I have this system that sends a signal wire from a controller to a multi-position switch. Every state this switch has run the signal into a resistor, the controller know what state was pressed by reading the voltage of its own signal wire.

I want to get rid of the switch(es) and use a digital pot to 'simulate' the same thing.

Problem is: The system is setup so that it has 2ma running in that signal wire. My digital pot can handle only a measely 1ma.

I need some way to cut that amperage down (i cant modify the exisiting system) without changing the voltage.



I figured that one way to do this (i'm not certain this will work) would be to place a small 1/4W resistor inline. Since the original resistors are 1/2W i think this will only let 1ma though there.

Am I even close to being right ?

 

you would need to change the value of the resistor to reduce the power flowing through, but then your digital pot would do this wen it changes value

 

The pot is dumb as hell and will gladly ask for more power then it can handel.

If i put a 1/4W resistor inline on the wire somewhere wont that prevent more then 1ma from ever getting to the pot ?

 

Let's go back to basics here: Ohms Law - V = IR

In your case, I is constant (2mA).

By putting a resistor inline, you are going increase the overall resistance, thereby forcing an increase of the signal voltage. Thus, you may get an incorrect reading of the state. Anyway, since you are dealing with a constant current output, you will ALWAYS have 2mA flowing (provided the output doesn't hit the supply rails)

Also, by having a fixed resistor inline, you are changing the zero offset of your switch-position/voltage curve.

The ideal way to solve your problem, is to re-program your system, so it only outputs 1mA. Failing this, maybe there is a solution using a shunt component, though the value of the shunt must be equal to the value of the pot at all times.

 

I can NOT modify the system.

If I put a 10ohm 1/4W resistor in there, and the system is expecting 5V, isnt is going to act like a 5V - 1amp signal ? (I know it would be SLIGHTLY over 5V)

Also, is a 1/4W resistor going to limit the max current in the line ?

 

Any thing that you put in series will raise the voltage at the source, so that won't work. One solution is to put a constant current sink in parallel with the digital pot. This will require a negative voltage in order to work down to ground.

__________________
see my website: www.geocities.com/russlk

 

right.

If the 1/4W thing didnt work, my plan was to use the pot and make an adjustable voltage sink...


Howver, I have NO IDEA of how to do that

Anyone want to stab at it ?

 

Voltage sink?? Not sure what you mean. Anyway, here's a fairly simple current sink for you. Not 100% accurate, due to opamp currents not flowing through "R", but should be good enough for your purposes.

So, if you have 2mA from your signal source, and set it so you have 1mA flowing into the sink, then arithmetically, you have 1mA flowing in your digital pot :P

Attached Images

I-Sink.jpg (6.7 KB, 525 views)

 

Well.... Thats looks nice although I have NO idea of how thats works


I was looking into this awhile ago and someone suggested use of a NPN transistor and an Op amp along with the pot I have to sink a specific voltage

Ie: If i need the existing system to see 2.5V of the 5V in its signal wire I set my voltage sink doodad to -2.5V

Any idea of that they were talking about OR can you explain what your diagram does... (I'm sure I could build it and it would work, BUT i really like to know WHY things i use work, not just that they do)

 

It's a constant current source - it's quite easy to understand how it works from ohms law.

If you have a constant resistance (R), and you keep the voltage across it constant - by simple application of ohms law, the current has to be constant as well.

In this circuit the opamp is used as a comparator, comparing the voltage across R with a reference voltage - if the voltage across R is less than the reference the comparator switches the transistor ON, causing the voltage to increase. If the voltage across R is greater than the reference it will turn the transistor OFF, causing the voltage to decrease.

This action results in a constant voltage across R, which means it has a constant current through it - by changing the value of R, or by altering the reference voltage, you can adjust the constant current drawn by the circuit.

Obviously this requires that the voltage on the emitter of the transistor is always going to be greater than the reference voltage - once it falls below this, the current will no longer be constant. Obviously a low reference voltage would be a good idea.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Well, I'm certain thats 100% accurate, but however makes little sense to me since its the circuit in the diagram i still dont undstand, not how ohms law works. I do understand some of it but...

If the op amp is just acting as a comparator why do I need it ? I ALWAYS want the current in the line to be 1ma of the 2.2 it usually is.

I have a pic with a comparator in it, can i just use that instead of the op amp ?

I (think) thats a PNP transistor in the diagram, is it possible to just use that if I dont need the compatator ?

All i want is to use this stupid digital pot without burning it up. It says in its datasheet that I can send 1ma max (although i tested it out and it handled 2ma just fine, prolly not for extrended time though)

Thanks for your help everyone

 

Here is another current source that you may be able to understand better:

Attached Images

CURRENT_SINK.jpg (7.3 KB, 470 views)

__________________
see my website: www.geocities.com/russlk

 

Well, I tried that first diagram, and..... it didnt work There is 2.45ma in the signal wire and still 2.45ma on the wiper of my pot. Thats 1.45 too much. Yea, I followed the diagram EXACTLY, used a 741 for the amp, and a PNP transistor.

Thanks Russlk, that does looks great. A few questin though... can you explain a little about how that works and how to adjust the resistors to sink a little more then 1ma (say 1.45 )... also, how am I going to come up with -5V ?? (dumb question i'm sure)

 

About that -5V... I just remembered seeing that my PIC has a -ref pin... Think I can use that ?

 

Un, answered my own question.... it looks like the Vref- is only for the pic's adc to read - voltages correctly. That sucks.

Any super duper easy way to get a -5V ?