Constant Current Load Circuit Issues

hp1729

Joined Nov 23, 2015
2,304
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.
View attachment 121516
I would have put OVLD to pin 5 of U2 instead of pin 7. Remove the 100 ohm resistor and jumper Q2 drain to pin 5.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
I would have put OVLD to pin 5 of U2 instead of pin 7. Remove the 100 ohm resistor and jumper Q2 drain to pin 5.
That seems to be the consensus solution and I agree. My only problem now is to figure out if Q2 is defective or are the Source & drain reversed.
 

LesJones

Joined Jan 8, 2017
4,511
Could Q2 be faulty ? (Source to drain short.) I think you need a resistor in series with U2.2 output (Pin 7) so that the overload signal from R6 can pull the input to Q1 low. Does the voltage on the drain of Q2 change when the gate voltage changes from a few mV to 3 volts ?

Les.
 

Kjeldgaard

Joined Apr 7, 2016
476
That seems to be the consensus solution and I agree. My only problem now is to figure out if Q2 is defective or are the Source & drain reversed.
If Q2 is reverse mounted, then the body diode will be in forward direction and you will then be measurering up to approximately 0.7 Volts the "Drain" terminal.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
If Q2 is reverse mounted, then the body diode will be in forward direction and you will then be measurering up to approximately 0.7 Volts the "Drain" terminal.
Yes, that is exactly what I was measuring, and that was the verification that the Drain/Source connections were reversed. Here is what happened. When I assigned the footprint for the BS170 on the PCB layout, I got the drain and source connections back-assward. My bad for interpreting the data sheet incorrectly.

So, I flipped Q2 around so that the drain and source connections are correct. I removed R6 and replaced it with a jumper to pin 5.

I powered up the circuit without the fan connected. I hooked up my source and cranked the current up to 2 amps. The heat sink temp rose to 63 deg C and the OL circuit kicked in and immediately dropped the current to 20 mA. When the heat sink cooled off, the load current went back up to 2 amps.

I then connected the fan and it comes on around 53 deg C. With the fan running the heat sink temperature stabilizes around 57-58 deg C at 2 amp load.

To make a long story short - everything is working the way it should. I'm a very happy camper!
 

Kjeldgaard

Joined Apr 7, 2016
476
Congratulations.

I have just a detail more:
My attention fell on Q2 and Q3 Gate circuits, where the drive voltage is only just over 3 Volt, barely above the Threshold voltage.
When BS170 has a maximum limit for the gate voltage of 20 Volts, I suggest removing R14 and R16.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Congratulations.

I have just a detail more:
My attention fell on Q2 and Q3 Gate circuits, where the drive voltage is only just over 3 Volt, barely above the Threshold voltage.
When BS170 has a maximum limit for the gate voltage of 20 Volts, I suggest removing R14 and R16.
The data sheet gives the following for BS170 Gate Threshold Voltage:
Min. - 0.8 volts
Typ. - 2.1 volts
Max. - 3.0 volts
When I measured the gate voltage today it was 2.77 volts. That's the reason for the voltage divider.
I don't understand the 20 volt threshold. What am I missing?
 

AnalogKid

Joined Aug 1, 2013
12,135
Th max threshold Vgs is the minimum voltage to start enhancement of the channel. In other words, 3 V isn't the maximum gate voltage allowed for operation. It is the minimum gate voltage needed to guarantee the start conduction under all variations of devices, production lots, temperature, etc. 20 V is the typical true maximum Vgs. Since the output of the 324 will pull the gate below threshold you don't need R14. To get greater enhancement, you should eliminate R13.

ak
 

Thread Starter

bertz

Joined Nov 11, 2013
327
Th max threshold Vgs is the minimum voltage to start enhancement of the channel. In other words, 3 V isn't the maximum gate voltage allowed for operation. It is the minimum gate voltage needed to guarantee the start conduction under all variations of devices, production lots, temperature, etc. 20 V is the typical true maximum Vgs. Since the output of the 324 will pull the gate below threshold you don't need R14. To get greater enhancement, you should eliminate R13.

ak
I'm a retiree and just getting back into electronics. This is my first project using MOSFETs so bear with me as I plod through this.

Going back to the data sheet I see upon closer inspection the Vgs values that I quoted earlier were for a specific set of conditions, namely Vds = Vgs and drain current is 1 mA.

So I went to the curves, specifically for the ON-Region Characteristics. I see a set of curves for Vgs values from 3.0 volts to 10.0 volts. However it looks like 4.0 volts is the highest you would want to go in order to keep drain current below 500 mA (the max rating for the device).

Now getting back to my application, if I remove the shunt resistor from the divider network Vgs will jump to 9.96 volts for Q2 and 10.99 volts for Q3. I don't see anything in the data sheet that indicates that the device can operate with 20 volts on the gate. It's clear from studying the data sheet that 2.77 volts is barely enough to open the channel. I get it. But wouldn't operating with 4.0 volts on the gate sort of act like a current limit? If so I can increase the value of the shunt resistors instead of removing them entirely, especially since the circuit seems to be working just fine as presently configured.

I'm probably missing something very obvious, but I'm learning a lot as we go along.
 

Kjeldgaard

Joined Apr 7, 2016
476
First, I would say that it's generally a really nicely done project.

The maximum Vgs are often found in the Absolute Maximum Ratings section.

The curves of Id/Vgs, Rds/Vgs, etc. are usually nominal values.

For current limitation is Vgs not a particularly good idea, often die transistor of heat stroke due to power dissipation, and there are also quite a large spread of Id/Vgs data.

A not entirely logical level transistor as the BS170, can be used on a 5 Volt CMOS logic output. But if there is possibility of something like 8 to 12 Volt Vgs, you are nicely between fully conductive and maximum Vgs.
 

Thread Starter

bertz

Joined Nov 11, 2013
327
First, I would say that it's generally a really nicely done project.

The maximum Vgs are often found in the Absolute Maximum Ratings section.

The curves of Id/Vgs, Rds/Vgs, etc. are usually nominal values.

For current limitation is Vgs not a particularly good idea, often die transistor of heat stroke due to power dissipation, and there are also quite a large spread of Id/Vgs data.

A not entirely logical level transistor as the BS170, can be used on a 5 Volt CMOS logic output. But if there is possibility of something like 8 to 12 Volt Vgs, you are nicely between fully conductive and maximum Vgs.
Excellent discussion on MOSFET characteristics inspired me to further understand how to read MOSFET data sheets. My first mistake was using the threshold Vgs values I referenced earlier to design the voltage divider circuitry. I now understand that these values are not intended for circuit designers. Vgs is a design parameter that defines the point where the device is at the threshold of turning on. It is an indication of the beginning, nowhere near the end. It also tells us that the gate voltage must be held below this value while in the off state to minimize leakage. By setting the voltage divider to deliver a gate voltage less than 3 volts, I am indeed fortunate that the MOSFETs turned on as they should. Sometimes it is better to be lucky than good.

To better understand how to turn on MOSFETs I looked at some of the curves in detail. The first curve I found that refers to the MOSFET turning on with increasing gate voltage is the Transfer Characteristic Curve. This plots drain-source voltage against drain-source current for a number of given gate voltages beginning with 3 volts and ending at 10 volts. What I gleaned from this is that the MOSFET should not operate with gate voltages less than 3 volts in order to fully turn on. I can see where this curve may have relevance if your operating in the linear region.

The curve that has data with the MOSFET fully on is the Output Characteristic Curve. The operating condition for the curve published in the data sheet is Id = 500 mA and Vds = 25 volts. Nevertheless, I see that for Vgs below 4.25 volts we are still in the linear region of the MOSFET. In order to fully turn on, we must be above this value. Again, for a Vgs of only 2.77 volts I am fortunate that my overload circuitry is working as well as it does.
I plan on replacing the 1.8k shunt resistors (R14 & R16) with 4.7k resistors to bring the Vgs up around 5 volts.

Thanks to Kjelgaard and AnalogKid for setting me on to the path of enlightenment.
 

Attachments

AnalogKid

Joined Aug 1, 2013
12,135
So I went to the curves, specifically for the ON-Region Characteristics. I see a set of curves for Vgs values from 3.0 volts to 10.0 volts. However it looks like 4.0 volts is the highest you would want to go in order to keep drain current below 500 mA (the max rating for the device).
That is true only if there is no impedance in the source or drain that limits the drain current. But there almost always is. Running the transistor "open loop" and depending on it's gain to set the output current rarely works because the gain varies significantly with temperature and among individual devices. For Q2, the U2.2 output stage limits the maximum current available to way under 0.5 A. You want Q2 to act as a saturated switch, so hit it with 5-10 V and let the circuit elements do their jobs. Q1 is inside the opamp feedback look, so the opamp corrects for device variations in the transistor.

ak
 

AnalogKid

Joined Aug 1, 2013
12,135
To better understand how to turn on MOSFETs I looked at some of the curves in detail. The first curve I found that refers to the MOSFET turning on with increasing gate voltage is the Transfer Characteristic Curve. This plots drain-source voltage against drain-source current for a number of given gate voltages beginning with 3 volts and ending at 10 volts. What I gleaned from this is that the MOSFET should not operate with gate voltages less than 3 volts in order to fully turn on. I can see where this curve may have relevance if your operating in the linear region.

The curve that has data with the MOSFET fully on is the Output Characteristic Curve. The operating condition for the curve published in the data sheet is Id = 500 mA and Vds = 25 volts. Nevertheless, I see that for Vgs below 4.25 volts we are still in the linear region of the MOSFET. In order to fully turn on, we must be above this value. Again, for a Vgs of only 2.77 volts I am fortunate that my overload circuitry is working as well as it does.
Absolutely correct., yunser.

ak
 

Plamen

Joined Mar 29, 2015
111
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:
View attachment 121461
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 3.1
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.

Petkan:
The driver headroom (max current) is defined by power supply voltage (V) minus voltage drop in current sense resistor divided by load resistance (neglecting MOSFET Rdson) i.e. I max = (V-I.Rcs)/Rload.. For instance with Rcs=1 Ohm (2V drop at 2A) and 12V power supply, a load resistance of 5 Ohm will exhaust the headroom. 2A will be enforced at lower Rload or Higher Vpower.

Note that this is an analog (continuously operating) current sink that will dissipate tremendous heat. You may consider switching mode (say Buck regulator with external error amplifier, comparing command current with actual (current feedback). If you need more details (send me a note to plamen_petko5@hotmail.com for LTSpice example. There are ready made Constant Voltage (CV), constant current (CC) switching mode power regulators
http://www.ebay.ca/sch/i.html?_from...+module.TRS0&_nkw=CV+CC+power+module&_sacat=0

They feature adjustable by two presets (voltage and current). With light load they regulate voltage as preset by pot. Once the current limit (set by preset) is exceeded - they enter current control (limiting) mode. Either way - switching mode operation and little heat dissipation.

If your goal is not to sink a particular current, originating from remove source (referred to your GND) this will be the simplest (you can remove the preset and replace with Pot.
However if you aim at say discharging a rechargeable battery at specific current i.e. you really need to sink (externally originating), not source current (load powered by the battery not by your circuit)- it will not be suitable.
 
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