Author Topic: Electronic constant current DC load  (Read 94067 times)

steve30

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Re: Electronic constant current DC load
« Reply #45 on: November 01, 2012, 10:17:22 AM »
I finally got round to building one of these yesterday. I'm sure it will help me with my power supply projects.

I used an IRF540 MOSFET and managed to get adjustment from a few milliamps right up to 3.5A when it was connected up to a 5V source.

I got a lot of oscillation though. There were some funky patterns on the oscilloscope :D. To fix this I moved the capacitor to the drain of the FET.

I think I might try some different MOSFETs as well as this is a bit of a learning experience for me as I have never used FETs before.

MJLorton

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Re: Electronic constant current DC load
« Reply #46 on: November 05, 2012, 07:41:17 AM »
I finally got round to building one of these yesterday. I'm sure it will help me with my power supply projects.

I used an IRF540 MOSFET and managed to get adjustment from a few milliamps right up to 3.5A when it was connected up to a 5V source.

I got a lot of oscillation though. There were some funky patterns on the oscilloscope :D. To fix this I moved the capacitor to the drain of the FET.

I think I might try some different MOSFETs as well as this is a bit of a learning experience for me as I have never used FETs before.

Hats off to you Steve and enjoy the tinkering until it does the job for you.

If you have a moment please post some pictures....we'd love to see it.

Cheers,
Martin.
Play, discover, learn and enjoy! (and don't be scared to make mistakes along the way!)

steve30

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Re: Electronic constant current DC load
« Reply #47 on: November 05, 2012, 08:26:21 AM »
I'll get the gear out later and do some photos of the oscilloscope. That would be cool.

I will note that the IRF540 requires higher voltages on the gate to be 'turned on', so I powered the circuit off a 24V PSU rather than just 5V.

Anyway, it worked nicely, so I thing I might try and make a nice hand drawn PCB for it.

btfdev

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Re: Electronic constant current DC load
« Reply #48 on: November 14, 2012, 12:13:25 PM »
Good Day Everyone,

Hi all, BTFdev is a newbie here.

First Martin, great job on providing further analysis and research on this subject. I found this very educational and valuable. I do have a few questions for the experts, please refer to the schematic diagram Circuit2.jpg posted by Martin.

1. With the combination of 5K/50K pots (R4/R2), can you please tell me what is the min and max voltage from R2 (Pin 2) ?

2. After passing the 1st Opamp, what is the min and max voltage at LM324 Pin1?

3. How about the 10K pot (R3)? What is the min and max voltage from R3?

4. After passing the 2nd Opamp, what is the min and max voltage at LM324 Pin7?

Thanks in advance.

Team BTF

jwrelectro

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Re: Electronic constant current DC load
« Reply #49 on: November 14, 2012, 05:01:18 PM »
OK,  I will try and answer your questions.  I am using a different opamp so Martin can jump in if I make any mistakes.  If you don't mind I will add some of my reasoning behind the answers.

1)  The output of the three port regulator LM7806 is +6-volts. so with R4 wiper at its pin 3 of the pot the full 6-volts is applied to R2.  So with R2's wiper (Pin 2 at Pin1 of that pot) the maximum voltage out of pin 2 to the opamp's pin 3 is 6-volts.   With R4's wiper at pin 1 of the pot the maximum voltage out is 6v (50 kohms / (5 kohms + 50 kohms)) = 5.45-volts.  The minimum voltage out of the R2 pot is 0-volts no matter where the wiper on R4 is set.

2)  The 1st opamp is a voltage follower with 100% negative feedback so the gain is unity (1).  Therefore  the voltage at pin 1 of the opamp is the same as the R2 wiper at pin 2 of the pot.

3)  The maximum voltage at R3's wiper is 6-votls and the minimum is 0-volts.  For this to be true for the 6-volts the wiper of R4 must be at pin 3 of that pot.

4)  The voltage at pin 7 is a little more difficult to determine because of VGS(threshold) which can vary from transistor to transistor.  I am using the same transistor as Martin and the VGS(threshold) is around 1.55 volts.  So the output voltage at pin 7 would be from around 0-volts to around 6 volts less the internal drop of the opamp.  You would not want to run the system at this level because what is happening is you are hitting the power supply rail which in this circuit is shown to be 6 volts.   Just as an example let's say the internal drop of the opamp is 200 mV.  So the max output would be 6v - 200 mV = 5.8 volts.  If the VGS(Threshold) is say 1.5 volts then the voltage across R5 would be 5.8v - (1.5v + 1.5 v) = 2.8 v and therefore equates to an ILoad of 2.8 Amps.  Please note that in this circuit arrangement you are not getting the full adjustment range out of R3 if you have the other pots set to obtain maximum control voltage.

Hope this helps... John
Good Day Everyone,

Hi all, BTFdev is a newbie here.

First Martin, great job on providing further analysis and research on this subject. I found this very educational and valuable. I do have a few questions for the experts, please refer to the schematic diagram Circuit2.jpg posted by Martin.

1. With the combination of 5K/50K pots (R4/R2), can you please tell me what is the min and max voltage from R2 (Pin 2) ?

2. After passing the 1st Opamp, what is the min and max voltage at LM324 Pin1?

3. How about the 10K pot (R3)? What is the min and max voltage from R3?

4. After passing the 2nd Opamp, what is the min and max voltage at LM324 Pin7?

Thanks in advance.

Team BTF
« Last Edit: November 14, 2012, 05:46:55 PM by jwrelectro »

steve30

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Re: Electronic constant current DC load
« Reply #50 on: November 17, 2012, 02:55:11 PM »
I finally got round to taking some pictures, and decided to write it all up in a web page.

Here it is:

http://stevecoates.net/cc_load/

MJLorton

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Re: Electronic constant current DC load
« Reply #51 on: November 19, 2012, 02:39:22 AM »
I finally got round to taking some pictures, and decided to write it all up in a web page.

Here it is:

http://stevecoates.net/cc_load/

Brilliant Steve. Thanks for the post.
Play, discover, learn and enjoy! (and don't be scared to make mistakes along the way!)

btfdev

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Re: Electronic constant current DC load
« Reply #52 on: November 23, 2012, 05:45:54 PM »
Thanks for the replies and inputs from our questions.

After even more Googling and Youtubing, we have more understanding of this dummy load concept. We also did build one prototype ourselves and now researching a better logic level MOSFET.

One thing we noticed from our research is that, some dummy loads are powered by separate battery but we also seen some powered entirely by the load. Any pros and cons for these two approaches? So in general for people who interested in building their own dummy load, here are a few things to consider:

- What to use to power your rail?
- What is the Voltage/Source to feed your opamp
- What is the voltage needed to feed your MOSFET to induce its function in ohmic and saturated region?

We saw some designs featuring voltage/current display and cooling fans, you also need to consider how are you going to supply power to these peripherals.

Looking forward for more info on this subject.

Team BTF.

SeanB

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Re: Electronic constant current DC load
« Reply #53 on: November 24, 2012, 12:29:37 AM »
Separate power means you can have a wide input voltage range, from the max of the pass element to where the sense resistor is dropping 90% of the voltage. Single source power limits you to the minimum needed for the controller to the max voltage rating of the control, typically 8-20V then. A separate power supply also allows you to have cooling fans to make the heatsink more efficient as a forced air cooler is both cooler running and a lot smaller.

dr_p

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Re: Electronic constant current DC load
« Reply #54 on: November 24, 2012, 08:03:31 PM »
Here is my take on Dave's CC load:

It has current and voltage displays and the base is doubled in plexi not to accidentally short the PCB.
I added a series 6A diode to the FET to protect against wrong polarity. Power resistor is 0R33, 5W, 10% with the worst thermal stability ever but I don't care. This means that 0.1A shows correctly on the panel meter but at 1.2A the resistor heats up and increases in value, dropping more voltage, thus the panel meter shows 1.6A and thins go downhill from there. Every part used is scrapped from somewhere else, so I used what I had.

Max current through it is short of 4A. Max drain current of the FET is 7A. Op-amp is power with 9V and is enough for 6.7A (displayed on the panel meter, tested on a lead-acid). The TO92 5V regulator is for the panel meters, but it was not necessary, they have their own LM1117-3.3.

http://www.ebay.com/itm/Mini-red-LED-Digital-Voltage-Panel-Meter-DC-4-5V-30V-gauge-/261031174906?pt=LH_DefaultDomain_0&hash=item3cc6aba2fa

These panel meters are very nice - I only have to remove a diode from the PCB and supply 4.5-30V for power and 0-30V to be measured. Internal resistor divider is 330K vs. 12K + 22K pot. On the one measuring current I replaced 330K with 50K pot and set it up to measure the voltage on the power resistor.  I was in a hurry so I didn't investigate how to move the the decimal dot, but it's doable. For 2$ shipped you can't go wrong with these.

It's powered from an external 20V wall adapter.

For a better one I would use 0R1 5% resistor 5W, 10-30A diode and higher current FET, use a PC CPU heat sink and fan.

SeanB

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Re: Electronic constant current DC load
« Reply #55 on: November 25, 2012, 12:23:05 AM »
Nice meters, I should buy a whole lot of them....... Nice setup there, nice and compact.

Fennec

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Re: Electronic constant current DC load
« Reply #56 on: November 30, 2012, 06:12:48 PM »
> and higher current FET, use a PC CPU heat sink and fan.

You should find a FET with a much lower RDS_on than more current. Your 2SK2749 has a RDS_on around 1,6Ω. At 2A you have 6,4W amount of heat to dissipated. That's much.

The IRF3710 for axample is a Monster with 23mΩ . At 2A you only have 92mW at the FET. 6,3 Watts under.
And this 6A reverse Diode you don't need, most MOSFETs have a build in high current reverse one. Much enough for your lill board.
This 0R1 5W resistor is not the best idea. At 2-3A it goes glow. Use 25W or more, like Martin. You feel better then, belive me.

The best, you don't need those large heatsinks or PC cooler. You can still use your old one. 

dr_p

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Re: Electronic constant current DC load
« Reply #57 on: December 01, 2012, 03:59:01 AM »
You should find a FET with a much lower RDS_on than more current.
not surprisingly these two go hand in hand. One 50A FET will have low Rds_on by design otherwise it will dissipate a lot of power.

The IRF3710 for axample is a Monster with 23mΩ . At 2A you only have 92mW at the FET.

you got this wrong: Rds_on is the FET is fully open, and the FET + resistor are the only components eating the power. There won't be 2 A flowing through there, but much more: 12V / ( 23mΩ+ 0.33Ω ) = 34A
The FET will never get to work in on-off state, it's Rds will never come close to those 23mΩ so it doesn't matter how low it is.
If you want to pull 2A from a 12V supply, you HAVE to drop 24W somewhere. You can choose a huge power resistor(also higher value) and a low Rds_on FET, thus dropping most power on the resistor. Alternatively you can use a lower power resistor (3-5W, but lower value) and a higher power(!) FET and drop more power on the FET. Either way it'll work, but I find FETs cheaper to come by - old ATX PSUs etc.

And this 6A reverse Diode you don't need, most MOSFETs have a build in high current reverse one.

The diode I used is in series with the FET and the power resistor, it's used to protect against reverse connection when the internal diode would be forward biased and subject to the whole power dissipation. It's not paralleled with the FET.

This 0R1 5W resistor is not the best idea. At 2-3A it goes glow. Use 25W or more, like Martin. You feel better then, belive me.

3A on a 0.1Ω dissipates 3*3*0.1=0.9W. What are you talking about?!

The best, you don't need those large heatsinks or PC cooler. You can still use your old one.

as I've said before, you have to dissipate the power somewhere. Either FET or resistor, so cooling stays the same, sorry.
« Last Edit: December 01, 2012, 04:05:13 AM by mihai.a.anghel@gmail.com »

SeanB

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Re: Electronic constant current DC load
« Reply #58 on: December 01, 2012, 09:05:10 AM »
The FET dissipates all of the power, the low on resistance allows it to be controlled into the linerar region, if the on resistance is dissipating all the load then the FETY is full on, and there is no longer control of the current.

As to power in the sense resistor you want the resistor to run as cool as possible, as the voltage drop across it changes with temperature. A hot resistor is not the same value as a cold one by a measurable amount.

Fennec

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Re: Electronic constant current DC load
« Reply #59 on: December 01, 2012, 12:53:40 PM »
So you don't have a PWM mode ?