Will IR3205 n MOSFET safely switch 32v DC

Hi All

I want to build a simple circuit that will involve using the Arduino to switch power on/off to a strip of LED lights powered by 32V DC supply from the mains via a digital out and an IR3205 n channel MOSFET.

I see Vdss is 55V, which suggests to me that using this MOSFET’s gate driven from the Arduino to switch 32V DC is within its limits. However, I also read somewhere recently that one should always use a MOSFET that is rated at twice the Vs i.e. in this case, 64V.

Is this the case? Or am I safe to use my IR3205 in this circuit?

Thanks all.

Did you mean IRF3205? www.irf.com/product-info/datasheets/data/irf3205.pdf If so, everything indicates you will need 10V on the Gate to turn it fully on into its very low resistance region.

CrossRoads: Did you mean IRF3205? www.irf.com/product-info/datasheets/data/irf3205.pdf If so, everything indicates you will need 10V on the Gate to turn it fully on into its very low resistance region.

Ah, sorry yes, my typo - I meant the IRF3205, thanks. And bummer... I made a mistake in not checking the Vg value. So, lo0oks like I now have to use some form of transistor device to up the 5V to 10V at the gate of the IRF3205. Drat!

HI,

This is a very high current Fet for what I suspect is fairly small current (correct me if I'm wrong). You will probably get by if you have a 5V arduino. Would not suggest this for a production design.

The 55V spec will be find, especially for an IR device. They are very tough and have voltage clamping built in. Again for a production design the 2 X Vcc rating is likely a good place to be.

If you have some time, eBay has 5 x IRLZ44N for $1.57 This would be a better choice as it is a logic level FET and you can draw serious amps even when driven from an arduino.

Good luck

JohnRob

JohnRob:
HI,

This is a very high current Fet for what I suspect is fairly small current (correct me if I’m wrong). You will probably get by if you have a 5V arduino. Would not suggest this for a production design.

The 55V spec will be find, especially for an IR device. They are very tough and have voltage clamping built in. Again for a production design the 2 X Vcc rating is likely a good place to be.

If you have some time, eBay has 5 x IRLZ44N for $1.57 This would be a better choice as it is a logic level FET and you can draw serious amps even when driven from an arduino.

Good luck

JohnRob

Thanks John!
I will certainly buy some IRLZ44N’s as you suggest - seem to be a much better option.
One quick qn if I may please - will the 5V from the Arduino digital pin actually be enough to switch the IRF3205 at all?

theMusicMan: One quick qn if I may please - will the 5V from the Arduino digital pin actually be enough to switch the IRF3205 at all?

It will, but that's in its linear region, so it'll likely get hot fairly quickly.

Cheers! Dirk

It most likely will. You could likely get away with it especially if there is no other load on the drive pin.

Like most specifications, the actual part is somewhere in mid range. You have a good chance of getting 5A or so. The risk is the MOSFET may not be fully "ON". Meaning you would like the drain end up near 0 volts. The good news is the MOSFET will get warm under these conditions and try to turn on more.

If you are buying FETs through eBay (China) you might look at a 2N7000 just to have on hand. They are a low current FET good for direct connection to the arduino and useful in a variety of places.

Good Luck.

Thanks again, John.

Can I ask now please, what item in a mosfet data sheet do I look for to determine if it is logic level? Is it Gate threshold? I ask, because the gate threshold I read for the 2N7000 is 3v max! Doesn't that mean the Arduino, providing 5v on digital pins, would cause the 2N7000 to burn out?

I obviously need a little guidance on reading and understanding datasheets

theMusicMan:
Can I ask now please, what item in a mosfet data sheet do I look for to determine if it is logic level?

Look at RDS(on) and the VGS associated with it. Logic-level MOSFETs will generally have a VGS of 5V specified for RDS(on). (It may have more than one VGS, but the RDS(on) value will likely be close to that of the one for 5V.)

theMusicMan:
Is it Gate threshold?

No. VGS(th) is the level (“threshold”) at which the MOSFET starts turning on (off for p-channel).

Cheers!
Dirk

dephwyggl:
Look at RDS(on) and the VGS associated with it. Logic-level MOSFETs will generally have a VGS of 5V specified for RDS(on). (It may have more than one VGS, but the RDS(on) value will likely be close to that of the one for 5V.)

No. VGS(th) is the level (“threshold”) at which the MOSFET starts turning on (off for p-channel).

Cheers!
Dirk

Great information, Dirk. Thank you sir :slight_smile:

The voltage rating overhead depends rather on the load - purely resistive or passive loads like heaters bulbs and LEDs are less of an issue, but stray inductance (such as from long wires which act as transmission lines) can lead to ringing, which can approch twice the supply in bad cases. Faster switching makes things worse as less inductance is needed to induce ringing.

Real inductive loads are worse and require some protection circuitry to control.

JohnRob: It most likely will. You could likely get away with it especially if there is no other load on the drive pin.

Like most specifications, the actual part is somewhere in mid range. You have a good chance of getting 5A or so. The risk is the MOSFET may not be fully "ON". Meaning you would like the drain end up near 0 volts. The good news is the MOSFET will get warm under these conditions and try to turn on more.

If you are buying FETs through eBay (China) you might look at a 2N7000 just to have on hand. They are a low current FET good for direct connection to the arduino and useful in a variety of places.

Good Luck.

Wrong. When a MOSFET warms up its junction resistance increases and will slowly turn off as the part gets hotter and hotter and the junction resistance increases more and more until it fails unless of course you have attached it to a heatsink.