hello guys, hope you all doing well please i would like to know if there is any resistors IC, which means instead of using 10 normal 1/2 resistors for example, i use that IC instead to save space or else ?

Thanks in advance for the help :)

Pelle

ok thanks great so the name is resistor net ! and which IC is used the most or it doesn't matter old new ? and does it get hot a lot or it's the same as normal 1/4, 1/2 w ... resistors ?

The data sheets give the maximum power to be dissipated by each element and the total power to be dissipated by the entire package.

groundfungus: The data sheets give the maximum power to be dissipated by each element and the total power to be dissipated by the entire package.

ok thanks but how to know now which exact IC to work with :S ? i need an IC alternative for 10 1K resistors 1/2W

First figure out what the power rating of the resistors really need to be in your circuit. For example if the supply voltage is 5 volts a 10k resistor will not dissipate more than 0.0025 watts (V^2/R = 25/10000) and any array of resistors with a power rating/resistor greater than that can be used. The 1/2 watt you mentioned may have been chosen more because of availability than necessity.

RoyK: First figure out what the power rating of the resistors really need to be in your circuit. For example if the supply voltage is 5 volts a 10k resistor will not dissipate more than 0.0025 watts (V^2/R = 25/10000) and any array of resistors with a power rating/resistor greater than that can be used. The 1/2 watt you mentioned may have been chosen more because of availability than necessity.

aha ok so the application that i wanted for is : 13 LEDs connected to shift register which connected to arduino, the current is 5V from a voltage regulator which is connected to a 12V battery

so you mean i should use a 1/4W resistor ?

You need to calculate how much current is flowing through each resistor and what voltage is being dropped across each resistor. Then the power that resistor is dissipating is the voltage times the current. Add them all up and compare that to the package's rating.

For one pack holding 13 resistors at 10K you can use this:- http://uk.farnell.com/bourns/4114r-2-103lf/resistor-network-10k/dp/9355774

Grumpy_Mike: You need to calculate how much current is flowing through each resistor and what voltage is being dropped across each resistor. Then the power that resistor is dissipating is the voltage times the current. Add them all up and compare that to the package's rating.

but i need something to start from an IC to start from, isn't the same thing IC and normal 13 resistors for each LED ?

but i need something to start from an IC to start from, isn't the same thing IC and normal 13 resistors for each LED ?

Sorry but I have no idea what those words mean, care to try again?

Grumpy_Mike:

but i need something to start from an IC to start from, isn't the same thing IC and normal 13 resistors for each LED ?

Sorry but I have no idea what those words mean, care to try again?

what i meant is an IC name to be based on for simple applications so i go search for it get it and learn more about how to use these ICs

These are not ICs these are resistor arrays. They do not have a name as such. They have a data sheet, there is not much you need to know.

I am still not sure what you don’t understand about calculating the dissipation for each resistor in the array and adding them up.

For one pack holding 13 resistors at 10K you can use this:-
http://uk.farnell.com/bourns/4114r-2-103lf/resistor-network-10k/dp/9355774

The data sheet is on the same page.

Grumpy_Mike: These are not ICs these are resistor arrays. They do not have a name as such. They have a data sheet, there is not much you need to know.

I am still not sure what you don't understand about calculating the dissipation for each resistor in the array and adding them up.

For one pack holding 13 resistors at 10K you can use this:- http://uk.farnell.com/bourns/4114r-2-103lf/resistor-network-10k/dp/9355774

The data sheet is on the same page.

ah ok so the IC doesn't have a code or something, so the current comes from 5V regulator which is up to 1A, then the current goes to arduino, on the other hand, the LEDs and 7segments are connected to Shift registers with every register can take up to 8, and these registers connected to arduino, between the LEDs, 7segments and the registers there is resisters 1K each (i need to see how much i should get a higher resistor for a lower brightness) , so i want to eliminate the resistors with a resistor array to save space, if that array let's say can hold up to 13 resistors (13 pins) means i need about many, so let's say i connected the LEDs to this array, and the array is connected to the shift register (how is it connected to shift register from the positive side of the LED ?!) i have searched a little about it, i see it's works fine for pull up and down, but how do we use it with the LEDs connected to shift register, how can i integrate the array in this case between them ?!

btw why do we use a resistor array other than for saving space ? or that's the only reason ?

ah ok so the IC doesn't have a code or something

Not a generic name like an IC would. However the device I linked to has a manufacturers part number of 4114R-2-103LF, there will be other almost identical offerings from other manufacturers with totally different numbers.

so the current comes from 5V regulator which is up to 1A

The maximum current capacity of the power supply is irrelevant.

so i want to eliminate the resistors with a resistor array to save space, if

eliminate is the wrong word, you mean replace.

how is it connected to shift register from the positive side of the LED

The common resistor connection in the array is connected to +5V then the positive (anode ) of the LED is connected to the resistor in the array and the negative (cathode ) of the LED is connected to the shift register output. This is known as current sinking and a logic zero will turn on the LED. That is the correct way to do it.

The other beginners way to do it is to connect the common resistor connection in the array is connected to ground, then the negative (cathode ) of the LED is connected to he resistor in the array . Then the the positive (anode ) of the LED is connected to the shift register output. This is known as current sourcing and a logic one will turn on the LED.

why do we use a resistor array other than for saving space ?

Because when you make the device you only have to make one placement of a component instead of 13 placements. This saves time and so saves money.

Grumpy_Mike:

so i want to eliminate the resistors with a resistor array to save space, if

eliminate is the wrong word, you mean replace.

lol you’re right

Grumpy_Mike:

how is it connected to shift register from the positive side of the LED

The common resistor connection in the array is connected to +5V then the positive (anode ) of the LED is connected to the resistor in the array and the negative (cathode ) of the LED is connected to the shift register output. This is known as current sinking and a logic zero will turn on the LED. That is the correct way to do it.

aha so same as the button pull down and up

Grumpy_Mike:
The other beginners way to do it is to connect the common resistor connection in the array is connected to ground, then the negative (cathode ) of the LED is connected to he resistor in the array . Then the the positive (anode ) of the LED is connected to the shift register output. This is known as current sourcing and a logic one will turn on the LED.

but isn’t that the resistor should be connected to the + of the LED ?

but isn't that the resistor should be connected to the + of the LED ?

The resistor works with the led for either lead. It is a series circuit.

retrolefty:

but isn't that the resistor should be connected to the + of the LED ?

The resistor works with the led for either lead. It's is a series circuit.

i meant for the LED, the current comes from the + (in fact it's the opposite for electrons) but usually the resistor is connected to the anode + of the LED from where the current comes to resist the number of current so how it would resist and limit the amount of current if it's connected from - of the LED ?

firashelou:

retrolefty:

but isn’t that the resistor should be connected to the + of the LED ?

The resistor works with the led for either lead. It’s is a series circuit.

i meant for the LED, the current comes from the + (in fact it’s the opposite for electrons) but usually the resistor is connected to the anode + of the LED from where the current comes to resist the number of current so how it would resist and limit the amount of current if it’s connected from - of the LED ?

Because in a simple series circuit the current is the same at any point in the circuit. It’s one of Kirchhoff’s circuit laws.

but usually the resistor is connected to the anode + of the LED from where the current comes to resist the number of current

No that is wrong and shows a total lack of understanding about electricity. It makes no difference which direction current flows. Beginners think that a resistor "uses up" electricity first before it gets to the LED. This is just rubbish.

Current flows through all parts of a series circuit.

Hi,

if that array let's say can hold up to 13 resistors (13 pins)

Sorry, it depends on what you want, resistors have 2 connections, an array of 13 resistors will have 26 pins , or if one end of the resistors is connected together then it will be 14pins.

We can see what you are trying to do, and the resistor networks are designed with this sort of job in mind.

I suggest you go to E14, farnell, RS or digikey online and use their search facility to look at their stock. What you will probably find arrays of 4 or 6 resistor with all their connections on pins, to 4 to 13 resistor arrays with a common connection pin.

Tom....... :)