Discussion in 'Power Electronics' started by tasty, Sep 14, 2019.

1. ### tasty

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Sep 14, 2019
Hello,

I'm trying to make an electronic load circuit like in this picture : https://learn-cnc.com/wp/wp-content/uploads/2018/04/ConstantCurrentExample.png . It works fine with low voltages like 8V but my supply is 55V. I use an IRF840 for the MOSFET and a 0.6 ohm shunt. I ajust the positive input with a potentiometer. I need to pull 5 amps on this 55V supply. So I need 3V on this shunt. But the MOSFET blow up at 0.8V. I don't undestand why. Can somebody help me?

Thank you.

2. ### Alec_t

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Jul 7, 2015
Welcome to EP!
An LM358 with a supply of 5V can provide a maximum output voltage of about 3.8V, so Vgs = 3.8V - 0.8V = 3V.
0.8V across 0.6Ω implies 1.33A current.
At 1.33A current the minimum drain-source voltage Vds of the FET is about 2V, provided Vgs is at least 4.5V (according to Fig 2 of the FET datasheet). However, with Vgs being only 3V then Vds is likely to be a lot higher than 2V. If Vds were, say, 5V then the FET would be dissipating at least 5V x 1.33A = 6.6W. Without a big heatsink it would get VERY hot. In the absence of any test load in series with the FET, Vds is 55V-0.8V = 54.2V. At 1.33A drain current the FET would have to dissipate 54.2V x 1.33A = 72W. So what size heatsink were you using when your FET fried?
To pull 5A from a 55V supply your poor old FET would have to dissipate (55V-3V) x 5A = 260W, which is way beyond its rating!
A 'logic-level' FET would be a better choice than the IRF840 unless you can use a higher supply voltage for the opamp.

Last edited: Sep 14, 2019
3. ### tasty

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Sep 14, 2019
Thank you for the answer. I forgot an information, sorry. The oamp have 10V not 5V for the supply. For the heatsink, it's not the issue because it's not enough hot to destroy the component (I can touch the MOSFET with my finger ). I think it's about 40-50 degre Celcius. I tested with an IRF1404 and the result was the same but when the voltage across the shunt was about 0.2-0.3V. May be I have to put a gate drive between the opamp and the gate of the MOSFET? What do you think?

Last edited: Sep 14, 2019
4. ### Alec_t

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Jul 7, 2015
Shouldn't be necessary if (1) the FET is a logic-level type and (2) the gate isn't being switched rapidly. A gate series resistor of, say, 330Ω would reduce the source current required from the opamp output.
Can you use a lower resistance shunt? 0.6Ω would dissipate 15W at 5A.

5. ### tasty

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Sep 14, 2019
Yes, actually I have 10 resistors of 1.5 ohm at 5W each one. I can use any value than I can make with all 10 resistors. The lowest resistance than I can put is connect all of them in parallel so 0.15 ohm. In this case I need 0.75 V to have 5A. Do you think that it will solve the problem? I can dissipate untill 50W with those resistances.

6. ### Alec_t

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Jul 7, 2015
0.15Ω would certainly reduce the heat in the shunt, but any FET is still going to roast at 260W!

7. ### tasty

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Sep 14, 2019
Ok but the resistors can take 50W (there are 10 of 5 W), even 15W is not an issue. I still don't understand why any MOSFET blow up with a supply of 55V. If it's not the heat the only option is the over current somewere in the circuit but where?

8. ### Alec_t

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Jul 7, 2015
The FET takes the brunt of the heat dissipation; not the shunt.

9. ### tasty

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Sep 14, 2019
But why? to much current ? to much voltage? because of oscillation in regulation system?

10. ### WHONOES

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May 20, 2017
Sounds as if the Fet is oscillating of its own accord. That can happen if you have a long wire between your opamp an the Fet gate. It is also possible that the opamp is unintentionally oscillating because of poor or non existent PSU decoupling. Trying one thing at a time, connect a 10μF capacitor across the supply pins of your opamp as close as you can get them. Then try connecting a 1K resistor between the gate of the Fet and the output of your opamp preferably attached directly to the Fet gate terminal.

11. ### tasty

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Sep 14, 2019
Thank you, I will test it tomorrow.

12. ### Alec_t

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Jul 7, 2015
Even if the FET isn't oscillating, its sole purpose as an electronic load is to dissipate energy. Energy = current x voltage x time. Most of the voltage is dropped across the FET drain-source (55V at the drain, ~3V max at the source); a much smaller voltage (~3V max) is dropped across the shunt. As the current through the FET and the shunt is the same, it follows that the energy loss in the FET is much greater than in the shunt.
If you were to use some resistance in series between the 55V rail and the drain, then that resistance would share some of the voltage drop and reduce the stress on the FET. However, you would then reduce the range of variation of the electronic load.

Last edited: Sep 15, 2019
13. ### tasty

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Sep 14, 2019
I tested and the result is the same, a beautiful explosion of the MOSFET. I put a fan to improve the cooling so the heat is definately no the issue. I added a 2200µF to the opamp supply and 1K between the out of opamp and the gate of the MOSFET. I used a 10N60C this time. Shoud I reduce the power dissipated by the MOSFET like Alec_t said?

https://imgur.com/a/g5rZcTN

PS: I saw a little drop of voltage on the opamp supply (like 0.2V) when I was increasing the voltage on positive input before the explosion.

Last edited: Sep 15, 2019
14. ### Alec_t

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Jul 7, 2015
Are you aware that the mounting tab of the FET is electrically connected to the FET drain (unless you have the T0-220F package version) ?

15. ### tasty

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Sep 14, 2019
Yes, I know that I soldered the positive of 55V directly on the drain pin. I tested with a TO-220 version (the IRF840) and it was the same result. And nothing else is connected to the heatsink. In theory, the 10N60C can handle almost 6kW am I right? But why this thing can't do it for just about 250W, considering that que heat is not the issue. Put a big heatsink and a 12V fan on it.

Last edited: Sep 15, 2019
16. ### Alec_t

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Jul 7, 2015
No. How did you arrive at that figure?
Here's what the datasheet says:

Absolute maximum power is 156W, provided you can keep the case temperature Tc down to 25C. If Tc is, say, 50C then the maximum power is derated to only 156-1.25*(50-25) = 125W.
You also have to consider the junction temperature inside the FET. If you crank up the current quickly the junction will heat up, possibly to a destructive level, long before the case can get even warm.

17. ### tasty

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Sep 14, 2019
Thank you, I will try to connect 6 MOSFET in parallel then. I will see.

18. ### hevans1944Hop - AC8NS

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Jun 21, 2012
@tasty You are conceptually incorrect in your application of the 10N60 MOSFET as a "variable resistance" electronic load. This device is intended to be used as a high-speed electronic switch, being either turned ON with low resistance from source-to-drain, or turned OFF with high resistance from source-to-drain. Either of these two states will dissipate minimal power.

It will be difficult to operate your MOSFET as a linear device, with source-to-drain resistance being controlled by the gate-to-source voltage, without electronic feedback, of which you have indicated none. So get with the program and begin posting schematics. For reasons I won't discuss here, it will be difficult to operate power switching MOSFETs without "blowing them up" even at modest power dissipation. For an excellent discussion and a sample circuit of "how to do it" please read Kerry Wong's web page here.

I was NOT going to jump into this thread because @Alec_t is competent to give you good advice. Problem is, you don't know WTF you are doing, so it's like trying to explain quantum mechanics to barnyard chickens. The chickens will nod their heads and pretend understanding, but they are just dumb chickens and will never understand quantum mechanics. You OTOH have some hope of eventually "getting it" although you are not there yet. Read on.

An electronic load is a nice thing to have handy during design, but realize that its function is to consume energy and dissipate power. Dissipating power means heat will be generated, no matter what device you use to dissipate power. Your job, should you choose to quit blowing up components, is to figure out a way to safely dissipate the power to the environment.
Okay, that means you need to dissipate 5 A x 55 V = 275 W. That's P = V x I as the "formula" for power dissipated in a resistive load with V volts across the load and I amperes flowing through it. That's almost as much as the heat generated by SIX 60 watt incandescent light bulbs (360 watts for six light bulbs versus 275 watts for your application). That's a lot of heat there! Common sense should tell you that the TO-220 package is not capable of this much power dissipation without a significant increase in semiconductor junction temperature, even with a "perfect" heat sink maintained (somehow) at 25C.

Since the current-measuring shunt is in series with the 55 V supply, that 5 A current will produce 5 A x 5 A x 0.6 Ω = 15 W power dissipation in the shunt. That's P = I2R as the "formula" for power dissipated in a resistive load. Fifteen watts is insignificant compared to two hundred seventy-five watts, but why do you need such a large-valued shunt resistor to measure five amperes of current? How did you decide a 0.6 ohm shunt was necessary?

But let's get back to power dissipating, electronically controlled, loads. For small amounts of power, up to about ten kilowatts or so, and low voltages, up to about a kilovolt or so, bi-polar junction transistors (BJTs) are almost ideal candidates. If higher current than a single device can safely pass is required, it is fairly simple to parallel two or more devices with "current sharing" resistors in series with their emitter leads.

Well, good luck with that then. Back to @Alec_t for further help...

19. ### Alec_t

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Jul 7, 2015
Fine in theory, difficult in practice when the FETs aren't fully on or off. Because of production spreads, even FETs from the same batch are unlikely to have exactly the same Vgs/drain-current characteristic. The result is that they won't share current equally and will probably self-destruct sequentially, domino style. Adding source resistors helps in current-sharing, but isn't a guaranteed cure.

20. ### tasty

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Sep 14, 2019
I already posted a link for the schematic. If you read ALL the posts you will understand better.

I know exactly want I want to do. I want to dissipate a certain power into an electronic circuit but with a CONSTANT current. If you read ALL the posts you will understand better.

The issue is not the heat like a said a lot of times. There is no time for the MOSFET to get warm, this is the issue. If you read ALL the posts you will understand better.

I have 10 resistances of 1.5 ohm with 5W each one like I already said. If you read ALL the posts you will understand better. I can make a lot of shunt value with them. The only constraint is to chose a value of resistance to have a voltage under then the maximum value that the output of the opamp can give.

This not a small amount at all for common electronics. The reason that I choose a FET is because most of the schematics that you can find in internet gives that. The BJTs can also do the work but the COMMON BJTs can take until 1A or 10A if you don't want a specific (and expensive) but most of the MOSFET can take a large amount of current.

Thank you for your post but I think that you are not in the good domain.

You can read this document from another person who already tried that. You can read the third paragraph to understant better if you want.