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R1 represents a resistive sensor in the figure below. In reality, it has also a capacitor in parallel. I'm trying to understand the effect of this capacitor. When a parallel C(is labelled as C2) is added, the output is saturated. What is the explanation of this? How can i tackle this C which negatively affect the output? enter image description here

enter image description here

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When a parallel C(is labelled as C2) is added, the output is saturated. What is the explanation of this?

At 2 kHz, the 5 nF cap has an impedance of about 16 kohm. Because it is in parallel with a resistor of value 300 kohm, the op-amp gain is increased about twenty times and it is much more likely that the op-amp output will saturate.

How can i tackle this C which negatively affect the output?

Put a 15 nF capacitor across R5 to "balance" the effect of the 5 nF across R1. However, the op-amp output may show signs of instability so, also try a 100 ohm resistor in series with V1.

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  • \$\begingroup\$ Thanks, acc. to my calculation the impedance of 5nF at 2kHz is about 16kohm. Nop. Thanks for your answer. \$\endgroup\$ – Berker Işık Oct 16 at 8:45
  • \$\begingroup\$ C2 is also variable with R1, so it's dependent to R1. It is not easy to adjust also C1 capacitor on the fly. \$\endgroup\$ – Berker Işık Oct 16 at 8:47
  • \$\begingroup\$ Oops yes. I calculated for 1 kHz. Fixed now. \$\endgroup\$ – Andy aka Oct 16 at 8:49
  • \$\begingroup\$ @BerkerIşık did you try adding a capacitor across R5 to counter the effect of the 5 nF? \$\endgroup\$ – Andy aka Oct 26 at 9:57
  • \$\begingroup\$ Yes, I've added 15nF in simulation. Beacuse of C , gain is 1/3 as expected. Thanks for this. But the problem was dynamically changing value of R1. In this time the question is, how to set C every time according to the R1 to maintain the ratio R5/R1. \$\endgroup\$ – Berker Işık Oct 26 at 11:33
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At 2kHz the impedance of C2 is 16kOhm, hence your amplifier acts like if R1 was equal to 300kOhm || 16kOhm = 15kOhm

Without the capacitor, your gain is -1/3 but with the capacitor the gain becomes -6.7 and hence you saturate the op amp (1V * 6.7 = 6.7 V pick > 5V pick)

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