Integrator : what is Integrator definition , formula , meaning circuit waveform transfer function

miller integrator transfer function , Integrator : what is Integrator definition , formula , meaning circuit waveform ?
Integrator
A circuit in which the output voltage waveform is the integral of the input voltage waveform is called integrator. Fig. 46 (a) shows an integrator circuit using op-amp.
Here, the feedback element is capacitor. the current drawn by op-amp is zero and the V₂ is virtually grounded.
Therefore,           I₁ = I_F and V₂ = V₁ = 0
V_IN – 0 / R = C d(0 – V₀) / dt
Integrating both sides w.r.t time from 0 to t, we get
V_IN / R dt = C d(-v₀) /dt = dt
= C (-V₀) + V₀| t = 0
If                        V₀| _{T = 0} = 0 V
Then,             V₀ = -1 /R     V_IN dt
The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RC.
If the input is the sine wave, the output will be cosine wave, if the input is a square wave, the output will be a triangular wave. for accurate integration, the time period of the input signal T must be longer than or equal to RC.
Fig. 46(b) shows the output of integrator for square and sinusoidal inputs.
Example 11. Prove that the network shown in figure below is a non-inverting integrator with V₀ = 2 / RC | V_S (T) dt.
Sol. The voltage at point A is V_0/2 and it is also the voltage at point B because different input voltage is negligible.
Therefore, applying node current equation at point B,
Or              V_B – V_A / R + V_B – V₀ / R + C dV_B / dt = 0
Or               2V_B / R – V_S / R – V₀ /R + C dV_B / dt = 0
V₀/R – V_S / R – V₀/ R + C/2 (dV₀ / dt) = 0
dV₀ / dt = 2V_S /RC
and                         V₀ = | 2/ RC V_S
Differentiator
A circuit in which the output voltage waveform is the differention of input voltage is called differentiator as shown is fig. 47(a).
The expression for the output voltage can be obtained from the kirchhoff’s current equation written at node V₂.
Since,                      I_IN = I_F
Therefore,            C d/dt (V_IN – 0) = 0 – V₀ / R
Thus. the output V₀ is equal to the RC times the negative instantaneous rate of change of the input voltage V_IN with time. a cosine wave input produces sine output.Fig. 47 (b) also shows the output waveform for different input voltages.
The input signal will be differentiated properly if the time period T of the input signal is larger than or equal to R_FC.
As the frequency changes, the gain changes. also at higher frequencies, the circuit is highly susceptible at high frequency noise and gets amplified. both the high frequency noise and problem can be corrected by additing, new components as shown in fig. 47 (c).
Voltage to Current Converter
Fig. 48 shows a voltage to current converter in which load resistor R_L is floating (not connected to ground).
The input voltage is applied to the non-inverting input terminal and the feedback voltage across R drives the inverting input terminal. this circuit is also called a current series negative feedback amplifier because the feedback voltage across R depends on the output current I_I and is in series with the input difference voltage V_d. writing the voltage equation for the input loop.
But V_d >> , since A is very large, therefore,
V_IN = V_F
V_IN = R I_IN
VIN = VIN /R
and since, input current is zero.
I_L = I_IN  = V_IN /R
The value of load resistance does not appear in this equation. therefore, the output current is independent of the value of load resistance. thus, the input voltage is converted into current, the source must be capable of supplying this load current.
Current to Voltage Converter
The circuit shown in fig. 49 is a current to voltage converter.
Due to virtual ground the current through R is zero and the input current flows through R_F.
Therefore,                         V_OUT = – R_F * I_IN
The lower limit on current measure with this circuit is set by the bias current of the inverting input.
Example 12. For the current converter shown in figure below, prove that
I_L / I_I = – (1 + R₁ / R₂)
Sol. The current through R₁ can be obtained from the current divider cicuit.
I₁ = I_L (R₂ / R₁ + R₂)
Since, the input impedance of op-amp is very large, the input current of op-amp is negligible.
or                          I_I = I₁ = – I_L (R₂ / R₁ + R₂)
Thus,                             I_L / I_I = – (R₁ / R₂)
Filters
A filter is a frequency selective circuit that passes a specified band of frequencies and blocks or attenuates signals of frequencies out side this band. filter may be classified on a number of ways.

1. Analog or digital
2. Passive or active
3. Audio or radio frequency

Analog filters are designed to process only signals while digital filters process analog signals using digital technique. depending on the type of elements used in their consideration, filters may be classified as passive or active.
Elements used in passive filters are resistors, capacitors and inductors, active filters on the other hand, employ transistors of op-amps, in addition to the resistor and capacitors. depending upon the elements the frequency range is decided.
RC filters are used for audio or low frequency operation LC filters are employed at RF of high frequencies.
The most commonly used filters are

1. Low-pass filter
2. High-pass filter
3. Band-pass filter
4. Band-pass filter
5. All-pass filter

Fig. 50 shows the frequency response characteristics of the five types of filter. the ideal response is shown by dashed line. while the solid lines indicates the practical filter response.
A low-pass filter has a constant gain from 0 Hz to a high cut-off frequency f_H. therefore, the bandwidth is f_H. At f_H the gain is down by 3 dB. after the the gain decreases as frequency increases. the frequency range 0 to f_H Hz is called pass-band  and beyond f_H is called stop-band.
Similarly, a high-pass filter has a gain from very high frequency to a low cut-off frequency f_L, below f_L the gain decreases as frequency decreases. At f_L the gain is down by 3 dB. the frequency range f_L Hz to infinity is called pass-band and below f_L is called stop band.
First Order Low-Pass Filter
Fig. 51, shows a fist order low-pass butter-worth filter that uses an RC network for filtering, op-amp is used in non-inverting configuration R₁ and R₂ decides the gain of the filter.
According to voltage divider rule, the voltage at the non-inverting terminal is
V₁ = -JX_C / R – JX_C = V_IN where X_C = 1 /2fC
-J 1 / 2fC
V₁ = 2fC / 2fCR – j = V_IN
2fC
-j / 2fCR – j = V_IN
1 / 1 + j2fCR = V₂
V₀ = ( 1 + R_f / R₁ ) V₁
= (1 + R_F / R₁) V_IN / 1 + j2fCR
V₀/V_IN = A_f / 1 + j(f/f_h)
Where, V₀/V_IN is the gain of the filter as a function of frequency.
A_F = (1 + R_F / R₁) = Pass-band filter gain
f = Frequency of input signal
f_H = Cut-off frequency of the filter = 1/2RC
Magnitude of the gain of low-pass filter
=| V₀ / V_IN| = A_f / 1 + (f/f_H)
and phase angle = – tan ^-1 (f/f_h)
At low frequencies = |V₀/V_IN| = A_f                
At                 f = f_H , = |V₀ / V_IN|= A_f / 2 = 0.707 A_F
and at                  f > f_H |V₀ /V_IN| <A_F
Thus, the low-pass filter has a nearly constant gain A_F from 0 Hz to high cut-off frequency f_H. At f_h the gain is 0.707 A_F and after f_H, it decreases at a constant rate with an increases in frequency. f_h is called cut-off frequency because the gain of filter at this frequency is reduced by 3 dB from 0 Hz.
A low-pass filter can be designed using the following steps:

1. Choose a value of high cut-off frequency f_h.
2. Select a value of C less than or equal to 1 uF.
3. Calculate the value of R = 1 / 2f_HC
4. Finally, select values of R₁ and R_F to set the desired gain using A_F = 1 + R_F/R₁

Example 13. Design a low-pass filter at a cut-off frequency of 1 KHz with a pass-band gain of 2.
Sol. Given f_h = 1kHz. Let C = 0.01 uF
Therefore, R can be obtained as
R = 1 / 2 x 10³ x 0.01 x 10^-6 = 15.9 k
A 20 k potentiometer can be used to set the resistance R. Since, the pass-band gain is 2, R₁ and R_F must be equal. Let R₁ = R₂ = 10 K
Second Order Low-pass Butter-Worth Filter
A stop-band response having a 40 dB / decade at the cut-off frequency is obtained with the second-order low-pass filter. A first order low-pass filter can be converted into a second order low-pass filter by using an addition RC network as shown in Fig. 52 (a).
The gain of the second order filter is set by R₁ and R_F, while the high cut-off frequency f_H is determined by R₂, C₂, R₃ and C₃ as follows:
f_H = 1 /2 R₂R₃C₂C₃
Furthermore, for a second-order low-pass butter-worth response, the voltage gain magnitude is given by
|V₀ / V_IN| = A_F / 1 +( f/f_H)⁴
A_F = 1 + R_F/R₁ = pass-band gain of the filter
where, f = frequency of the input signal
Except for having the different cut-off frequency, the frequency response of the second order low-pass filter is identical to that of the first order type as shown in fig. 52 (b).
First Order High-Pass Butter-Worth Filter
Fig. 53(a) shows the circuit of first order high-pass filter. this is formed by interchanging R and C in low-pass filter.
The lower cut-off frequency is f_L. this is the frequency at which the magnitude of the gain is 0.707 times its pass-band value. all frequencies higher than f_L are pass-band frequencies with the highest frequency determined by the closed-loop bandwidth of the op-amp.
V₀ = (1 + R_F/R₁)* V₁
= (1 + R_F /R₁) * J2fRC/1 + J2fRC  = * V_IN
V₀ / V_IN = A_F |f/f_L) / 1 + J(f/f_L)|
The magntiude of the gain of the filter is
|V₀ /V_IN| = A_F (f/f_L) / 1 + (f/f_L)
at               f < f_L  |V₀/V_IN|< A_F
at              f = f_L |V₀/V_IN|< A_F 2
at              f > f_L |V₀ /V_IN|= A_f
If the two filters (high and low) band-pass are connected in series it becomes wide band filter whose gain frequency response is shown in fig. 53(b).