Author Topic: How does this circuit function?  (Read 769 times)

0 Members and 1 Guest are viewing this topic.

Offline jeffjmrTopic starter

  • Regular Contributor
  • *
  • Posts: 144
  • Country: us
How does this circuit function?
« on: December 03, 2022, 01:28:49 am »
Since JU1 is in parallel and a dead short bypassing R55 and R56, how does this circuit function? Any resistor in parallel with 0 ohms is 0 ohms.

But the circuit does work. It calibrates this power supply’s ammeter. I don’t understand how.

Can someone enlighten me?

Jeff
 

Offline BillyO

  • Super Contributor
  • ***
  • Posts: 1665
  • Country: ca
Re: How does this circuit function?
« Reply #1 on: December 03, 2022, 01:34:06 am »
1) is the jumper in place?

2) doe the calibration pot make a difference?
Bill  (Currently a Siglent fanboy)
--------------------------------------------------
 

Offline jwet

  • Frequent Contributor
  • **
  • Posts: 519
  • Country: us
Re: How does this circuit function?
« Reply #2 on: December 03, 2022, 02:56:25 am »
My best guess is that its a two point calibration to get rid of an offset that might be on the signal feeding the meter.  With the current set to zero, you close the jumper and adjust the mechanical zero on the meter for zero- having the jumper closed, gives you a magnified response.  This cancels any offset in the circuit and any meter mechanical offset due to mounting (gravity).  You then open the jumper, go full scale on the current and adjust the pot for full scale.  You then sweep between zero and full scale on the meter even with some gain and offset errors on the input.  The jumper is only used for cal.  The meter resistance is usually around 50 or 100 ohms.  If you meausre across the pot and the 3.6K, you should get something like 4k when the pot is centered.
 
The following users thanked this post: tooki

Offline jeffjmrTopic starter

  • Regular Contributor
  • *
  • Posts: 144
  • Country: us
Re: How does this circuit function?
« Reply #3 on: December 03, 2022, 04:44:42 pm »
‘Doh!  My bad, Billy.

I shouldn’t have asked until I checked. The jumper may indeed not be closed.

Jwet, your response is intriguing but the calibration instructions do not call for changing nor address the jumper.

I need to unstack my power supplies so I can take the cover off of one of these (I have two) and check the status of that jumper. These are BW 1743 supplies. They have digital meters.

The reason I am looking closely at this circuit, is one of these has fairly accurate and linear meters, and the other does not. The instructions call for calibration at mid scale. I am trying to see what I can do to minimize the errors at the extremes of the scale. It reads high at low values and low at high values so I think?? it may be addressable in the circuit as opposed to just errors internally to the digital meters.

Am I on the right track?

Thanks for your inputs.

Jeff

 

Offline jwet

  • Frequent Contributor
  • **
  • Posts: 519
  • Country: us
Re: How does this circuit function?
« Reply #4 on: December 03, 2022, 06:37:53 pm »
If the meters have a zero adjust, you could follow my procedure anyway.  You seem to have a span and zero type problem.

This whole thing could be a left over from a time when they had mechanical meters?
 

Offline jwet

  • Frequent Contributor
  • **
  • Posts: 519
  • Country: us
Re: How does this circuit function?
« Reply #5 on: December 03, 2022, 07:37:23 pm »
Looked it up on line, the current meter has a zero and a span spot.

Check out page 24 -
https://bkpmedia.s3.amazonaws.com/downloads/manuals/en-us/1743_manual.pdf
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf