I am designing a DC solid-state relay using an FQP30N06L (a MOSFET with 32 A, 60 V, Rds(on) = 0.035 Ω @Vgs = 10 V) to control using a Raspberry Pi.

I am trying to control an electric door lock with 6-12 V 1200 mA; the nominal resistance at 12 V is 9 Ω. I have seen many articles on the Internet, but all are confusing.

So, as per my specification:

The power dissipated by the MOSFET is P = I2 X R; I is the load current of electric door = 1.2 A; R is the Rds(on) of the MOSFET = 0.035 Ω.

This gives me a power dissipation of P = 1.2 A X 1.2 A X 0.035 Ω = 0.0504 W.


In the datasheet you've linked, there is a table labelled Thermal Characteristics:

enter image description here

The value that is relevant is \$\mathrm{R}_{\Theta\mathrm{JA}}\$ - this is the thermal resistance between the junction (i.e., the active semiconductor region) and the surroundings with no heatsink. You can calculate (to a first approximation) the temperature rise using this value:

\$T_\mathrm{J} = T_\mathrm{A} + \mathrm{R}_{\Theta\mathrm{JA}}\times P\$

where \$T_\mathrm{A}\$ is the ambient temperature and \$P\$ is the power you've calculated. Using these figures, assuming \$T_\mathrm{A} = 25~^\circ \mathrm{C}\$, gives you a temperature of ~\$28~~^\circ \mathrm{C}\$. This means you are unlikely to need a heatsink for this application.

  • \$\begingroup\$ so so have you comapared this value ~28 ∘C with the 125°C which is often assumed as a maximum temperature for silicon, so we're safe. is that so? \$\endgroup\$
    – Sabha
    Nov 19 '21 at 10:38
  • \$\begingroup\$ You will barely feel any heat coming off this at 28 C - so no need for a heatsink. 125 is high, and look at Spehro's answer to see how raising the junction temperature can feedback causing further temperature rises. \$\endgroup\$
    – awjlogan
    Nov 19 '21 at 10:41

If you drive the MOSFET gate with 10V you will not need a heatsink. Or even 5V. Power dissipation is less than 0.1W when conducting, and even if you switch it a bit slowly the spikes in power dissipation during switching won't add up to much for infrequent switching. It will barely feel warm. At around 0.5W or 0.6W you need to start worrying about it (for a TO-220 case), especially if the environment might be hot (the Rds(on) goes up as the die gets hotter too).

However, if you attempt to drive this particular part, which is not rated for 3.3V drive, directly with the Raspberry Pi 3.3V output or if you omit the flyback diode from the coil you may well burn it out. You're probably not doing that, but better to mention it.

  • \$\begingroup\$ Thank you @Spehro Pefhany \$\endgroup\$
    – Sabha
    Nov 19 '21 at 10:43
  • \$\begingroup\$ From raspberry pi, I am using PC817 optocoupler and then the optocoupler emitter is connected to gate terminal of MOSFET. in between raspberry pi and optocoupler I have added a 220 k ohm resister and LED. I have also added a 1N4001 flyback diode from the coil. 12V supply is connected to optocoupler collector, and between optocoupler collector and 12V supply their is 2.kohm resistor and led. Also I have added a pull down resitor of 10K ohm at the MOSFET gate to ground. \$\endgroup\$
    – Sabha
    Nov 22 '21 at 8:16

for both calculations & where is it usefull or not, depends on each datasheet of each component you choose, then the type oh heatsink depends on both datasheet indications, & component hardware material

If you have to ask yourself about that need consider you must check for those informations on datasheets each time you are over 0.75A in one standalone componant as a minimum.

In general you know you need one from the component form factor also, it is made for that so it need it by default so calculate its size from datasheet informations

in your use case you say you have a quite low 0.05A but this mosfet is build for using a heatsink so use one by default a small one will be enough mayb just a metal plate behind it with a small size.


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