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Please be gentle as I am a total newbie with electronics.

I recently bought an Arduino beginners kit. In that kit was a 9V motor with the proper transistor (I think these are also called MOSFET?). The printing on the transistor reads IRF520N IOR P533B D5A2 (see picture left). I have started my own little project and bought a 12V pump and hooked it up just as I previously did with the motor from the kit and it worked perfectly (see picture). Since then I decided I wanted to buy more transistors of the same type and I found one which I thought was the same. It is labeled IRF520N IOR P702D AP9Q (see picture right) but when I use that transistor the pump doesn't start pumping.

Apparently the two transistors are not completely identical. I found the datasheet to the newly acquired transistor (listed on the page of the online store I bought it from) but I don't really understand anything it's trying to tell me. Also I couldn't find the datasheet for the one delivered with the Arduino.

So my questions are:

  • Why is that transistor not working?
  • Could you point out which values in the datasheet are important to me so I can pay attention to those when I look for another transistor?
  • What does the transistor actually do which I bought and can I still use it for something (I actually bought quite a lot of them..)?

If I missed any important information please ask in the comments and I will include it in my post. I included a schematic at the bottom.

Edit #1:

As joren vaes suggested I used the new transistors with the old motor and a 9V power supply. To my surprise that actually made the motor spin. So it looks like the new transistor can't handle the 12V?

Edit #2:

I did some experimentation with pump, 9V and older/newer transistor. Using the 9V batterie with the 12V pump works fine but only with the old transistor. With the new transistor the pump moves super slow and the transistor got worryingly hot. The old transistor doesn't heat up noticeably. As suggested by joren vaes i made another photo of the setup which shows the wiring properly and photos (first page, second page) of the instructions which came along in the kit. There isn't really any more valuable information about the transistors which came along if you were hoping for that. Also the Arduino supplies 5V and the code I uploaded just sets the transistor pin HIGH when the button is pressed. I know that I wouldn't need an Arduino for that ;)

Edit 3: Included schematic. I hope I didn't do anything wrong..

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Have you tried using the newly bought transistor with the original motor? Also, learning to read schematics is essential for doing electronics work. It's really simple - the only difficult part for a beginner will be understanding the symbols. Feel free to ask me more specific questions! \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:55
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    \$\begingroup\$ I would very much doubt that the transistors are very different. The transistor is a IRF520N - all of the other stuff that follows will have to do with package type, ROHS compliance, lotnumbers, ... and shouldn't influence the performance by any significent amount \$\endgroup\$ – Joren Vaes Apr 11 '17 at 11:58
  • \$\begingroup\$ I would not have thought about it, but I did try it and you are right! It works with the original motor and a 9V battery (which runs on 9V). Ill include that in the post. Also: thats what I thought. But the original transistor totally works with 12V and the pump but the new transistors dont (I tried 3 of them, so it can't be one which shipped broken) do the transistors actually only work with 9V and the one I got from the kit just has a higher tolerance? \$\endgroup\$ – Neuron Apr 11 '17 at 12:02
  • \$\begingroup\$ I can't believe that any of the two is. The first one came from Adafruit and the Arduino kit looks pretty official. The other comes from the biggest electronics store here in Germany and Austria (conrad.de). They also have physical stores and seem really professional in what they do \$\endgroup\$ – Neuron Apr 11 '17 at 12:04
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    \$\begingroup\$ Just as a note to you about Conrad: They do not practise any quality control or in-store stock tracking, so you cannot always be fully certain the device you got "offline" fits the datasheet, as batches may be mixed. I'm not even certain they keep track for the on-line supply. From way back when I remember quality of even LEDs being questionable. \$\endgroup\$ – Asmyldof Apr 11 '17 at 13:44
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I will not repeat RobhercKV5ROB's nice summery of the important specs.

I would just like to add that another very important bit of information in the datasheet is the pinout - in our case the Gate is pin 1, Drain is pin 2, Source is pin 3. The big metal tab is also connected to the Drain terminal, internally (so watch out when working with the transistor, since this metal piece is at the same voltage as part of your circuit!). This pinout is different for different parts and different manufacturers, so it's a good idea to verify this from the manufacturers' datasheet every time you are using a new part.

The IRF520N is rated for a \$V_{GS}\$ of +-20V. This means that you can apply up to 20V difference between pin 1 and 3 on the device - as you mention you were using a 9V battery or a 12V supply, neither of which come close to this maximum rating. This cannot be our problem.

I also checked your picture, and it seems like you wired the transistor up correctly. I would say, take some more pictures of the different setups so we can have a look at those, to try and find out what's going on here. Perhaps also point to the location of the tutorial you started out with, telling you how to wire up the mosfet in the original experiment.

What does the IRF520N actually do:

The IRF520N is (in full fancy words) an "Enhancement-mode n-channel metal-oxide-semiconductor field-effect transistor"*. Most people just write N-MOS (and pronounce it as "en-moss") or N-Channel MOSFET. I would usually link to the wikipage, but unfortunatly the one on MOSFETs is really quite complicated and I fear it will do more harm than good.

In very simple terms, as you already discovered, a MOSFET (just like any other transistor) is a electronic switch. The switch is located "between" the "Drain" and the "Source" terminals. The voltage difference between the "Gate" and the "Source" pin determines how hard this switch is turned on. If this difference is smaller than the "threshold voltage", found in the datasheet, the switch is "off". If we go above the threshold, the transistor is turns on. Take note that this is not a very "binary" behavior - it's not like a physical switch that's either all the way on or all the way off - it's a more gradual transition.

These are very usefull parts, and people use them all the time. In fact, I have a bag of almost 50 IRF540N's (a slightly bigger version of your transistor) laying around here in my lab...somewhere...

*Enhancement mode means that the transistor is "off" when no voltage is applied between gate and source, and turns on when this is increased. There also exist "depletion mode" transistors - here the transistor is already turned on at 0V! applying a positive voltage will turn it on even harder, and to turn it off, we need to apply a negative voltage. Depletionmode MOSFETs are very uncommon outside of specialized scenarios - so much so that most people in the last years of the EE bachelors won't even know they exist, but I felt it was worth mentioning them here, for completeness sake.

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  • \$\begingroup\$ Thanks a lot for the detailed post. I did include the things you were asking for. I also included some info about an experiment I did with pump and 9V. \$\endgroup\$ – Neuron Apr 11 '17 at 13:22
  • \$\begingroup\$ IIRC, aren't all JFETs depletion mode? I know depletion mode sounds a bit odd to those of us who have only used BJT & MOSFET enhancement-mode circuits, but if there are a whole category of FETs that are entirely depletion mode; hard to call it quite so uncommon. -- +1 for fixing my error in omitting the pinout from my answer. \$\endgroup\$ – Robherc KV5ROB Apr 11 '17 at 19:08
  • \$\begingroup\$ You are correct, I should have phrased this differently, and explicitly say that depletionmode MOSFETs are uncommon. As JFET's are always depletion, and almost all MOSFETs are enhancement mode. I know a lot of people that finished a bachelor's in EE and are halfway through their master's and haven't heard of depletion mode MOSFETs. - I'll edit my answer to clarify that it's depletionmode mosfets that are uncommon. \$\endgroup\$ – Joren Vaes Apr 11 '17 at 19:13
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  1. I agree with JorenVaes that for most purposes (yours included) an IRF520N "is an IRF520N". HOWEVER, there is a slight possibility that the one you bought has different pin assignments than the original, so you might have it wired wrong.

  2. On a datasheet, the values that you'll be looking at the most (for an N-channel MOSFET, like the IRF520Ns we're talking about here) are:

    • Id "Continuous Drain Current" -- how much current you can put through the transistor before things start melting or blowing up
    • Vds "drain-source voltage" how much voltage the transistor can effectively block. hitting it with more voltage than this will cause it to fail in one of 3 ways (let power through, but not enough to harm it {very rare}, melt {done it}, or explode {done it})
    • Rds(on) -- How much resistance the transistor causes when it's "turned on" ... this tells you how much voltage you're going to lose across the transistor per amount of current you ask it to pass, and how much power it's going to be dissipating (which heats it up)
    • Vgs(max) -- Same as Vds, except that it tells you how much voltage you can send to the gate pin (the one that controls the transistor)
    • Vgs(threshold) -- Tells you how much voltage you have to send to the gate before the transistor will start letting current flow through it.
    • Ptot "Power Dissipation" -- tells you how much total power (heat) the transistor can handle before it dies/melts/blows up (NOTE: this isn't how much power can "pass through" it, just how much power can be "lost" in it. i.e. it's the limit for current * voltage drop across the transistor.)
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  • \$\begingroup\$ Thanks for the reply! It was really helpful. But I couldn't figure out the problem. You got an upvote but I'll see if someone can answer all questions before I pick the "right" answer \$\endgroup\$ – Neuron Apr 11 '17 at 13:28

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