Second Order Op-amp Circuit

In summary: As soon as the input is changed from 0 to v_g , then \frac{dv_o}{dt} is no longer 0.In summary, the rate of change of the output of both op-amps is zero when the initial energy stored in the circuit is zero and the input signal jumps instantaneously from 0 to Vg. However, as soon as the input changes, the rates of change will no longer be zero. This may initially be confusing, but it is a natural consequence of the circuit's behavior.
  • #1
paulmdrdo
89
2
Homework Statement
the rate of change of the output of both op-amps when the initial energy stored in the circuit is zero and the signal jumps instantaneously from 0 to Vg.
Relevant Equations
SEE ATTACHED IMAGE
242722


I was reading about this 2nd order op-amp circuit which is essentially a cascaded integrator and got confused with the explanation of the book regarding the rate of change of the outputs. The book said that when the initial energy stored in the circuit is zero then

242715


this rate of change is zero when Vo1(0) = 0(initial energy is zero). Now going to the rate of change of the first of amp,
242716

would this also be equal to zero? The reason I asked this is when I derived for this rate of change I ended up with this

242717

the input Vg is a non zero quantity therefore this rate of change can not be zero. But the initial energy stored is zero so this should not be the case. This got me confused. Please enlighten me. TIA!
 
Last edited:
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  • #2
paulmdrdo said:
Problem Statement: the rate of change of the output of both op-amps when the initial energy stored in the circuit is zero and the signal jumps instantaneously from 0 to Vg.
Relevant Equations: SEE ATTACHED IMAGE

View attachment 242722

I was reading about this 2nd order op-amp circuit which is essentially a cascaded integrator and got confused with the explanation of the book regarding the rate of change of the outputs. The book said that when the initial energy stored in the circuit is zero then

View attachment 242715

this rate of change is zero when Vo1(0) = 0(initial energy is zero). Now going to the rate of change of the first of amp,
View attachment 242716
would this also be equal to zero? The reason I asked this is when I derived for this rate of change I ended up with this

View attachment 242717
the input Vg is a non zero quantity therefore this rate of change can not be zero. But the initial energy stored is zero so this should not be the case. This got me confused. Please enlighten me. TIA!
Your equations look correct to me. :smile:

You are correct that the rates of change of voltages, [itex] \frac{dv_{o1}}{dt} [/itex] and [itex] \frac{dv_o}{dt} [/itex] are zero when the input is 0 V, and both capacitors are uncharged (zero initial energy).

But it doesn't stay that way. As soon as the input is changed from 0 to [itex] v_g [/itex], then [itex] \frac{d v_{o1}}{dt} [/itex] is no longer 0.

You can make a similar statement about the next stage too.
 

What is a Second Order Op-amp Circuit?

A Second Order Op-amp Circuit is a type of electronic circuit that uses an operational amplifier (op-amp) to amplify and manipulate an input signal. It is called "second order" because it has two reactive components (such as capacitors or inductors) in its feedback loop, which allows it to have a second-order response to changes in the input signal.

What are the advantages of using a Second Order Op-amp Circuit?

One advantage of using a Second Order Op-amp Circuit is its ability to provide a more precise and stable output signal compared to a first order circuit. This is because it has a higher order response, which means it can better filter out noise and unwanted signals. Additionally, second order circuits can also have a wider bandwidth and faster response time.

What are some common applications of Second Order Op-amp Circuits?

Second Order Op-amp Circuits are commonly used in audio and signal processing applications, such as active filters, oscillators, and amplifiers. They are also used in control systems, where precise and stable amplification is required.

What is the difference between a Second Order Op-amp Circuit and a First Order Op-amp Circuit?

The main difference between these two types of circuits is the number of reactive components in their feedback loops. A first order circuit has only one reactive component, while a second order circuit has two. This means that a second order circuit has a higher order response and can provide more precise and stable amplification compared to a first order circuit.

What are some common challenges when designing a Second Order Op-amp Circuit?

One common challenge when designing a Second Order Op-amp Circuit is ensuring stability. The addition of a second reactive component can introduce more complexity and potential for instability in the circuit. Another challenge is selecting the right values for the reactive components to achieve the desired frequency response and gain. Additionally, noise and interference can also be a challenge in these circuits, so proper shielding and grounding techniques may be necessary.

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