Current Sense

Go To Last Post
22 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hi all ,
This circuit is wrong apparently (Regarding +12V) ,
How could I sense the current ?
Is +12V to +5V conversion the only way ?

Thanks in advance

Attachment(s): 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Looks like a differential amp with a gain of 3. 10 amps should read 1V across rsense, 3V at output of amp?

Imagecraft compiler user

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

bobgardner wrote:
Looks like a differential amp with a gain of 3. 10 amps should read 1V across rsense, 3V at output of amp?

Hi bobgardner , glad to see you
Yes it's a differential amp with that gain but the problem is +12v(I think so at least)
because the + and - pins of the opamp would see high voltage and this may cause problem , don't you think like that ?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

The op amp inputs must always be within the amp's supply voltage rails; in this case +5V and GND. The circuit shown will try to make the inputs be at +9.5 volts. This won't work.
You could swap the position of R29 and the load to get a workable circuit. Note that you will have to use an op amp that has its inputs and output rated for rail-to-rail operation.

Letting the smoke out since 1978

 

 

 

 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

The LM358 will run at 12V, so using that supply instead of 5V might be an option.

If I recall, the 358 doesn't want to see an input pin at its positive supply voltage, although it is OK with an input pin at GND.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

or something like this

  rsense
-+\/\/\+---
 |     |
 []    []39k
 |     |
-+-   -+-
 |     |
 []    []10k
 |     |
 V     V

This divides both 12V ends of the rsense down to 3V. Now the gain of the diff amp (connected in the middle of the divider) can be 10 or whatever to amplify the IR drop across the rsense up to something sensible. Maybe add caps from the center to gnd for a filter.

Imagecraft compiler user

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

why do you not put the load from +12V to the Rsense resistor and that again to ground?????

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Currently the circuit is non-sense.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Good to see an Australian style diagram for once!!

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Thanks for the replys

meslomp wrote:
why do you not put the load from +12V to the Rsense resistor and that again to ground?????

I need high side sense

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

If you happen to also need to switch your load, you might want to look at the AUIR3316 from IR.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

you might want to look at the ACS712 current senseors. they have the advantage that the input is isolated from the output, so you can put them anywhere in the circuit.
http://www.allegromicro.com/en/Products/Part_Numbers/0712/0712.pdf

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
The LM358 will run at 12V, so using that supply instead of 5V might be an option.

In theory that would work.

The main reason to keep the op-amp at +5 V & Gnd, however, is that the Vout will not exceed +5 V.

The input the the AVR ADC is limited to +5 V, (assuming running the micro on +5V/Gnd).

If one were to run the op-amp on +12 V, then one would have to either scale the output, or clamp the input or the output so as to prevend the output from exceeding 5 V.

Bob's second diagram feeding the op-amp will work.

Note, also, that if one uses a pot for one of the legs, then one can adjust the delta V per input amp, and essentially have a gain control, albeit not in the normal sense.

The pot would replace the lower resistor, or be placed between two resistors, so as once again to insure that the input to the op-amp could never exceed +5 V regardless of the pot setting.

JC

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Chris-Mouse wrote:
you might want to look at the ACS712 current senseors. they have the advantage that the input is isolated from the output, so you can put them anywhere in the circuit.
http://www.allegromicro.com/en/Products/Part_Numbers/0712/0712.pdf

I agree with this, otherwise you are going to need a 10W resistor, presuming you want to measure 10A. They come in 5, 20 and 30A ratings.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You COULD connect the op amp to 12v and then connect the ground side through a zener. Might have to be a little more complicated.

Then the rsense volage should be in the correct area.

You will get a screwy output range that you will have to level shift.

hj

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Thanks for the new replys
I Ordered ZXCT1008 (Thanks to dalpilot)

bobgardner wrote:
or something like this

  rsense
-+\/\/\+---
 |     |
 []    []39k
 |     |
-+-   -+-
 |     |
 []    []10k
 |     |
 V     V

This divides both 12V ends of the rsense down to 3V. Now the gain of the diff amp (connected in the middle of the divider) can be 10 or whatever to amplify the IR drop across the rsense up to something sensible. Maybe add caps from the center to gnd for a filter.

Is This Circuit Correct ?

Attachment(s): 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

What is you avversion about drawing GROUND and VCC the correct way around?

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Other than than the position of vcc and gnd, I the think the idea is: divide the 24V down to less than 5V, then amplify the small differential voltage up so you can use most of the 5V range of the a/d converter. Might be good to try it on a white prototype board before building a circuit board.

Imagecraft compiler user

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Lostisland,

The problem I see with your approach is that you are using a quasi difference amplifier whose performance is almost entirely dependent upon excellent matching of the resistors around the op-amp stages. Yes, you have specified 1 percent resistors but you are aiming for a gain of 100 and I fear that you will, with its present design, be disappointed with the results.

Could I suggest that you add a small preset pot to each of the stages so that with zero current flowing through the current shunts you can trim out the effects of resistor and op-amp mismatches? Hopefully the stages will then behave satisfactorily for other signal inputs.

Cheers,

Ross

Ross McKenzie ValuSoft Melbourne Australia

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Sorry for the position of Gnd & Vcc

valusoft wrote:
Lostisland,
The problem I see with your approach is that you are using a quasi difference amplifier whose performance is almost entirely dependent upon excellent matching of the resistors around the op-amp stages. Yes, you have specified 1 percent resistors but you are aiming for a gain of 100 and I fear that you will, with its present design, be disappointed with the results.

Could I suggest that you add a small preset pot to each of the stages so that with zero current flowing through the current shunts you can trim out the effects of resistor and op-amp mismatches? Hopefully the stages will then behave satisfactorily for other signal inputs.

Cheers,

Ross

What if I use the circuit at the start of this thread with these 2 changes :
1- +5v would be +12v
2- with 2 resistors at the output of the opamp I make one third of the output for the ADC

Which circuit is better ?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

UP Please