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Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.

In your case, we have

(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass charcteristiccharacteristic), and

(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.

Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).

Hence, the output is as expected and, therefore, correct.

Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.

In your case, we have

(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass charcteristic), and

(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.

Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external elements and lowpass property of the opamp).

Hence, the output is as expected and, therefore, correct.

Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.

In your case, we have

(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and

(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.

Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).

Hence, the output is as expected and, therefore, correct.

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source | link

Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.

In your case, we have

(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass charcteristic), and

(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.

Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external elements and lowpass property of the opamp).

Hence, the output is as expected and, therefore, correct.