Constant Current Load Circuit Issues

Thread Starter

bertz

Joined Nov 11, 2013
327
I am in the process of building a constant current load. The specs are 0-2 amps with an input voltage not to exceed 25 volts. Before getting into the discussion, here is the circuit:
Constant Current Load Circuit.jpg
The problem is that I am not getting the full 2 amp load as I increase R3. Here is a table that shows what is happening:
Pin 3 Pin 5 Pin 6 Pin 7
Volts Volts Volts Volts
0.00 0.00 0.05 2.30
0.50 0.20 0.20 2.90
1.00 0.40 0.40 3.13
1.50 0.60 0.53 3.25
2.00 0.80 0.54 3.25
2.50 1.00 0.51 3.23
3.00 1.20 0.52 3.23
3.50 1.40 0.53 3.23
4.00 1.60 0.53 3.23
4.50 1.80 0.53 3.23
5.00 2.00 0.53 3.23
The pins refer to the LM324. What I would expect to happen is that as the non-inverting input (pin 5) increases, the Op-amp would do its thing and adjust the output so that the inverting input (pin 6) matches the non-inverting input. However we seem to hit a brick wall around .53 volts, and consequently the current maxes out at .53 amps.

Here is some additional info:
The fan runs all the time, as soon as we power up the circuit.
The voltage drop across R6 is 2.53 volts which indicates a 25 mA current when there should be NONE!
LOAD is 11.90 volt, 2.2Ah LiPo

Could it be that Q2 and Q3 are installed backwards?
All comments and suggestions welcome.
 
Last edited:

wayneh

Joined Sep 9, 2010
17,496
Your load at J2 is wired wrong. The MOSFET controls the current to ground, and yet you're showing one pole of the load is grounded.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Your load at J2 is wired wrong. The MOSFET controls the current to ground, and yet you're showing one pole of the load is grounded.
Of course one side is grounded. The load is either a battery or a power supply. How could it work otherwise?
Current flows from the load to the MOSFET drain, down through the source, through the current sense resistor and then to ground. The other side of the load is connected to ground to complete the circuit.
 

Kjeldgaard

Joined Apr 7, 2016
476
Some thoughts and observations:
What is the purpose of the fairly large value of R1 and C4?
What is the supply voltage measured directly at U2 pin 4?
OVLD signal from Q2 looks mysterious? - I would pull U2 pin 5 to ground instead.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Some thoughts and observations:
What is the purpose of the fairly large value of R1 and C4?.
I threw R1 into the circuit as a current limiter to protect the IC, completely ignoring the voltage drop.
The purpose of the capacitor is to stabilize the output and prevent oscillation. I don't think 100Nf is that high.
[QUOTE="What is the supply voltage measured directly at U2 pin 4?.[/QUOTE]
The supply voltage at pin 4 is only 6.4 volts!! Obviously this chops off the output at pin 7.
Good catch! Thank you! The 220 ohm resistor must go!
[QUOTE="OVLD signal from Q2 looks mysterious? - I would pull U2 pin 5 to ground instead.[/QUOTE]
When Q2 is turned, the idea is to force the control voltage at pin 7 to ground thus shutting down the current source and protecting the circuit. I see what you mean about pulling pin 5 to ground instead, but why wouldn't pulling pin 7 to ground accomplish the same thing? Just asking.

While were on this discussion, why is R6 showing a voltage drop if Q2 is not turned on?

Finally, I was pondering the circuit diagram vs. the PCB layout and I found that Q3 is indeed reversed. Would that cause the fan to be on all the time?
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Huh? Your label is confusing. Those are current sources, not loads (as long as you're not charging the battery). If there is a battery or PS at J2, isn't R5 your load?
Nothing confusing here. The terminology is pretty standard. Check it out mate.
 

wayneh

Joined Sep 9, 2010
17,496
Nothing confusing here. The terminology is pretty standard. Check it out mate.
Sorry, it's not standard to refer to the current source as the load. Your active circuit turns R5 and the MOSFET into the adjustable load. Your guy refers to the item at J2 as the PSU, which is indeed standard terminology.

But it's moot. For your application, the wiring at J2 is fine. Only the label is off.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Huh? Your label is confusing. Those are current sources, not loads (as long as you're not charging the battery). If there is a battery or PS at J2, isn't R5 your load?
I will concede however that "SOURCE" would be a more appropriate label.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Have you tried disconnecting the overload circuit? Divide and conquer to isolate the problem.
I believe Kjeldgaard put his finger on the primary issue. The 220 ohm resistor was chopping my supply voltage to the opamp rail in half. That in turn was what was limiting the current to around .5 amp.
Tomorrow I will remove the 220 ohm resistor and reverse Q3 to see if the fan will stop running. I believe the OVERLOAD portion of the circuit will require some tweaking as well.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
I believe Kjeldgaard put his finger on the primary issue. The 220 ohm resistor was chopping my supply voltage to the opamp rail in half. That in turn was what was limiting the current to around .5 amp.
Tomorrow I will remove the 220 ohm resistor and reverse Q3 to see if the fan will stop running. I believe the OVERLOAD portion of the circuit will require some tweaking as well.
OK, I removed the 220 ohm resistor and swapped the drain and source on Q3. Here are the voltages on the opamp pins:
Pin3 Pin5 Pin6 Pin 7
0.00 0.00 0.05 2.54
0.50 0.20 0.20 2.90
1.00 0.40 0.40 3.15
1.50 0.60 0.60 3.35
2.00 0.80 0.80 3.56
2.50 1.00 1.00 3.76
3.00 1.20 1.20 4.00
3.50 1.40 1.40 4.17
4.00 1.60 1.60 4.36
4.50 1.80 1.80 4.55
5.00 2.00 2.00 4.74

The bottom line is the circuit is now working as it should with one exception which I'll get to in a bit. I now have full range of 0-2 amps load current across the full range of R3. In addition, the fan only comes on when the heat sink temperature reaches 55 deg C. So far so good.
Next, I disconnected the fan and set R3 for full 2 amps to test the overload circuit. At 66 deg C the OL kicked in as indicated by the LED. The only problem is the load current did not cut back. So I disconnected the Source and allowed the MOSFET to cool down. Then made a couple of voltage measurements across R6. These were the results:
1.79 volts across R6 @ NO LOAD
2.45 volts across R6 @ 0.4 amps
2.65 volts across R6 @ 0.6 amps
It seems to me that Q2 is sinking current from U2.2, pin 7 since the voltage reading across R6 are identical to the voltages on pin 7 less 0.7 V drop across Q2.

Here are my questions:

1. Why is Q2 allowing current to pass through it if it is turned OFF? Are the Drain and Source connections reversed as was the case with the fan MOSFET?

2. If I connect R6 to the non-inverting input (pin 5) of U2.2 as Kjeldgaard suggests, won't I just be seeing the output of pin 1 as divided by R2-R4?

3. Is R6 really necessary? Again, I just threw it in there to limit current in this loop.

Attached is the revised schema.
Constant Current Load Circuit.jpg
 

wayneh

Joined Sep 9, 2010
17,496
1. Why is Q2 allowing current to pass through it if it is turned OFF? Are the Drain and Source connections reversed as was the case with the fan MOSFET?
That's the first thing to check, yes. Also check the gate voltage to make sure it's what you expect.
2. If I connect R6 to the non-inverting input (pin 5) of U2.2 as Kjeldgaard suggests, won't I just be seeing the output of pin 1 as divided by R2-R4?
Not if the OVLD pulls the pin to ~ground via Q2.

3. Is R6 really necessary? Again, I just threw it in there to limit current in this loop.
No. You could eliminate R6 in this arrangement, since there is no need for current limiting.
 

AnalogKid

Joined Aug 1, 2013
10,987
Huh? Your label is confusing. Those are current sources, not loads (as long as you're not charging the battery). If there is a battery or PS at J2, isn't R5 your load?
R5 is the current sense resistor. It is part of the load, but the intent is that the majority of the power is dissipated in the MOSFET. I've done this, but I had R5 at 0.1 ohm (with other circuit changes to adjust the gain).

ak
 

AnalogKid

Joined Aug 1, 2013
10,987
The overload circuit tries to put a 100 ohm load to GND on U2.2, and U2.2 tries to close its loop anyway. If you want the overload circuit to shut down the output, delete R6, put a 1K resistor in series with the Q1 gate and tie the Q2 drain directly to the Q1 gate. This will remove gate drive from Q1 without damaging U2.

ak
 

bertus

Joined Apr 5, 2008
22,270
Hello,

Is there a fault in the J2 connector?
Pin 1 is connected to the mosfet and pin 2 to ground.
Should pin2 be connected to the powersupply?

Bertus
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Hello,

Is there a fault in the J2 connector?
Pin 1 is connected to the mosfet and pin 2 to ground.
Should pin2 be connected to the powersupply?

Bertus
No, this part of the circuit is functioning exactly as it should. See Post #12.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Move "the top of R6" from U2 pin 7 to U2 pin 5, and the Overload Protection circuit will work.
Yes, I see how this would work to bring the current down in an OVERLOAD condition IF, and it's a big IF, Q2 is turning on and off in response to a temperature rise in the heat sink.
As I stated previously, the triggering mechanism appears to be working correctly. The LED comes on when the heat sink temperature hits around 66 deg C and the Q2 gate voltage goes from a few millivolts to around 3 volts. However this has no apparent effect on the Current Source circuit..
Because I see a voltage drop across R6 that is proportional to the output voltage on pin 7, I am inclined to believe that Q2 is conducting all the time!
1.79 volts across R6 @ NO LOAD
2.45 volts across R6 @ 0.4 amps
2.65 volts across R6 @ 0.6 amps
So once again, I ask the question: Why is Q2 conducting when it is not triggered?
 

Thread Starter

bertz

Joined Nov 11, 2013
327
The overload circuit tries to put a 100 ohm load to GND on U2.2, and U2.2 tries to close its loop anyway. If you want the overload circuit to shut down the output, delete R6, put a 1K resistor in series with the Q1 gate and tie the Q2 drain directly to the Q1 gate. This will remove gate drive from Q1 without damaging U2.

ak
I see exactly what you are saying and there is no doubt that this would work, except for two issues:
  1. Q2 appears to be conducting all the time and
  2. The circuit board is already manufactured and does not lend itself readily to modification. Kjeldgaard's solution is doable without butchering the board if I can figure out the issue with Q2.
 
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