# Source potential for MOSFET with grounded substrate

Considering the configuration below, the source's potential should be invariant with respect to Vg (as long as Vg>Vt). So Vs is equal to Vdd. But if the substrate is connected to the source, as shown here, when Vs reaches Vg-Vt, the transistor switches off, and Vs remains at Vg-Vt. Is this correct? I am asking because I read in a material that in the first situation Vs depends on Vg.

EDIT (quote from a lab guide): If the M1 transistor is in cut-off stage, for the case in figure 4.2b, the output voltage will depend on VGG and the threshold voltage, following a relation that contains also elements (parameters) specific to the intrinsic structure of the basis substrate.

• Can you post the material's name or link? Nov 17 '14 at 12:52
• I posted a quote from the material which is actually a lab "guide". Am I right? Thank you. Nov 17 '14 at 14:04

In your first circuit, there exist "body effect". But if you can omit that effect $$v_{GS} = v_{DS} = V_{dd}-V_{s}\\ v_{DS} > v_{GS} - V_{TN}$$

It should work in saturation region. Unless $v_{GS} < V_{TN}$, $v_{DS}$ can't be zero.

• Ok, but the potential difference that has the most significance in the channel's width is the one between the gate and the substrate, no? And if we don't consider the Body Effect, Vs should equal Vdd in the first configuration. Please correct me if I am wrong, these questions arise from the fact that I have just started studying MOSFETs. Nov 17 '14 at 14:24
• When channel formed completely, it will be "resistive", if $I_{D}$ is 0, then $V_{DS}=0, V_{D}=V_{S}$, but $V_{s}$ is unknown in your first circuit. Nov 17 '14 at 14:45
• It can be considered that Vs is left floating (or connected to the drain of a closed MOSFET). So in the first configuration Vs = Vd and in the second Vs = Vg (=Vdd) - Vt. Nov 17 '14 at 14:50
• In the first circuit, you can only make sure $V_g=5V$, and your substrate $V_p=0$, and if omit body effect, $V_{g} - V_{p} > V_{TN}$, if you can also make sure $I_{D} = 0$, then $V_{S}=V_{D}$. Actually, $I_{D}$ is determined by $V_{GS}$ and $V_{DS}$. Nov 17 '14 at 14:58
• That is what I wanted to know. So if $I_D=0$ then $V_S=V_D$. Thank you very much. Nov 17 '14 at 15:04

Effectively, your circuits show a diode connected mosfet. As the name suggests, it will behave similar to a diode. The source voltage will depend on the current flowing through your transistor, so with an unconnected source it is hard to say what will be the voltage at the node Vx.

In the first circuit, where the mosfet body is not connected to the source, we need to consider the parasitic diodes that are caused by the way a mosfet is produced. In the circuit below (center of the figure) I drew the parasitic diodes (sorry for the bulk connection in the figure, the circuit drawing tool doesn't have a mosfet where the bulk is not connected to the source, but assume it is connected to gnd).

The parasitic diode between the source and bulk will short the source to ground, forcing the transistor on. So in that case you will be shorting your power supply... simulate this circuit – Schematic created using CircuitLab

When you connect the bulk of the transistor to the source, you have a regular diode connected transistor.

A source that describes mosfet circuits very well (in my opinion) is "Design of analog cmos integrated circuits" by Razavi. Page 53 of the book shows the small signal model for a diode connected transistor. From the small signal circuit we find that the mosfet impedance (and thus Vx) depends on the current flowing through the diode connected transistor. $$\frac{V_x}{I_x} \approx \frac{1}{g_m + g_{mb}}$$ When not considering the body effect, the voltage Vx will also be dependent on the current through the transistor, however the body effect (g_mb) makes the effect bigger.

Indeed, when the voltage Vx rises above Vg-Vt, the transistor will turn off.