**Active Filters**

In the RC Passive Filter tutorials, we saw how a basic first-order filter circuits, such as the low pass and the high pass filters can be made using just a single resistor in series with a non-polarized capacitor connected across a sinusoidal input signal.

We also noticed that the main disadvantage of passive filters is that the amplitude of the output signal is less than that of the input signal, ie, the gain is never greater than unity and that the load impedance affects the filters characteristics.

With passive filter circuits containing multiple stages, this loss in signal amplitude called “Attenuation” can become quiet severe. One way of restoring or controlling this loss of signal is by using amplification through the use of **Active Filters**.

As their name implies, **Active Filters** contain active components such as operational amplifiers, transistors or FET’s within their circuit design. They draw their power from an external power source and use it to boost or amplify the output signal.

Filter amplification can also be used to either shape or alter the frequency response of the filter circuit by producing a more selective output response, making the output bandwidth of the filter more narrower or even wider. Then the main difference between a “passive filter” and an “active filter” is amplification.

An active filter generally uses an operational amplifier (op-amp) within its design and in the Operational Amplifier tutorial we saw that an Op-amp has a high input impedance, a low output impedance and a voltage gain determined by the resistor network within its feedback loop.

Unlike a passive high pass filter which has in theory an infinite high frequency response, the maximum frequency response of an active filter is limited to the Gain/Bandwidth product (or open loop gain) of the operational amplifier being used. Still, active filters are generally much easier to design than passive filters, they produce good performance characteristics, very good accuracy with a steep roll-off and low noise when used with a good circuit design.

The most common and easily understood active filter is the **Active Low Pass Filter**. Its principle of operation and frequency response is exactly the same as those for the previously seen passive filter, the only difference this time is that it uses an op-amp for amplification and gain control. The simplest form of a low pass active filter is to connect an inverting or non-inverting amplifier, the same as those discussed in the Op-amp tutorial, to the basic RC low pass filter circuit as shown.

This first-order low pass active filter, consists simply of a passive RC filter stage providing a low frequency path to the input of a non-inverting operational amplifier. The amplifier is configured as a voltage-follower (Buffer) giving it a DC gain of one, Av = +1 or unity gain as opposed to the previous passive RC filter which has a DC gain of less than unity.

The advantage of this configuration is that the op-amps high input impedance prevents excessive loading on the filters output while its low output impedance prevents the filters cut-off frequency point from being affected by changes in the impedance of the load.

While this configuration provides good stability to the filter, its main disadvantage is that it has no voltage gain above one. However, although the voltage gain is unity the power gain is very high as its output impedance is much lower than its input impedance. If a voltage gain greater than one is required we can use the following filter circuit.

The frequency response of the circuit will be the same as that for the passive RC filter, except that the amplitude of the output is increased by the pass band gain, A_{F} of the amplifier. For a non-inverting amplifier circuit, the magnitude of the voltage gain for the filter is given as a function of the feedback resistor ( R_{2} ) divided by its corresponding input resistor ( R_{1} ) value and is given as:

Therefore, the gain of an active low pass filter as a function of frequency will be:

- Where:
- A
_{F}= the pass band gain of the filter, (1 + R2/R1) - ƒ = the frequency of the input signal in Hertz, (Hz)
- ƒc = the cut-off frequency in Hertz, (Hz)

Thus, the operation of a low pass active filter can be verified from the frequency gain equation above as:

- 1. At very low frequencies, ƒ < ƒc
- 2. At the cut-off frequency, ƒ = ƒc
- 3. At very high frequencies, ƒ > ƒc

Thus, the **Active Low Pass Filter** has a constant gain A_{F} from 0Hz to the high frequency cut-off point, ƒ_{C}. At ƒ_{C} the gain is 0.707A_{F,} and after ƒ_{C} it decreases at a constant rate as the frequency increases. That is, when the frequency is increased tenfold (one decade), the voltage gain is divided by 10.

In other words, the gain decreases 20dB (= 20log 10) each time the frequency is increased by 10. When dealing with filter circuits the magnitude of the pass band gain of the circuit is generally expressed in *decibels* or *dB* as a function of the voltage gain, and this is defined as:

Design a non-inverting active low pass filter circuit that has a gain of ten at low frequencies, a high frequency cut-off or corner frequency of 159Hz and an input impedance of 10KΩ.

The voltage gain of a non-inverting operational amplifier is given as:

Assume a value for resistor R1 of 1kΩ rearranging the formula above gives a value for R2 of

then, for a voltage gain of 10, R1 = 1kΩ and R2 = 9kΩ. However, a 9kΩ resistor does not exist so the next preferred value of 9k1Ω is used instead.

converting this voltage gain to a decibel dB value gives:

The cut-off or corner frequency (ƒc) is given as being 159Hz with an input impedance of 10kΩ. This cut-off frequency can be found by using the formula:

where ƒc = 159Hz and R = 10kΩ. |

then, by rearranging the above formula we can find the value for capacitor C as:

Then the final circuit along with its frequency response is given below as:

If the external impedance connected to the input of the circuit changes, this change will also affect the corner frequency of the filter (components connected in series or parallel). One way of avoiding this is to place the capacitor in parallel with the feedback resistor R2.

The value of the capacitor will change slightly from being 100nF to 110nF to take account of the 9k1Ω resistor and the formula used to calculate the cut-off corner frequency is the same as that used for the RC passive low pass filter.

An example of the new **Active Low Pass Filter** circuit is given as.

Applications of **Active Low Pass Filters** are in audio amplifiers, equalizers or speaker systems to direct the lower frequency bass signals to the larger bass speakers or to reduce any high frequency noise or “hiss” type distortion. When used like this in audio applications the active low pass filter is sometimes called a “Bass Boost” filter.

As with the passive filter, a first-order low-pass active filter can be converted into a second-order low pass filter simply by using an additional RC network in the input path. The frequency response of the second-order low pass filter is identical to that of the first-order type except that the stop band roll-off will be twice the first-order filters at 40dB/decade (12dB/octave). Therefore, the design steps required of the second-order active low pass filter are the same.

When cascading together filter circuits to form higher-order filters, the overall gain of the filter is equal to the product of each stage. For example, the gain of one stage may be 10 and the gain of the second stage may be 32 and the gain of a third stage may be 100. Then the overall gain will be 32,000, (10 x 32 x 100) as shown below.

Second-order (two-pole) active filters are important because higher-order filters can be designed using them. By cascading together first and second-order filters, filters with an order value, either odd or even up to any value can be constructed. In the next tutorial about filters, we will see that Active High Pass Filters, can be constructed by reversing the positions of the resistor and capacitor in the circuit.

Error! Please fill all fields.

good work

The non-inverting lpf does not have the specified input impedance of 10k ohms, but rather a much higher impedance. In fact the 10k resistor doesn’t do much if the particular amp used has a typically high input impedance. To lower the impedance to 10k, , the resistor should go to ground and the input applied directly to the input pin .

Excellent tutorial. It is very useful for beginners.

It is explained very simple way.

Thanks

need help..

i’m using a simplified non-inverting amplifier filter circuit..with R1=6.8k, R2=680k, C=100n and input voltage, Vin=20mV..when i try this circuit in multisim, i got Vout= 81.15mV and its mean i have gain around 4.. how can i prove it by calculation?

You have a DC gain of 101 and a high frequency gain of 1 as R2 is shorted by the capacitor. Your gain of 4 is therefore at some frequency value in between. For a gain of 4 you need R1 = 6k8 and R2 = 20k4. This will give Xc as 21031 ohms, then find f by substitution in Xc = 1(2pi.f.C), which should give 75.7Hz. Use the formulas above.

excellent tutorials u have given. i learned a lot thank u.

I have a question regarding the magnitude frequency response of the “Simplified non-inverting amplifier filter circuit”.

Does the magnitude approach (R1+R2)/R1 at very low frequencies and approach “1” at very high frequencies?

This is what I found when I derived the transfer function. I was surprised by this because the low pass active filter we studied in class (which is the one you call “Equivalent inverting amplifier filter circuit”) approaches zero as frequencies get really high.

thanks for your help in advance.

Hello RM, Both LPF’s are the same except one is a non-inverting amplifier while the other is an inverting amplifier. As calculated above, the voltage gain of the circuit is 10 at low frequencies (dc to 159Hz), reducing at -20dB/decade at the corner frequency, 159Hz to a gain of 1 (unity) at higher frequencies, (around 10kHz and above) for the non-inverting amp.

This is because the gain of a non-inverting amplifier is 1+(R2/R1). Then at low frequencies R2 = 9k1ohms and at high frequencies resistor R2 is shorted by the capacitor as Xc becomes very small, so the op-amps gain simply becomes 1+(0/1k) = 1. Then your calculations are correct. 🙂

A clear explanation.well done.

[spam deleted]

very easy to understand

Great work. Easy explaination