2 added 324 characters in body edited Jul 4 '16 at 12:55 Sagie 17211 silver badge88 bronze badges 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\$$. 1 answered Jun 24 '16 at 13:41 Sagie 17211 silver badge88 bronze badges 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).