Why is my LTspice simulation so different from my calculation by hand?

Question

This is the circuit:

$$V_{in} g_{m1} + \frac{V_{in}} {r_{o1}} = (V_{in} - V_o) s C - \frac{V_o} {r_{o2}}$$

$$\Rightarrow \frac{V_o} {V_{in}} = \frac{s C - g_{m1} - 1/ r_{o1}} {s C + 1/r_{o2}}$$

The pole frequency is when C = 1e-13 = \$ 10^{-13}\ \mathrm{F}\$ (3 dB loss)

$$ f_p = \frac{\omega_p} {2 \pi} = \frac{1} {2 \pi r_{o2} C} \approx 20.37\ \mathrm{MHz} $$

The graph (green line) shown is around 33.29 MHz.

Formula correction according to @Puk. (Shout out to Puk!)

$$V_{in} g_{m1} + \frac{V_{o}} {r_{o1}} = (V_{in} - V_o) s C - \frac{V_o} {r_{o2}}$$

$$\Rightarrow \frac{V_o} {V_{in}} = \frac{g_{m1} - s C} {s C + 1/r_{o2} + 1/r_{o2}}$$

For analysis in feedback loop.

Open loop DC gain:

$$V_{in} g_{m1} + \frac{V_{o}} {r_{o1}} = \frac{V_o} {r_{o2}}$$

$$\Rightarrow \frac{V_o} {V_{in}} = - g_{m1} (r_{o1} \parallel r_{o2})$$

But I don't know how to calculate the \$\beta\$ for this feedback network.

Answer 1

In the small-signal circuit, \$r_{o_1}\$ and \$r_{o_2}\$ appear in parallel between the output node and ground, so your hand calculation isn't correct (note that it also predicts the wrong DC gain). Specifically, the \$V_{in}/r_{o1}\$ term should be \$V_o/r_{o1}\$. The correct pole frequency is $$f_p = \frac{1}{2\pi(r_{o1} \parallel r_{o2})C}.$$

I'm not going to attempt to calculate the exact values \$r_{o1}\$ and \$r_{o2}\$, but using \$r_o\propto \lambda^{-1}\$ and assuming you've calculated \$(2\pi r_{o2}C)^{-1}\$ correctly,

$$f_p = \frac{r_{o1} + r_{o2}}{r_{o1}}\frac{1}{2\pi r_{o2}C}.$$ $$= \frac{0.1^{-1} + 0.14^{-1}}{0.1^{-1}}\times 20.37\text{ MHz}$$ $$= 34.9\text{ MHz}$$

which is much closer to what your simulation is giving.