How to provide "bypassing" or "decoupling" capacitors in the remote sensor? - Page 1
Forum section: electronics / general

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 12, 2011, 06:07:12 pm
content: Let's say there is a remote board with LEM LA 55-P (with "current" output) current transducer. The board draws power (+/-15V) from a local board as shown in figure below. I wish to install decoupling/by-passing capacitors on the remote board. How should I connect the capacitors? Across +15V and -15V?Cheers.

author: Unknown
author_rank: Unknown
timestamp: September 12, 2011, 06:30:21 pm
content: If the data sheet didn't give any info, I would connect it across the power pins, as you would do with an op-amp. You don't want to connect a capacitor between a power rail and the output unless explicitly suggested by the manufacturer, since this will load the transducer.

author: IanB
author_rank: Super Contributor
timestamp: September 12, 2011, 07:24:28 pm
content: I might be barking up the wrong tree here, but I'm puzzled by your diagram. In the "Long Wires" I only see three wires. Where is the ground wire to provide the 0 V reference for the +/- 15 V supply and to provide the return path for the current loop output? Is it just an accidental omission from the drawing?

author: Unknown
author_rank: Unknown
timestamp: September 12, 2011, 07:31:26 pm
content: Why would it need a ground connection? Current can return via V+ or V-, depending on polarity (sinking or sourcing). I agree that labeling the supplies as symmetrical is a bit misleading, but people do it all the time with opamps, which don't have any notion of ground either.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 12, 2011, 07:48:34 pm
content: Quote from: IanB on September 12, 2011, 07:24:28 pmIt's not an accidental omission... The sensor does not require a "0V"... It's output is in the form of "current"...

author: onemilimeter
author_rank: Frequent Contributor
timestamp: September 12, 2011, 07:51:21 pm
content: Quote from: alm on September 12, 2011, 07:31:26 pmThe sensor can sense primary current in directional. The datasheet recommends that the sensor needs +/-15V supplies.

author: IanB
author_rank: Super Contributor
timestamp: September 12, 2011, 07:51:56 pm
content: Quote from: alm on September 12, 2011, 07:31:26 pmOK, I think I get it. I'm not particularly familiar with instrumentation and signalling and that's why I thought I might be barking up the wrong tree. But on thinking about it I can see how it works.(As an aside, I have never really got my head round dual rail vs single rail supplies on op amps either. That one continues to cause me a lot of head scratching.)

author: IanB
author_rank: Super Contributor
timestamp: September 12, 2011, 07:54:02 pm
content: But I still don't think the sensor sees +/- 15 V supplies. From the sensor's perspective it sees a 30 V potential difference that might just as well be labelled 0 V and +30 V.OK, it is really +15 V and -15 V relative to the measurement line which is tied to 0 V via Rm. Never mind.

author: Unknown
author_rank: Unknown
timestamp: September 12, 2011, 09:12:27 pm
content: Quote from: IanB on September 12, 2011, 07:51:56 pmThat's because people started coming up with the nonsense about symmetrical power supplies and virtual grounds. An opamp just has a positive and a negative supply, and the inputs and outputs have to be somewhere between those rails (with some extra margin). It can just sink or source currents from those rails to drive the output to any voltage in between. Ground is irrelevant as far as the opamp is concerned, as long as your signal stays within the rails (so it makes sense to define ground to be somewhere within this range). Same with this transducer.