0
\$\begingroup\$

I was planning on making a voltage controlled resistor using a power MOSFET in order to load one 12V lead acid battery. I tried implementing the circuit below; everything was going well for a few seconds until the power supply suddenly switched off only for me to find out that the MOSFET was completely shorted.

schematic

simulate this circuit – Schematic created using CircuitLab

The MOSFET drain was attached to a heatsink by a screw and some thermal paste; in addition, a 12V fan was supplied through the same 12V supply and was cooling off the MOSFET. The MOSFET wasn't anywhere near its limit temperature when I checked it. The DAC output voltage was set to 2.39V in order to obtain approximately 6A of current as per the VGS to ID curves. I do realize that this open loop system is unreliable and subject to the variations of temperature, which is why I plan on making a feedback loop using a current sensor in the future. My question is, why did this happen?

Note: I have not surpassed any maximum ratings here as far as I know.

\$\endgroup\$
16
  • 4
    \$\begingroup\$ 72W is a tall order for a TO-220; at that level, you're mostly out of the realm of air-cooled heatsinks and into water/TEC setups. The junction-to-heatsink thermal resistance of 1.25 C/W gives a delta-T of 90C on top of your heatsink-to-ambient thermal resistance. Your max Tj is 175, which means your heatsink needs to be better than 1 C/W; possible but it's a lot of work to run your device at maximum Tj. I'm thinking the negative Vth temperature coefficient and constant Vgs caused you to exceed 6A; were you measuring current? \$\endgroup\$
    – vir
    Apr 28, 2022 at 19:33
  • 3
    \$\begingroup\$ You should not rely on typical characteristics. Current measurement and feedback are necessary. Also most MOSFETs don't handle linear operation well. Make sure SOA includes DC. \$\endgroup\$
    – Lars Hankeln
    Apr 28, 2022 at 19:53
  • 2
    \$\begingroup\$ @A.H.Z Saturation is the risky region for many FETs (especially those optimized for switching), due to various effects like localized thermal runaway. \$\endgroup\$
    – nanofarad
    Apr 28, 2022 at 20:10
  • 2
    \$\begingroup\$ @A.H.Z You may have the same thermal runaway issue with parallel FETs and they may thus need individual source and/or drain resistors (not sure which offhand) of some small value, OR independent control/feedback to counteract this. \$\endgroup\$
    – nanofarad
    Apr 28, 2022 at 20:19
  • 2
    \$\begingroup\$ This is not how to design an Active Load. Try this >>step 1) define design limits for V,I,P,$ 2) consider PWM or constant current with storage series tank load. Allocate Pd loss budget for each component with total < 15%. 3) Determine max temp rise and DCR, ESR, RdsOn, Rbat to achieve specs. 4) only use air coils and very low ESR caps. Compare BJT switches with IGBT and FET's choices. 5) compute Q of the load and target low Q = 1 at max load. \$\endgroup\$
    – Tony Stewart EE75
    Apr 28, 2022 at 23:30

1 Answer 1

Reset to default
1
\$\begingroup\$

The DAC output voltage was set to 2.39V in order to obtain approximately 6A of current as per the VGS to ID curves.

Like this: -

enter image description here

  • So, when 2.39 volts are first applied (at ambient of 25°C) the current drawn will be about 0.7 amps (blue circle).
  • The MOSFET will be dissipating approximately 12 x 07 watts i.e. 8.4 watts
  • But it will be warming rapidly and, as it warms it draws more current and warms even faster
  • After a very short length of time (possibly less that 1 ms), the MOSFET junction has warmed to 175°C and it's now dissipating about 12 volts x 7 amps (green circle) = 84 watts.

And maybe, at this point you believed it might stop warming?

After all there are no graphs above 175°C so, what really happens?

Answer: it gets rapidly warmer and warmer until it reached a temperature of around 600°C and then the device suffers melt down.

Why does this happen?

Answer: because you are controlling the device in a region where thermal runaway will occur.

If you were controlling the device with gate-source voltages above about 4 volts then the MOSFET will naturally be in the region where there can be no thermal runaway.

At the absolute minimum, you need a feedback mechanism that can rapidly reduce the gate-source voltage should current start to rise due to thermal runaway. But, it's worse because many modern MOSFETs comprise of thousands of parallel MOSFETs internally and, each have their own characteristic and, although you may think the current is held at (say) 6 amps, it will be flowing in only a small part of the overall area of those thousands of transistors. Result is the same.

Remedy: use a MOSFET designed for linear applications and use a feedback mechanism that is lightning fast.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ By linear operation I assume that you are referring to the saturation region ( Vds>Vgs-vth) correct? \$\endgroup\$
    – A.H.Z
    Apr 29, 2022 at 15:46
  • 1
    \$\begingroup\$ @A.H.Z linear operation means applications that are are not switching applications i.e. applications that seek to control the drain current by regulating the gate-source voltage. \$\endgroup\$
    – Andy aka
    Apr 29, 2022 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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