Battery discharger - need some help.

[Andrea87]

I'm building a constant current load to discharge my 18650's and measure some capacity, based on this schematic, but without good results. I've replaced the MAX480 with any single supply opamp I had laying there, tried TL082, LM358N, NE5532, OPA2228 without getting any regulation. Since I've used dual opamps, I've tried both to leave the second OP connections free or setting it to a voltage follower pointed to 0V.

The mosfet is allways being driven at full +V voltage by the opamp, saturating it and getting all the available current into it, I have been measuring 3 amps off a single AA cell. Can anyone tell me what I've done wrong? traces on the pcb have made using eagle following the schematic both of the first and the second driver, and then double checked before building the circuit, so the problem is not in the circuit board. Some photos and schematics about the second driver built yesterday:

First, the second schematic redrawn in eagle to produce directly the pcb.
http://i44.tinypic.com/2m6sw86.png

second, the pcb artwork drawn following the schematic posted up there.
http://i41.tinypic.com/2wd9gfb.png

and then some construction pics. The driver has been build on a double sided board, any track is perfectly insulated, no shorts have been found anywhere. Top layer is connected to gnd, all needing pins have got a large area of copper scraped off to avoid involontary groundings.



Can anyone please help me to figure out why this circuit is saturating my mosfet and won't get in any way to regulate? I've tried anything. pulling up down the output with two 100k resistors connected to v+ and gnd, setting the second opamp as a voltage follower getting 0V imput to avoid oscillations, but nothing got my circuit into proper regulation. It would be a pretty good circuit, costs like $10 and can deliver a stable current load to measure any capacity... I just don't mind why it's not working :thinking:

 

[PeAK]

I'm building a constant current load to discharge my 18650's and measure some capacity, based on this schematic, but without good results.
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The mosfet is allways being driven at full +V voltage by the opamp, saturating it and getting all the available current into it, I have been measuring 3 amps off a single AA cell. Can anyone tell me what I've done wrong?
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Can anyone please help me to figure out why this circuit is saturating my mosfet and won't get in any way to regulate?
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Andrea,
Part of the problem is that the schematic is wrong. The circuit works for a power supply where there is a hidden (i.e. implied) connection from the return leg of the supply to "circuit ground" in the schematic.

This is not the case for a battery where the negative terminal is truly "floating". To correct this, just add a ground connection (ie. a wire) to the circuit ground without any other changes to your circuit. I've diagrammed it in an updated schematic for you. Note: You do not have to remove the ground connection from the leg of "Rc" that I diagrammed...it works just as well with the connection...just add the new ground connection from the leg of the supply to ground.

P.S. Trying it figure out what it is doing with your
present connection is a bit more difficult, I fill you in when I get it sorted. What is the value of V+ ?

PeAK

 

[Andrea87]

yes, the return leg is connected to the circuit ground. in the second pic there's a thin wire going from the negative pin to the third screw of the mammuth, going to ground. as V+ I am using one or two 4-AA NimH packs, so about 5 or 10V.
I'm getting the right return signal to the inverting pin of the opamp. About 0.3V at 3A, which sounds good for 0.1 ohm sense resistor. Also reference voltage is good, in this case setted to 0.1V to get a simple 1A fixed load. I have tried many operationals there, none of them worked. just need to understand why

 

[Mr Happy]

Andrea,
Part of the problem is that the schematic is wrong. The circuit works for a power supply where there is a hidden (i.e. implied) connection from the return leg of the supply to "circuit ground" in the schematic.

This is not the case for a battery where the negative terminal is truly "floating". To correct this, just add a ground connection (ie. a wire) to the circuit ground without any other changes to your circuit. I've diagrammed it in an updated schematic for you. Note: You do not have to remove the ground connection from the leg of "Rc" that I diagrammed...it works just as well with the connection...just add the new ground connection from the leg of the supply to ground.

I do not follow the necessity of this change to the schematic?

My analysis is this: In the original schematic the bottom of RC is connected to B1 negative. Since essentially no current will be flowing through RA, RB and RC in the configuration as shown, the voltage at the top of RA will be the same as at the bottom of RC and the circuit alteration will make no difference.

 

[Mr Happy]

Andrea:

What do the voltages at pins 2 and 3 of the op amp do when you adjust R6? That should give a clue as to why the op amp output is doing what it is doing.

Is the MOSFET undamaged? As you know they are very sensitive to static. Might it have been damaged in handling?

 

[PeAK]

yes, the return leg is connected to the circuit ground. in the second pic there's a thin wire going from the negative pin to the third screw of the mammuth, going to ground. as V+ I am using one or two 4-AA NimH packs, so about 5 or 10V.

The circuit looks fine. Are the 4 large resistors setting the current all equal to 0.1 ohms ?
It would be better to set the reference to 1V and the external resistors to 1 ohm to give a 1 amp discharge. A 0.1V drift on a 0.1V reference can mean either twice the current or no current while the same drift on 1V would result in only a 10% change.

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I'm getting the right return signal to the inverting pin of the opamp. About 0.3V at 3A, which sounds good for 0.1 ohm sense resistor.

The value at the inverting terminal of the opamp should be equal to the reference value of 0.1V on the non-inverting node.
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Also reference voltage is good, in this case setted to 0.1V to get a simple 1A fixed load. I have tried many operationals there, none of them worked. just need to understand why

The value of 0.1V is too low for the op-amp to work properly. It must be at least 2.5V above ground and/or greater for a TL082.

The Max480 is fine at 0.1V but the pin out is different for the op-amp output pin (Pin 6 instead of Pin 1) and the power supply pin (Pin 7 instead of Pin 8). The existing board would need to make some cut to some traces and add some jumpers to accomodate this opamp.

EDIT: The red lines indicate traces to cut. The green lines indicate joins to make using jumpers:

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In the original schematic the bottom of RC is connected to B1 negative
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As Andrea said a thin wire was physically added (dotted red line) to connect "Supply under Test" (aka 18650 supply battery) negative to the bottom of resistor RC (gnd)...In the original schematic (in the article), this connection does not exist making for 6 ground connnection points (vs 5). See Figure:

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PeAK

 

[Mr Happy]

As Andrea said a thin wire was physically added to connect B1 negative to the bottom of resistor RC (gnd)...In the original schematic (in the article), this no connection does not exist.

I don't understand this comment. In the original schematic the foot of RC is shown with a ground symbol, is it not? This means it is connected to the other four locations in the circuit with a ground symbol, including battery negative.

 

[PeAK]

I don't understand this comment. In the original schematic the foot of RC is shown with a ground symbol, is it not? This means it is connected to the other four locations in the circuit with a ground symbol, including battery negative.

There are two batteries. One is B1 and the other is the 18650 termed "supply under test" in the diagram. Andrea and I are referring to negative terminal of the 18650 not having a ground connection. See previous post where an updated schematic was added with a dotted red line. Hope this clarifies things.

cheers,
PeAK

 

[Mr Happy]

There are two batteries. One is B1 and the other is the 18650 termed "supply under test" in the diagram. Andrea and I are referring to negative terminal of the 18650 not having a ground connection. See previous post where an updated schematic was added with a dotted red line. Hope this clarifies things.

No, I think you misunderstand. The supply under test is not meant to have a common ground connection with the dummy load, it is meant to be completely isolated apart from the two connections to the constant current circuit. This is explained in the circuit description:

Battery operation [of the circuit] provides isolation while eliminating grounding problems.

In addition, it makes no difference to the operation of the circuit whether you move that ground connection or not since, as I said above, the negative terminal of the supply under test and the negative terminal of the battery B1 are already at the same potential, being connected as they are through the resistor chain RA ... RC which is carrying no current. There is no mistake there.

 

[PeAK]

No, I think you misunderstand. The supply under test is not meant to have a common ground connection with the dummy load, it is meant to be completely isolated apart from the two connections to the constant current circuit. This is explained in the circuit description:
In addition, it makes no difference to the operation of the circuit whether you move that ground connection or not since, as I said above, the negative terminal of the supply under test and the negative terminal of the battery B1 are already at the same potential, being connected as they are through the resistor chain RA ... RC which is carrying no current. There is no mistake there.

If Ra...Rc carry no current, then they can be removed to simplify the analysis. The circuit you end up with has an infinite number of solutions and no feedback around the op-amp. To prove this just assume any level of current and the source of MOSFET can assume any solution. I'll give another proof tomorrow but the circuit is very standard...until then, be Mr Happy. Cheers.

 

[Andrea87]

@ Peak: I am using a TL082 or NE5532, so the schematic is ok.

@MR happy: using 0.1R resistor, pin 2 stays steady at 0.3V, pin 3 floats between 0 and 0.5V adjusting the 500K pot.

Battery negative is common with circuit negative, allowing feedback.

 

[Mr Happy]

@MR happy: using 0.1R resistor, pin 2 stays steady at 0.3V, pin 3 floats between 0 and 0.5V adjusting the 500K pot.

In spite of his strange obtuseness, I believe PeAK gave you the correct answer. The MAX480 in the original circuit has some special characteristics including low voltage, single rail operation, and the ability for inputs and output to operate right down to the negative supply rail. For this circuit to operate successfully you will need an op amp with similar characteristics. I do not believe the TL082 can substitute since it is a dual rail amplifier that needs a supply voltage of at least 12 V (+/- 6 V) and does not operate right down to the negative rail.

To make this circuit work as designed you will need the MAX480 or an equivalent substitute.

Edit: note that another characteristic of the MAX480 is micro-current supply, and the calculated drain of the whole circuit is given as only 18 µA. Consequently they expected it could run for years on a single battery.

 

[Mr Happy]

If Ra...Rc carry no current, then they can be removed to simplify the analysis. The circuit you end up with has an infinite number of solutions and no feedback around the op-amp. To prove this just assume any level of current and the source of MOSFET can assume any solution. I'll give another proof tomorrow but the circuit is very standard...until then, be Mr Happy. Cheers.

Yes, remove those resistors and replace them with a piece of wire. The ground connection will then be in the same place that you put it. When I said "no" current I clearly meant negligible current as I first explained in post #4. Do I have to spell everything out in detail? This is all explained in the circuit description, if you would take the time to read it:

Bias current for the op amp (3 nA maximum) flows through the series combination of RA, RB, RC, and R10. Multiplying this current by the sum of the four series resistors (100 Ω) yields a 300-nV error. This error voltage is small compared to the voltage presented by R6 at the op amp's noninverting terminal. Therefore, the resulting output-current error is insignificant.

I am really quite sure the circuit will operate as Maxim originally published it and your modification is not needed.

 

[Andrea87]

My distributor doesn't have max480 available, but I do have some similar ops there... They're OPA177's, would you think these would work? They look pretty similar to maxim's max480, and would fit well with peak's circuit mod.

Edit: I've contacted my local distributor, he can get both max4236/4237 and max4238/4239. They look good, in about ten days I should be able to get some cheap. Which one should I choose?

 

[PeAK]

Yes, remove those resistors and replace them with a piece of wire
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We finally agree that we need to add a piece of wire from the "supply under test" negative terminal to circuit ground :twothumbs This is quite different from what you "meant" earlier. To be even clearer, your earlier stance/statement would imply that the circuit would not care if the values of Ra...Rc doubled or tripled and in the limit approach the circuit that was incorrectly drawn in the EDN article.

 

[PeAK]

My distributor doesn't have max480 available, but I do have some similar ops there... They're OPA177's, would you think these would work? They look pretty similar to maxim's max480, and would fit well with peak's circuit mod.

Edit: I've contacted my local distributor, he can get both max4236/4237 and max4238/4239. They look good, in about ten days I should be able to get some cheap. Which one should I choose?

There are two issues. One is finding an op-amp that is pin compatible with your PCB/TL082 and the other is finding an op-amp that is electrically compatable with the max480. Some of the newer op-amps may not be available in the DIP package that you are using and may require attaching wire jumpers to a surface mount package. On the electrical front look for a description along the lines of

"The input common-mode range extends below the negative supply range, and the output swings rail-to-rail "
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Direct replacements for the max480 are offered by Analog Devices in the form of the OP193 and OP90 and these will work provided you hack your PCB with the cuts and joins mentioned earlier. Electrically, of your suggestions, the max4236 would work but I'm not sure about the pinout or the packaging available.

P.S. The OP90 is in current production and offered in a DIP

PeAK

 

[Andrea87]

Pinout aren't a problem, smd chips either. In my current pcb, pins 5, 6and 7 are floating, and can be connected to anything, originally I had left them floating, then configured the second opam as a voltage follower. I can produce pretty good pcbs at nearly no price, using toner transfer paper method. I can even build up a ±5V powered system, I have a ton of AA rechargeables around and using two packs won't be a problem. I will have a couple of max4236 coming there next week. op90 and op193 aren't available at my reseller.

I will try something with the OP177 later, using a dual power supply, I should have another 2 or 3 pcbs ready to build.

 

[MatajumotorS]

You can look at this schematic: http://kaspars.id.lv/projects/electronicload/load.pdf
from here:http://www.geocities.com/hagtronics/load.html

It is 100% working, i have built one, some info at my homepage under "other projects"

Ofcourse You can build Bigger or smaller, depending on needed power rating

 

[Mr Happy]

We finally agree that we need to add a piece of wire from the "supply under test" negative terminal to circuit ground :twothumbs This is quite different from what you "meant" earlier. To be even clearer, your earlier stance/statement would imply that the circuit would not care if the values of Ra...Rc doubled or tripled and in the limit approach the circuit that was incorrectly drawn in the EDN article.

Let me just say one more time that the original circuit will work as published and contains no mistake. The circuit was published by Maxim and the engineers at Maxim have a slight possibility of knowing how to design circuits. There is in addition a reader comments section below the article, and nobody has questioned the correctness of the circuit or asked why it doesn't work. The obvious conclusion: it works as designed.

 

[Andrea87]

I've rebuilt a new circuit using the OP177 precision opamp, a 0.5ohm sense resistor, IT WORKS WELL! I'll post some pics soon