I intend to use the mosfet module shown for switching large LEDs. Do I need to ground the negative of the DC in for the load?
Thank you in advance.
- Show us a schematic of your module so we can answer effectively.
If you don’t have one, draw it from the foil trace connections.
Yes. Or, put another way, you need a common ground between the Arduino and the DC power supply for the LEDs.
Please post a link to the spec of these LEDs. If, by "large", you mean high power, this type of LED needs a constant-current power supply, and you can't use a MOSFET to dim them.
Thanks for your reply, just need to switch them on and off, no dimming required. I will ensure that I have a common ground !
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Ok, but they aren't the type that require a constant-current power supply, right?
No it's not necessary
If yours is like the one I looked at the - input is connected to the power ground. There is no opto isolator to eliminate this connection. Just connect your load to the output section power and ground and your power supply on the input. Look at this schematic. Connecting the grounds will just turn it on and keep it on.
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Many thanks for that, I'm assuming that as I am supplying LEDs and no inductive load, that the Flyback diode is not required?
The LEDs will be supplied from a LiPo battery.
Thanks for your reply!
You are welcome.
That should be just fine. Be sure you get at least 3V from the liPo as less then that may damage the MOSFETs. If you can read them what is the number of the MOSFET.
- Move the top of R2 to the left side of R3.
YES!
You need to connect to the DC- and OUT-, because this is where the actual switching takes place. This is a so called "low side" switch. You do not need to connect the DC+ and OUT+. You can connect the positive power directly to your LED! If you check the bottom side of the PCB you will notice that the DC+ and OUT+ are connected straight through. The negative DC in is connected to signal ground on the module.
NOTE!!! You need to have an input signal of more than 8V, preferably at least 10V with the MOSFETs on that board, or you will have a large RDSon, resistance in the MOSFETs, and they will consume power and get really, too, hot.
The MOSFETs will start conducting with a 3.3V input signal, but in practice they will not start working like a switch until the input signal is above 8 volt. At lower VGS they will be more like a current source/limiter, for which they where not built and therefore may not work well at all.
Well, just a small warning there.
EDIT: Depending on the exact type of MOSFETs on your module, the signal voltage required for proper operation may vary, but certainly not below 4 volts and preferably not below 5, at least not for large loads.
I think you are looking at the wrong MOSFET
The flyback diodes are internal to the MOSFETs, they're not something you can remove or opt-out.
The intrinsic diode is not a flyback diode.
There may very well be many different types of MOSFETs on generic boards like these, so speaking out of own experience and from what I could tell from the photo this was the same. In any case 3.3V is a very low VGS for most power MOSFETs and is likely to cause excessive loss and heat dissipation in the MOSFET.
Correct, the MOSFET diod isn't actually a flyback diode, even if often called so. I didn't notice there was a dashed flyback diode added up in the corner of the schematic and assumed the ones in the MOSFETs was referred to.
If the load isn't inductive, the flyback diode shouldn't be needed.
Not what I asked at all... I was asking about the LEDs.
It isn't? Please explain.
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The only ground in @SierraGolfMike's diagram is GND and since DC- it's already connected to GND it's not necessary.
I must have missed that somewhere. I still don't see it. What makes me think that might not be true is the original question: