# Will this MOSFET switch circuit work?

I should point out that the LED I'm trying to switch is actually an LED strip with a one meter length and a total of 60 LEDs. Total power consumption is 4.8W at 12V. By the way, 'IC1' is the 7805, which converts the 12V DC into 5V DC.

This is the MOSFET I plan on using: http://panda-bg.com/datasheet/860-064009-Transistor-IRF530-ST.pdf (IRF530-ST)

I'm new to the field of electronics so I am not quite sure if it will work so any kind of feedback is very much appreciated :)

Correction to the schematic: your IRF530 is installed backwards. You don't want the body diode "pointed" in the same direction as your LED string. The N-FET source terminal should be connected to the negative terminal of your 12V supply.

While using a LMx24 as a comparator isn't the best, since its input common mode range and its output range includes the negative/ground supply and the threshold voltage of the IRF530 is at most 4V, this circuit (with the corrected FET direction) will mostly work if you use a LM7805 or higher voltage (accounting for dropout, 5 to 9V should be a reasonable choice of 78xx). From the comments: remember that the IRF530 has its on resistance rated at a $V_{GS}$ of 10V. Acceptable performance may be found for this application at lower gate voltage, but choosing a different FET with a "logic level" gate drive capability would help if a 7805 is used.

However, your op-amp (being used as a comparator) has no hysteresis. This means that at certain light levels, you will have a flickering LED string because your threshold potentiometer and the LDR/resistor divider have almost the same outputs. For some good information on why you want hysteresis and how to implement it, check out TI's TIDU020A document. The difference in their circuits is:

• Note 5V won't turn the MOSFET fully on, its on resistance is specified at Vgs=10V. (The threshold, 2-4V, is where it conducts 0.25 mA, not very bright). You could happily run the whole thing off 12V if it's a stable supply, or a 9V regulator should be good enough otherwise. For hysteresis, all you need is a resistor from opamp VOUT to IN+, much higher (say 20-50x) than the potentiometer R3 - say 220K or 470K. Aug 7, 2016 at 15:46
• @BrianDrummond But if I switch directly to the 12V supply, would the LM124-324 handle the 12 Volts? Aug 7, 2016 at 15:51
• Check its datasheet... Aug 7, 2016 at 16:03

• The IRF530 might have a threshold voltage as high as 4 V. If you drive it with 5 V, that's only one volt above threshold and it won't be very strongly on. You will not achieve the low Rds(on) specified in the datasheet (that value is specified with a Vgs of 10 V).

This could lead to excessive heating in the MOSFET.

I would recommend using a comparator chip rather than an op-amp. If you choose one with an open-collector output (which is common) and appropriate ratings for maximum output voltage (12 V), and provide a pull-up to 12 V, the FET gate can be pulled to 12 V when "on".

Or choose a FET with a guaranteed Rds(on) at 5 V Vgs. The IRF530 is really overkill for your design since it is specified for 14 A and you are only switching about 400 mA (12 V / 4.8 V).

• Leaving the three unused op-amps in the '324 package unconnected could lead to oscillations and affect the behavior of the op-amp you are using, or overheat the '324 chip. Typically, unused devices in op-amp packages can be connected as voltage followers (with the outputs not connected to anything) to avoid this issue.

• I understand that this particular MOSFET is a bit overkill but I plan on adding more LED strips to the circuit in the future so I don't really feel the need to change it. It just makes it a lot more future proof I guess :D On a side note can you please elaborate on how to connect the 3 other op amps as voltage followers?? :) Aug 7, 2016 at 15:56
• Google "op amp voltage follower" and look at the image results. Aug 7, 2016 at 16:04
• If you're going to stick with the IRF530 then find a way to drive the gate to 10 V for on. Aug 7, 2016 at 16:06
• Good note on the unused op-amps. I was thinking about that, then forgot to mention it. If possible, connect the voltage follower to the middle of a voltage divider as best-practice. Aug 7, 2016 at 16:22
• @The Photon I got it. Thank you very much for your feedback dude and I'm sorry if I wasted your time with my thoughtlessness. Have a very nice day! :) Aug 7, 2016 at 16:25