can someone help me with my problem?

Hi everyone, i would like to ask you guys 1 more question about a arduino project that i’ve encountered some problems with…

May i know what rating resistors(How many Ohms) should i use for the voltage dividers for the Resistors? ( so that the 24 volts won’t burn my arduino board( my arduino board will be sensing HIGH OR LOW SIGNALS )?

http://www4.snapfish.com.sg/snapfishsg/slideshow/AlbumID=2777091019/PictureID=103601264019/a=651875019_651875019/otsc=SHR/otsi=SPIClink/

as you could see…
INPUT:
the voltage from the input is 24.5V +/-
the current it supplies is 0.000004A (4uA)

should the resistors be just

R1: 4 ohms
R2: 21 ohms

so output will be 4v at arduino… with 1 v +/- for free play…

power rating for both resistors : 1/4 watts
?

or should it be more? like 400 ohms and 21000 ohms?

if it should be more… why?

what power rating resistors should i use?

sorry… but my theory is not very good… trying to learn as much as possible here… =)
:-[

What is the 24V input coming from?

The ratio of resistors for the voltage divider is correct. The current flow through the resistor is defined by Ohm's law: V = I * R. With a voltage drop of 4 volts, and a resistance of 4 ohms, I = 1A.

Power is amperage * voltage, so a 1A current flow at 4 volts requires the resistor to dissipate 4 watts. That's a really healthy resistor you need.

A 4000 Ohm resistor would have 0.001A and would dissipate 0.004 watts.

I assume you are thinking of powering and arduino through a voltage divider. Don't even think about it, it is not the way to do it. The results are unbelievably poor and you waste a tone of power. Use a linear voltage regulator or switching regulator module.

What I hear is that you want to sense the existence of a 24.5 volt 4 microamp signal. Is this to be an on-off signal that you want to input on a digital input pin?

4 microamps is not a lot to work with. At 24.5 volts, the current across a 6 megaohm resistor is 4 microamps. You could build a voltage divider with a 1.2 megaohm and 4.7 megaohm resistors to yield around 5V in the centre. Since you only have 4 microamps, I doubt this would make a useful input to the arduino.

For this small a signal, I suspect you would need an amplifier to deal with it. If it were me, I think I would use the signal to turn on a JFET, or feed it into a comparator with JFET inputs.

PaulS

What is the 24V input coming from?

The ratio of resistors for the voltage divider is correct. The current flow through the resistor is defined by Ohm's law: V = I * R. With a voltage drop of 4 volts, and a resistance of 4 ohms, I = 1A.

Power is amperage * voltage, so a 1A current flow at 4 volts requires the resistor to dissipate 4 watts. That's a really healthy resistor you need.

A 4000 Ohm resistor would have 0.001A and would dissipate 0.004 watts.

the 24.65V is coming from a sensor.. DC signal.. hmm.. so the voltage divider could work with arduino?
meaning that it could sense the 4V as high?

Grumpy_Mike

I assume you are thinking of powering and arduino through a voltage divider. Don't even think about it, it is not the way to do it. The results are unbelievably poor and you waste a tone of power. Use a linear voltage regulator or switching regulator module.

hi there, nope.. im not gonna power my Arduino through this voltage divider.. i want the input pins to sense it as a "HIGH" or "LOW"

gardner
What I hear is that you want to sense the existence of a 24.5 volt 4 microamp signal. Is this to be an on-off signal that you want to input on a digital input pin?

4 microamps is not a lot to work with. At 24.5 volts, the current across a 6 megaohm resistor is 4 microamps. You could build a voltage divider with a 1.2 megaohm and 4.7 megaohm resistors to yield around 5V in the centre. Since you only have 4 microamps, I doubt this would make a useful input to the arduino.

For this small a signal, I suspect you would need an amplifier to deal with it. If it were me, I think I would use the signal to turn on a JFET, or feed it into a comparator with JFET inputs.

Hi there, yes.. i do.. you got me correctly.. but i think i measured the existing current wrongly..
I'm currently doing a project for my friend.. his SENSOR has a output voltage of 24.65 DC voltage.. BUT i wasn't sure about the current.. so i used my multimeter to check.. it displayed a reading of "4microamps"..
i think i got it wrongly as i remember that without a LOAD(resistor) there won't be current flowing through it.. ( from what i learnt from school, i've probably given the knowledge back to the teacher after my final exams. :cry: )

i've read the datasheet of the sensor.. it says OUTPUT: 30V 1A
Does that mean that i could draw a max of 30V and 1A out of the sensor signals?

if so.. how do i design my circuit so it does not burn out my resistors, damage the sensors and my arduino input pins?

Thank you guys for replying to this thread.
Really appreciate it man.. =)

Does that mean that i could draw a max of 30V and 1A out of the sensor signals?

Yes, it does. However, the ACTUAL current draw will be a function of the voltage divider resistors you use. Large values = small current.

The sensor can not PUSH 1A to your Arduino. The Arduino must PULL the current, and a voltage divider made of anything but two equal length pieces of copper will not pull much current.

BUT i wasn't sure about the current.. so i used my multimeter to check.. it displayed a reading of "4microamps"... i think i got it wrongly as i remember that without a LOAD(resistor) there won't be current flowing through it

If you just put a multi meter set to current from ground to 30V you have in effect put a short circuit across it. This, if you are lucky, should have blown the fuse in your meter and you will need to replace it. If it really did show 4uA then your sensor is not working.

One amp is a lot of current for a sensor to give, this current has to come from somewhere so what are you powering it with? That power supply has to be capable of supplying an amp at 30V.

Use a R2 as 1K in this calculator:-

For best results put a 5v1 zenner diode across R2 to protect against any over voltage.

If you just put a multi meter set to current from ground to 30V you have in effect put a short circuit across it. This, if you are lucky, should have blown the fuse in your meter and you will need to replace it. If it really did show 4uA then your sensor is not working.

One amp is a lot of current for a sensor to give, this current has to come from somewhere so what are you powering it with? That power supply has to be capable of supplying an amp at 30V.

Use a R2 as 1K in this calculator:-
Electronics 2000 | Potential Divider Calculator
For best results put a 5v1 zenner diode across R2 to protect against any over voltage.

Hi yes.. I think I've solved my problem.. The sensor wasn't spoilt.. I think my multimeter is.. =)

thanks anyway for your help! =)