The compensator is an essential electronic filter that enhances the speed and stability of a control system during dynamic operations. In most studies, the compensator is depicted as an active circuit placed around an operational amplifier (op amp), with its characteristics assumed to be ideal. When used in low-bandwidth systems, today's converters can achieve crossover frequencies exceeding 100 kHz even with minimal output capacitance, ensuring rapid transient responses to minimize voltage drops. However, considering the operational amplifier as a perfect model no longer suffices in these applications, leading to significant gain and phase distortions. By examining the open-loop gain and the influence of the low and high-frequency poles of the selected op amp on the overall response, appropriate components can be chosen to avoid disrupting the gain and phase characteristics needed for the crossover. In this first part, we will focus on the effects of open-loop gain, deliberately ignoring low and high-frequency poles. The second part will explore the impact of these additional poles and demonstrate how improper selection can undermine the final outcome.
Different types of compensators are commonly categorized as Type 1, Type 2, and Type 3 in switching converters. All three models feature a pole at the origin to provide the maximum available quasi-static gain (S = 0), ensuring precise output regulation. The Type-1 compensator is a straightforward integrating circuit that offers no phase boost. Type 2 builds upon Type 1 by incorporating a pole/zero pair to provide a phase boost of up to 90°. Lastly, the Type-3 circuit includes another pole-zero pair, boosting the phase by 180°. Figure 1 illustrates the frequency response (both amplitude and phase) of the three compensators along with their respective transfer function expressions. For further insights into these circuits, refer to [1].
Figure 1: The choice of compensator depends on the desired phase boost.
The Type-2 compensator is frequently encountered in current-mode power supplies, offering a maximum phase increase of 90° to provide substantial compensation. Figure 2 displays its application around an op amp. A resistor divider monitors the controlled variable (V_out, which represents the output voltage in this case), and several passive components form the filter. To determine the transfer function of the converter, we first examine the open-loop gain A_OL of the op amp and assess how it impacts the final expression. The transfer function G of the circuit relates to the excitation signal V_out and the output response V_FB through a mathematical relationship.
Figure 2: In this compensator, we consider the op amp to have a finite open-loop gain but ignore its internal pole.
Introduction to Rapid Analysis Techniques:
Various methods can determine the dynamic response of filters. This article employs Fast Analysis Circuit Technology (FACTs) as detailed in [2] and [3]. The core principle of these FACTs is to determine the circuit time constant under two distinct conditions: when the excitation signal vanishes (V_out drops to 0 V) and when the response clears (V_FB = 0). Using this approach, you'll discover how quick and intuitive it is to determine a specific transfer function.
As referenced, the transfer function for a first-order system with non-zero quasi-static gain can be expressed as:
Equation (1)
The first term G_0 represents the gain exhibited by the system when S = 0. The item carries units, as applicable. Here, since we're discussing gain [V]/[V], there are no units, making G dimensionless. The numerator N(s) governs the zero of the transfer function. Mathematically, the zero is a specific point s_z where the response becomes null. Theoretically, considering that the excitation signal spans the entire s-plane (not merely the vertical axis in harmonic mode), adjusting the input signal to the zero frequency s_z causes the zero to appear as the output response zeroing. Specific impedance combinations in the circuit block signal propagation, producing an output of 0 V, despite the presence of an excitation source. The zero is the root of the numerator. This is a convenient mathematical abstraction that significantly aids in identifying zeros without needing to write a series of algebras. For more details on this method, consult [4].
The denominator D(s) comprises the natural time constants of the circuit. These time constants t = RC or t = L/R are obtained by setting the excitation signal to zero and determining the impedance of the capacitance or inductance "indicated" in this configuration. By "observing," I mean imagining placing an ohmmeter temporarily on a removed capacitor or inductor to read the resistance it shows. This is a relatively straightforward exercise. Examining the passive circuit in Figure 3, you see an injection source-excitation-bias network on the left side. The input signal propagates through the mesh and nodes to form the observed response of a resistance R_3.
Figure 3: Determining the time constant of the circuit requires setting the excitation to zero and observing the resistance provided by the energy storage element temporarily removed from the circuit.
To determine the time constant of this example circuit, we set the excitation to 0 (a 0V voltage source is replaced by a short circuit, and the 0A current source is open) and remove the capacitor. Then, we mentally connect an ohmmeter to determine the resistance provided by the capacitor terminal. Figure 4 guides you through these steps.
Figure 4: After changing the 0V source to a short circuit, you determine the resistance at the capacitor terminal.
If you follow the steps in Figure 4, you "see" R_1 in parallel with R_2 and in series with R_4, all in parallel with R_3 and in series with r_C. The time constant of this circuit can only be calculated by R and C_1:
Equation (2)
We can prove that the pole of a first-order system is the reciprocal of its time constant. Therefore:
Equation (3)
Now, what is the quasi-static gain of this circuit when s = 0? Under DC conditions, the inductor is shorted, and the capacitor is open. Applying this concept to the circuit in Figure 3, plotted as shown in Figure 5, you imagine cutting the connection before R_4 and see a resistor divider with R_1 and R_2. The Thévenin voltage on R_2 is:
Equation (4)
The output resistance R_th is a value where R_1 and R_2 are connected in parallel. The complete transfer function thus involves a resistor divider consisting of R_4 in series with R_th and R_3 loaded. r_C is off because capacitor C_1 is removed in this DC analysis. So you can write:
Equation (5)
Figure 5: You disconnect the capacitor in the DC circuit and calculate the transfer function for this simple resistor arrangement.
Basically, this is the case, but we are missing a zero. How do we know if it has reached zero? Well, this is a useful technique: Consider the circuit in Figure 3 in your mind, shorting the capacitor C_1. Now, suppose you excite a circuit with a short-circuit capacitor. Can you observe the response of V_out based on the oscilloscope? Of course, r_C is shorted to R_3, and although the amplitude may be low, the input signal will still propagate and respond. If the answer to this exercise is "Yes, although C_1 short circuit, but still responding," then there is about zero and C_1. If you are dealing with a circuit with an inductor L_1, then do the same exercise, but use an open inductor. If you still have a response in this mode, there are zero related to L_1.
We say in the preamble that the zero point is represented in the circuit by blocking the propagation of the excitation signal and produces an output that is empty. If we consider a deformation circuit – where C_1 is replaced – as shown in Figure 6, when the excitation skewing network, what specific conditions mean response is empty? Having an empty response only means that the current in the R_3 cycle is zero. If the resistor has no current, no voltage is applied and V_out is 0 V, which is not a short circuit but a virtual ground.
Figure 6: In this morphing circuit, when r_C and C_1 are converted into a short circuit in series, the response disappears.
If R_3 has no current, then r_C and 1/sC_1 are connected in series to create a converted short:
Equation (6)
The root s_z is the zero position we need:
Equation (7)
Thus there are:
Equation (8)
Now we can combine all of these results to form the final transfer function characterized by the circuit of Figure 3:
Equation (9)
This is called a low-entropy expression, and you can immediately identify the gain, pole, and zero. High-entropy expressions are obtained by applying large-scale external forces to the original circuit when considering the impedance divider, for example:
Equation (10)
Not only can you make mistakes when deducing expressions – I will! – But formatting the results to something like (9) requires more effort. Also, please note that we did not write a line of algebra when writing (9). If we later find an error, it is easy to go back to a separate drawing and fix it separately. The correction of (9) will be easy. Now try the same correction in (10) and you might start from scratch. Your check, in the expression (9) Mathcad® table and plotted (10) are the same frequency response as shown in Figure 7.
Figure 7: Fast Mathcad® tells you whether the expressions derived with FACTs match the responses returned by the original expression.
The introduction to FACTs is intended to illustrate how easy and efficient it is to use them on simpler and more complex circuits. By breaking down a complex architecture into simple, separate circuits, you can quickly write transfer functions, sometimes just by checking, as we did. Now that we have introduced the tool, let’s apply it to our Type 2 compensator.
FACTs applied to Type-2 compensators:
To efficiently use FACTs into the circuit of Figure 2, we began by calculating the energy storage components: C_1 and C_2. Considering that their independent states are variable – as they are not in series or in parallel – this is a second-order system. Considering a non-zero quasi-static gain, such a system can be expressed in the following form:
Equation (11)
For second-order systems, we can prove that the denominator follows the following formula:
Equation (12)
The coefficient s is simply the sum of the time constants that determine the zero point excitation. The S2 coefficient is slightly more complicated because it introduces a new symbol: . This symbol means that the impedance you "see" the end of the C_2, C_1 and replaced by a short circuit. At first glance, it is a bit difficult to understand, but there is no insurmountable. We will understand it in a few words.
Solution according to route the circuit of Figure 3, we can study the system s = 0, as shown in FIG. 8. During the analysis, V_ref is a perfect source and its dynamic response is 0 (ignoring the modulation of our application, its voltage is fixed). Therefore, it naturally does not exist in the small signal circuit, and the short circuit is used in the AC analysis.
Figure 8: Disconnect all capacitors under DC conditions: The op amp operates in an open loop configuration.
The op amp provides a voltage equivalent to e times the open loop gain A_OL. The voltage at the inverting pin is related to the low side impedance R_lower, in which case e is a non-zero value:
Equation (13)
In this circuit we have two capacitors and therefore have two separate time constants. To determine the first time constant associated with C_2, we set the excitation signal to 0, we determine the impedance from C_2, C2 connects the terminals, and C1 is removed from the circuit. Figure 9 sketches.
Figure 9: First time constant associated with capacitor C_2: What is the impedance you see between its terminals?
If the previous example checks well, the presence of a voltage control source – the operational amplifier – does not work with this simple method. To determine the impedance provided by the C_2 terminal, we can connect the test generator I_T and we will determine the voltage V_T across it. Then V_T / I_T will give us the desired impedance. Sketch relates to a current source 10 as shown in Figure. The first simple equation you can write is related to e. The voltage between the input pins of the op amp is the negative value of the voltage applied to R_1 and R_lower in parallel:
Equation (14)
Figure 10: You install a test generator to determine the impedance across C_2.
The output of the op amp is e times the open loop gain A_OL. Therefore:
Equation (15)
Substituting (14) into (15) yields:
Equation (16)
V_T is the voltage of the current source. There is a negative e at its left end and a V_FB at the right:
Equation (17)
If we extract V_FB from (17), combined with the result of (16), we have:
Equation (18)
Our impedance is simple:
Equation (19)
Therefore the first time constant t_2 is expressed as:
Equation (20)
Second time constant related with C_1, the schematic to be updated, as shown in Figure 11. We did not install the current generator because the result is obvious: the resistance across C_1 is the determined resistance of C_2 in series with R_2:
Figure 11: Determine the second time constant immediately because it is the resistor that drives C_2 in series with R_2.
We have two time constants and we can do the second order term. We said that we need to evaluate , where C_2 is shorted, and we look at the resistance from the terminal of C_1. Figure 12 shows the new sketch. Since we have a Frank short circuit in the loop associated with R_2, then the resistor R is R_2:
Equation (21)
Therefore, if we combine the time constants according to (12), we derive the denominator D(s):
Equation (22)
This second-order form can be rearranged, assuming that the quality factor Q is much less than one. In this case, the two poles are completely separated: one controls the low frequency and the second is at the top of the spectrum. From (12) we can prove that the two poles are defined as:
Equation (23)
Equation (24)
Equation (25)
If we apply these definitions to (23), simplify and rearrange, we get:
Equation (26)
Equation (27)
Since we have a denominator, do we have zero points in this circuit? We can use the techniques shown before: If we imagine a short circuit, C_1 or C_2 then C_1 and C_2, are these three configurations responsive? If C_1 is shorted, we have a simple inverter with R_2 and other resistors: there is a zero associated with C_1. If C_2 is shorted, the op amp is 0: C_2 has no zero. If both capacitors are shorted, of course, there is no response. To determine the zero position, what does in Figure 13 prevent the propagation of the excitation and make the response empty? If the impedance provided by C_1 and R_2 becomes a shorted transition, the response disappears:
Equation (28)
then
Equation (29)
then
Equation (30)
Which gives the zero point at:
Equation (31)
We now have the final transfer function
Equation (32)
and
Equation (33)
and
Equation (34)
and
Equation (35)
Compare the response between circuits:
Now it makes sense to compare the dynamic response brought by the type-2 circuit (where we consider the open-loop gain), and the type 2 perfect transfer function is given below:
Equation (36)
among them
Equation (37)
Equation (38)
Equation (39)
For example, we compare the ideal op amp and open-loop gain 50 dB op amp (such as TL431). When the compensator must achieve the following goals: f_c = 10 kHz and 20 dB gain compensation at this frequency, the phase boost must be 65°. R_1 and R_lower are calculated for the 12V output and the 2.5 V reference voltage. (31) and two (36) dynamic response as shown in Figure 14. The deviation of cross gain and phase boost is negligible. However, the gain at (31) is 35 dB at 120 Hz and 45 dB at (36). Finally, the finite static gain of the limited A_OL option is only 36.4 dB(»66), while the infinite time is the perfect op amp. What is the impact of these numbers? Insufficient gain at twice the power frequency will affect the ability of the control system to reject the rectified ripple. Output variables may be affected by this component, especially under voltage mode control. In addition, control variables may have significant static errors if the implant gain is low. If you now select the operational amplifier has a higher A_OL, for example 80 dB, the deviation disappears, two curves are very close to each other, as shown in Figure 15.
Figure 14: In the Bode plot of type 2, we consider that the open-loop gain A_OL and the low-side resistance R_lower are not much different from the original perfect equation.
Figure 15: When the open loop gain A_OL increases, the two curves are well superimposed. The quasi-static gain is increased to 66.3 dB compared to 36 dB with 50 dB A_OL gain.
To sum up:
This first part demonstrates the effect of open-loop gain in a compensator with a non-ideal op amp. When the op amp is no longer considered perfect, you can see the effect of the weak open loop gain in the low frequency range of the dynamic response and evaluate the performance degradation caused by this situation. In this first part, we only consider the effect of open loop gain. In the second part, we will add a complication analysis, adding two low and high frequency poles, and the IC designer naturally places it in an operational amplifier to ensure its stability.
References:
1. C. Basso, “Designing Control Loops for Linear and Switching Converters – A Tutorial Guideâ€, Artech House 2012, ISBN 978-1-60807-557-7
2. C. Basso, “Linear Circuit Transfer Functions – An Introduction to Fast Analytical Techniquesâ€, Wiley 2016, ISBN 978-1-119-23637-5
3. V. Vorpérian, “Fast Analytical Techniques for Electrical and Electronic Circuitsâ€, Cambridge University Press 2002, ISBN 978-0-521-624428
4. C. Basso, “Fast Analytical Techniques at Work with Small-Signal Modelingâ€, APEC Professional Seminar, Long Beach (CA), 2016, http://cbasso.pagesperso-orange.fr/Spice.htm
Accessores for electric components , like connector caps , rail cable clips , cable gland , pin remove tools , crimping tools.
Pg7 Cable Gland,Emc Cable Gland,M23 Connector Cap,Rail Cable Clips
Kunshan SVL Electric Co.,Ltd , https://www.svlelectric.com