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Figure 1-2 shows a simplified equivalent circuit for an op-amp. As you see, it is composed of multiple MOSFETs. For a CMOS op-amp to work properly, these MOSFETs need to operate in the saturation region. Figure 1-3 shows the saturation region of the MOSFET.

In this region, the MOSFET operates as follows:

- As the gate-source voltage increases, the drain current increases.
- As the drain-source voltage increases, the drain current increases slightly.

A slight change in drain current causes a considerable change in drain-source voltage.

The portions of the op-amp provide the following functions:

- Differential input pair: Amplifies a difference in voltage between the V
_{IN (+)}and V_{IN (-)}inputs - Current mirror: Provides an equal amount of current to Q
_{p1}and Q_{p2}comprising the differential input pair. The current mirror acts as load resistance for the differential input pair. Typically, the output of the current mirror (i.e., the drain terminal of the differential input pair) has high impedance, which is difficult to obtain with a typical resistor. As a result, the first-stage differential amplifier has a high gain. Such a resistive load composed of transistors is called an active load. - Current source: Determines the amount of current that flows to the differential input pair and the common-source amplifier. The current source acts as an active load for the common-source amplifier.
- Common-source amplifier: Provides the drive current for an external load connected to the output and compensates for the gain of the first-stage differential amplifier

Before going into the operation of an op-amp, let’s discuss the drain voltage of Q_{n1} in the current mirror.

The drain-source voltage (V_{DS_n1}) and the drain-gate voltage (V_{DG_n1}) of Q_{n1} are equal. Figure 1-4 plots the conditions under which V_{DS} = V_{DG} is satisfied. Since the resulting curve looks like the I_{F}-V_{F} curve of a diode, the connection of Q_{n1} is called a diode connection. In Figure 1-4, the drain current is large because it is the I_{D}-V_{DS} curves of a discrete N-channel MOSFET with a large channel area. The internal MOSFETs of an IC have a drain current two to three orders of magnitude lower than this.

As Figure 1-4 indicates, after the drain current exceeds a certain point (at a V_{DS} of 1.5 V or higher), a slight change in the drain current hardly affects the drain-source voltage.

Next, let’s consider how the current source works. First, let’s consider a circuit without a current source as shown in Figure 1-6. The subsequent common-source amplifier is identical to that of the previous op-amp.

An equal voltage (V_{DD} – V_{IN}) is applied to the differential inputs, V_{IN(+)} and V_{IN(-)}. Hence, V_{SG} = V_{IN}. At this time, when the drain current (I_{D_p1}) is conducted, the drain voltage of Q_{p1} settles to a voltage at which V_{SD_p1}＋V_{DS_n1}＝V_{DD}. Since I_{D_p1} is copied by the current mirror, the circuit composed of Q_{p2} and Q_{n1} has the same voltage relationship as this.

Suppose that the voltage applied to V_{IN(+)} and V_{IN(-)} increases by ΔV to (V_{DD} – V_{IN} + ΔV). Since the circuit of Figure 1-6 has a current mirror, the same amount of current flows to the differential input pair. However, without a current source, the currents flowing to the differential input pair decrease by the same amount. As a result, the drain-source voltage of Q_{n2} connected to the common-source amplifier also decreases.

This is equivalent to a decrease in the gate-source voltage of Q_{n3} (V_{GS_n3}) of the common-source amplifier. The common-source amplifier has a current source (Q_{p4}), which raises the drain-source voltage (V_{DS_n3}) to oppose a decrease in V_{GS_n3}, keeping the current constant. In other words, the output voltage (V_{OUT}) increases even though the V_{IN(+)} and V_{IN(-)} inputs have the same voltage and phase. It is essential that the op-amp have a constant output when a common-mode input (same input voltage) within the range shown in the datasheet is applied to V_{IN(+)} and V_{IN(-)}. The circuit of Figure 1-6 cannot satisfy this requirement.

Next, let’s consider the circuit of Figure 1-2 with a current source (Q_{p3}). Suppose, for example, that the input voltage applied to V_{IN(-)} and V_{IN(-)} increases by ΔV to (V_{DD} – V_{IN} + ΔV). Since this circuit has a current source, the current flowing to the differential input pair remains unchanged. Therefore, the drain-source voltage of Q_{n1} (V_{DS_n3}) remains unchanged. Likewise, V_{DS_n2} remains unchanged. Therefore, the output voltage is constant for the common-mode input voltage.

(The V_{SD_p3} of Q_{p3} compensates for ΔV. The current flowing to the differential input pair changes because the source-drain voltage of the current source changes. Since the drain-source voltage of the current source changes, the drain current (ID) changes. However, ID changes only slightly with VDS. Therefore, ID does not change significantly.)

So, the role of the current source is to keep the output voltage constant when the common-mode input voltage is applied to V_{IN(+)} and V_{IN(-)}.

Next, let’s consider the case in which different voltages are applied to V_{IN(+)} and V_{IN(-)}.

- Suppose that V
_{IN(+)}and V_{IN(-)}initially have the same voltage (V_{DD}– V_{IN}) and then the V_{IN(-)}voltage increases by ΔV. - V
_{SG_p1}decreases, causing I_{D_p1}to decrease by ΔI_{p1}. However, as explained above, Q_{n1}has a diode connection. Therefore, V_{DS_n1}remains unchanged. So, the drain voltage of Q_{p1}remains constant. - The current mirror copies the decreased I
_{D_p1}to the drain current of Q_{n2}(I_{D_n2}). - This is contradictory since the drain current of Q
_{n3}(I_{D_p3}) in the current source remains unchanged. Therefore, the drain voltage of Q_{n2}(V_{DS_n2}) increases to increase the current flowing through Q_{n2}. - You might think that an increase in V
_{DS_n2}causes V_{SD_p2}to decrease, causing I_{D_p2}to decrease. Note, however, that the current from the current source (I_{D_p3}) remains unchanged. Since I_{D_p1}has decreased by ΔI_{p1}, I_{D_p2}should increase, not decrease. Therefore, the source voltage of Q_{p2}increases. - The source-gate voltage of Q
_{p1}(V_{SG_p1}) increases, causing its drain current (I_{D_p1}) to increase. - I
_{D_p1}is copied to the drain current of Q_{n2}(I_{D_n2}). Then, the operation returns to Step 3.

Eventually, the drain voltage of Q_{n2} (V_{D_n2}) increases from the initial voltage.

The increased V_{D_n2} is transferred to the subsequent common-source amplifier.

The V_{GS_n3} of the common-source amplifier increases, causing I_{D_n3} to increase. However, the increase in I_{D_n3} is constrained by Q_{p4} of the current source. Since the increase in V_{GS_n3} does not lead to an increase in I_{D_n3}, the drain-source voltage of Q_{n3} (V_{DS_n3}) decreases.

This means that when the V_{IN(-)} voltage increases, the V_{OUT} voltage decreases.

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1. What is an op-amp?

1.1. Characteristics of op-amps (What is the ideal op-amp?)

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