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 V_{2} is virtually grounded.

Therefore, I_{1} = I_{F} and V_{2} = V_{1} = 0

V_{IN} – 0 / R = C d(0 – V_{0}) / dt

Integrating both sides w.r.t time from 0 to t, we get

V_{IN} / R dt = C d(-v_{0}) /dt = dt

= C (-V_{0}) + V_{0}| t = 0

If V_{0}| _{T = 0} = 0 V

Then, V_{0} = -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 V_{0} = 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} – V_{0} / R + C dV_{B} / dt = 0

Or 2V_{B} / R – V_{S} / R – V_{0} /R + C dV_{B} / dt = 0

V_{0}/R – V_{S} / R – V_{0}/ R + C/2 (dV_{0} / dt) = 0

dV_{0} / dt = 2V_{S} /RC

and V_{0} = | 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 V_{2}.

Since, I_{IN} = I_{F}

Therefore, C d/dt (V_{IN} – 0) = 0 – V_{0} / R

Thus. the output V_{0} 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_{F}C.

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 + R_{1} / R_{2})

**Sol. **The current through R_{1} can be obtained from the current divider cicuit.

I_{1} = I_{L} (R_{2} / R_{1} + R_{2})

Since, the input impedance of op-amp is very large, the input current of op-amp is negligible.

or I_{I} = I_{1} = – I_{L} (R_{2} / R_{1} + R_{2})

Thus, I_{L} / I_{I} = – (R_{1} / R_{2})

**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.

- Analog or digital
- Passive or active
- 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

- Low-pass filter
- High-pass filter
- Band-pass filter
- Band-pass filter
- 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 R_{1} and R_{2} decides the gain of the filter.

According to voltage divider rule, the voltage at the non-inverting terminal is

V_{1} = -JX_{C} / R – JX_{C} = V_{IN} where X_{C} = 1 /2fC

-J 1 / 2fC

V_{1} = 2fC / 2fCR – j = V_{IN}

2fC

-j / 2fCR – j = V_{IN}

1 / 1 + j2fCR = V_{2}

V_{0} = ( 1 + R_{f} / R_{1} ) V_{1}

= (1 + R_{F} / R_{1}) V_{IN} / 1 + j2fCR

V_{0}/V_{IN} = A_{f} / 1 + j(f/f_{h})

Where, V_{0}/V_{IN} is the gain of the filter as a function of frequency.

A_{F} = (1 + R_{F} / R_{1}) = 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

=| V_{0} / V_{IN}| = A_{f} / 1 + (f/f_{H})

and phase angle = – tan ^{-1} (f/f_{h})

At low frequencies = |V_{0}/V_{IN}| = A_{f }

At f = f_{H} , = |V_{0} / V_{IN}|= A_{f} / 2 = 0.707 A_{F}

and at f > f_{H} |V_{0} /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:

- Choose a value of high cut-off frequency f
_{h}. - Select a value of C less than or equal to 1 uF.
- Calculate the value of R = 1 / 2f
_{H}C - Finally, select values of R
_{1}and R_{F}to set the desired gain using A_{F}= 1 + R_{F}/R_{1}

**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 10^{3} 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, R_{1} and R_{F} must be equal. Let R_{1} = R_{2} = 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 R_{1} and R_{F}, while the high cut-off frequency f_{H} is determined by R_{2}, C_{2}, R_{3} and C_{3} as follows:

f_{H} = 1 /2 R_{2}R_{3}C_{2}C_{3}

Furthermore, for a second-order low-pass butter-worth response, the voltage gain magnitude is given by

|V_{0} / V_{IN}| = A_{F} / 1 +( f/f_{H})^{4}

A_{F} = 1 + R_{F}/R_{1} = 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.

V_{0} = (1 + R_{F}/R_{1})* V_{1}

= (1 + R_{F} /R_{1}) * J2fRC/1 + J2fRC = * V_{IN}

V_{0} / V_{IN} = A_{F} |f/f_{L}) / 1 + J(f/f_{L})|

The magntiude of the gain of the filter is

|V_{0} /V_{IN}| = A_{F} (f/f_{L}) / 1 + (f/f_{L})

at f < f_{L} |V_{0}/V_{IN}|< A_{F}

at f = f_{L }|V_{0}/V_{IN}|< A_{F} 2

at f > f_{L} |V_{0} /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).