connection - the center - they they also have surrounding metal often as a ground, which is also used as the shield (balanced lines also have their shield as the ground reference).  I've read the thread, and its quite hard to tell if you're joking, or you've managed to "do" electronics all this time without realizing that voltage is a potential difference. 

Are you talking about that thing on some computer mother boards that looks like a battery but is actually a security device? I always thougt that wasn't really one pin but the case was grounded. How can they possibly do that; the output pin can go above and below 0 effectively making 1 and 0 over one wire: 1t would be +5v for 1 and 0 would be -5 volts. But how would that get power?, a cap inside that recharges when you first turn it on?

I have never seen such a security device on a PC motherboard.  "One-Wire" could I suppose be used on PC motherboards, but that's just the spec.  I think you are referring to these devices? https://en.wikipedia.org/wiki/1-Wire#/media/File:I-button.jpg  That use the one wire protocol: https://en.wikipedia.org/wiki/1-Wire  .  I have never seen one on a PC motherboard, but I have seen coin batteries used for bios/RTC backup, could that be what you're mistaking for "not a battery" ?

As I said, I can't tell if you're serious... so apologies if I come across a bit rude.
 

Offline GeoffreyF

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If there isn't a circuit, no current will flow, and no signal will travel. Doesn't matter what you do at the input or output pins if there's only one wire connected and no ground path to complete the circuit.

This is practically the first thing you learn in any electricity tutorial/book.

Yes I know that but my question was how to do it. You often see in designs to build things with one output but I'm guessing they are omitting the ground because its common sense.

For instance your cable TV runs off one conductor to a station miles away. Unless that works off RF where they are sending RF down a cable instead of an antenna but that would be a signal with no return path.

Actually, on what you state with Cable TV, NO.  The shield is a conductor, it is grounded, it is the return path.  It doesn't come from miles away, it comes from an amplifier or other equipment hundreds of feet away.   Although the return of a circuit is not always a ground, there is always a return. That  is why they are called "Circuits". Even old fashioned Telegraph circuits which uses the actual ground for the return, still has a return path.  You say you come here for clarity but it seems like you turn from it or introduce speculations instead of the explanation which you sought.  The concept of "Potential" or voltage is based on the idea of a circuit or a path of return for charge carriers, such as electrons.
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Offline tooki

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There are no one-wire circuits.
WHAT?!? You mean Dallas Semi was lying to us all this time?!?  :palm:

Are you talking about that thing on some computer mother boards that looks like a battery but is actually a security device? I always thougt that wasn't really one pin but the case was grounded. How can they possibly do that; the output pin can go above and below 0 effectively making 1 and 0 over one wire: 1t would be +5v for 1 and 0 would be -5 volts. But how would that get power?, a cap inside that recharges when you first turn it on?
Uhhhh… no. IDEngineer was referring to the 1-Wire standard, which was designed by Dallas Semi.

As for the alleged security device: Ummm, are you sure you aren't confounding various conspiracy theories and other technologies? Right now, there's the Bloomberg story about alleged covert security implants in SuperMicro server motherboards, which is probably false. (I am incredibly excited to see how that clusterfuck resolves!!)
Dallas Semi made battery-backed RAM modules that were commonly used in 70s/80s equipment. They're common in test gear for storing calibration constants and user settings.

Do you have an example of or reference to the "battery [that] is actually a security device"?


How can they possibly do that; the output pin can go above and below 0 effectively making 1 and 0 over one wire: 1t would be +5v for 1 and 0 would be -5 volts. But how would that get power?, a cap inside that recharges when you first turn it on?
What are you referring to?

FYI, 1-Wire is basically a 3 or 5V bus (provided by a pull-up resistor) that transmits ones and zeros by pulling the bus low (0V, not negative!!) for different periods of time. (A capacitor in the device keeps it powered while it does this.)
 

Offline IDEngineer

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Can't believe this thread is still going.

"Ground" is just a fancy term for "reference". Seems like this discussion comes up every once in a while. Here's an experiment that sometimes helps.

You have a 12V battery and a DC voltmeter. You want to know the voltage potential at the + terminal. So you take the red lead of the meter and touch it to the + terminal. You leave the black lead touching nothing, because "you don't care about the - terminal, just the +". What does the meter read? Zero. Why? Because voltage measures the DIFFERENCE between two points. A measurement - almost by definition - has to be relative to something, some agreed-upon baseline reference point. When you stick the positive lead on the + terminal, the - lead is connected to nothing... and there is no voltage between the + terminal and nothing, so the meter correctly reads zero.

Now, you touch the black lead to the - terminal of the battery and magically, you see 12VDC. Why? Because now the meter sees a difference in voltage between its two leads.

Next you touch the black lead to the + terminal along with the red lead. They're both touching the same terminal. The meter reads... zero, because there's no DIFFERENCE between the two leads, they're both touching the same thing.

Now let's grab a second 12V battery. And let's connect Battery A's - terminal to Battery B's + terminal. Now you effectively have a three-terminal battery: A + terminal at one end, a "middle" terminal, and a - terminal at the other end. If put both of the leads on the + terminal, what will the meter display? Zero, because there is no DIFFERENCE between the leads when they're touching a shared conductive object. If put both of the leads on the - terminal, what will the meter display? Zero again, because there is no DIFFERENCE between the leads when they're touching a shared conductive object.

Now put the black lead on the - terminal and the red lead on the "middle" terminal. The meter reads 12V... even though there's a whole second battery right there! Why? Because the DIFFERENCE the leads see is just the difference between the terminals of one battery. Since the other battery is not "in the circuit", it's as if it doesn't exist to the meter. You can connect it, and disconnect it, and wave it around, and carry it outside, and the meter won't know or care.

Put the read lead on the + terminal while leaving the black lead on the - terminal. The meter reads 24V. Why? Because you're measuring across BOTH batteries. The first battery has a DIFFERENCE of 12V between its terminals, the second battery also has a DIFFERENCE of 12V between its terminals, and since you've effectively "stacked" the two batteries, the meter "sees" both differences and sums them.

Now we come to the heart of the matter: Move the black lead from the - terminal to the "middle" terminal. What the heck? The meter now reads 12V! But the red lead is still on the + terminal, the one that was giving us 24V just a moment ago!?! How can the same + terminal yield two voltages? The answer is, because the + terminal - by itself - is meaningless. The + terminal only has a "voltage" RELATIVE TO SOMETHING ELSE. The meter reads the DIFFERENCE between two points, remember? So yes, you left the red lead on the + terminal but you moved the black lead to a different place on the battery... and so the difference between that new place and the + terminal is not the same voltage as the difference between the two places you were measuring just a moment ago.

We can call the - terminal "ground", and then we have two different voltages we can measure that are both positive relative to our agreed-upon ground point. Or, we can agree to call the "middle" terminal "ground" and then we have two different voltages, one of which is positive relative to "ground" (the + terminal) and one of which is negative relative to "ground" (the - terminal). Heck, we could even agree to call the + terminal "ground" and then we'd have two different voltages that are both negative relative to "ground".

If you've followed along here, you should now understand that voltage is a measurement of the DIFFERENCE in potential between two points. We can arbitrarily refer to some voltage as "ground" and then measure everything relative to that point.

Hopefully now you recognize that a single contact between two systems won't convey any information. It's just like the very first step above, touching just the red lead to the + terminal... the meter read zero because it could not discern any DIFFERENCE between its two leads. That + terminal might be at 1V, 12V, 1000V, could be AC or DC, could be anything, but the meter cannot SEE any of that until it can compare its red lead to its black lead.

Likewise with your two Arduinos. Connecting a single wire between two otherwise isolated systems doesn't give them any way to measure DIFFERENCE on that wire. They need a shared reference point, an agreed-upon baseline.

To avoid confusion: There are many ways - some intentional, others not! :-DD - to provide this reference point. Connecting a second "data" wire might, in fact, work under some circumstances because now you've established a two-point connection and under a magic set of conditions each might provide a reference (really, a current path) for the other. But it's unwise to rely on "magic" in electronics. Hopefully, someone's earlier and excellent example of a balanced differential pair (without a formal ground) will now also make sense because the receiving system does, in fact, have TWO points and can measure the difference between them.

Bottom line: You really do need to provide a common reference. Ground is almost always that common reference except in special cases like differential pairs. For your Arduinos, it's best to connect their grounds - with the caveat that you need to watch out for ground loops and the like, which is an entirely separate topic.

Hope this helps.

« Last Edit: October 21, 2018, 05:59:37 pm by IDEngineer »
 

Offline Buriedcode

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Can't believe this thread is still going...

To be fair, some are just coming late to the first post and adding their two cents - like myself.  I don't think its being dragged on so much as, everyone adding their own explanations.
 

Offline Nusa

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Can't believe this thread is still going...

To be fair, some are just coming late to the first post and adding their two cents - like myself.  I don't think its being dragged on so much as, everyone adding their own explanations.

Including replies by the original poster, so it's not like the thread has been highjacked. Yet.
 
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Offline alsetalokin4017

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Well. Without overcomplicating things, the answer in general to Beamin's original question is "yes".

But... for those who do like complexity and who are arguing about whether one-wire "circuits" are possible... the answer would have to be "it all depends on what's going on in the wire."

The easiest person to fool is yourself. -- Richard Feynman
 


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