Should gain of this differential circuit be absolute value?

Question

Content is from the book Design of Analog CMOS Integrated Circuits (https://electrovolt.ir/wp-content/uploads/2014/08/Design-of-Analog-CMOS-Integrated-Circuit-2nd-Edition-ElectroVolt.ir_.pdf) on Page 112.

Due to a manufacturing error, in the circuit of Fig. 4.19, \$M_2\$ is twice as wide as \$M_1\$. Calculate the small-signal gain if the DC levels of \$V_{in1}\$ and \$V_{in2}\$ are equal.

The following is my analysis. Since it's symmetrical, we can use virtual ground to analyze this circuit:

First, we only analyze the left part with KCL:

$$ -\frac{V_{out2}} {R_D} = V_{in} g_{m1} + \frac{V_{out2}} {r_{o1}}$$ $$\Rightarrow \frac{V_{out2}} {V_{in}} = -g_{m1} (R_D \parallel r_{o1})$$

Now we only analyze the right part with KCL:

$$ -\frac{V_{out1}} {R_D} = -V_{in} g_{m2} + \frac{V_{out1}} {r_{o2}}$$ $$\Rightarrow \frac{V_{out1}} {V_{in}} = g_{m2} (R_D \parallel r_{o2})$$

We combine this two equation into one:

$$\frac {V_{out2} - V_{out1}} {2 V_{in}} = -\frac{1}{2} (g_{m1} + g_{m2}) R_D$$

Since \$g_{m2} = 2 g_{m1}\$ $$\frac {V_{out2} - V_{out1}} {2 V_{in}} = -\frac{3}{2} g_{m1} R_D$$

But the book said:

$$|A_v| = \frac{2 R_D} {1/g_{m1} + 1/ g_{m2}} $$ $$|A_v| = \frac{4} {3} g_{m1} R_D$$

Answer 1

The following is my analysis.

Firstly, we analyse the left part of this circuit with the use of KCL.

$$- \frac{V_{out2}} {R_D} = (V_{in1} - V_p) g_{m1} + \frac{V_{out2} -V_p} {r_{o1}}$$

$$\Rightarrow - V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) = V_{in1} g_{m1} - V_p (g_{m1} + \frac{1} {r_{o1}})$$

$$\Rightarrow V_p = \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1} + \frac{1} {r_{o1}}}$$

Secondly, we analyse the right part of this circuit with the use of KCL.

$$ \frac{V_{out1}} {R_D} = -(V_{in2} - V_p) g_{m2} + \frac{V_p -V_{out1}} {r_{o2}}$$

$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) = - V_{in2} g_{m2} + V_p (g_{m2} + \frac{1} {r_{o2}})$$

We could substitute \$\frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m2} + \frac{1} {r_{o2}}}\$ for \$V_p\$ from what we previously got.

$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) = - V_{in2} g_{m2} + \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1} + \frac{1} {r_{o1}}} (g_{m2} + \frac{1} {r_{o2}})$$

Since we already assume \$g_{m1} r_{o1} \gg 1\$ and \$g_{m2} r_{o2} \gg 1\$

$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) = - V_{in2} g_{m2} + \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1}} (g_{m2})$$

$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) = - V_{in2} g_{m2} + V_{in1} g_{m2} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}$$

$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) - V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}= V_{in1} g_{m2} - V_{in2} g_{m2} $$

We also assume \$r_{o1} \gg R_D\$ and \$r_{o2} \gg R_D \$

$$\Rightarrow V_{out1} (\frac{1} {R_D}) - V_{out2} (\frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}= V_{in1} g_{m2} - V_{in2} g_{m2} $$

$$\Rightarrow (\frac{1} {R_D}) (V_{out1} - V_{out2} \frac{g_{m2}} {g_{m1}}) = V_{in1} g_{m2} - V_{in2} g_{m2} $$

For third equation.

$$- \frac{V_{out2}} {R_D} = \frac{V_{out1}} {R_D}$$

$$\Rightarrow - V_{out2} = V_{out1} $$

Substitute \$-V_{out2}\$ for \$V_{out1}\$

$$\Rightarrow (-\frac{1} {R_D}) V_{out2} (1 + \frac{g_{m2}} {g_{m1}}) = g_{m2} (V_{in1} - V_{in2}) $$

The gain will be

$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}} = \frac{2 V_{out2}} {V_{in1} - V_{in2}} = \frac{2 g_{m2}} {-\frac{1} {R_D} (1 + \frac{g_{m2}} {g_{m1}})} $$

$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}} = - \frac{2 R_D g_{m2}} {(1 + \frac{g_{m2}} {g_{m1}})} $$

As we all know \$g_{m2} = 2 g_{m1}\$

$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}} = - \frac{2 R_D g_{m2}} {(1 + 2)} =- \frac{4 R_D g_{m1}} {3} $$