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AN2653 Datasheet(PDF) 8 Page - STMicroelectronics |
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AN2653 Datasheet(HTML) 8 Page - STMicroelectronics |
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8 / 22 page ![]() Out-of-the-loop compensation method AN2653 8/22 3 Out-of-the-loop compensation method 3.1 Theoretical overview A simple compensation method, using only one extra component, consists in adding a resistor in series between the output of the amplifier and its load (see Figure 13). It is often referred to as the out-of-the-loop compensation method because the additional component (ROL) is added outside of the feedback loop. The resistor isolates the op-amp feedback network from the capacitive load. From Figure 14, the loop gain with this compensation method is: This compensation introduces a zero in the loop gain, just after the pole caused by the capacitive load, at: This pole is also unfortunately shifted to lower frequencies at: However, due to the zero, the effect of the pole is minimized and the stability is improved. To obtain a good level of stability, ROL must be chosen such that the frequency of the zero occurs at least one decade before unity-gain frequency. It then allows a significant shift of the phase and therefore increase phase and gain margins. The previous equation shows that if ROL >> Zo, then -Vr/Vs= A(ω ), and the circuit is stable. In that case, pole and zero occur at the same frequency. However, the value of ROL is limited by the load impedance, ROL and RL acting as a a divider bridge from the operational amplifier output. Therefore, in order to minimize the error on Vout, ROL must be very small compared to RL (for example, a maximum of 1%, but this criterion depends on the required accuracy). Finally, this compensation method is effective, but the drawback is a limitation on the accuracy of Vout depending on the resistive load value. Figure 13. Out-of-the-loop compensation schematics Figure 14. Out-of-the-loop equivalent schematics for loop gain calculation V r V s ------ – A ω () 1 R OL R L ----------- jR OLCLω ++ ⎝⎠ ⎛⎞ ⋅ 1 R OL Z o + R L ------------------------ jZ o R OL + ()C Lω ++ --------------------------------------------------------------------------------- = 1 2 π R OL ⋅ R LCL || ------------------------------------------------ 1 2 π Z o R OL + () ⋅ R LCL || ------------------------------------------------------------------ |
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