next topic »

[on:]

hi,
on my current project im trying to monitor two motors using transducers what im trying to achieve is this, if the motor fails i loose a 0v to my ic. alarm sounds.

data sheet of tranducer.
http://docs-europe.electrocomponents.com/webdocs/06c4/0900766b806c43de.pdf


the way im trying to so it is this, set the motor to its lowest speed setting then use a trimming pot to adjust when the transistor turns on at the lowest speed setting, then if the motor fails and goes below the pre-set level i loose my 0v to my ic.



just wondering if this is going to be a very accurate way to achieve this or is there another more accurate way i could do this, (id really struggle to program this obviously this would be the best way)

thank you

[Reply #1 on:]

Since you didn't see fit to add your sketch to the post and you didn't clearly describe what the LT????? thing is... It looks like it should work. The 'sensor' has an output at some unstated condition... and you are using a part of that output to trigger the FET... OK as far as it goes however the V_gth of a fet is generally about 2 volts. If you need sensing at a volt or less (> .8V) you should use an NPN type transistor, a 2N3904 should work very well and you are missing one part, Fet or Bipolar transistor THERE must be a pull-UP on the drian or collector lead (the one that goes to the '328 input) unless you have set the pin as an input and set it true (high) without that it cannot work and without your sketch there is no way to tell. IMO

Doc

[Reply #2 on:]

hi doc
the LT in my diagram is the transducer im using LTSR-25, its output is 2.5v to 4.5v.

im asuming  2.5v at min current 0.001a and 4.5 at max current 25amps. thats the range of the transducer.

yes sorry i should have said , the input is in my sketch has internal pull up resistor, when a low is present an i get an output to an alarm.

thank you

[Reply #3 on:]

Sorry I read your post before I had my morning coffee... The device you chose is a Bi-lateral current sensor so its output voltage will be 1/2 of the supply voltage and swing positive for current flow shown by the arrow on the sensing device. The Arrow points to the load... If the current flow was the other way (from the load to the source) then the voltage will drop... That's the EASY part. You chose a 25A sensor and on the data sheet there is a formula for determining the voltage represented by a specific current... READ the Data sheet first and work through the calculations first. It is very likely that you will need either an Op-Amp or better a comparator as the voltage change is going to be small, on the order of 10mV/A and likely the sensor isn't going to be very workable for small current changes < 1A.. There are other devices that are probably more suitable. All those devices are hall effect type devices and you should be able to find one better suited to what you need. Match the sensor to the MAX expected load current. You WILL however have to make the output ground referenced or sense the increase from mid point that represents the load and of course the sense pin Must be analog, or use a comparator [better] because then you can use a 'digital' input as the comparator output will swing from Vsource to ground) when your load changes. IMO

Doc

[Reply #4 on:]

hi doc
many thanks on your input really appreciate it, for my purpose im really not looking for very accurate current readings, all im interested in is if the current falls below the pre-set level then i loose my low input then alarm,   
 
im going to knock up a board in the next couple of days puts some loads on it and have a play.

could you please expain a bit more on what you mean by "output ground reference" i dont quite understand that

thank you

[Reply #5 on:]

Yes, I can. If you had read the data sheet you would have realized that the sensor is Bi-Lateral, that is that the sensor will sense current in either direction... to or from the source and that the sensor is radiometric, that is that it will center itself to 1/2 of the sensor power source - Not the load voltage but the current that is to be measured as the sensor is a hall effect device and therefore isolated from it's load.... The sensor will swing about 1/2 of the sensor supply voltage in response to the applied load. This is done so the sensor can sense both drain and charge to the load power source. It will go 'above' the 'center' for loads applied and 'below' the 1/2 point for sources of 'current' applied. What you need to do is either write your sketch to sense or respond to one 'direction' (polarity) of change in voltage or apply the sensor output to a level shifter, typically made from an op-amp and some other parts to get a "Ground Referenced Output". The real issue here is that I don't think from  what you have written so far that you understand very much of what I have said so far. I feel that I know what it is that you need to do, however explaining it to you is difficult because you don't have the understanding of general electronics that I do and it is more work than I really want to do. Not because I am lazy... far from it I looked at the information for your sensor and had the answer immediately... the issue is that it too difficult to explain in words and goes somewhat beyond my understanding of the scope of this forum. I have the tools to draw the diagrams and simulate it for that matter. However doing so would mean several more hours of my time and I have spent far too long at these replies as it stands now. MY take of this forum is that it is a place of learning not a place to go to in order to get your work done for you. If you carefully follow what I have said so far, I think you have enough clues except one and that is you might have to 'amplify' the swing to get it to the point where the Arduino can differentiate or accurately sense your "Trouble Condition" as the Arduino can either sense a voltage in reference to it's supply voltage or to an internal 1.1V reference Both are 10 bit measurements so at the 5V reference your "granularity is 4.88 mv / step or measurable change  or 1.1/1024 if you use the internal reference. Again YOU are the engineer here not me... it is YOU WHO MUST LEARN, not me. Further if I make a mistake I have to "Fix" it... and I am NOT being compensated for my time or my responsibility in becoming what is in effect "The Project Engineer". I am happy to give advice but not to engineer the whole project. I Do Hope that what I have written so far has been helpful and if you need further help then you will have to create accurate drawings of what you need and what you have done to create this project so far.   IMO

Doc

[Reply #6 on:]

thanks for sharing your knowlage doc, really appreciate it as i said i wouldnt be able to solve my problem using code so my aim was to solve it using components im going to try to use a comparator circuit to do this.
thank you