2 added 324 characters in body
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Edit: as user "Virange" suggested, the \$r_{oc}\$ resistor probably represents the internal resistance of the current source marked as \$i_{SUP}\$.

While there is indeed, as another responder mentioneduser "Bhuvanesh N" indicated, a very small leakage from the body/bulk to the drain and the source terminals, I don't think \$r_{oc}\$ representscan't represent this leakage, as this leakage is absolutely meaningless for the small signal equivalent circuit.

The leakage current is similar toessentially the leakage current of a reverse-biased diode (since there are effectively two diodes in the device   - athe diode's P being the bulk-drain diode of the FET, and a bulk-source diode - which are reverse-biased)the diode's N being the FET's source, drain, and whilen-channel. While one might want to take itthis leakage current into consideration while solving the DC-equivalent circuit, I feel it isits effects are completely negligible for the small-signal model.

What is relevant for the small-signal model is the body effect. The body effect means \$V_{SB}\$ changes \$V_{TH}\$, and therefore \$I_{DS}\$. In this specific circuit, the input signal changes \$V_{SB}\$, which introduces another effect on the amplified signal via body effect.

In the small-signal model circuit provided in the question, \$r_{o}\$ represents the channel length modulation (due to \$V_{DS}\$ changes), and I believe \$r_{oc}\$ represents the body effect (due to \$V_{SB}\$ changes) is repesented by the controlled current source marked as \$-g_{mb}v_s\$.

While there is indeed, as another responder mentioned, a very small leakage from the body/bulk to the drain and the source terminals, I don't think \$r_{oc}\$ represents this leakage.

The leakage current is similar to the leakage current of a reverse-biased diode (since there are effectively two diodes in the device - a bulk-drain diode and a bulk-source diode - which are reverse-biased), and while one might want to take it into consideration while solving the DC-equivalent circuit, I feel it is negligible for the small-signal model.

What is relevant for the small-signal model is the body effect. The body effect means \$V_{SB}\$ changes \$V_{TH}\$, and therefore \$I_{DS}\$. In this specific circuit, the input signal changes \$V_{SB}\$, which introduces another effect on the amplified signal via body effect.

In the small-signal model circuit provided in the question, \$r_{o}\$ represents the channel length modulation (due to \$V_{DS}\$ changes), and I believe \$r_{oc}\$ represents the body effect (due to \$V_{SB}\$ changes).

Edit: as user "Virange" suggested, the \$r_{oc}\$ resistor probably represents the internal resistance of the current source marked as \$i_{SUP}\$.

While there is indeed, as user "Bhuvanesh N" indicated, a very small leakage from the body/bulk to the drain and the source terminals, \$r_{oc}\$ can't represent this leakage, as this leakage is absolutely meaningless for the small signal equivalent circuit.

The leakage current is essentially the leakage current of a reverse-biased diode   - the diode's P being the bulk of the FET, and the diode's N being the FET's source, drain, and n-channel. While one might want to take this leakage current into consideration while solving the DC-equivalent circuit, its effects are completely negligible for the small-signal model.

What is relevant for the small-signal model is the body effect. The body effect means \$V_{SB}\$ changes \$V_{TH}\$, and therefore \$I_{DS}\$. In this specific circuit, the input signal changes \$V_{SB}\$, which introduces another effect on the amplified signal via body effect.

In the small-signal model circuit provided in the question, \$r_{o}\$ represents the channel length modulation (due to \$V_{DS}\$ changes), and the body effect (due to \$V_{SB}\$ changes) is repesented by the controlled current source marked as \$-g_{mb}v_s\$.

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While there is indeed, as another responder mentioned, a very small leakage from the body/bulk to the drain and the source terminals, I don't think \$r_{oc}\$ represents this leakage.

The leakage current is similar to the leakage current of a reverse-biased diode (since there are effectively two diodes in the device - a bulk-drain diode and a bulk-source diode - which are reverse-biased), and while one might want to take it into consideration while solving the DC-equivalent circuit, I feel it is negligible for the small-signal model.

What is relevant for the small-signal model is the body effect. The body effect means \$V_{SB}\$ changes \$V_{TH}\$, and therefore \$I_{DS}\$. In this specific circuit, the input signal changes \$V_{SB}\$, which introduces another effect on the amplified signal via body effect.

In the small-signal model circuit provided in the question, \$r_{o}\$ represents the channel length modulation (due to \$V_{DS}\$ changes), and I believe \$r_{oc}\$ represents the body effect (due to \$V_{SB}\$ changes).