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why we use 0 ohm resistor

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karthigowri

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why we use 0 ohm resistor in electronic circuits
 

for may reason : to have a jumper if there is no enought space for the layout, and to have a resistor which can be modifed/changed later,
 
0 ohm resistors are used specially in SMD circuits, where the pin geometric of the ICs used, prohibit connections when the board are mounted. If you have unused pins you can pull them out and terminate them with a 0 ohm resistor, to make it possible to have access to this pin in the future, or during prototyping

Some places it is convenient to use 0 ohm resistors as programming shunts, or for places where a single resistor may be split into serial or parallel combinations in the future, to get the correct value/power rating.

Or you can use the 0 ohm resistor as a link if the board need it.
 
why we use 0 ohm resistor in electronic circuits
Click to expand...
Hi karthigowri

In fact they can be used in two purpose ( better to say i use them for these two purposes perhaps the other guys have better application for them .

1st: as a jumper but it's dealing with a good shape .
2nd : they can tolerate only 0.25W ! 0.25W jumpers so i use them as protection ( i use 0.25W zero ohm resistors )

That's all . but if there is any other application for them i'll be glad if anybody tell me too .

Best Wishes
Goldsmith
 
Another useful achievement of SMD 0R shunts is the possibility to get rid need to assemble by hand, but once such device fits into a standard package pattern, allow employ the same pick and place machine used for the other devices on the board.




+++
 

Gorgon said:
Since V=0 and R=0, 0/0 = 1 current :razz: Bad joke!
Click to expand...

Yes that was my point, I don't see how the 0 ohm resistor can work as a protection device , for all I know the PCB track may melt before the resistor is damaged.
Assuming a few mOhms the resistor may take 10 Amps or so with no problem.
 
I used 0 ohm resistors many times so separate ground signals or power signals when routing PCB. For example if I have a circuit with analog ground and digital ground, on schematic use a 0 ohm resistor to connect both grounds at unique point.
 
And how much current does this translate to?
Click to expand...

Don't see the issue like this , consider , you want implement over voltage protection you can use that and then use a varistor in order to short the line so it will be destructed .
As another method , it can be burned sooner than the PCB or the other elements .

As a simple test , try to do what i suggested , you'll see the result is amazing .


Best Wishes
Goldsmith
 

goldsmith said:
Don't see the issue like this..
Click to expand...

So this is a practical method you use that is not based on any theoretical calculations.
In other words you expect the resistor to blow off before any tracks are damaged without taking into consideration how many amps the 0.25W translates to and how many amps the pcb track can handle.
 

goldsmith said:
Don't see the issue like this , consider , you want implement over voltage protection you can use that and then use a varistor in order to short the line so it will be destructed .
As another method , it can be burned sooner than the PCB or the other elements .

As a simple test , try to do what i suggested , you'll see the result is amazing .


Best Wishes
Goldsmith
Click to expand...

Dear Goldmith,
You are confused, power don't blows resistors. Energy does. When you use 0 ohm resistors is the energy that blows your resistor. Checking resistors datasheet, you will see that 2A current for 5 seconds on a 0805 chip resistor will blown it.
If you try to use power, there is not sollution, because 0 ohm resistors theoretically don't disipates power. P = V x I. Supposing R = 0 ohm, V = I x R, put any value to I, V will be allways 0, then P will be always 0. Of course, R isn't really 0. Thats why manufacturer specifies blown energy, and it's expressed in energy units (square amperes per second or I^2t=A^2xSec)

Check attached datasheet, there is maximun current specified for jumpers resistors or 0 ohm resistors.
 

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So this is a practical method you use that is not based on any theoretical calculations.
Click to expand...
Hi Alex
Of course i don't use theoretical calculations for that . most of the times i design it by some rules of thumb and some trial and error .
In other words you expect the resistor to blow off before any tracks are damaged
Click to expand...
Indeed

without taking into consideration how many amps the 0.25W translates to and how many amps the pcb track can handle.
Click to expand...
Certainly i take care of this issue .
For such a purpose i take care that my tracks be able to handle the inrush current .

Regards
Goldsmith

- - - Updated - - -

You are confused, power don't blows resistors. Energy does. When you use 0 ohm resistors is the energy that blows your resistor. Checking resistors datasheet, you will see that 2A current for 5 seconds on a 0805 chip resistor will blown it.
If you try to use power, there is not sollution, because 0 ohm resistors theoretically don't disipates power. P = V x I. Supposing R = 0 ohm, V = I x R, put any value to I, V will be allways 0, then P will be always 0. Of course, R isn't really 0. Thats why manufacturer specifies blown energy, and it's expressed in energy units (square amperes per second or I^2t=A^2xSec)

Check attached datasheet, there is maximun current specified for jumpers resistors or 0 ohm resistors.
Click to expand...


Hi penrico
Thanks for your description but i'm familiar with things you said but as i told before i use some rules of thumb and trial and error to design it ( to make it easy but if i want to do it completely theoretically of course i know the rules and i know how to do it but it waste more time ( first we must know the theorem behind everything and 2nd we must try to use simplest methods . )


Best Luck
Goldsmith
 
You are confused, power don't blows resistors. Energy does. When you use 0 ohm resistors is the energy that blows your resistor. Checking resistors datasheet, you will see that 2A current for 5 seconds on a 0805 chip resistor will blown it.
If you try to use power, there is not sollution, because 0 ohm resistors theoretically don't disipates power. P = V x I. Supposing R = 0 ohm, V = I x R, put any value to I, V will be allways 0, then P will be always 0. Of course, R isn't really 0. Thats why manufacturer specifies blown energy, and it's expressed in energy units (square amperes per second or I^2t=A^2xSec)
Click to expand...
Resistors have both power and pulse energy maximum rating, in so far resistor can be "blown" by exceeding the power limit.

For 0-ohm resistors you need to know the residual resistance (a low milliohm number) to refer to power or energy limits or specify the rating in A and A²s. I don't however see an I²t rating in the linked datasheet

By the way, I²t isn't an energy quantity, I²Rt is.
 
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