0
\$\begingroup\$

I am designing a dummy load, and it should be capable of sinking around 75 A @ 80 V (not always, varies from 10 to 75 A). I want to choose MOSFET for the application as shown in the image.

I would like to ask a doubt regarding the MOSFET power dissipation. Is it the max power going through it or the current^2 * Rds(on)? From what I understood the dissipated power is current^2 * Rds(on) (not for fast switching applications). In short, what is the power rating for the MOSFET, and why? Also, how does heat generated differ from the overall power rating?

enter image description here

\$\endgroup\$
1
  • 2
    \$\begingroup\$ For a linear application, you will never operate at rated RdsON, but some much higher value unless you saturate your regulation. Look up the SOAR curve for your MOSFET instead of RdsON. \$\endgroup\$
    – winny
    Commented Sep 27, 2023 at 10:19

3 Answers 3

4
\$\begingroup\$

Your circuit will operate in the linear region and this means that the gate-source voltage will be moderate to low in value and nowhere near the value to obtain lowest \$R_{DS(on)}\$ (a parameter associated with switching applications) and, unfortunately you can't use it to calculate power dissipation. However, you can use your simulator to calculate the true power dissipated namely \$I_D \times V_{DS}\$.

This always equates to the MOSFET power dissipated (although it doesn't account for the small amount of power used by the gate). However, yours is a linear application so the gate power is largely irrelevant.

You should also be aware of thermal runaway problems with MOSFETs in linear applications and try and focus on choosing a MOSFET that is designed to be less problematic under these circumstances. IXYS make a few parts that are specified for linear applications. Be also aware that several parallel MOSFETs with individual op-amp control is the way to go.

Treat each parallel arrangement of MOSFET and control op-amp separate i.e. don't try and share the same current sense resistor.

what is the power rating for the MOSFET

Data sheets tell you a figure on their front page about power rating but this won't help you. The figure quoted is for switching applications i.e. it is a "summary" value for short pulses that counts for nothing in your type of linear application. Hence, a single data sheet value for the power rating will not help.

You have to dig deeper and look at the safe operating area curve in the data sheet. Then you look for the curve that implies continuous operation. In your previous question on this subject I showed this graph for an IGBT and, we can use that to demonstrate what I mean: -

enter image description here

Previously, your requirement was for 50 amps at 85 volts and that meant that the peak power is 4.25 kW but, where those lines cross on the graph, the device can only sustain this power for 3 or 4 ms maximum.

So, you look at the DC curve (continuous operation) and see that the device can only sustain about 8 amps maximum when \$V_{DS}\$ is 85 volts. That's a power of 680 watts and not 4.25 kW.

This implies to me that the minimum number of parallel devices needed is seven i.e. 7 devices with their associated current sense resistors and op-amps. Of course this was for an IGBT but, it won't be much different when using MOSFETs. For your new requirement (circa 6 kW) you might be looking at ten parallel circuits and, if you choose smaller devices than the IGBT I targeted in the previous question, you might be looking at 12 to 15 parallel circuits.

\$\endgroup\$
2
\$\begingroup\$

It's quite simple:
If your circuit has to sink 75 A @ 80 V it will dissipate \$P_{tot}=75\text{A} \cdot 80\text{V} = 6\text{kW}\$.

The power dissipated by R_shunt is \$P_{shunt} = R_{shunt} \cdot I^2\$.

In short, what is the power rating for the MOSFET, and why?

Obviously the rest of the power \$P_{tot} - P_{shunt}\$ needs to be dissipated by the MOSFET.

Also, how does heat generated differ from the overall power rating?

If there is not other way for the power to go (and in your circuit there isn't) the overall power rating equals the power going into heat (in the resistor and in the MOSFET).

\$\endgroup\$
1
\$\begingroup\$

I'll treat this question as "how to select a dummy load design", not necessarily concerning a specific implementation. (In other words, heading off the X-Y problem, in case it applies here.)


For typical load applications -- testing supplies, batteries, etc. -- one must define the load range V and I, and P if less than Vmax * Imax, the accuracy and precision, the response time, and any other details if applicable (is switching noise tolerable? -- how much, in what frequency bands?).

Consider some basic options:

  • Load Resistors

    We can just wire up some dumb ass resistors and get cooking. It's hard to argue with passive chunks of metal and ceramic if it gets you where you need to go!

    Obvious downsides: the V/I ratio is fixed by configuration, is not very adjustable (at best selectable by wiring or switches, or using pricey rheostats -- which in these values, are probably pretty chunky themselves, more of a selector than a continuous adjustment), and can only be made to approximate an arbitrary (CC, CV or other) curve through continuous adjustment. Upside: resistors can be very compact, as the operating temperature can be considerably higher than semiconductors -- if you don't mind that 250°C air is rising off of them! Which are probably cheaper than a semiconductor and heatsink assembly, too (which needs to operate below maybe 100°C -- considerably more area required).

    Gotcha: heatsink type resistors (aluminum-case and thin film bolted types), which have all the downsides of semiconductors: limited operating temperature, heatsink required. Well, maybe not all, you still get the pulse capacity; well sort of, the thin-film ones less so. As you can see, there are quirks to understand in all corners of available components as well as design approaches.

  • Transistors and Heatsinks

    The standard approach, more or less, as it's readily controllable, easy to design and build, and scalable I guess as long as you don't mind it's 0% efficient by definition. (Actually negative I suppose, because you won't usually have the controls powered by the load, but, that's a tiny distinction.)

    The downsides, as mentioned, are the heatsinks must be quite large and heavy to keep operating temperatures down. You might even find water cooling is desirable. A lot of transistors are required for this kind of power level. Pay careful attention to SOA (safe operating area), which must not be violated for any given transistor.

    We can make improvements, or compromises, upon the basic semiconductor load dump: consider putting a resistor in series with a transistor. Depending on drain current, drain voltage is reduced; this restricts the operating area, so is an acceptable option if you don't need full current capability below some minimum operating voltage.

  • Resistors and Transistors

    We can combine the best of both worlds, by using an array of shunt (parallel) resistors to carry the brunt of the load, and a much smaller semiconductor load to take up the difference. We set the resistors such that, for a given load current, slightly less than the total is taken up by the enabled resistors, and so as current setting increases, the difference rises until the next resistor pulls in, and so on.

    This has the downside that, as resistors switch in and out, say due to load voltage varying (as it will due to mains ripple on a linear supply or active PFC, or while ramping the current setpoint), a blip is generated where a given resistor turns on or off, and the linear sink takes a moment to catch up. The resistor switching can be made intentionally slow so that the linear sink has time to react, but there will always be some residual blip as the control loop acts to compensate for the event.

    I've personally made such a design, which uses three 200W rated MOSFETs in the linear sink, and has a total capacity of 100-500V 0-4A. Response time is in the 10s of µs, good enough for what I'm using it with (mains supplies with quite slow compensation). The resistor-switching blips are measurable, but easily averaged out of such measurements (again, the supply's loop compensation is much slower). Unfortunately, I don't have a BOM cost handy -- it was a one-off design, and I don't really know for sure if it's cheaper than an all-transistor version, or at what point one would break even.

  • Power Converter and Battery

    This solution is perhaps most obvious when cycle testing batteries. You cycle energy between a pair of batteries, as long as you start with average charge below 50% or thereabouts, and have a small external supply available for "topping up" overall system charge, to compensate for charge and converter losses over multiple cycles. More practically, you might have a larger reservoir battery, operating nominally in the say 50-80% charge range, so it has ample capacity to absorb charge from the BUT (battery under test), and it might be several times oversized; it could also be used for multiple simultaneously, so long as the total charge is conserved (i.e. you mostly have one BUT discharging into another that's being charged, and so on, with a small remainder going in/out of the reservoir).

    Well, it also works well for power supply testing, and as a general-purpose power supply itself. For PS testing, the converter should have a CC/CV/R characteristic (whatever is appropriate for testing), and the reservoir battery must have a voltage limiter on it -- since continuous PSU testing would eventually overcharge it. In terms of motor drives: a braking resistor.

    The main downsides to a power converter, are the difficulty in building it (an SMPS is quite a step up in challenge from a linear supply/sink/amplifier), and the relatively slow response time -- the control can't respond any faster than the output filters permit, generally in the ~kHz.

  • Grid Tied Load

    Same idea as above, but with a grid-tie inverter returning excess power to the mains supply. Even more difficult, of course -- I certainly wouldn't recommend building one from scratch, not without experience in all of the circuits and standards involved for such a device -- but it's very much an attractive option at high power levels, for sustained testing. If it's a power supply under test, then the overall effect is simply to return power to its inlet; which might be done in an easier way elsewise (e.g. return to its internal DC link supply), but returning it to the grid is more general-purpose: you aren't bypassing anything so can test the supply entirely by itself, and you don't care what it uses internally (does it use a 160V or 320V or 400V or 650V or... rail?), if you have access to such connections at all.

    It also works well for batteries, where the grid is used effectively as a battery itself, whether by way of momentarily offsetting your own energy consumption, or fully reversing mains power flow (assuming permission from the utility, of course!). In that case, any regular battery charger can be used for the charge cycle, and a grid-tied load to discharge.

Regarding noise: obviously, the switching methods have some potential for EMI and RFI emissions. They can be filtered arbitrarily well, but this comes at the expense of response time, if that should be a concern. The combined "resistor power DAC" and linear sink method exhibits some transition noise, as mentioned in that section.

Also, kind of as a side to the power converter and battery option: instead of a braking resistor and battery, it can be switched directly into the braking resistor, with minimal energy storage (i.e. just EMI filters), no battery. Or other load elements could be used (like TVS diodes or MOVs*, giving more stable clamping voltage -- perhaps better for peak load operation?), but resistors are cheapest. Basically, a buck or boost (or both) converter into a load resistor.

*Given that MOVs are never rated for continuous duty; such a design would need to be evaluated to prove feasibility.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.