is it possible to achieve a phase margin of \$ 45^\circ \$ while only adjusting \$C_1\$ alone

Question

Schematics above shows a two-stage fully differential amplifier with differential input and output signals. Neglect the body effect and parasitic capacitances of the transistors.

$$ V_{dd} = 3.3\,\text{V} $$ $$ \mu_n C_{ox} = 134\,\mu \text{A}/\text{V}^2, \quad \mu_p C_{ox} = 45\,\mu \text{A}/\text{V}^2 $$ $$ \lambda_n = 0.1\,\text{V}^{-1}, \quad \lambda_p = 0.2\,\text{V}^{-1} $$ $$ I_1 = 0.45\,\text{mA}, \quad I_2 = 0.6\,\text{mA} $$ $$ L = 0.35\,\mu\text{m} $$ $$ W_1 = W_2 = 50\,\mu\text{m}, \quad W_3 = W_4 = 50\,\mu\text{m} $$ $$ W_5 = W_6 = 70\,\mu\text{m}, \quad W_7 = W_8 = 60\,\mu\text{m} $$ $$ C_2 = 1.35\,\text{pF} $$

Assume the two-stage amplifier in unity gain feedback configuration. Determine the value of \$C_1\$ to obtain a phase margin of \$45^\circ\$. Assume that the unity gain frequency is 400 MHz.

(b): Phase Margin and Feedback Configuration

Determine the value of \$ C_1 \$ to achieve a phase margin (\$ PM \$) of \$ 45^\circ \$ for the two-stage amplifier in a unity-gain feedback configuration, given that the unity-gain frequency (\$ \omega_t \$) is 400 MHz.

Given:

• Unity-gain frequency: \$ f_t = 400\,\text{MHz} \$
  $$ \omega_t = 2\pi f_t = 2\pi \times 400 \times 10^6\,\text{rad/s} = 2.513 \times 10^9\,\text{rad/s} $$

• Total DC gain: $$ A_v = 736 $$ (from Part (a))

Understanding Phase Margin in Two-Pole Amplifiers:

In a two-pole system, the open-loop gain is given by:

$$ A(s) = \frac{A_v}{(1 + s/\omega_{p1})(1 + s/\omega_{p2})} $$

In unity-gain feedback configuration, the loop gain $$ T(s) $$ is:

$$ T(s) = A(s) $$

Phase Margin (\$ PM \$) Calculation:

The phase margin is:

$$ PM = 180^\circ - \angle T(j\omega_t) = 180^\circ - \left[ \tan^{-1}\left( \frac{\omega_t}{\omega_{p1}} \right) + \tan^{-1}\left( \frac{\omega_t}{\omega_{p2}} \right) \right] $$

Calculating the Poles:

1. First Pole (\$ \omega_{p1} \$) at the Output of the First Stage:

$$ \omega_{p1} = \frac{1}{R_{o1} C_1} $$

2. Second Pole (\$ \omega_{p2} \$) at the Output of the Second Stage:

$$ \omega_{p2} = \frac{1}{R_{o2} C_2} $$

From Part (a):

• $$ R_{o1} = 14.815\,\text{k}\Omega $$
• $$ R_{o2} = 5.555\,\text{k}\Omega $$
• $$ C_2 = 1.35\,\text{pF} $$

Calculate \$ \omega_{p2} \$:

$$ \omega_{p2} = \frac{1}{R_{o2} C_2} = \frac{1}{5.555 \times 10^3\,\Omega \times 1.35 \times 10^{-12}\,\text{F}} = \frac{1}{7.5 \times 10^{-9}\,\text{s}} = 133.33 \times 10^6\,\text{rad/s} $$

Convert \$ \omega_{p2} \$ to frequency:

$$ f_{p2} = \frac{\omega_{p2}}{2\pi} = \frac{133.33 \times 10^6}{2\pi} \approx 21.22\,\text{MHz} $$

Assessing the Phase Margin with Given \$ \omega_t \$:

Given that \$ \omega_t = 2.513 \times 10^9\,\text{rad/s} \$, we can calculate:

1. Phase Shift Due to the First Pole:

Since \$ \omega_t \gg \omega_{p1} \$ (because \$ A_v \$ is large):

$$ \frac{\omega_t}{\omega_{p1}} = A_v = 736 $$

Thus:

$$ \theta_1 = \tan^{-1}\left( \frac{\omega_t}{\omega_{p1}} \right) \approx \tan^{-1}(736) \approx 89.92^\circ $$

2. Phase Shift Due to the Second Pole:

$$ \frac{\omega_t}{\omega_{p2}} = \frac{2.513 \times 10^9}{133.33 \times 10^6} \approx 18.85 $$

$$ \theta_2 = \tan^{-1}(18.85) \approx 86.97^\circ $$

3. Total Phase Shift and Phase Margin:

$$ \angle T(j\omega_t) = \theta_1 + \theta_2 \approx 89.92^\circ + 86.97^\circ \approx 176.89^\circ $$

$$ PM = 180^\circ - \angle T(j\omega_t) = 180^\circ - 176.89^\circ \approx 3.11^\circ $$

1. Desired Phase Shift from Second Pole:

$$ PM = 180^\circ - [ \tan^{-1}(A_v) + \tan^{-1}\left( \frac{\omega_t}{\omega_{p2}} \right) ] = 45^\circ $$

Given that $$ \tan^{-1}(A_v) \approx 90^\circ $$, we have:

$$ 45^\circ = 180^\circ - 90^\circ - \tan^{-1}\left( \frac{\omega_t}{\omega_{p2}} \right) $$

$$ \tan^{-1}\left( \frac{\omega_t}{\omega_{p2}} \right) = 45^\circ $$

$$ \frac{\omega_t}{\omega_{p2}} = \tan(45^\circ) = 1 $$

Thus:

$$ \omega_t = \omega_{p2} = 133.33 \times 10^6\,\text{rad/s} $$

But this contradicts the given \$ \omega_t = 2.513 \times 10^9\,\text{rad/s} \$.

**Therefore, it's impossible to achieve a phase margin of \$ 45^\circ \$ with \$ \omega_t = 400\,\text{MHz} \$ and \$ C_2 = 1.35\,\text{pF} \$.

Solution: Adjust \$ C_2 \$ to Increase \$ \omega_{p2} \$:

Since \$ C_2 \$ is limiting the phase margin, we need to reduce \$ C_2 \$ to increase \$ \omega_{p2} \$.

Calculating Required \$ \omega_{p2} \$:

Given \$ PM = 45^\circ \$ and \$ \theta_1 \approx 90^\circ \$:

$$ PM = 180^\circ - \theta_1 - \theta_2 = 180^\circ - 90^\circ - \theta_2 = 90^\circ - \theta_2 = 45^\circ $$

Thus:

$$ \theta_2 = 45^\circ $$

$$ \frac{\omega_t}{\omega_{p2}} = \tan(\theta_2) = \tan(45^\circ) = 1 $$

So:

$$ \omega_{p2} = \omega_t = 2.513 \times 10^9\,\text{rad/s} $$

Calculating Required \$ C_2 \$:

\$ \omega_{p2} = \frac{1}{R_{o2} C_2} \$ implies \$C_2 = \frac{1}{R_{o2}} \$

Answer 1

To get a phase margin of \$ 45^{\circ} \$ in a two-pole system, all you need to do is place the second pole on the gain crossover frequency, \$ \omega_{p_2} = \omega_t \$. Thus, you need to increase \$ C_1 \$ in your case so that \$ A_v\cdot \omega_{p_1}=\omega_{p_2} \$. In other words, following a single-pole approximation until we hit \$ \omega_{p_2} \$, the roll off slope is \$20\,\text{dB/dec}\$. Working back from the gain crossover frequency, you can get the required value of \$ \omega_{p_1} \$ and hence \$ C_1 \$.