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Consider the Dynamic Response of Operational Amplifier in Type-2 Compensator

Considering the dynamic response of operational amplifier in type-2 compensator (Part 2)

03

Fact applied to type-2 compensator

In order to efficiently apply fact to type-2 compensator, we first consider energy storage elements C1 and C2. Considering that their independent states are variable - such as they are not series or parallel - this is a second-order system. Considering the non-zero quasi-static gain, the system can be expressed as:

ï¼ˆ11ï¼‰

For second-order systems, we can prove that the denominator follows the following formula:

ï¼ˆ12ï¼‰

The coefficient s is only the sum of the time constants for determining the zero excitation. S Â² The coefficient is slightly more complex because it introduces new symbols. This symbol means that you "Imagine" the impedance at both ends of C2, while C1 is replaced by a short circuit. At first glance, it's a little difficult to understand, but we'll understand it with a little explanation. According to the way of solving the circuit in Fig. 2, we can study the system with S = 0 (Fig. 7). In the process of analysis, VREF is a perfect source and its dynamic response is 0 (ignoring the modulation we apply, its voltage is fixed). Therefore, it naturally does not exist in small signal circuits and is in the form of short circuit in AC analysis.

Figure 7: disconnect all capacitors under DC conditions: the op amp operates in an open-loop configuration.

The voltage provided by the operational amplifier is equivalent to e times the open-loop gain AOL. The voltage of the inverting pin is related to the low side impedance rlower. At this time, e is a non-zero value:

ï¼ˆ13ï¼‰

This circuit has two capacitors, so it has two separate time constants. To determine the first time constant related to C2, we set the excitation signal to 0, determine the impedance of C2, C2 connects the terminal, and C1 is removed from the circuit. To determine the impedance provided by the C2 terminal, we can connect the test generator it and determine the voltage Vt at both ends. Then VT / it will provide the impedance we want. The first simple equation that can be written is related to E. The voltage between the input pins of the operational amplifier is:

ï¼ˆ14ï¼‰

The output of the operational amplifier is:

ï¼ˆ15ï¼‰

Substituting (14) into (15) yields:

ï¼ˆ16ï¼‰

VT is the voltage of the current source:

ï¼ˆ17ï¼‰

If VFB is extracted from (17), combined with the results of (16), we have:

ï¼ˆ18ï¼‰

The impedance is:

ï¼ˆ19ï¼‰

Therefore, the first time constant T2 is expressed as:

ï¼ˆ20ï¼‰

The second time constant is related to C1 (Fig. 8). We did not install the current generator because the result is obvious: the resistance at both ends of C1 is the determined resistance after C2 and R2 are connected in series:

ï¼ˆ21ï¼‰

Figure 8: immediately determine the second time constant because it is the driving resistance of C2 and R2 in series.

We have two time constants for the second-order term. We need to evaluate, where C2 is replaced by a short circuit, and we look at the resistance at C1. Since we have a frank short circuit in the circuit involving R2, the resistance R is R2:

ï¼ˆ22ï¼‰

Therefore, if we combine the time constant according to (12), we get the denominator D (s):

ï¼ˆ23ï¼‰

This second-order form can be rearranged, assuming that the quality factor Q is much less than 1. At this time, the two poles are completely separated: one controls the low frequency, and the second is located in the upper part of the spectrum. From (12), we can prove that the two poles are defined as:

ï¼ˆ24ï¼‰

ï¼ˆ25ï¼‰

If we apply these definitions to (23), simplify and rearrange, we get:

ï¼ˆ26ï¼‰

ï¼ˆ27ï¼‰

If we imagine short circuiting C1 or C2 or C1 and C2, are these three configurations responsive? If C1 is short circuited, we have a simple inverter with R2 and other resistors: there is a zero related to C1. If C2 is short circuited, the op amp is 0: C2 has no zero point. If both capacitors are short circuited, of course, there is no response. If C1 and R2 are short circuited, the response disappears (Figure 9):

ï¼ˆ28ï¼‰

then

ï¼ˆ29ï¼‰

Figure 9: if the serial connection between R2 and C1 is converted to short circuit, there is no signal response: this explains how the zero point is generated.

It is given that the zero point is located at:

ï¼ˆ30ï¼‰

Now there is the final transfer function

ï¼ˆ31ï¼‰

and

ï¼ˆ32ï¼‰

ï¼ˆ33ï¼‰

ï¼ˆ34ï¼‰

ï¼ˆ35ï¼‰

04

Compare the response between circuits

Now it is meaningful to compare the dynamic response brought by type-2 circuit (where we consider open-loop gain). The perfect transfer function of type 2 is:

ï¼ˆ36ï¼‰

among

ï¼ˆ37ï¼‰

ï¼ˆ38ï¼‰

ï¼ˆ39ï¼‰

For example, we compare an ideal operational amplifier with an open-loop gain of 50dB (such as TL431). At this time, the compensator must meet the following objectives: FC = 10kHz, gain compensation of 20dB at this frequency, and phase lift must be 65 Â°. R1 and rlower are calculated for 12V output and 2.5V reference voltage. The two dynamic responses of (31) and (36) are shown in Figure 10. The deviation between crossover gain and phase boost is negligible. However, at 120 Hz, the gain of (31) is 35 dB and (36) is 45 dB. Finally, the quasi-static gain of limited AOL is only 36.4db (Â» 66), while it is a perfect operational amplifier at infinity. When the gain is less than twice, the power frequency will affect the ability of the control system and suppress the rectifier ripple. The output variable may be affected by this element, especially under voltage mode control. In addition, if the implantation gain is low, the control variable may have significant static error. If you choose an op amp with higher AOL, such as 80dB, the deviation disappears and the two curves are very close to each other.

Figure 10: in the baud diagram of type 2, we believe that the open-loop gain AOL and low side resistance rlower do not greatly affect the original perfect equation.

05

summary

In the first two chapters of this paper, the influence of a compensator using a non ideal operational amplifier on the open-loop gain is introduced. If the op amp is not perfect, you can see the effect of weak open-loop gain in the low frequency range in the dynamic response to evaluate the performance degradation caused by this situation. get in touch with us
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