7 The opamp and stability

Q 7.1

(a)

We now define an (angular) frequency ${\omega }_1=\frac{1}{R_2\cdot C_1}$

(b)

An answer:
For a phase margin of $45°$, the phase shift of each low pass transfer is $-67.5°$. With $A_o\gg 1$ this cannot occur for $|A_{\mathit{loop}}=1$. Hence: NO.

(c)

An answer:
For a phase margin of $45°$, the summed phase shift of the two low pass transfers is $-135°$. With ${\omega }_0\gg {\omega }_1$ this can occur for for an $\omega \gg \mathit{omeg}{a}_1$, for a loop gain magnitude 1. Hence: YES.

(d)

An answer:
For a phase margin of $45°$, the summed phase shift of the two low pass transfers is $-135°$. With ${\omega }_1\gg {\omega }_0$ this can occur for for an $\omega \gg \mathit{omeg}{a}_0$, for a loop gain magnitude 1. Hence: YES.

(e)

An answer:

For the case that ${\omega }_1\gg {\omega }_0$:

$$\begin{array}{rcll}& \omega & \gg {\omega }_0& \\ & & \to \angle \frac{A_o}{1+\mathit{j\omega }∕{\omega }_0}\approx -90°& \\ & & \to & \\ & & \angle \frac{1}{1+\mathit{j\omega }∕{\omega }_1}\approx -45°& \\ & & |\frac{1}{1+\mathit{j\omega }∕{\omega }_1}|\approx \frac{1}{\sqrt{2}}& \\ & \omega ={\omega }_1& =\frac{A_0}{\sqrt{2}}\cdot {\omega }_0\approx A_0\cdot {\omega }_0& \end{array}$$

Using an asymptotic Bode plot, the $\sqrt{2}$ is easily lost. A graphical “derivation” would be similar to the plot below.

Q 7.2

(a)

An answer:
From the description above: $A_v(\mathit{j\omega })=\frac{A_{\mathit{DC}}}{1+\mathit{j\omega }∕{\omega }_0}$ with $A_{\mathit{DC}}=10^5$, ${\omega }_0=2\pi f_{\mathit{UG}}∕A_{\mathit{DC}}$

Indicate if the following statements are true or false. Motivate your answers with a proper explanation or calculation.

(b)

An answer:
False. $$\begin{array}{rcll}& z_{\mathit{in}}& \frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=\frac{v_{-}-v_{\mathit{OUT}}}{R_1}& \\ & & v_{\mathit{OUT}}=-A_v(\mathit{j\omega })\cdot v_{-}& \\ & z_{\mathit{in}}& =\frac{R_1}{1+A_v(\mathit{j\omega })}& \\ & & =\frac{R_1}{1+\frac{A_{\mathit{DC}}}{1+\mathit{j\omega }∕{\omega }_0}}& \end{array}$$

For (radian) frequencies lower than ${\omega }_0$, $A_v$ is almost real and almost $10^5$ and hence there $z_{\mathit{in}}\approx \frac{R_1}{A_{\mathit{DC}}}$ which shows that the statement is false.

(c)

An answer:
False. $$\begin{array}{rcll}& z_{\mathit{in}}& =\frac{R_1}{1+A_v(\mathit{j\omega })}& \\ & & =\frac{R_1}{1+\frac{A_{\mathit{DC}}}{1+\mathit{j\omega }∕{\omega }_0}}& \end{array}$$

For all frequencies, the numerator of the equation for $z_{\mathit{in}}(\mathit{j\omega })$ has a real part larger than or equal to 1, and a non-zero imaginary part. The norm of the denominator is hence larger than 1; then $|z_{\mathit{in}}|<R_1$ for all frequencies.

(d)

An answer:
True. The opamp has 1 pole. Now an extra pole is added in the feedback path. This results in a phase shift that may be close to $-180°$ at (magnitude of the) loopgain values equal to 1. This then corresponds to a low phase margin.

For nicely behaved amplifiers, the phase margin is usually assumed to be $60°$ or larger.

(e)

An answer:
False. The loopgain has a second order low pass transfer function. The gain at a phase shift equal to $-180°$ is zero. The gain margin is therefore large, $\infty$.

(f)

An answer:
False.

$$\begin{array}{rcll}& z_{\mathit{in}}& =\frac{R_1}{1+\frac{A_{\mathit{DC}}}{1+\mathit{j\omega }∕{\omega }_0}}& \end{array}$$

This equation may be rewritten in a standard form, but you may also just plug in some extreme values:

$$\begin{array}{rcll}& z_{\mathit{in}}& =\frac{R_1}{1+A_{\mathit{DC}}}\mathrm{\text{(for }}\omega =0)& \\ & z_{\mathit{in}}& =R_1\mathrm{\text{(for }}\omega \to \infty )& \end{array}$$

At lower frequencies, $A$ is higher, resulting in a lower $z_{\mathit{in}}$. There is a frequency range where the $z_{\mathit{in}}$ shows a (first order) frequency dependency, roughly between ${\omega }_0$ and the unity gain (radian) frequency.

(g)

An answer:
False.

Q 7.3

(a)

An answer:
A straight forward derivation (using e.g. $v_{-}=v_{+}$ which holds for stable systems with $A\to \infty$) yields

$H(\mathit{j\omega })=\frac{v_{\mathit{out}}}{v_{\mathit{in}}}=\frac{R_2+Z_{L1}}{R2}=1+\frac{\mathit{j\omega }{L}_1}{R_2}$

Note that this yields some kind of high-pass characteristic.

(b)

An answer:

$$\begin{array}{rcll}& H(\mathit{j\omega })& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=A\left(v_{\mathit{in}}-\frac{R_2}{R_2+\mathit{j\omega }{L}_1}{v}_{\mathit{out}}\right)& \\ & & =A\cdot v_{\mathit{in}}\frac{1}{\left(1+\frac{A}{1+\mathit{j\omega }{L}_1∕R_2}\right)}& \\ & H(\mathit{j\omega })& =\frac{A}{1+\frac{A}{1+\mathit{j\omega }{L}_1∕R_2}}& \end{array}$$

For a Bode plot, rewriting into a standard form is the easiest. The pole(s) and zero(s) and (here) the DC-voltage gain follow directly.

$$\begin{array}{rcll}& H(\mathit{j\omega })& =\frac{A}{1+\frac{A}{1+\mathit{j\omega }{L}_1∕R_2}}& \\ & & =\frac{A\left(1+\mathit{j\omega }{L}_1∕R_2\right)}{1+A+\mathit{j\omega }{L}_1∕R_2}& \\ & & =\frac{A}{1+A}\cdot \frac{1+\mathit{j\omega }{L}_1∕R_2}{1+\mathit{j\omega }\frac{L_1∕R_2}{1+A}}& \end{array}$$

(c)

An answer:

$$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =\frac{A_0}{1+\mathit{j\omega }∕{\omega }_1}\cdot \frac{1}{1+\mathit{j\omega }{L}_1∕R_2}& \end{array}$$

Conclusion: at the second pole ($f=\frac{10^6}{2\pi }$) the phase shift of the loopgain is $-135°$. At the frequency where the loopgain equals 1, the phase margin is larger than $45°$.

Q 7.4

(a)

An answer:

$$\begin{array}{rcll}& z_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}& \\ & & i_{\mathit{out}}=\frac{v_{\mathit{out}}+A{v}_{\mathit{out}}}{R_1+\mathit{j\omega }{L}_2}=\frac{1+A}{R_1+\mathit{j\omega }{L}_2}{v}_{\mathit{out}}& \\ & z_{\mathit{out}}& =\frac{v_{\mathit{out}}}{i_{\mathit{out}}}=\frac{R_1+\mathit{j\omega }{L}_2}{1+A}& \\ & & =\frac{R_1}{1+A}\cdot \left(1+\mathit{j\omega }\frac{L_2}{R_1}\right)& \end{array}$$

(b)

An answer:

(c)

An answer:
Stability is about loop gain; for this circuit the loop gain with an infinite load impedance is $$\begin{array}{rcll}& A_{\mathit{loop}}& =\mathit{A\beta }& \\ & & =A=\frac{A_0}{1+\mathit{j\omega }{\tau }_1}& \end{array}$$

From which it follows that the phase of the loop gain is between $0°$ and $-90°$, yielding a phase margin $>90°$. The phase margin is hence also larger than $60°$.

(d)

An answer:

$$\begin{array}{rcll}& A_{\mathit{loop}}& =\mathit{A\beta }& \\ & & \beta =\frac{R_L}{R_1+R_L+\mathit{j\omega }{L}_2}& \\ & & =\frac{R_L}{R_1+R_L}\cdot \frac{1}{1+\mathit{j\omega }{L}_2∕(R_1+R_L)}& \\ & & =\frac{R_L}{R_1+R_L}\cdot \frac{1}{1+\mathit{j\omega }∕{\omega }_0}& \\ & & A=\frac{A_0}{1+\mathit{j\omega }∕{\omega }_1}& \\ & A_{\mathit{loop}}& =\frac{A_0}{1+\mathit{j\omega }∕{\omega }_1}\cdot \frac{R_L}{R_1+R_L}\cdot \frac{1}{1+\mathit{j\omega }∕{\omega }_0}& \end{array}$$

The pole at ${\omega }_0$ can be at any (radian) frequency, depending on (among others) the value of the load resistor $R_L$. If ${\omega }_0<<<{\omega }_1$ or if ${\omega }_0>>>{\omega }_1$ the phase margin can be sufficient (check!). For ${\omega }_0\approx {\omega }_1$ the phase margin will be very small for large DC opamp gain values $A_0$.

Q 7.5

(a)

An answer:
Assuming stable behavior, for an ideal opamp $v_{+}=v_{-}$ which yields directly $$\begin{array}{rcll}& & V_{\mathit{REF}}=\frac{R_1}{R_1+R_2}{v}_{\mathit{OUT}}& \\ & & v_{\mathit{OUT}}=\left(1+R_2∕R_1\right)V_{\mathit{REF}}& \end{array}$$

(b)

An answer:
At the unity-gain frequency, the modulus of the gain is 1 by definition, leading to $$\begin{array}{rcll}& A(\mathit{j\omega })& =\left|\frac{A_0}{(1+\mathit{j\omega \tau })}\right|\equiv 1& \\ & & \Rightarrow & \\ & & |1+\mathit{j\omega \tau }|=A_0& \\ & & \Rightarrow & \\ & & A_0=\sqrt{1^2+{\omega }^2{\tau }^2}& \\ & & \Rightarrow & \\ & & f=\frac{1}{2\pi }\sqrt{\frac{A_0-1}{{\tau }^2}}\approx \frac{A_0}{2\mathit{\pi \tau }}& \end{array}$$

Another approach is by drawing a Bode plot. Asymptotically, the (angular) corner frequency lies at $1∕\tau$ at a gain of $A_0$, and rolls-off first order for higher frequencies. It crosses the zero dB axis if the (modulus of the) gain is reduced by a factor $A_0$. Being a first order system that must be at a frequency factor of $A_0$ higher than the corner frequency.

(c)

An answer:
The loop consists of (in arbitrary order):

• the transistor, which translates into a transconductance $g_m$ for small signals
• the resistive divider with $C_{\mathit{out}}$
• the opamp (via the non-inverting input

The loop gain is then

$$\begin{array}{rcll}& A_{\mathit{loop}}& =-g_m\cdot \frac{R_1+R_2}{1+\mathit{j\omega }(R_1+R_2)C_{\mathit{out}}}\cdot \frac{R_1}{R_1+R_2}\cdot \frac{A_0}{1+\mathit{j\omega \tau }}& \\ & & =-g_m{R}_1{A}_0\cdot \frac{1}{1+\mathit{j\omega }(R_1+R_2)C_0}\cdot \frac{1}{1+\mathit{j\omega \tau }}& \end{array}$$

(d)

An answer:
It was stated that the unloaded regulator is stable, with a sufficient phase margin. Furthermore the loop gain at DC is probably much larger than unity because $A_0>>1$. This implies that the (magnitude part of the) Bode plot of $A_{\mathit{loop}}$ is as shown below:

• $A_{\mathit{loop}}(0)>>0$
• there are two poles, where the second one is located at frequencies where $|A_{\mathit{loop}}|<1$ (for sufficient phase margin)

A higher $I_{\mathit{OUT}}$ increases the $g_m$ of the transistor. That shifts the curve in the Bode plot upwards, which decreases the phase margin.

Q 7.6

(a)

An answer:
From the Bode plot, we can see that the op-amp transfer function has a DC gain of $A_0=80\mathit{dB}=10^4$, and three poles. The first pole is at $f_1=1\mathit{kHz}$, the second one at $f_2=100\mathit{kHz}$, and the third one at $f_3=1\mathit{MHz}$. After the first pole, the slope is 20 dB/decade. After the second pole it is 40 dB/decade and after the third pole it is 60 dB/decade.

This shows that the op-amp transfer function can be written as the cascade (for gain: multiplication) of three first-order transfer functions:

$$\begin{array}{rcll}& A_{\mathit{opamp}}(\mathit{j\omega })& =A_0\cdot \frac{1}{1+j\frac{\omega }{2\pi f_1}}\cdot \frac{1}{1+j\frac{\omega }{2\pi f_2}}\cdot \frac{1}{1+j\frac{\omega }{2\pi f_3}}& \end{array}$$

(b)

An answer:
Configuration (a) has positive feedback and configuration (b) is inverting, which leaves configuration (c) and (d) to choose from. Now, let’s observe the op-amp transfer function again:

Suppose we would apply unity feedback ($\beta =1$). In that case $A_{\mathit{loop}}=A_{\mathit{opamp}}(\mathit{j\omega })\beta =A_{\mathit{opamp}}(\mathit{j\omega })$, so the loop transfer is simply the transfer of the op-amp itself. For unity gain feedback, we can read the gain margin by checking the gain at a phase of -180 degrees. This turns out to be -20 dB, the resulting amplifier using $\beta =1$ would be unstable. We can also check the phase margin by checking the phase at a gain of 1 ($=$ 0 dB). This turns out to be $-225°$, so the phase margin (distance to $-180°$) is $-45°$. This also indicates instability.

In other words, the op-amp is not “unity gain stable”: If we apply a feedback network with a transfer of $\beta =1$, the closed-loop circuit is unstable. For unity gain stability, at least $+45°$ phase margin is required.

For the circuit in (d), the feedback network $\beta$ has a gain of $R_1∕(R_1+R_2)=1∕10=-20$dB. So the open-loop loop transfer $A_{\mathit{loop}(\mathit{j\omega })}=A_{\mathit{opamp}}(\mathit{j\omega })\beta$ for the circuit is (d) is simply the opamp voltage gain, downshifted by 20 dB. This is shown below:

We can see that the circuit has (close to) zero gain and phase margin, so it is still unstable or at best very close to instability.

For the circuit in (c), the feedback network $\beta$ has a gain of $R_1∕(R_1+R_2+R_4)=1∕57=-35$dB.

We can read a gain margin of 15 dB and a phase margin close to $45°$. So, the closed-loop transfer of this circuit is stable. Also, for the desired frequency range of $0\dots 10$kHz, the circuit has (close to) 20 dB closed-loop gain.

(c)

An answer:
This has been done graphically in (b).
The phase margin can be calculated by evaluating the (angular) frequency ${\omega }_x$ for which $\left|A_{\mathit{loop}}(j{\omega }_x)\right|=1$ and checking $\arg <!--\; FUNCTION\; APPLICATION\; -->(A_{\mathit{opamp}}(j{\omega }_x)\beta )$. That is however computationally a lot of work: after maximum simplification of the loop gain, we are still left with a high order function of $\omega$... $$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& \approx A_{\mathit{loop}0}\cdot \beta \cdot \frac{1}{j\frac{\omega }{2\pi f_1}}\cdot \frac{1}{1+j\frac{\omega }{2\pi f_2}}\cdot \frac{1}{1+j\frac{\omega }{2\pi f_3}}& \end{array}$$

Calculating the gain margin is a lot easier. The above approximation for $A_{\mathit{loop}}$ can be used to get the (angular) frequency ${\omega }_x$ at which $\mathit{Arg}(A_{\mathit{loop}}(j{\omega }_x))=-\pi$. Then that ${\omega }_x$ can be used to get the gain margin $|A_{\mathit{loop}}(j{\omega }_x)|$. Straight forward algebra (or plain symmetry) shows that

$$\begin{array}{rcll}& & \mathit{Arg}(\frac{1}{1+j\frac{\omega }{\omega 0}}\cdot \frac{1}{1+j\frac{\omega }{N\cdot \omega 0}})=\frac{-\pi }{2}\to {\omega }_x=\sqrt{N}{\omega }_0& \end{array}$$

which can be used to calculate $|A_{\mathit{loop}}(j{\omega }_x)|$ and thereby to calculate the gain margin.

(d)

An answer:
From 0 to 10 kHz, we only have to consider the first pole of the op-amp.

$$\begin{array}{rcll}& A_{\mathit{opamp}}(\mathit{j\omega })& \approx A_{\mathit{loop}0}\cdot \frac{1}{1+j\frac{\omega }{2\mathit{\pi f}1}}& \\ & & =A_0\cdot \left(\frac{1}{1+j\frac{\omega }{2\pi \cdot 1\cdot 10^3}}\right)& \end{array}$$

The output resistance of this type of non-inverting amplifier circuit with a voltage $A$ can be derived to be:

$$\begin{array}{lll}\hfill r_{\mathit{out}}=\left(R_1+R_2\right)||\frac{R_4}{1+\mathit{A\beta }}& & \hfill \end{array}$$

Plugging in the frequency-dependent gain of the opamp:

$$\begin{array}{lll}\hfill z_{\mathit{out}}=\left(R_1+R_2\right)||\frac{R_4}{1+A_0\cdot \left(\frac{1}{1+j\frac{\omega }{2\pi \cdot 1\cdot 10^3}}\right)\cdot \left(\frac{R_1}{R_1+R_2+R_4}\right)}& & \hfill \end{array}$$

Of course, this can be written in more convenient forms. And in more inconvenient forms.