Dear Team,

May I know the name of the below circuit and it's purpose.

I know for DC it will be open and ac it's act as inverting amplifier.

But what is the use of this circuit

 

Hi hoy,
Where is the complete circuit where this clip from taken from?

I looks incomplete.
E

 

I got it this is a proportional controller
https://www.researchgate.net/figure...er-using-operational-amplifier_fig2_338104251

 

hi hoy,
It appears to part of a ramp oscillator loop, whose amplitude is controlled by the feedback loop formed by that lower left INA OPA configuration N4, R32 which controls the current flowing into the PT100 sensor. So keeping the PT100 ramp drive current at the set level by N5.

E

Update: your post #3, confirms this.

 

It performs the (inverting) proportional-integral function.

 

Ignore resistor R26 for the moment.
This is a lowpass filter or op amp integrator circuit.



The problem with this ideal integral function is that any DC input will be integrated to infinity. In other words, the op amp output will saturate to the supply rails. You need a resistor in parallel with the capacitor in order to create a “leaky” integrator.

 

It is an integrator. R26 adds a break frequency at the R26, C63 frequency. The voltage gain decreases as an integrator at a single-pole slope until this frequency is reached, then much above it, C63 is about a short and the gain flattens out to -R26/R21.

Thank you.
Do I need to keep a large resistor parallel with C63?

 

The problem with this ideal integral function is that any DC input will be integrated to infinity. In other words, the op amp output will saturate to the supply rails. You need a resistor in parallel with the capacitor in order to create a “leaky” integrator.

When the opamp stage - with either a C or a series R_C in the feedback path - is used as a controller (as mentioned by the TO in his post#3) it is NOT necessary to have an additional resistor in parallel to the feedback capacitor.
No DC problem because of the closed DC loop with negative feedback.

 

When C63 is fully charged,it will act as open circuit.Then there is no feedback path.Please correct me if I am wrong

 

When C63 is fully charged,it will act as open circuit.Then there is no feedback path.Please correct me if I am wrong

I wrote " When the opamp stage... is used as a controller...."
In this case, C63 will not be fully charged because the whole stage is part of a negative DC-stabilizing feedback loop.

The same applies - for example - when an opamp-based integrator stage is used as one unit within a state-variable filter circuit with overall feedback (KHN or Tow-Thomas)

 

Dear Team,

May I know the name of the below circuit and it's purpose.

I know for DC it will be open and ac it's act as inverting amplifier.

But what is the use of this circuit

View attachment 332772

It is a basically a Low Pass Filter with the Time Constant R26 * C63.

 

..... Without the feedback resistor, at 0 Hz (DC) the gain is the open-loop gain of the op-amp and that will cause the output to be unstable - not oscillatory, but amplifying anything at the input that departs from exactly 0 V, including the offset voltage of the op-amp.

Yes - but only in case the circuit is used as a stand-alone gain stage (see my post#9).