I am working on a design of a control system that controls a few small solenoids, each drawing about 200 mA of current and operating at 15-18 V. These solenoids are activated by NMOS transistors. Due to the size constraint that I have for the PCB containing these transistors, I would ideally like to use SOT-23 NMOS transistors.

Now, if one does the math, the power dissipated by each transistor activating a solenoid is P = (0.2 A)(18 V) = 3.6 W. However, it seems that most of the SOT-23 package NMOS transistors can only dissipate a maximum of about 1 W.

My question now is this - is the maximum power dissipation rating primarily due to the heat build-up that could damage the device if it exceeds its rated power dissipation? Or is it the limit of the internal materials that the transistor is made of?

Each solenoid in this system is typically never activated for longer than 20 seconds, and at the very extreme it might be activated for 1 minute, which happens almost never. The average on-time for these solenoids are between 1 second and 10 seconds. Will this relatively short on-time (dissipating 3.6 W of power) have a detrimental effect on a transistor that can only dissipate a maximum of 1 W? Or can one safely assume that the average on-time of the solenoid is short enough that the transistor will not heat up too much to cause any damage?

I have used the 2N7002 NMOS transistor (which has a very low max. power dissipation rating) in a prototype for this system with no problems at all and no evident heat build-up by the transistors. Should I rather get a properly rated transistor (3.6 W or higher rated), or can I safely use a 1 W rated transistor for this system?

Your comments and suggestions will be highly appreciated. Thanks!


4 Answers 4


What you have calculated is the power dissipated by the solenoid and transistor together. The transistor will maybe drop 0.1 volts across it when passing 0.2 amps hence its power dissipation is 20 milli watts.

In your case, the 2N7002 is maybe a little closer to the limit. See the graph below: -

enter image description here

What this tells you is that if you drive the gate with 5 volts and you are passing 0.2 amps through the drain, the volt drop from drain to source might be 0.3 volts. Power dissipated is 60 mW.

However, if you are driving the gate with 3V logic signals it is likely to fry because at 0.2A there is no resolution to the 3V curve. In reality the circuit will take about ~50mA with 75% of the voltage being across the transistor so power would be ~675 mW and too much for a puny 2N7002.

As an aside, having read the data sheet, ~0.1A (Fairchild) is the absolute limit for the 2N7002 so, you should not consider using this device for driving a 0.2A solenoid. On the other hand, for NXP, the limit is 300mA (SOT23 package) so you need to check what the source is for the device. It can be annoying when different suppliers do this. Supertex state 115mA (same as FC) and I won't go into any others but, hopefully you see the issue.

You'll also need a flyback catch diode across the solenoid because, when you turn the FET off, the current thru the solenoid has created a magnetic field and hence has stored energy - this energy turns into a big voltage spike that will easily damage transistors when they try to deactivate the solenoid.


You're calculating the power dissipation of your transistor wrong.

18V * 0.2A is the power of the solenoid. The power of the NMOS is the voltage drop across the transistor times the current through it.

Typically for a completely switched MOSFET you will have an ohmic value given in the datasheet as rds(on), which is the on-state resistance. So you can calculate the power as if the transistor were a resistor (P = I²R).

Often the datasheet also quotes a peak pulse current and the duration of those peaks, which would be what you assume. It will take quite more power for a short time, but it needs time for the heat to dissipate and the part will not overheat and get damaged. Those pulses are typically a lot shorter than seconds or minutes.


You make a common mistake. You should be using the NMOS transistor as a switch meaning that it is either conducting or non-conducting. When it is non-conducting, no current flows so P = V * I = 18 V * 0 = 0 W, Good, no problem.

When the NMOS is conducting and the solenoid is activated current flows but the 18 V should not fall across the transistor, it should fall over the solenoid ! If you look in the 2n7002 datasheet you should find that it has an on resistance of around 4 ohms, this value depends on the Vgs you apply and the current. The current will be 0.2 A * 4 ohms = 0.8 Volts So the dissipated power will be: P = 0.8 * 0.2 = 0.16 W This is low enough !

Conclusion: this will work absolutely fine !

  • 2
    \$\begingroup\$ For completeness, the FET will dissipate more power at the moment of switching : between ON and OFF it may have up to 18V and up to 0.4A (worst case, 9V and 0.2A) momentarily. For a solenoid the switch rate is low enough this doesn't matter which is why @FakeMoustache didn't mention it. But in fast switching like PWM, it would matter... \$\endgroup\$
    – user16324
    Oct 9, 2015 at 11:18
  • \$\begingroup\$ Very true, I was assuming very infrequent switching, not more than a few times per second or so. \$\endgroup\$ Oct 9, 2015 at 11:19
  • \$\begingroup\$ @BrianDrummond the use case is given in the opening post their average on time is given as between 1s and 10s, so they won't be switched very often. \$\endgroup\$
    – Arsenal
    Oct 9, 2015 at 11:39
  • 1
    \$\begingroup\$ @Arsenal - very true. All I was thinking is, if the asker - or someone searching "NMOS transistor power dissipation" - reads this correct answer and uses it in a different context, it might be worth adding a spoiler for their surprise... \$\endgroup\$
    – user16324
    Oct 9, 2015 at 12:28
  • \$\begingroup\$ @BrianDrummond sound reasoning there, happens more than often that answers are used in different context and then people come back crying... +1 so it doesn't get lost. \$\endgroup\$
    – Arsenal
    Oct 9, 2015 at 12:44

You're using the MOSFET as a switch. When the MOSFET is ON, it will appear as a low value resistance between drain and source. This is called Rds ON. For a 2N7002 from NXP this is about 2.8 Ohm (assuming Vgs=10V, Id=0.5A and Tj=25C).

With 200mA of drain current the voltage drop is 0.56V. The power dissipation of the MOSFET is 0.56V * 0.2A = 112mW. This means that with 18V supply, (18V - 0.56V) is across the solenoid.

The moment the voltage of your load is important is when the switch is OFF. In that case the MOSFET should not conduct current (maybe small leakage), and MOSFET needs to "block" the full supply voltage across drain and source. This is the drain-source voltage rating, which is 60V for the 2N7002.

The reason why your prototype appeared to work fine is because we only had 112mW power dissipation. The NXP datasheet specifies 0.83W max power dissipation at 25C ambient. Power limits are primarily set by the max junction temperature and thermal resistance. A SOT-23 has typical thermal resistance values of 100 - 150C/W (depends on PCB, layout, etc. - often is a bit of a guess).

Assuming worst case the junction of the device heats to 0.112W * 150C/W = 16.8C above ambient. Above ambient is important because it means you may need to derate a device if your product needs to work at higher ambient temperatures. E.g. if the circuit needs to work at an ambient temperature of 80C, the junction temperature would be 96.8C. The limit is 150C, so the circuit should be OK to use.

It is dangerous if specifications for the generic parts are different between manufacturers. In that case it may be wise to select a different MOSFET or stay within the worst-case specifications.

  • \$\begingroup\$ With regards to the thermal resistance (junction-to-ambient), the datasheet for the 2N7002 gives a value of 625 ºC/W. Assuming the transistor operates at room temperature (25 ºC) with a power dissipation of (0.2 A)*(0.56 V) = 0.112 W, would that then mean the temperature at the junction is (625 ºC/W)*(0.112 W) + 25 ºC = 95 ºC? How long would it take for the device to reach this temperature for the given power dissipation? I have not felt any significant rise in temperature when testing the prototype. \$\endgroup\$
    – wave.jaco
    Oct 13, 2015 at 5:35
  • \$\begingroup\$ Hard to say because it depends on thermal mass. You probably need to determine the length of a "heat soak" test yourself. If the PCB is likely to be heated up be external sources as well, you may need to wait for them to settle first. 625C/W seems a bit high, however reviewing my figures 200-250 C/W is not uncommon. But it depends on many things; how much copper there is on the board, 2vs4 layer boards, airflow, etc. Also notice that this is the internal junction temperature being calculated. The case will be cooler than that! So I'm not all surprised you didn't notice hot spots. \$\endgroup\$
    – Hans
    Oct 13, 2015 at 10:52
  • \$\begingroup\$ Thanks for the answer Hans. You mention that factors such as the amount of copper on the PCB that could have an influence of the heating-up of the transistor. Can one generally assume that the more copper there is on the board, the better the heat dissipation will be? Or is this not always the case? Also, how would the amount of layers affect the heat dissipation? \$\endgroup\$
    – wave.jaco
    Oct 14, 2015 at 6:44
  • \$\begingroup\$ From my understanding, copper on outer layers dissipate heat to the environment. So more copper = better dissipation. More copper on inner layers is often found to have better dissipation for power devices in many appnotes as well, probably because it can better transport the heat to a wide surface area, for again better cooling. However, if you really think you're pushing too much power through a SOT-23, then consider upgrading the device or package. Thermal design is often a guess work where you hope the worst-case scenerarios are "worst" enough. \$\endgroup\$
    – Hans
    Oct 14, 2015 at 8:06

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.