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Electronic load module 100V, 10A, 75W, commissioning, test

TechEkspert 10170 8

TL;DR

  • An $8 electronic load module lets you test power supplies by forcing a controlled current instead of using bulbs, resistors, or a bare transistor on a heatsink.
  • The board uses an IRFP250N, TL431, LM258, LM393, a bent-wire shunt resistor, a 5-20k potentiometer, and a 12V supply.
  • Auction specs claim 75W maximum power, 50W continuous power, 10A maximum current, and 100V maximum voltage.
  • To start it, mount the transistor on a heatsink with fan, connect the load terminals in series with an ammeter, and feed 12V into J3.
  • The 50W continuous rating is lower than the 75W peak rating, so cooling is critical during use.
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  • Electronic load module 100V, 10A, 75W, commissioning, test
    When testing, for example, power supplies, an electronic load is useful to, for example, force the flow of a specific current. In practice, light bulbs are often used quickly (which is a bad solution due to the low resistance of the cold fiber), resistors, or a random transistor on the heatsink, the base or gate of the transistor is controlled by an operational amplifier, or often without a feedback loop with an ordinary potentiometer. It is available on auction portals electronic load module priced ~ $ 8. In the description of the auction, we see the following parameters: maximum power 75W, continuous power 50W, maximum current 10A, maximum voltage 100V. On the board, we see a measuring resistor (in the form of a bent wire), a transistor IRFP250N, TL431, LM258, LM393. To start the artificial load, mount the transistor on a heat sink (preferably equipped with a fan), attach a 5-20k potentiometer ensuring current regulation, and connect a 12V power source.

    The V- V + connector is used to connect the cables connecting the tested device,
    in series in this circuit it is worth to turn on the ammeter to control the set current.
    The 12V power supply is supplied to the J3 connector, the device consumes a current

    Cool? Ranking DIY
    About Author
    TechEkspert
    Editor  
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    TechEkspert wrote 6988 posts with rating 5427, helped 16 times. Been with us since 2014 year.
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  • #2 17412247
    Anonymous
    Level 1  
  • #3 17412253
    TechEkspert
    Editor  
    There is no scheme, the description of the auction is usually tragic, this is a common feature of these cheap modules ...
    It is important that it works :) and does not generate oscillations ...

    The artificial load that you have presented has turned out to be very compact.
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  • #4 17412495
    tplewa
    Level 39  
    @TechEkspert

    Let's take a look at the IRFP250N and take into account that in this system it works in DC mode ... unfortunately, the documentation does not often provide SOA for DC mode ... but even looking at it:

    Electronic load module 100V, 10A, 75W, commissioning, test

    anyway, I wrote about this transistor in another thread regarding the artificial load, only they went there even more, i.e. 150W (the same transistor, but better cooling) ;)

    https://www.elektroda.pl/rtvforum/topic3462045.html#17222762

    These 75W are even temporary, and 50W are with good cooling :) for a transistor ... at 100V, it's not even to mention ...
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  • #5 17413738
    _johnny_
    Level 10  
    IGBT transistors are best suited for dissipating DC power. They have a much wider ch-ke SOA, so you can lose up to 300W in a housing like the transistor above.
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  • #6 17414054
    tplewa
    Level 39  
    _johnny_ wrote:
    IGBT transistors are best suited for dissipating DC power. They have a much wider ch-ke SOA, so you can lose up to 300W in a housing like the transistor above.


    They don't necessarily have to be IGBTs. Anyway, it's better to give more transistors - in fact, apart from the transistor, we have few elements, in total an operational amplifier and a measuring resistor ... we distribute the current nicely to the transistors, so we don't have to force them. However, with forcing it is so often that it will not die right away, but the service life is drastically shortened by excessive overheating of the structure and suddenly it flies from time to time ...

    By adding some CPU + DAC + ADC, you can make a pretty cool artificial load with various interesting functions, e.g. for discharging batteries (counting ampere hours etc.).
  • #7 17422472
    maciej_333
    Level 38  
    TechEkspert wrote:
    There is no schema

    Found this pattern in less than a minute: Link . As you can see, someone took the trouble to draw it from a PCB. Who will now solve the puzzle related to the purpose of using the D1 diode and the threshold at which the overvoltage protection will work?
  • #8 17422925
    jaroslawk
    Level 21  
    Hello,
    If I am not mistaken, the output of the R13 resistor should be connected to GND and not to V- and the system works from about 0.7V.
  • #9 17424782
    TechEkspert
    Editor  
    This is the scheme, but in a place where someone did reverse engineering ... and not at the seller / manufacturer.

    As for D1, this is a primitive way to get virtual mass?
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Topic summary

✨ The discussion revolves around the use of an electronic load module with specifications of 100V, 10A, and 75W for testing power supplies. Users share experiences with DIY implementations, highlighting the importance of proper cooling for components like the IRFP250N transistor. Concerns are raised about the lack of detailed schematics in auction listings for these modules, with suggestions for improvements such as using multiple transistors to distribute current and enhance longevity. The conversation also touches on the potential of integrating microcontrollers for advanced functionalities in artificial loads, including battery discharging capabilities. Additionally, there are inquiries about circuit design and component connections, particularly regarding overvoltage protection.
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FAQ

TL;DR: A $8, 75 W load module delivers only about 50 W continuous, “These 75 W are even temporary” [Elektroda, tplewa, post #17412495] IRFP250N’s DC SOA drops below 20 W at 100 V.
Why it matters: Knowing the real limits prevents MOSFET failure and saves test time.

Quick Facts

• Module price: USD ≈ 8 [Elektroda, TechEkspert, post #17412176] • Advertised limits: 100 V, 10 A, 75 W peak, 50 W continuous [Elektroda, TechEkspert, post #17412176] • Mandatory supply: 12 V/≥200 mA for control circuit [Elektroda, TechEkspert, post #17412176] • IRFP250N RDS(on): 0.085 Ω typ. @10 Vgs [IRFP250N Datasheet] • Typical SOA at 25 °C: 40 W @ 50 V DC [IRFP250N Datasheet]

1. What is the safe continuous power for the IRFP250N on this board?

Stay near 50 W with strong airflow. The DC safe-operating-area curve shows the transistor falls below 40 W at 50 V and <20 W at 100 V [IRFP250N Datasheet][Elektroda, tplewa, post #17412495]

2. Can the module run the advertised 75 W indefinitely?

No. Contributors report 75 W is only short-term; long tests should not exceed 50 W even with a fan [Elektroda, tplewa, post #17412495]

3. Why must I use a heat sink and fan?

At 50 W the MOSFET’s junction rises ≈62 °C above heat-sink temperature (RθJC 1.25 °C/W) [IRFP250N Datasheet]. Forced airflow keeps case temperature under 100 °C and prevents thermal runaway.

5. At what voltage does the load begin to conduct?

Because R13 ties to ground, regulation starts around 0.7 V—the TL431 threshold plus op-amp input offset [Elektroda, jaroslawk, post #17422925]

6. How do I wire the potentiometer for current control?

  1. Connect the wiper to the op-amp non-inverting input. 2. Tie one end to 12 V, the other to ground. 3. Use 5 – 20 kΩ to cover 0–10 A range [Elektroda, TechEkspert, post #17412176]

7. How can I stop oscillations?

Keep wiring short, add a 100 nF ceramic across V+ / V−, and place a 1 µF capacitor across the sense resistor to slow the loop—“important that it works and does not generate oscillations” [Elektroda, TechEkspert, post #17412253]

8. Is an IGBT a better choice than the IRFP250N?

Yes. IGBT devices show a wider DC SOA; users report up to 300 W dissipation in similar packages [Elektroda, johnny, post #17413738]

9. What if I parallel multiple MOSFETs instead?

Paralleling two MOSFETs halves current per device, doubling reliability. Balance with 0.1 Ω source resistors to avoid current hogging [Elektroda, tplewa, post #17414054]

10. What happens at 100 V and 1 A (100 W)?

The transistor exceeds its DC SOA; thermal stress can cause sudden die cracking—an observed failure mode in high-voltage loads [IRFP250N Datasheet].

11. How do I insert an ammeter correctly?

  1. Break the V+ line from the supply. 2. Wire the meter in series between supply and V+ terminal. 3. Keep leads short to reduce inductance [Elektroda, TechEkspert, post #17412176]
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