Unwind di Analogue Integrated Circuit Design - 3 di 5

Analysis of the full-differential 5-transistor operational transconductance amplifier (OTA)

Assumptions

• M1, M2: same Vtn, Kn’, W1=W2 and L1=L2;
• M7, M3, M4: same Vtp, Kp’, L3=L4=L7, W3=W4=W7*(N/2);
• M5, M6: same Vtn, Kn’, W5=N*W6, L5=L6;
• VI1=VI2=VIC, VO1=VO2=VOC;
• All MOSFETs are in saturation region.

Current mirror assumptions

Let's consider the previous N/2 factor and the M6-M5 current mirror:

ID iN IREF IDG5 a5

M1-M2 differential pair

Considering the M1-M2 differential pair:

iIs IREIsa TI e

Looking at the circuit, the current that flows in M1 must be the same that flows in M3, so the 2 current mirrors must be built such that they respect this KCL:

Ida IDIds REFe iIdaIn IDREFe4 if

Small signal behaviour

Let's analyse now the small signal behaviour at low-frequency (parasitic capacitances are open circuits) considering only a differential input: vi1=vid/2, Vi2=-vid/2. In particular, it's considered an ideal model of the MOSFET and parasitic elements are added:

ros 704 Ma144µ ra Ma2oz RID VID22 205145

The gates of the devices M7, M3, and M4 are part of a current mirror that generates a constant current every time, so the gate voltage must be constant in time. For this reason, in an AC analysis, the gates must be connected to an AC ground. The same thing happens to the device M5.

Looking at the circuit, it's clear that there is some type of anti-symmetry in the circuit, evidenced by this line. There is some anti-symmetry because if vid grows up, then the gate voltage of M1 grows up, and the gate voltage of M2 decreases. Due to this, the potential vs is constant in time, so in the AC analysis, it becomes an AC ground.

Equivalent circuit analysis

The analysis of the circuit can be developed considering only half of the circuit. Considering only the right part of the circuit, the analysis must be done on the following equivalent circuit:

Noiµ ra raVid2Voi 8mi ViTgsTgs Toi D203 ee ViIm DVoi 701 203 i

Analysing now the left side of the circuit, the analysis must be done on the following equivalent circuit:

Noa Ma2oz704 vid2ImaNo Ta 20h ivid

Due to the anti-symmetry of the circuit and due to the equality between the devices, it can be assumed:

iIm 9miIo 202 i103 204e aVoiVoi

So the differential output is:

ImaNoiVod ViImNoi TaToiD 20h2032e 2vidIVidGm Toi 203

Differential and common gain

Now can be defined the differential gain of the amplifier:

ImitaVodAdama 2032Vid

It can be noted that the result is similar to the gain obtained in the case of a common source biased by a current mirror.

It can also be defined the common gain of the amplifier due to the differential inputs, which is:

Alda iVoc Voi 0Norta 2radVid

This gain is equal to 0 because when the inputs are 2 opposite signals, the outputs are equal in modulus but opposite in sign.

Common-mode input

Let's consider now the opposite input, the common-mode input:

VideoVia Vicvii e

The circuit becomes:

704RosMi MrraraVic Vic205

To simplify the analysis, it's useful to divide the resistor r05 in 2 equivalent resistors in parallel which value is double of the original one:

704RosMi rara MrVic Vic2205 2205

In this case, the input voltages increase equally, so the source voltage of M1 and M2 cannot be considered an AC ground anymore.

Analysis of the left half of the circuit

Considering only the left half of the circuit:

NoVic Imbitosi io roNgs Ngs Innings2205

In this situation, the source terminal is not connected to AC ground, so it must be considered body effect and the factor gmb’:

tie Vs Vbi Usb VbVsNgs Vb 0a

Considering the KCL at this node:

iVoi Fm_vsI Imi 0Vic1 I 9ms e201201 203

While at this node:

iGm8mi _Vie 0Imbit Voi atzio to

Summing the 2 equations, the result is:

Voii0 vs Tos2e rzio te 703