# 2N7000 behaviour explanation

I recently bought a 2N7000 transistor and though it was a cheaper replacement MOSFET compared to the STP55NE06 as I didn't need to handle high voltages and high currents. However it acted very differently. Usually with the STP55NE06, when I connect like this:

simulate this circuit – Schematic created using CircuitLab

The LED starts as OFF (assuming the MOSFET gate was previously set to ground).

If I connect LINK 1, the LED turns ON and when I disconnect LINK 1, it remains ON as the MOSFET gate is floating towards the positive. If I then connect LINK 2, the LED turns OFF and when I disconnect LINK 2 it remains OFF.

HOWEVER

If I replace it with the 2N7000:

simulate this circuit

This time the LED starts off being ON. If I connect LINK 1, nothing happens. If I disconnect LINK 2, nothing happens. Then when I connect LINK 2, the LED turns OFF. If I then disconnect LINK 2, the LED turns back ON.

It honestly feels like the 2N7000 behaviour is like the STP55NE06 but with a pull up resistor on the gate...

simulate this circuit

Also how come I can power the LED like this (albeit quite dimly):

simulate this circuit

Wait this means that my theory with the 2N7000 having a internal pull-up resistor could be true.

Anyways, can someone tell me why 2N7000 behaves so differently compared to the STP55NE06 despite both being Mosfets?

Edit: I measured the resistance between the GATE and DRAIN of the 2N7000 and I'm surprised to see a 6.5k resistor between it. The STP55NE06 had NONE! additionally the 2N7000 has A DIODE from SOURCE to GATE. The STP55NE06 also had NONE! It's like this:

simulate this circuit

Can someone confirm that there is a pull up resistor and a diode inside the 2N7000?

• What resistance does the LED series resistor have in your schematic? Comparing the datasheets, the 2N7000 has much higher Rdson of 5 ohms compared to STP55's 0.022 ohms. That might explain what you see. – Fizz Apr 28 '16 at 6:25
• Oh you can just ignore that. I know some people on this site are skeptical of not putting a resistor in front of any LED. Funny thing, the LED I use is 12V compatible because it has an internal resistor. – Bradman175 Apr 28 '16 at 6:27
• No, you can't ignore its value (relative to MOSFET Rdson) even if it's built-in the LED module. – Fizz Apr 28 '16 at 6:30
• I got another 2N7000 and that one worked perfectly fine. Either ESD magic or packaging error. – Bradman175 Apr 28 '16 at 7:41
• @Fizz is right: leaving the gate unconnected is not how you're supposed to use a MOSFET, though most of them will work as you expect. It's like removing the case from a BJT and using it as phototransistor: most will work, but characteristics are not guaranteed. – Dmitry Grigoryev Apr 28 '16 at 7:51

One thing you haven't considered is that the particular 2N7000 sample you're experimenting with may be faulty (you could easily damage it with an ESD for example). Or perhaps you have got an STP55NE06 with a far better gate insulation than the spec requires, and an 2N7000 which just barely meets the spec.

I suggest you repeat your experiments with a different 2N7000 sample. Chances are the weird behavior will go away.

• See my edit I made. – Bradman175 Apr 28 '16 at 7:17
• Honestly that looks like a dead MOSFET to me. – Dmitry Grigoryev Apr 28 '16 at 7:22
• Ok this is interesting, I have another 2N7000 and it works much more like expected. You are probably right Is it possible to get the wrong transistor and package it with the wrong name by accident? – Bradman175 Apr 28 '16 at 7:29
• @Fizz I'm sorry to hear you're a mere mortal and not a child of Thor like the rest of us on this blog. I have killed (and diagnosed and had to replace) at least 5 FETs while doing hobby projects at home without any ESD equipment. Not at once, of course, that happened over several years. – Dmitry Grigoryev Apr 28 '16 at 7:31
• @Fizz OK, I removed that statement from the answer, since it's not relevant after the last question edit anyway. Thanks. – Dmitry Grigoryev Apr 28 '16 at 7:43

Ok, so (to summarize what you said) the only difference is in the floating gate behavior. I guess the 2N7000 accumulates gate charge by itself much easier somehow. In general, MOSFETs with low Rdson (such as the STP55, 0.022 ohms) need much higher gate charge to turn on, whereas MOSFETs with higher Rdson (such as 2N7000, 5 ohms) need lower gate charge. This happens because MOSFETs with low Rdson are constructed by paralleling many more cells. The Fairchild datasheet for 2N7000 gives a ballpark figure for the (on) gate charge of 2N7000 of 4nC in Figure 10. The STP55 needs about 80 to 105nC gate charge.

I'm not enough of a MOSFET expert to tell you how the 2N7000 gets its floating gate charge in this case. So you basically need to make sure "link 2" (to ground) is connected to keep the 2N7000 off.

And regarding the accepted answer, according to Infineon (p. 6) soft ESD damage results in an increase in leakage current, not the other way around. The reason for this is that (soft) ESD damage is characterized by a deterioration of the (edges) of the gate oxide layer. So the 2N7000 that worked for you may be one that's actually more ESD damaged than the one that made you unhappy. However, I stated in a comment, since floating gate use is off-datasheet use (because there's no minimum or typical leakage current promised), manufacturing variation is actually enough of an explanation for why two 2N7000s behaved differently in your circuit with their gates floating.

And as I as I said in another comment, your finger may have enough charge to turn on a MOSFET and even leave it on for quite a while. The less charge a MOSFET needs, the easier is to do this and the lower its leakage current is, the longer it will retain gate charge. That demo is for a 5nC (typ.) to 30nC (max) gate charge MOSFET. Again there's no minimum leakage specified, so no max discharge time can be inferred. The datasheets for MOSFETs in general only seem to give a max leakage current, which puts a lower (but no upper) limit on the discharge time with the gate floating, i.e. they promise the gate doesn't lose charge by itself faster than a certain rate, which is what you care about in typical usage (i.e. with gate connected). Floating gate usage is not something that most MOSFET manufacturers care about.

• Well I need to use a MOSFET with a floating gate charge characteristic similar to the STP55NE06. I know I can't use the 2N7000 and your theory is to do with the RDS-ON. Then how about the fact I can power the LED from the gate? – Bradman175 Apr 28 '16 at 7:01
• @Bradman175: I'm not sure I understand your followup question. You probably can't use the 2N7000 with a floating gate in the way you want it. I don't know why you can't connect its gate to ground. Perhaps you should explain that. – Fizz Apr 28 '16 at 7:06