The equivalent input circuit of a typical op-amp consists of a differential input pair, a current source, and a current mirror (active load) as shown in Figure 1. The current source established by the reference bias (VB1) determines the amount of current that flows to the differential input pair. Because of the current mirror circuit, the differential input pair basically provides Qn1 and Qn2 with the same current. As a result, VDS_qn2 is transferred to the output stage of the op-amp.
Suppose that VIN(+) and VIN(-) have dropped by ΔV. This causes the source-gate voltage (VSG) of Qp2 and Qp3 to increase, which in turn causes their drain current to increase. Since the current source provides more current, the drain-source voltage of Qp1 (VDS_qp1) also increases. As a result, the source-gate voltage (VSG) of Qp2 and Qp3 decreases back to the previous level.
Therefore, the output of an op-amp remains unchanged when common-mode signals are applied to its differential input pair.
The assumption of the above operations is that the MOSFETs are in the saturation region.
Next, let's consider the conditions under which the MOSFET enters the saturation region. Figure 2 shows the I D-VDS curve of an N-channel MOSFET (SSM3K16). The drain current remains almost constant (ΔV/ΔI = high impedance) over the drain-source voltage (VDS) range in the saturation region. The following relationship must be satisfied in order for the MOSFET to operate in the saturation region, where Vth is the gate-source voltage (VGS) at which the drain current begins to flow.
VDS > VGS - Vth (1)
Here, we write an equation for VIN(+) with respect to GND to determine its minimum value, VIN(+)_min.
VIN(+) = VGS_qn1 + VSD_qp2 – VSG_qp2 (2)
At VIN(+)_min, the source-drain voltage of Qp2 decreases to a level called the pinch-off voltage at which Qp2 is about to transition from the saturation region to the linear region. Let this voltage be VSD_qp2_min and the threshold voltage of Qp2 at which the drain current begins to flow be Vth_qp2. Then, the following equation is obtained from Equation 1 that represents the saturation condition:
VSD_qp2_min = VSG_qp2 – Vth_Qp2 (3)
Substituting Equation 2 into Equation 3 gives:
VIN(+)_min = VGS_qn1 – Vth_qp2 (4)
Up until this point, we have regarded MOSFETs as three-terminal devices. In actuality, however, there is one more terminal called the back gate. (In the case of typical discrete MOSFETs, the back gate is internally connected to the drain terminal.)
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