PDA

View Full Version : current monitor?



stbrnrd
05-24-2006, 10:05 AM
could the stamp monitor 4-20mA, without the need of (asides form a resistor or capacitor) external components? if so how would the comand be structured ... i need the current loop monitored, if it is within range (4-20mA) then a pin would do (either h or l), but if it isn't...another pin to do another action.
make sense?

Chris Savage
05-24-2006, 10:29 AM
Without external components, no.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Chris Savage
Parallax Tech Support
csavage@parallax.com (mailto:csavage@parallax.com)

stbrnrd
05-24-2006, 11:16 AM
OOOkay...i guess i couldn't have my cake and eat it...
if you so incline would you be able to help on this ordeal? as far as needed components is concerned ....i'll take the fifth.
but i would greatly appreciate the help http://forums.parallax.com/images/smilies/smile.gif

Chris Savage
05-24-2006, 09:55 PM
Measuring current will require at minimum a few external components...If you want any accuracy you will need an ADC as well.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Chris Savage
Parallax Tech Support
csavage@parallax.com (mailto:csavage@parallax.com)

T Chap
05-28-2006, 03:33 PM
if all you need to do is set a threshold and output a high at "X" current, here is the schematic for it. You can order the AD628AR off digikey, the comparator is an LM 339. This can be set extremely sensitive. There is a tool to adjust ot you own needs below the link:

note that I used a 12v supply because I am putting 24 volts into it. You can use 5v if needed, see the tool.

http://melrosemagazineonline.com/currentmonitor.jpg

http://www.analog.com/Analog_Root/static/techSupport/designTools/inter
activeTools/ad628/ad628.html?diffamp=AD628%20+%2F-%2015V

T Chap
05-28-2006, 03:41 PM
oops, not shown is a " shunt" resistor, just a regular 1 ohm, 1 watt resistor that is inserted in series between the load(led, motor, etc) and the positive voltage feeding it.

IN+(pin1) is on the side feeding into the resistor, IN-(pin8) is the side of the resistor going into your load. The shunt resistor has a voltage drop when there is a load present, the op amp converts the voltage drop(difference) into a positive going voltage directly proportional to the voltage drop(set by gain of course, so it could be larger or smaller).

Any differential op amp will work OK, but this chip are designed specifically for the purpose. google current sensing IC for other ideas, but this works incredibly well.

stbrnrd
06-23-2006, 06:53 AM
Thank you for the help http://forums.parallax.com/images/smilies/smile.gif)

giving it a shot in a couple of days when the "samples" get here .