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Topic: Ground connection between 2 electronic devices needed? (Read 298 times) previous topic - next topic

Lemonbonbon

Oct 11, 2018, 02:11 pm Last Edit: Oct 11, 2018, 02:15 pm by Lemonbonbon
Hi,

I am not that advanced in electronics, so i am asking this question:

The pins on the arduino (A) do not deliver enough current for a seperate electronic component (B). Therefore I did connect this other device (B) with ground and Vcc to a laboratory power supply. To share data/information between my arduino (A) and the device (B) I use jumper wires to connect the pins, but I have no connection between the arduino (A) ground pin to the ground pin of the other device (B).

Do I have to connect a wire between the arduino ground pin and the device ground pin? By doing this, I would also connect the arduino ground pin to the power supply minus-output. Is this okay? I noticed that there is a voltage difference of 0,8 V between the ground of the device (B)/ minus-pin of power supply and the arduino ground (A).

Thanks

Perehama

F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Lemonbonbon

But is there not already an implicit ground connection, because the ardunio should be grounded through its USB-cable and the PC-powercable, while the PC uses the same ground as the power supply?

Furthermore, is the minus-output of the power supply connected the the ground or floating? The supply has just plus and minus output.

Perehama

But is there not already an implicit ground connection, because the ardunio should be grounded through its USB-cable and the PC-powercable, while the PC uses the same ground as the power supply?

Furthermore, is the minus-output of the power supply connected the the ground or floating? The supply has just plus and minus output.
Pretty sure ground on a USB port is not tied to earth ground on a 3-pin AC power cord. If you are measuring a difference in voltage potential, there is one.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Perehama

Instead of telling us device B, tell us what the device is, with model number etc. Tell us what power supply you are using. A picture can also help explain your setup.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Perehama

Yes, connect the ground pin on all 3.
DO NOT connect DC - on the power supply to Earth Ground on the power supply.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Lemonbonbon

I have attached a little picture to ilustrate the situation. The possible ground wire is in also included.

Perehama

I have attached a little picture to ilustrate the situation. The possible ground wire is in also included.
Yes, make that connection.
In not doing so, you do not have any ground reference that is common between A and B so data signals sent (Voh) may not meet the requirements data signals received (Vih).
There is no danger to making that connection.
Without knowing more about component B, I can't offer additional input, but the ground link is a YES.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Paul__B

Pretty sure ground on a USB port is not tied to earth ground on a 3-pin AC power cord.
And I am pretty sure it is on virtually all desktops as they all have a  3-pin power cord as well as virtually all laptops and USFF machines whose power cord has a 3-pin plug.  Those with a 2-pin plug are obviously the exception.

If you are measuring a difference in voltage potential, there is one.
Possibly, which is why the grounds must be tied together to remove that difference.  If something burns, then one of your power supplies is dangerously faulty and you found out the hard way.

Perehama

And I am pretty sure it is on virtually all desktops as they all have a  3-pin power cord as well as virtually all laptops and USFF machines whose power cord has a 3-pin plug.  Those with a 2-pin plug are obviously the exception.
Possibly, which is why the grounds must be tied together to remove that difference.  If something burns, then one of your power supplies is dangerously faulty and you found out the hard way.
In USB, there are 4 grounds related to this discussion. Earth Ground, Chassis Ground and Supply Ground and Signal Ground. Earth Ground is the longest round pin of a 3-prong power plug that gets connected to a wall outlet that is connected to a rod buried in the actual ground. Chassis Ground is the shielding around the USB wires, the metal connector housing, etc. that connect all metal surfaces to ground static charges. Supply Ground, which is one of 4 pins inside the USB connector, and one of 4 wires inside a USB cable, is the negative pole of the 5 volt rail of the power supply and is commonly decoupled from Earth Ground as can be seen in this schematic of an ATX power supply. The symbol for Earth Ground is a vertical line with 3 horizontal lines beneath it in decreasing size from top to bottom. The symbol for Supply Ground is a vertical line with a single thick horizontal line. Chassis Ground looks like and upside down TV antenna. Signal Ground is and upside down triangle. Supply Grounds must be tied as Signal Ground references Supply Ground.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Paul__B

Well, if you think that the black wires on your ATX power supplies are not connected to the chassis and ground pin of the IEC connector, I suggest you go test a few and be incredulous!  :smiley-roll:

Perehama

Well, if you think that the black wires on your ATX power supplies are not connected to the chassis and ground pin of the IEC connector, I suggest you go test a few and be incredulous!  :smiley-roll:
You are right. A simple continuity test between the USB ground pin and the ground pin would make this discussion moot.
F=C/V=(A*s)/V=J/V^2=(W*s)/V^2=(N*m)/V^2=C^2/J=C^2/(N*m)=(s^2*C^2)/(m^2*Kg)=s/Ω=1/(Ω*Hz)=s^2/H

Southpark

You are right. A simple continuity test between the USB ground pin and the ground pin would make this discussion moot.
Did you do such a measurement?

Paul__B

We will see.

I did before posting, and on the two I tested, the black output wires were grounded.  I shall be most interested to hear if anyone does not find it so.

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