Response time needed for the circuit shown

[patan.gova]

Hello,
How to calculate the response time(time it takes to give the output) of the circuit shown below.



Thanks in Advance.

 

[Tunelabguy]

The passive HPF has no effect on initial response time (to a step function from the sensor). The LPF is first order, and so it reaches 63% of its ultimate (gain=101) value in 68 msec. (which is 680K x 100nF). But by this time the HPF is starting to fall off too, so it starts heading back toward zero with a time constant of 220 msec, so the output of the amplifier never does reach its gain=101 value. Calculating what the response time is for this circuit depends on how you want to define the response time. Normally for a first order system, the response time is the time to reach 63% of the eventual steady-state value. But since this circuit never does reach the 100% value, the meaning of the 63% point is somewhat ambiguous.

 

[LvW]

Just one comment/correction:
The lowpass time constant is lower by a factor of 100:

T=C*(680k||6.8k)~C*6.8kohms=860 µsec.

 

[patan.gova]

Tunelabguy "Calculating what the response time is for this circuit depends on how you want to define the response time."

I am collecting the samples in a 1 sec interval of the signal resulted from the output of this circuit.So, I want to know the resposne time of the circuit so that it will used(helps) in the post processing of the signal for the final result analysis.

Sorry LvW,
In this formula T=C*(680k||6.8k)~C*6.8kohms, I didn't got about this 680k||6.8k..

And can I assume the response time of the overall circuit as 860usec or is there anything more to add.

 

[patan.gova]

Hello, could someone explain a but clearly about the response time needed by the circuit in post[1].

 

[BradtheRad]

Here is a simulation of the response to 1Hz AC square waves.

The output immediately jumps to the input voltage plus the charge on the capacitor.
Then it starts heading toward zero.

 

[patan.gova]

Thanks BradtheRad.

and What is the formula used or the expression to calculate the response time for the circuit shown in the post1.

 

[BradtheRad]

What is the formula used or the expression to calculate the response time for the circuit shown in the post1.

My simple simulation only demonstrated the time constant for the first stage (RC network). However you will not apply square waves to your device.

I think you are interested in the cutoff frequencies for the LP and HP filters?

https://en.wikipedia.org/wiki/RC_time_constant

1.

The high-pass section blocks slow voltage changes.

The cutoff frequency is calculated as f = 1 / ( 2 * Pi * R * C )

The values are 4.7 uF and 47K.

The cutoff frequency works out to 0.7 Hz.

2.

The low pass section consists of an op amp whose feedback circuit is designed to attenuate rapid voltage changes.

The 100 nF capacitor charges and discharges through the 680k resistor. This works out to a cutoff frequency of 2.34 Hz.

 

[FvM]

The lowpass time constant is lower by a factor of 100:

T=C*(680k||6.8k)~C*6.8kohms=860 µsec.

Due to the non-invertiging topology, it's no pure low-pass, the transfer characteristic is "1 + LP", a pole-zero combination.

The low-pass time constant (pole) is 68 ms as said by Tunelabguy, 680 us is an additional zero.

 

[BradtheRad]

I was off base about which resistor the capacitor charges and discharges through. It is in fact the 6.8k resistor, because the op amp output provides a low-impedance connection to the supply rails.

As for the question, I would not be surprised if it is really about the response curve to a sine sweep.

Simulation of entire circuit shown in OP. The center frequency is 2.2 Hz. The effective gain turns out to be around 80 (not 100 as explained by Tunelabguy, post #2).



To show the LP response only, I switched out the first capacitor. The rolloff frequency appears to be about 2.2 Hz. The gain is 100.