Context: I'm studying basic electronics at an undergraduate level.

One of the questions in past exams was to explain the usefulness of the following MOSFET wiring. Judging from the picture I assume that the question refers to the fact that Vo is wired back to the gate of the MOSFET.

enter image description here

I skimmed through my book and the closest thing I could find was under the paragraph "reference voltages and current sources", but it hardly mentioned anything. Would anyone be so kind as to explain things a little bit ? What's so special about the above circuit and what exactly does a reference circuit do ?

I googled "reference circuit" and got the following from Wikipedia:

"A reference circuit is a hypothetical electric circuit of specified equivalent length and configuration, and having a defined transmission characteristic or characteristics, used primarily as a reference for measuring the performance of other, i.e., real, circuits or as a guide for planning and engineering of circuits and networks".

Is this relevant to my case ?

I tried to solve the circuit given \$K=1.5mA/V^2, V_T=2V, R=10KΩ\$. I came up with the following two equations:

\$I_D = K(V_{GS}-V_T)^2=K(V_G-V_S-V_T)^2=K(V_0-V_T)^2\$, since the gate and Vo are short-circuited. The second equation is \$V_o = 20 - I_D R=20-10I_D\$.

Therefore combining the two equations, I get: \$(20-V_o)/10=1.5(V_o-2)^2\$. From this equation I get \$V_0=\{0.87, 3.03\}\$, accepting the \$V_o = 3.03\$ so that \$V_{GS} = V_G = V_o = 3.03 > V_T = 2\$.

A colleague of mine though calculated \$V_o=9V\$. Does anyone see where I'm mistaken ?

Another colleague pointed me at this link [pdf] with a MOSFET diode-connected circuit whose description paragraph sheds some light.

  • \$\begingroup\$ Can you not put "EDIT FOO" in your question? The edit history is at the bottom of your post, at the link that currently says "edited 5 minutes ago". There's no need to put that information again in the question: that just makes it harder to read for people like me who are reading it for the first time. \$\endgroup\$ – Phil Frost Jan 21 '14 at 16:59
  • \$\begingroup\$ Ok... is it customary at this SE site? In others the exact opposite is required, because if you heavily edit your question, you make the answers look somewhat irrelevant or deficient. \$\endgroup\$ – stathisk Jan 21 '14 at 17:02
  • \$\begingroup\$ hopefully you edit it to make it more clear, not to make it a different question, and it was coherent enough to start that good answers were possible. If good answers aren't possible, then the question should be closed, then edited, then reopened. \$\endgroup\$ – Phil Frost Jan 21 '14 at 17:03

Consider the equivalent BJT circuit, which may be more familiar:


simulate this circuit – Schematic created using CircuitLab

This holds provided the input voltage is >0.65V, with of course some variation based on temperature, output current, manufacturing variation, etc. However, as a first approximation this circuit outputs a constant 0.65V.

The MOSFET circuit is no different, but instead of the 0.65V from a forward-biased silicon PN junction, we get the threshold voltage of the MOSFET. This parameter varies between models of MOSFETs, but is usually some volts. If the output voltage, which is also the gate voltage, is above the threshold voltage, the MOSFET turns on more, shunting more current to ground, increasing the current through the resistor, lowering the ouput/gate voltage such that an equilibrium is reached:

$$ V_{GS} = V_{out} \approx V_{GS(th)} $$

This sort of circuit would be useful as a reference voltage, for example, to implement a voltage regulator, because the output voltage is relatively unaffected by the input voltage. A single transistor such as this isn't necessarily a good voltage regulator on its own, but it could be the basis for something better. A good regulator starts with a reference such as this which might vary based on other parameters (output current, supply voltage, temperature), then isolating or compensating for those parameters from the reference.

  • \$\begingroup\$ Ok, I got the logic behind the circuit (it helped to study again how the MOSFET functions internally). Also, since \$V_{out} \approx V_{th}\$ (I assume that \$V_{GS(th) \equiv V_{T}=2V}\$) isn't 3.03V a more physically compatible value than 9V ? \$\endgroup\$ – stathisk Jan 21 '14 at 19:12
  • 1
    \$\begingroup\$ @Zet If your friend did not ask for the threshold voltage of the MOSFET, he could not possibly be right. Unlike BJTs, where the 0.65V is determined by the nature of all silicon PN junctions, it's possible to construct MOSFETs with a wide range of threshold voltages. However, 3V is probably more common than 9V, though either is possible. Even within specimens of the same model, there may be significant variance. For example, 2N7000 lists 0.8V (min), 2.1V (typ), 3V (max). \$\endgroup\$ – Phil Frost Jan 21 '14 at 19:17

I've never seen the circuit before, but at a first-pass guess it looks like a sort of voltage reference. In a typical resistive divider (replace the FET with a resistor), if the input voltage or the resistance change over time, the output of the divider also changes. In this case, though, the properties of the FET fight that change. It's negative feedback.

So say we're in a steady state. Some current flows through the FET, resulting in a voltage across the FET. That voltage is also on the gate of the FET, causing the FET to turn "on" to some degree. If the FET were all the way on, the voltage across it would be very low, so it would turn itself off again. If it were all the way off, the voltage on the gate would be the full 20V, so it would turn itself on. We're somewhere in between, in an analog mode. The FET goes to some voltage, depending entirely on the transfer characteristics of the FET.

So there's some amount of current flowing, depending on the impedance of the FET in that state. Say something changes to force more current to flow; the resistor warms up and its resistance goes down, or the voltage source rises. More current flows through the FET drain-source. The voltage across the FET drain-source rises. But that means the voltage on the FET gate-source also rises, meaning the impedance of the FET goes down. So more current flows through the FET, but through a lower impedance. So the output voltage won't rise as much as it would if the FET were a resistor.

The inverse applies if something causes less current to flow through the FET. This could be the voltage source falling, the resistance increasing, or a load being applied to the output terminal! If you have a resistive divider, adding a load changes the output voltage. This way, it doesn't, so long as the load isn't so great it eats all the available current from the resistor.

So it's not just a voltage reference, it's a voltage regulator.

  • 3
    \$\begingroup\$ BUT a regulator whose voltage is dependant on Vgsth - the gate to source threshold voltage - the voltage at which the device JUST turns on with a predefined Ids (usually low or very low). Vgsth (or just Vth) may vary by around 2:1 in some devices. Actual regulated voltage will be somewhat higher that Vgsth as FET has to turn on harder to lower Vd to the gate voltage that will support I = (Vdd-Vd)/10k. \$\endgroup\$ – Russell McMahon Jan 21 '14 at 9:20

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