1. The Basic Feedback Loop#
The simplest control loop can be conceptualized as a desired output value relating to some input value (setpoint) through both a feedforward gain path and a negative feedback gain path:
Where:
- $V_i$ = input signal
- $V_e$ = error signal
- $A$ = feedforward (open-loop) gain
- $\beta$ = feedback gain
- $V_o$ = output signal
- $V_f$ = feedback signal
Deriving the Closed-Loop Gain#
We will now derive the expression for the closed loop gain
$$G_{CL} \equiv \frac{V_o}{V_i}$$To do this, simply multiply around the loop and treat the summing junction as just that; a summing operation, noting the sign convention of the signals. We arrive at the following 3 expressions:
$$\begin{align} &V_o = A V_e \tag{1.1} \\ &V_e = V_i - V_f \tag{1.2} \\ &V_f = \beta V_o \tag{1.3} \end{align}$$Substituting $(1.2)$ into $(1.1)$ $\quad \Rightarrow \quad V_o = A(V_i - V_f) \quad \Rightarrow \quad$ Plug into $(1.3)$
$$ \Rightarrow \quad V_o = A(V_i - \beta V_o) $$Now solve for $\frac{V_o}{V_i}$:
$$ \Rightarrow \quad V_o = A V_i - A \beta V_o $$$$ \Rightarrow \quad V_o + A \beta V_o = A V_i $$$$ \Rightarrow \quad V_o(1 + A \beta) = A V_i $$$$ \Rightarrow \quad \boxed{G_{CL} \equiv \frac{V_o}{V_i} = \frac{A}{1 + A \beta}} \tag{1.4} $$This is typically where most application notes end; with an expression for closed loop gain that doesn’t really tell you much in terms of how to actually design a control loop. This is not most application notes; we want to understand how to design a proper control loop, so we continue on. So, with this classic expression for $G_{CL}$, we will modify it by multiplying the expression by the quantity $\frac{1}{\beta}$. It seems odd as to why we would do this, so allow me to explain. First, the expression for $G_{CL}$ becomes:
$$ \boxed{G_{CL} = \frac{1}{\beta} \frac{A \beta}{1 + A \beta}} \tag{1.5} $$The reason as to why this is such an elegant expression has two parts; the first of which being the fact that we now have a common term $A\beta$ in both the numerator and denominator. If you picture the loop in your mind, this expression starts at the summing junction, moves through both the feedforward and the feedback paths, and returns to the summing junction. Because of this, we commonly refer to this quantity as the Loop Transmission, and we designate it as the quantity $T$:
$$ \Rightarrow \qquad \boxed{T = A \beta} \tag{1.6} $$This quantity is leveraged in our stability analysis when we are finding the crossover frequency and phase margin (more on this later). The second reason why this expression is so powerful is in relation to the concept of an ideal control loop. Recall that in an ideal control loop, the error signal $V_e$ approaches zero ($V_e \to 0$). Derive the expression for our closed loop gain
$$ G_{CL} = \left. \frac{V_o}{V_i} \right|_{V_e \to 0} $$Using equation $(1.2)$, we apply the constraint $V_e \to 0$
$$ \cancel{V_e}^{\to 0} = V_i - V_f $$$$ \Rightarrow \quad V_i = V_f $$Substitute into eqn. $(1.3)$ and solve for $\frac{V_o}{V_i}$:
$$ V_i = \beta V_o $$$$ \Rightarrow \quad \boxed{\left. \frac{V_o}{V_i} = \frac{1}{\beta} \right|_{V_e \to 0}} \tag{1.7} $$We see that $\frac{1}{\beta}$ is the ideal gain of our control loop as the error signal approaches zero! To really hammer this point home, consider the case of a non-ideal, noninverting operational amplifier shown below:
In the op amp above, it is easy to see that the governing equations for this loop are exactly the same as the simple loop we started with (equations 1.1 - 1.3). We can ignore the output impedance of the op amp $r_o$ because it becomes vanishingly small in closed loop ($r_o \to 0$). The only difference here is that when multiplying around the loop from $V_o$ to $V_f$, we have a voltage divider, meaning
$$ \beta = \frac{R_{bot}}{R_{top} + R_{bot}} $$Given that we derived the ideal gain under the condition that $V_e \to 0$ to be $\frac{1}{\beta}$, we find the ideal gain of the noninverting op amp to be
$$ \boxed{\frac{1}{\beta} = 1 + \frac{R_{top}}{R_{bot}}} $$The equation with which we are all familiar.
Recalling equation $(1.5)$, we can write the expression for the nonideal op amp as
$$ G_{CL} = \left(1 + \frac{R_{top}}{R_{bot}}\right) \left(\frac{T}{1 + T}\right) $$It is now appropriate to introduce a new term. Notice how the above expression contains the ideal gain and some factor of loop transmission $\frac{T}{1 + T} < 1$. This tells us that there is a small discrepancy in the actual op amp output compared to the ideal gain that forces the output to be slightly less than ideal. Naturally, this quantity is commonly referred to as the Discrepancy Factor
$$ \boxed{D \equiv \frac{T}{1 + T}} \tag{1.8} $$To best illustrate the relationship between $G_{ideal} = \frac{1}{\beta}$, $A_{ol}$, and $G_{CL}$, we will graph it out
We note the difference between $A_{ol}$ and $G_{CL}$ to be $1 + A \beta$, the difference between $A_{ol}$ and $G_{ideal}$ to be $T = A \beta$, and lastly the aforementioned discrepancy factor separating the ideal gain $\frac{1}{\beta}$ and $G_{CL}$. It is important to recognize here that we are dealing with a double log plot, meaning multiplication and division in the linear domain become addition and subtraction in the log domain.
The illustration above shows the simplest possible case where there are no poles or zeroes (everything is flat with frequency), so what if we added some frequency dependence? Say, for instance, that the open loop gain block $A$ has a single real pole?
This representation of our loop grounds us in reality, as voltage feedback op amps are typically dominant pole compensated around 10Hz to allow for unity gain stability. In principle, all op amps, including current feedback op amps, have some dominant pole regardless anyway (infinite bandwidth is not a thing).
As a quick refresher, we are defining our real pole in this instance as $\frac{1}{1 + s/w_p}$ to assert a left-half-plane pole.
$$ \frac{1}{1 + s/w_p} = \frac{w_p}{s + w_p} $$$$ \Rightarrow \quad w_p \mathcal{L}^{-1} \{ \frac{1}{s + w_p} \} = w_p e^{-w_p t} $$$$ = \boxed{w_p e^{-\frac{t}{\tau_p}}} \tag{1.9} $$Left half plane poles in the time domain are represented by negative exponentials and therefore correspond to stable natural modes, while their counterpart right half plane poles in the time domain correspond to unstable natural modes due to the positive exponential
$$ w_p \mathcal{L}^{-1} \{ \frac{1}{s - w_p} \} = \boxed{w_p e^{\frac{t}{\tau_p}}} \tag{1.10} $$Back to our control loop which is now first order (single pole). We will start by deriving the closed loop gain $G_{CL}$ algebraically:
$$ G_{CL} = \frac{1}{\beta} \frac{A\beta}{1 + A\beta} $$$$ \Rightarrow \quad A \equiv \frac{A_{ol}}{1 + \frac{s}{\omega_p}} $$$$ \Rightarrow \quad G_{CL} = G_{ideal} \frac{ \frac{A_{ol}\beta}{1 + \frac{s}{\omega_p}}}{1 + \frac{A_{ol}\beta}{1 + \frac{s}{\omega_p}}} \quad \Rightarrow \quad T = A \beta $$$$ \Rightarrow \quad G_{CL} = G_{ideal} \frac{ \frac{T}{1 + \frac{s}{\omega_p}}}{1 + \frac{T}{1 + \frac{s}{\omega_p}}} $$$$ \Rightarrow \quad G_{CL} = G_{ideal} \frac{ \frac{T}{1 + \frac{s}{\omega_p}}}{1\left[\frac{1 + \frac{s}{\omega_p}}{1 + \frac{s}{\omega_p}}\right] + \frac{T}{1 + \frac{s}{\omega_p}}} = G_{ideal} \frac{ \frac{T}{\cancel{1 + \frac{s}{\omega_p}}}}{ \frac{1 + T + \frac{s}{\omega_p}}{\cancel{1 + \frac{s}{\omega_p}}}} $$$$ \Rightarrow \quad G_{CL} = G_{ideal} \frac{T}{1 + T + \frac{s}{\omega_p}} = G_{ideal} \frac{T}{1 + T + \frac{s}{\omega_p}} \frac{ \frac{1}{1 + T}}{\frac{1}{1 + T}} $$$$ \Rightarrow \quad G_{CL} = G_{ideal} \cancel{\frac{T}{1 + T}}^{\to D} \frac{1}{1 + \frac{s}{(1 + T)\omega_p}} $$$$ \Rightarrow \quad \boxed{G_{CL} = G_{ideal} D \frac{1}{1 + \frac{s}{(1 + T)\omega_p}}} \tag{1.11} $$We have arrived at an algebraic expression for the closed-loop gain $G_{CL}$, though it is not immediately apparent how we would use it to perform our stability analysis. The next section will introduce two practical methods to analyze the loop.
2. Phase Margin and Stability#
Understanding Phase#
Before diving right into stability analysis, it is worth first having a healthy understanding of phase and phase margin. What is phase, and how does it relate to frequency (and time)? Consider a vector with unity length rotating around in a circle as shown below:
A unit-length vector rotating around the unit circle is moving at an angular velocity $\omega$, where
$$ \omega = d\theta/dt $$$$ \int_0^t \omega,dt = \int_0^{\theta} d\theta $$where $\omega = 2\pi f$. Therefore:
$$ \boxed{\theta = 2\pi f t} $$Mapping this back to our control loop, we can designate the variables as
$$ \boxed{\theta_c = 2\pi f_c t_d} \tag{2.1} $$where $f_c$ is the crossover frequency, $t_d$ is the propagation delay through the loop, and $\theta_c$ is the loop phase at crossover with units $[rad/s]$. We will mention now that the crossover frequency $f_c$ is the point at which the loop gain is unity. For an input signal with an amplitude of $1Vpp$ and a frequency of $f_c$, for example, the output of the loop will be $1V_pp$ regardless if the loop gain is greater than 1 for lower frequencies as we will address in short order.
Back to phase. The phase of the loop at crossover determines the physical propagation delay through the loop for LTI systems with constant group delay, neglecting physical propagation delay from any pure time delay perspective e.g. phase velocity of the signal (non-minimum phase components). A phase shift of $\theta_c = -180^\circ$ yields a pure oscillation at the output of the loop with an oscillation frequency of $f_c$.
To illustrate this, we can solve Equation $2.1$ for $t_d$ and plug in a phase value of $\theta_c = -180^\circ$, but first we must express the sign convention for the phase response for a pure delay of an LTI system being negative:
$$ \theta_c(\omega) = -\omega t_d $$Now our phase requirement of $\theta_c = -180^\circ$
$$ \theta_c = -180^\circ\frac{\pi}{180^\circ} = -\pi $$Equating the two and solving for t_d is exactly Equation $2.1$ but with the proper sign convention
$$ t_d = \frac{\theta_c}{2\pi f_c} = \frac{\cancel{\pi}}{2\cancel{\pi} f_c} $$$$ \Rightarrow \quad \boxed{t_d = \frac{1}{2f_c}} \tag{2.2} $$This tells us for a phase lag of $\theta_c = -180^\circ$, we get a time delay of half of the crossover period, yielding a pure signal inversion at the summing junction of the control loop, which gives rise to a pure oscillation at the output of our loop at an oscillation frequency of $f_c$.
This oscillation at $\theta_c = \pm 180^\circ$ gives rise to our stability metric known as phase margin. It tells us effectively how close our loop is to instability, or more succinctly, how much phase lag we have in hand before hitting that $180^\circ$ threshold. Thus, our expression for phase margin is:
$$ \varphi_m = \theta_c - (-180°) $$$$ \Rightarrow \quad \boxed{\varphi_m = \theta_c + 180^\circ} \tag{2.3} $$Naturally, a phase shift of $-180^\circ$ will yield a $\varphi = 0^\circ$.
Finding Crossover Frequency and Phase Margin#
To analyze the loop, break it and insert a test source at the break point, setting all independent sources (e.g. $V_i$) to zero:
ADD BD FOR BROKEN LOOP FOR CONSTRUCTING A$\beta$ ANALYSIS (and A = $\frac{1}{\beta})$
Find the loop transmission $T=AB$ by multiplying around the broken loop, in factored pole-zero form:
$$ T = AB = \frac{A_0 B}{1+s/\omega_p} $$The easiest way to find crossover is to sketch the asymptote plots for magnitude and phase of $T$, and read off where the magnitude crosses 0 dB — the phase there gives the phase margin.
Phase/Gain Behavior of Poles and Zeros#
- LHP pole: phase lag, down to $-90°$ through the pole
- LHP zero: phase lead, up to $+90°$ through the zero (hence “lead”)
- RHP pole/zero: phase inverted relative to their LHP counterparts — notoriously hard to compensate
Phase shifting begins a decade before a pole/zero and totals $\pm90°$ per pole/zero by a decade after it.
First-Order System Rule#
For a single-pole-dominated loop, the maximum phase shift is $\theta_c=-90°$, so:
$$ \varphi_m = -90° - (-180°) = 90° $$A single-pole-dominated loop is guaranteed stable.
Worked Example: Voltage-Feedback Op Amp#
Voltage-feedback op amps are typically dominant-pole compensated at a low frequency (e.g. 10 Hz), rolling off like the single-pole example above.
Given open-loop gain $A_0 = 100,\text{dB}$ and desired non-inverting gain $G_{ideal}=10,\text{dB}$:
$$ B\,(\text{dB}) = A_0 - G_{ideal} = 100 - 10 = 90\,\text{dB} \;=\; T\,(\text{dB}) $$Crossover frequency: with a pole at 10 Hz rolling off at $-20,\text{dB/decade}$, dropping 90 dB takes:
$$ \frac{90\,\text{dB}}{20\,\text{dB/decade}} = 4.5 \text{ decades} \quad\Rightarrow\quad F_c = 10\,\text{Hz}\times10^{4.5} \approx 316.2\,\text{kHz} $$Phase margin: the pole’s phase settles to $-90°$ about a decade above 10 Hz — well below $F_c$ — so:
$$ \theta_c \approx -90° \quad\Rightarrow\quad \varphi_m = -90°-(-180°) = 90° $$Graphical Stability Check: $A$ and $1/B$ Intersection#
At the intersection, $A = 1/B$, i.e. $AB=1$ (0 dB) — this is the crossover frequency. The rate of closure is the difference in slopes of the two curves at that point.
Rule: a rate of closure of $-20,\text{dB/decade}$ at the intersection, called the Rate-of-Closure (ROC) criterion, satisfies the Nyquist Stability Criterion for minimum phase systems i.e. the loop is guaranteed stable. This sketch is also the visual picture of the gain-bandwidth tradeoff, giving rise to the gain-bandwidth product (GBW) spec in voltage-feedback op amps.
Coming Soon#
- Input & output impedance derivations of the 4 fundamental loop topologies (permutations of current/voltage inputs & outputs)
- Loop noise analysis