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When simulating the 2N7000 in CircuitLab, I was surprised to see that the current was lower than expected for my circuit (90mA at the drain) when the gate is at 3.3 V. So, I built a quick breadboard version and saw that the current was actually 98mA.

I suppose the ideal circuit would drive the gate hard on, but I'm wondering why CircuitLab is simulating around 10mA lower than my breadboard.

Is it perhaps because I've make a mistake in my schematic? Maybe circuit simulators aren't perfect and shouldn't be expected to replicate real world results? Do I have an 2N7000 made above manufacturing spec? Or something else?

Edit: I was hoping to get the spice model from the manufacturer. Turns out that for some reason I had purchased this particular part via eBay instead of a proper component supplier (probably to avoid the handling fee), so I have no idea who is the manufacturer because eBay sellers rarely/never include a datasheet.

For comparison, I added the other 2 from the top 3 choices of NMOS in CircuitLab. Curiously, the 3rd performs quite differently.

schematic

simulate this circuit – Schematic created using CircuitLab

DC sweep

DC sweep snip

Breadboard

2N7000

NMOS choices

Datasheet

Source: OnSemi NDS7002A datasheet (not actually the manufacturer of my part)

Edit: According to the datasheet from OnSemi, Vgs(th) seems to be at most 3 V, so I'd expect anything over that would be full on. V_TO (threshold voltage) in CircuitLab is 2.2 V.

Nick Bolton
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    What temperature were you running the simulation at? What temperature was the device on the breadboard? – Hearth Nov 05 '22 at 14:44
  • Did not know you could change the simulation temperature! How do I change that? A quick search didn't give me anything useful. On the breadboard, about 30°C (I have a radiator under my bench). Air temp is about 20°C. I didn't change the temp in CL so I guess it's whatever the default would be. – Nick Bolton Nov 05 '22 at 14:58
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    I have no idea how to do it in circuitlab. But my point was to point out that, even if you had a perfect model of your specific device, several characteristics are strongly dependent on temperature. – Hearth Nov 05 '22 at 14:59
  • Ah, I see your point. Good point. – Nick Bolton Nov 05 '22 at 14:59
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    You want your Mosfet to be turned in then why are you looking at the Vgs threshold voltage when it is almost turned off? It is fully turned on with a Vgs of 10V, turned on fairly well with a Vgs of 4.5V and guaranteed to conduct at least 1mA when its Vgs is 3V at 25 degrees C. – Audioguru Nov 06 '22 at 01:41
  • Regardless of temperature, etc, real devices have variations and datasheets are typical. – copper.hat Nov 07 '22 at 04:01

3 Answers3

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Why do you think there is a problem? The simulator uses some model it is programmed to, and the physical component has manufacturing tolerances.

The 2N7000 is barely useful at 3.3V gate voltage to begin with as it may not properly turn on. So don't expect it to work properly as a switch.

Justme
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One of the basic principles of circuit design with semiconductors is their properties vary. Their properties vary from part to part, their properties vary from manufacturer to manufacturer, their properties vary from lot to lot, and their properties vary from temperature to temperature.

Part of learning circuit design is getting a feel for what properties you can trust and what properties you can't. In the case of a FET's threshold voltage (which is what determines its behavior when it's not getting enough gate voltage to go into the \$\mathrm{R_{DS_{on}}}\$ region), the variation can be quite high.

If you were operating a FET in the saturated region (i.e. the constant current, with a healthy drain to source voltage), and you told me that you had two FETs whose drain current was within 50% of each other I'd say "huh, pretty close match".

You didn't say what your goal was with that circuit, but if the intent is to pull the drain to ground and put 100mA through the resistor, while using an I/O that gives 3.3V out, then you need to either buy a FET that is _specifically designed to fully turn on at 3.3V \$\mathrm{V_{GS}}\$, or you need to use a bipolar transistor (and remember that you need a resistor between input and base).

TimWescott
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  • Yeah, I know, but I'm still a little confused because of how CircuitLab is simulating it. The threshold is 2.2 V for the spice model in CircuitLab. Is that because this is the start of the threshold or the maximum threshold? – Nick Bolton Nov 05 '22 at 15:36
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    @NickBolton The threshold voltage is a single, well-defined point (with some temperature dependence), not a range; the only range is the range of values that individual devices will fall within. Most SPICE models use typical values that you can generally expect most individual FETs to be close to. Note also that the threshold voltage is only the voltage where the FET *barely* begins to conduct; to turn a FET on you need to drive the gate to several volts higher than the threshold voltage. Exactly how much higher depends on another device characteristic, the FET's transconductance. – Hearth Nov 05 '22 at 15:49
  • Got it. If not Vgs(th), what value or values do I look for in the datasheet to meet this criteria? > "specifically designed to fully turn on at 3.3V VGS" – Nick Bolton Nov 05 '22 at 15:50
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    You look for a part that has an \$\mathrm{R_{DS_{ON}}}\$ rating at a \$\mathrm{V_{GS}}\$ of 3.3V. At root, a datasheet is a promise, from the manufacturer to you, that a part will meet specifications. If they don't specify it in their datasheet, then there's no promise. – TimWescott Nov 05 '22 at 16:18
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According to the datasheet from OnSemi, Vgs(th) seems to be at most 3 V, so I'd expect anything over that would be full on. V_TO (threshold voltage) in CircuitLab is 2.2 V.

3 volts is the threshold of slight conduction (usually about 1 mA): -

enter image description here

If you want to know when the MOSFET is "full on" look at the graphs in the data sheet (same as NDS7002) and, bear in mind these are typical graphs only. There will be considerable variation from one device to another: -

enter image description here

So, "full-on" is when the gate-source voltage is 10 volts but this only means that if your drain current is 1 amp, the drain-source voltage drop is going to be about 1.2 volts. Hardly what some folk would call "full-on". However, if you were only interested in what the volt drop of the device was at 100 mA then, drain-source would drop about 0.1 volts.

I built a quick breadboard version and saw that the current was actually 98mA.

That doesn't surprise me and, the MOSFET would have about 0.9 volts across it because of the 100 Ω resistor you are using from the 10 volt supply to the drain. The MOSFET is in the saturation region and dissipating 90 mW. Did you notice it might be getting a little warm? Not the best region to be in for using the MOSFET as a switch; you need more gate voltage or, find a MOSFET with a much lower \$V_{GS(TH)}\$.

Andy aka
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    The resistor on the breadboard is 100 Ω. Same as the schematic. But yes the resistor was hot as expected (but thankfully not on fire). – Nick Bolton Nov 05 '22 at 16:12
  • I don't remember the FET being too hot to touch. But it probably was because it can handle 200mA, and I was passing 100mA through it without it being full on. – Nick Bolton Nov 06 '22 at 07:31
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    @NickBolton for future reference: the 200 mA quoted in the datasheet is often a value with an additional heatsink, you have to read the datasheet very closely when you try to get close to the maximum current ratings of components. Not sure for this one, but just a reminder. – Arsenal Nov 07 '22 at 13:08