Bidirectional Current Monitor

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

Hi guys

I'm designing a bidirectional current monitor circuit for my SLA battery fuel gauge

I found 2 designs for this in two application notes of National & Liner

I modified these designs as follows:
1- RO1 & RO2 are pots to scale the output.
2- D1 & D2 to feed the A/B part to zero when B/A measures high current.

The circuit works fine under simulator

In real life..
1- Both outs gives output when no load, It may be the amp offset (It's not a problem).
2- When a load attached the corresponding out decreases while it should increase as it do in simulator !!

I don't know why it doesn't run as in simulator

Can anybody help ?

Thanks in advance.

Attachment(s): 

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

Don't reinvent the wheel - it will come ugly and useless. Use this IC (or similar) instead.

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

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

No, you need a LEM DHAB http://www.lem.com/hq/en/content...

Imagecraft compiler user

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

MBedder thnx for replay ,,
If I have choice I'll chose this chip as a second choice after a dedicated MCU or one of MAXIM's charge counter.
But I'm bounded to a tight budget, limited time, and the parts where readily available in the local market & these specialized chips are not available her.

Is there a way to know where's the bug within this circuit ?!

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

bobgardner:
I think this device is more suited to current ranges of 100 to 1000 Amps
While my current range is 0.25 to 12.0 Amps.

I'm tring to implemint the methods mentioned here

http://www.google.com/url?sa=t&source=web&cd=3&ved=0CCYQFjAC&url=http%3A%2F%2Fwww.linear.com%2Fad%2F16-high_speed.pdf&ei=BN3ITbfYG4bMgQeY5YX8BQ&usg=AFQjCNGsTcGYe-JSfPJTfmUkSLT41i_i9g

but using the common opamps like LM358N.

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

Building a 80+ dB CMR differential amplifier with LM358 and general purpose resistors is a nonsense - you will just waste your time and money. Get the proper parts from Farnell/DigiKey/Mouser etc. and don't f.ck the brains anymore.

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

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

Sorry about your brain ..

I'm not that good in analog stuff, so I'm asking for help.

I think the idea of forums is to discus about what we do including the problems..

But It seems that as time pass some guys forget the origin of the idea.

Any how , many thanks for help, I'll search for help somewhere else .. again .. many thanks for all.

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

PortiaCyberLogic wrote:
Sorry about your brain ..
Not mine but yours.

Quote:
I'm not that good in analog stuff, so I'm asking for help.
You've got it.

Quote:
But It seems that as time pass some guys forget the origin of the idea.
No we don't.

Quote:
Any how , many thanks for help, I'll search for help somewhere else .. again .. many thanks for all.

Fare thee well,
and if forever,
still forever,
fare thee well

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

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

Its just a series R that wont get too hot with your max amps running thru (like .1 ohms?) it with a diff amp to step the small voltage drop across the R to the range 0-5V so you can read it with the a/d. I think an LM358 used as a diff amp with about a gain of 10 would work. Try it on a white plastic prototype board.

Imagecraft compiler user

Last Edited: Sun. May 15, 2011 - 03:54 PM
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

bobgardner:

As shown in the schematic I did exactly what you say..
Now it works very stable, & gives proper readings,
But only amp B that is work up with current in a direction & down in the reversed direction, the amp A don't change & gives a fixed stable value equals to no current value.
The shunt res. is 0.1 ohms & it don't heat that much.

I'm trying to know why amp A doesn't work..

Any help will be appreciable..
Many thanks for your answer.

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

The Linear tech part isn't quite the same as the LM358 so you were hoping for magic for it to work. The thing to remember with op-amps, is that many of them don't like their inputs near the power rails and the outputs don't swing from rail to rail. Have a look at the LM358 datasheet and see what it says. Some op-amps will specify rail to rail inputs and/or outputs. MBedder was helping you but the delivery was somewhat terse. We think he has Asperger's as he is quite smart but lacks a little in social skills.

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

Thanks Kartman..

This is the real problem,
I redesigned the circuit, implementing another concept using the same opamp, and it works fine now.

To not f**k more brains ..
I may post the new design if someone rely interested.

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

PortiaCyberLogic wrote:
I may post the new design if someone rely interested.
Sure ... post it.

Ross McKenzie ValuSoft Melbourne Australia

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

You only need one opamp for a bidirectional current sensor. If the opamp is running from 5V, bias it at 2.5V, and rig the gain so that 12A across your sense R is 2.5V, -12A goes down 2.5V to 0 (use a rail to rail opamp). That's just Algebra, not Calculus.

Imagecraft compiler user

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

bobgardner:

This is the concept I implemented without the rail to rail opamp, cause I power the opamp from the BAT 12V & the output is within 5 volt.

Also I'm biasing at 1.0V cause the charging current is 2.1A for max. while the load current can reach up to 12.5A
so I whited the bias toward the discharge to maximize the current resolution.

Here is the schematic..

Attachment(s): 

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

I would suggest that once you have those two pots set and you know what values you want for them, replace them with normal resistors. Pots tend to degrade over time - dirt and humidity and corrosion and stuff will affect them, which can have a big impact on your circuit. If this is just a short-term plaything it doesn't matter, but if you hope to use this thing for the next 5 years, you might want to consider it.

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

@PortiaCyberLogic
What's the voltage of the battery?

______
Rob

Scattered showers my arse -- Noah, 2348BC.
Rob Gray, old fart, nature photographer, embedded hardware/software designer, and serial motorhome builder, www.robgray.com

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

frankvh:
These 2 pots are multi turn cermet trimmer they are more reliable than pots.
Also how can I find a fixed resistors that matches the exact pot value after adjustment ?!

Graynomad:
The battery is 12V SLA I think the voltage range is (13.4V - 10.0V) & up to 13.8 under charging.

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

PortiaCyberLogic wrote:
frankvh:
These 2 pots are multi turn cermet trimmer they are more reliable than pots.
Also how can I find a fixed resistors that matches the exact pot value after adjustment ?!

Pretty easily. Measure the final pot value with your multimeter, then use one or more normal resistors in series or parallel to replicate its value (for each half of the pot).

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

RV2 for example:
After adjustment reads 9.786 K from exact 10.14 K in room temperature.
To do what you say:
1 * 9.1 K
1 * 680
2 * 3
And all must be exact in values or to be picked from a large set & reconstruct according fractions
Or to use 0.05% tolerance res. in this case it will be more practical to convert the whole set to 0.05%
and no need to trimming
Also I think more res. in signal pass = more noise
and I think small values has different thermal response of the large values ..

I think there is more simpler solution for this issue
is to use resistor chips 1% or 0.5% and compensate the output in software.

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

As you were already told before, everything can be easily done with a dedicated current sensing differential amplifier IC which contains the precisely matched resistors on chip. Even if you manage to get the 0.05% discrete resistors, your monitor will be temperature dependent, not even speaking of the pots.

Warning: Grumpy Old Chuff. Reading this post may severely damage your mental health.

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

You can get any ratio within 2% with a voltage divider with 1% Rs. I remember seeing a program that would pick the nearset standard R values when you input a voltage, but I can't remember where I saw it.

Imagecraft compiler user

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

Hi all ..

I rely appreciate all of your efforts to help.

I develop this device for a co. working in cine supplies & gadgets.
They already make a battery packs for off-studio shooting equipments such as sun-guns, cine cameras, audio systems, etc.
They just need to upgrade there battery packs with a facility to tell how much time left before battery die, & to give information about the battery quality.

And they do that upgrade while they are the top seller, i.e. they don't rely need to upgrade !.

Also they will add this upgrade with no extra charge to there clients !.

So there specifications & priorities are:
1- Upgrade cost: The lowest possible.
2- No critical components.
3- Stuck to the local available components.
4- Size: The smaller.
5- Acceptable error margin: Up to 10%.

Someone expert told me before:
When you design something you aim for the best.
But professional designer just make it work & reliable according specifications.

This device will not be used for medical, military, scientific, or surveillance purpose & the technicians who deal with these battery packs already have there myths about batteries in general.

What I mean is:
I'm not that new, I already know about the dedicated components that specialized for this kind of work & I already worked with such components before, but I gotta stuck to the specifications ,,
And this is why my 1st question was "Why this simple circuit doesn't work ?"
cause I just wanna make it work or something a like, but not too far solution.

Now I finished the beta prototype & they examined it.
They found it OK for there needs, & now we try to minimize the power consumption as it draws about 10 mA

Anyhow, I hope to not push on any one ask for something that may look stupid,,
Sometimes these stupid things may be the optimal solutions, or just facts that you gotta deal with.

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

You also have to factor inthe cost to design it. How much time have you spent trying to solve this problem? Would using a dedicated chip albrit more expensive, be cheaper in the long run?

As for your specific problem, you need to remember the output of the lm358 may not swing as close to the power rails as you might like. Have you measured the voltages at the opamp inputs and outputs? What are they? What do you expect the opamp to do in these conditions and does this line up with the datasheet specs?

Sounds like what you want is a 'gas gauge' or coulomb counter. There are specific chips to do this. To reach your goal, you may have to re-evaluate your constraints otherwise you might spend excessive time on something that would never fly or fly poorly at best.

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

Kartman:
You are right about time factor,,
Solving the problem took about 2 days, waiting for the specific parts to be delivered will take about 5 days + shipping + the clients specified for no special parts.

BTW before I take a decision of building a discrete solution, I & there engineer discussed in details all of the possibilities & solutions, & the conclusion was to go with the discrete solution.

I changed the opamp to be LM324A to reduce parts number & cost.

About measurements:

Supply voltage (Battery input): 11.83 V

Ideal (no load):
OpAmp In: 9.16 V, OpAmp Out: 1.00 V

Loaded: 10 A
OpAmp In: 8.15 V, OpAmp Out: 4.42 V

I think in this configuration, the opamp inputs will never reach power in, cause the resistor network divides input voltage.
Please refer to the last schematic.

And according the res. values & working params, the output is always up to 5V max.

I think it works.