IRLZ44N has VGS between 1V and 2V.

That’s another one I’ve seen suggested a lot. That could work, voltage divider off the arduino or battery.

 

AO3400:


They're surface mount, but you can mount them on through hole adapters. The adapters are two sided, so you can mount a gate pull up/down resistor on the other side.

The adapters are around a dime each from Ali Express. The leadframes also run about a dime per pin; when you can find them. You could also use 0.025" square male header pins. Right angle would probably be more convenient than the straight ones.

Make sure you teach your students how to handle devices properly. Most MOSFETs don't have any protection. 2N7000 are very easy to damage.

 

That’s another one I’ve seen suggested a lot. That could work, voltage divider off the arduino or battery.

Why do you need a voltage divider?
VGS(th) of 2V max means that the MOSFET will turn on with a gate voltage greater than 2V. It will work fine with 3.3V.

 

The irlz44 has a range of 1-2v, voltage divider for that.

 

The irlz44 has a range of 1-2v, voltage divider for that.

No. You are misinterpreting that data sheet.

VGS(th) min 1V means some devices will turn on with VGS at 1V (or it could mean all devices will turn off with VGS below 1V).
VGS(th) max 2V means all devices will turn on with VGS at 2V.

Thus, all devices will turn on with VGS at 3.3V.

VGS absolute min and max is -16V to +16V.

 

The irlz44 has a range of 1-2v, voltage divider for that.

Unless the voltage used to drive the gate is larger than the maximum allowed, you don't need to use a voltage divider. Driving the gate with a higher voltage will give you a lower on resistance.

 

Thanks for clearing that up.

 

So handling MOSFETs. I really just want a general purpose transistor, probably a FET, maybe a JFET would be better for kids. I like the bigger TO-220 package, seems easier to handle.

 

So handling MOSFETs. I really just want a general purpose transistor, probably a FET, maybe a JFET would be better for kids.

All JFETs are depletion mode devices. MOSFETs come in depletion or enhancement mode, but enhancement mode are much more common. So common that enhancement mode is assumed.

You're not going to find power JFETs to be readily available.

I like the bigger TO-220 package, seems easier to handle.

SOT-23 aren't difficult to handle if you put them on SIP adapters. But they do require ESD considerations.

Integrated circuits on Arduino boards are mainly CMOS, so they can also be damaged from abuse.

The leads on TO-220 devices are too large for solderless breadboards. You can force them in, but doing that can damage the clips. I've been twisting them 90 degrees to avoid abusing breadboards.

 

All these mosfets we’ve looked at are usually 50-100v. I’m assuming I couldn’t use it as a switch for something that runs off 110vac?

 

All these mosfets we’ve looked at are usually 50-100v. I’m assuming I couldn’t use it as a switch for something that runs off 110vac?

1) You use a thyristor (SCR, triac, or quadrac) to switch AC.
2) For classroom experiments you need to stay away from 110VAC. It is way too dangerous.

 

Never hear of a thyristor. The tube amp I built has a MOSFET. it’s directly after the input transformer and rectifier. Looks like it’s controlling the VRM (voltage regulation module), so must be acting like an amplifier to raise or lower the voltage and thus the volume.

 

So handling MOSFETs. I really just want a general purpose transistor, probably a FET, maybe a JFET would be better for kids. I like the bigger TO-220 package, seems easier to handle.

For an introductory course in electronics you don’t need to complicate things. There is no need to introduce FET and MOSFET.

Even PNP and NPN is complicated enough. Stick with NPN BJT.

 

All these mosfets we’ve looked at are usually 50-100v. I’m assuming I couldn’t use it as a switch for something that runs off 110vac?

The tube amp I built has a MOSFET. it’s directly after the input transformer and rectifier. Looks like it’s controlling the VRM (voltage regulation module), so must be acting like an amplifier to raise or lower the voltage and thus the volume.

Amplifying AC is different than trying to switch it.

You can't use a BJT to do that either.

 

Ok npn bjt, graphic I found on electronics.stackexvhange.com

R1 should probably be higher, like 1k or 3k, etc. 470ohm gives 20mA!!!

But I was looking at the 2N3904 data sheet. It says emitter-base breakdown voltage is 6v. Does that mean I can’t put a 9v across it?

R2 then allows about 1mA to the base. It says the DC current gain at 1mA is 70? I’m not sure what I’m looking at.

No maybe I just need the base current above 50nA for it to turn on.

 

Let's look at some fundamentals.

Ignore D1 and Q1 for a moment.
The maximum current through R1 is
V1 / R1 = 9V / 470Ω = 19mA
Hence you will never get more than 19mA flowing through D1 and Q1.

The maximum current through R2 is
V1 / R2 = 9V / 10kΩ = 0.9mA
You will never get more than 0.9mA flowing through the base-emitter junction of Q1.
The emitter-base breakdown voltage is 6V. Note that this is emitter-base and not base-emitter.
This is reverse voltage applied to the base-emitter junction.

The base-emitter junction is a P-N junction. It will start to conduct at about 0.6V
Beyond 2V you would have destroyed the transistor.
With R2 limiting the current to 0.9mA, the base-emitter voltage will not exceed about 0.75V.

In this application, Q1 is acting as a switch, pulling the cathode of D1 towards ground.
We want the resistance of Q1 to be as low as possible, i.e. Q1 is in saturation mode.
For this, we derate the current gain of Q1 to 10 (instead of 70).

With D1 in place, the max collector current is now,
(9V - 2V)/ 470Ω = 15mA

With a current gain of 10, we want a base current of at least 15mA/10 = 1.5mA
We are only supplying 0.9mA. Hence the value of R2 = 10kΩ is too high.
Change R2 to 4.7kΩ and now we can supply close to 2mA at the base.

 

The base-emitter junction is a P-N junction. It will start to conduct at about 0.6V
Beyond 2V you would have destroyed the transistor.
With R2 limiting the current to 0.9mA, the base-emitter voltage will not exceed about 0.75V.

Ok I get most of that, but please explain where you got the 0.6v and 0.75v.

 

Ok I get most of that, but please explain where you got the 0.6v and 0.75v.

A transistor starts turning on before the base-emitter voltage is 0.7V. That's usually not a consideration because you'll be more concerned with a steady state condition (the calculations for operating points is often iterative, but you can skip the initial turn on condition unless that's where you want the steady state operating point to be).

The 0.75V is not data based. You can see from the datasheet information you posted that the manufacturer is stating that it could be as high as 0.95V at an emitter current of 55mA.

You should also note that \( V_{EBO} \) is specified as 6V for 2N3904. If sufficient reverse current flows, current gain (beta) for the transistor will be permanently reduced.

 

Here’s part of what you were talking about. Vbe(sat) goes from .65-.8 at 20mA, which is max current through collector as we calculated

 

This is complicated.
Once the switch is ON, R1 and Q1 have a voltage drop based on Ic. And D1 has a voltage drop based on its color, 2-3v. And then there’s the whole gain factor, also based on Ic, which goes back to the base current, voltages.
This is probably beyond the class level and my ability to teach it. Although we should be able to simulate it in Eagle spice.