# Itis Strange

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## Itis Strange

's contributions- 01.31.2011
- Design second- and third-order Sallen-Key filters with one op amp
- Below it should have said, Capacitor value ratios must be less than the square of the damping factor.

- 01.31.2011
- Design second- and third-order Sallen-Key filters with one op amp
- bryman 79, I missed your question below. The first order section in Figure 2 is unconditionally stable. Its second order section is the same as that in Figure 1. Equation 28 evaluates the stability of both. The Oscillation frequency calculated in the Okawa tool is the imaginary part of the pole with the lowest damping factor. You can see that these two are identical. Thanks for referencing the Okawa tool at http://sim.okawa-denshi.jp/en/Sallenkey3Lowkeisan.htm . It has a much nicer user interface than the article's spreadsheet. The Okawa tool implements designs for which op amp gains are exactly 1. These have very low sensitivities to resistor and capacitor values, a good thing. But this is also a limitation in that the ratio of capacitor values in sections with high Q (low damping ratio) poles can be very far from 1. Capacitor value ratios must be greater than the square of the damping factor. In some cases, this can cause problems. The article's spreadsheet lets you trade off a bit of low sensitivity for lower capacitor ratios by specifying op amp gains of a little more than 1. Or you can specify gains of 1 and get the lowest possible sensitivity designs as with Okawa and live with the capacitor ratios. The proper strategy is to select a maximum capacitor ratio that you can tolerate and then experiment with the lowest gain that gives you a realizable filter.

- 01.31.2011
- Design second- and third-order Sallen-Key filters with one op amp
- HI All, the link to the spreadsheet has been restored.

- 01.31.2011
- Design second- and third-order Sallen-Key filters with one op amp
- Figure 1 corresponds to equation 28 and Figure 3 to equation 38. I'll see if I can get EDN to fix the link.

- 08.23.2017
- In defense of the current-feedback amplifier
- I’d like to propose a simple means for determining whether current or voltage is the predominant form of feedback in any op amp-based circuit. The method requires no knowledge of the inner workings of the op amp. Only simple impedance measurements are required. All op amps have a non-zero, finite differential input impedance Z as seen at their inverting inputs. This means that their output levels are proportional to the signal levels of either voltage or current sources connected across their inputs. Therefore, there is no basis for claiming that any op amp is completely unresponsive to either voltage or current feedback. But there is a reasonable criterion for determining which type is dominant. The impedance Z sees a feedback network whose Thevenin equivalent is a voltage source V in series with a resistance R. The Norton equivalent is a current source I in parallel with R. Once we constrain V = I x R, the two equivalents are indistinguishable from the point of view of the impedance Z. Regardless of which equivalent you apply, if Z is less than R, it is more accurate to say that Z sees a current source I = V / R than a voltage source V = I x R. The reverse is true if Z is greater than R. As such, no op amp is inherently predominantly responsive to one type of feedback; it is the combination of the inverting input and the feedback network impedances which makes this determination. Simulations of representative op amps dubbed CFA’s by the industry reveal their values of Z. Consulting their data sheets shows that the manufacturer’s recommended feedback networks for most (but not all) gain and power supply combinations are such that Z is less than R. In those cases, it is more accurate to state that the entire circuit is more responsive to current than voltage feedback. And that is the justification for the industry referring to them as CFA’s.

- 08.23.2017
- In defense of the current-feedback amplifier
- Consider any commercially available CFA IC, or one built out of discrete transistors. Configure it for closed loop operation. Excite the non-inverting input with a voltage Vi. Measure the inverting input current In. Measure the differential input voltage Vnp. Measure the output voltage Vo. Calculate rn = Vnp/In and z = Vo/In. Simulate Sergio Franco’s Figure 9 model using the calculated values and the value of Vi. You’ll discover excellent agreement between the CFA and the model.

- 08.23.2017
- In defense of the current-feedback amplifier
- 4% error is "great accuracy"? On what planet? You're evading and ignoring the the sim I sent you showing agreement to better than 0.2% between your discrete model and Sergio Franco's Figure 3.

- 08.23.2017
- In defense of the current-feedback amplifier
- Moo Koo, the 4% disagreement between your simulation and your equation is suspicious. I sent you some results by email. I removed the 22K from the model you were working with, something not found in a CFA. I measured your model’s inverting input impedance and transimpedance. I populated Sergio Franco’s Figure 3 model with it and developed the equations for it using only Kirchoff’s Laws (all that was needed.) I then simulated your discrete circuit and Sergio’s model. The equation and Sergio’s sim agreed to four decimal places (I didn’t check for more.) They also agreed with the sim of your discrete circuit without the 22k to 3 ½ decimal places. If your criterion for vindication is the degree to which a model matches a simulation, then you must discard your own in favor of Sergio’s. This thing the industry calls a CFA can be well explained by Sergio’s model… and current feedback.

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