# Active Filter Design

Discussion in 'Electronics Homework Help' started by shaneyj, Jun 15, 2017.

1. ### shaneyj

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Jun 10, 2017
Hello All,
We have started the unit on active filter design. The intro was short and sweet and this being an "introduction" to filters, band pass, and other jargon of which I am slightly unfamiliar, I am left wanting.
I understand the basic principle behind filters; block unwanted frequencies and allow others through.
I don't think our professor expects much of us, probably satisfied with following the textbook examples, but I really want to understand what I am doing.

This is what I have so far on the first part of the assignment...

Requirements: Low pass filter with a critical freq of 10kHz. (I assume critical to mean cutoff freq.)

We are instructed to use the design procedures from "Op Amps for Everyone" 4th Edition by Bruce Carter.

The values were calculated using formulas from Chapter 6 of the text. I was able to get a hint from another guy in the class about using a bode plotter to look at the results, but this is the first time I have used a bode plotter... Seems simple enough- dB over freq.

Can you help me decipher what I am looking at?

Does this look like the plot of a low pass filter? Thanks in advance to all the gurus...

24
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Jun 10, 2017

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3. ### OBW0549

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Jul 5, 2016
Some of your component values look rather... odd.

R1 and R2 are awfully small; remember, the 741 has to drive R1 (through C2) against whatever the input signal source is putting out, and 110Ω is a VERY heavy load. R1 and R2 should both be at least 10 times larger, or even more.

C2 looks like a mistake. Did you really mean 0.2 Farads? 0.2 microfarads would be more like it.

4. ### LvW

604
146
Apr 12, 2014
Yes - as OBWO549 has recommended - use C2=2*C1 and make a rescaling of the components.
Fore example: R1=R2=1100 Ohms and C1=10nF, C2=20nF.
However, please realize that the used opamp (741) has a slew rate that does not fulfill the requirements for a 10kHz cut-off frequency. You need a better opamp (higher slew rate).

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5. ### shaneyj

24
4
Jun 10, 2017
Good catch on C2. It should be 0.2uF.
As far as component values, the design instructions per the text say to start at 0.1uF for C1, then move forward from there. R1=R2=1/(2*sq rt 2*pi*C1*f) (I don't have the text in front of me, this is from memory).
It then says if resistor values are too small or large then pick another value for C1 and recalculate.
Being that I'm still learning and gaining an understanding, I don't know what too small or large is.

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6. ### shaneyj

24
4
Jun 10, 2017
Thank you for the advice on the op amp. I'll redo the circuit with that in mind.

7. ### OBW0549

159
118
Jul 5, 2016
That works, though for that high a frequency my first guess for C1 would be something closer to 0.01μF or even 1000pF; that's just my particular "gut feel", though.

As a general rule of thumb, I try to make sure that any loads on an op amp's output are no smaller than 1.0kΩ; the lower the load resistance, the harder it is for the op amp to drive and in most cases, the lower the op amp's open-loop gain.

"Too large" usually ends up being a resistance value which will create significant input offset errors due to the op amp's input bias current flowing through that resistance. For FET input op amps, I normally put that upper limit at about a megohm, and for bipolar op amps 100kΩ; but it all depends on the input bias current. Maximum resistance also is influenced by the environment, particularly the possibility of picking up stray interference; the more resistance, the more susceptible the circuit will be.

8. ### Audioguru

3,002
673
Sep 24, 2016
You should use a modern opamp that works at all audio frequencies. The lousy old 741 opamp was introduced 49 years ago, but not for audio (maybe AM radio sound quality?).

Your lowpass filter is simple so it does not block frequencies, instead it reduces frequencies. If the cutoff is at 10kHz then it is at -3dB which sounds a little less than lower frequencies. 20kHz will be reduced a little more to -12dB and 40kHz will be reduced 12dB more to -24dB which still has plenty of output level.

9. ### shaneyj

24
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Jun 10, 2017
Thanks for the response Audioguru... What does it mean that the gain is at -3dB? Can you articulate?

10. ### Audioguru

3,002
673
Sep 24, 2016
You have a professor and are studying active filters so I assumed you were taught about decibels.

11. ### LvW

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Apr 12, 2014
Clarification: It doesn`t reduce frequencies. Instead, it is correct to say "...it reduces the amplitudes of higher frequency componenets..."

12. ### Electrical Engineer David

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May 29, 2017
Decibels are another means of stating the ratio of the output to input. If the output is greater than the input then the term "gain" is used. If the output is smaller than the input then the term "attenuation" is used. For example, "The attenuation of the filter at 1 MHz is 100 dB".

The difference is the ratio is expressed in exponential terms and it is also weighted. The difference in the weighting between comparing voltages and power (i.e Watts) is such that regardless of whether you are referring to voltage or power the number of decibels will be the same. An example will help.

To convert between decibels and the ratio of voltages use

1) dB = 20 X log (Vout/Vin)

To convert between decibels and a ratio of power (i.e. Watts) use

2) dB = 10 X log (Pout/Pin)

If you have trouble with the term "log" read it as "the power of 10 that equals". For instance if Vout = Vin then Vout/Vin = 1. So the above would read 20 X the power of 10 that equals 1? Notice we use equation 1) since we are dealing with voltages not power levels. Well, since the power of 10 that equals 1 is 0, this becomes 20 X 0 = 0. So if the output voltage equals the input voltage this is 0 dB.

If Vout = 10 X Vin then Vout/Vin = 10. So the above would read 20 X the power of 10 that equals 10. Well, since the power of 10 that equals 10 is 1 this becomes 20 X 1 = 20 dB's.

If Vout = 100 X Vin then Vout/Vin = 100. So the above would read 20 X the power of 10 that equals 100. Well, since the power of 10 that equals 100 is 2 this becomes 20 X 2 = 40 dB's.

Note that if the output signal is smaller than the input signal the equation will result in a negative value. For example:

Vout = Vin/10 then Vout/Vin = 1/10th. So the above would read 20 X the power of 10 that equals 1/10th. Well since the power of 10 that equals 1/10th is -1 this becomes 20 X (-1) = -20 dB's.

In such cases you will often hear it stated as some number of dB's of attenuation. At other times it is stated as negative dB's. Still others will leave this out altogether if the speaker assumes the listener is aware it must be attenuation.

Every +20 dB increase is a ten-fold increase in the ratio of output voltage to input voltage (i.e. 20 dB means Vout = 10 X Vin, 40 dB means Vout = 100 X Vin, 60 dB = 1000 X Vin. Likewise every -20 dB's is a one-tenth decrease in the ratio of output voltage to input voltage (i.e. -20 dB means Vout = Vin/10, -40 dB means Vout = Vin/100, -60 dB means Vout = Vin/1000).

Another interesting and often used decibel increment is 6 dB. If Vout = 2 X Vin then the number of decibels is 20 X the power of 10 that equals 2. Since the power of 10 that equals 2 is 0.3 this becomes 20 X 0.3 = 6 dB. So every 6 dB increase corresponds to a further doubling of the ratio of output to input voltage.

From the above you should be able to recognize that +18 dB's is an 8-fold ratio between Vout and Vin since every 6 dB increase is a doubling of the voltage ratio (i.e 6 dB is X 2, 12 dB is X 4, and 18 dB is X 8 .

Likewise +26 dB means Vout is 20 times Vin. (Hint 26 is 20 + 6).

One final thing to note is 20 dB is a 10-fold increase in voltage. But it is a 100 fold increase in power since this will result in 10 times the current and power is Voltage X Current. Using equation 2) 10 dB's is a 10-fold increase in power. 20 dB's is a 100-fold increase in power. Likewise -10 dB's means Pout = Pin/10 and -20 dB's means Pout = Pin/100. Power level is actually the more common concern and hence what was used to define the units of decibels.

So when considering power the output to input power ratio increases 10-fold for every 10 dB increase and it doubles for every 3 dB increase. The +3 dB and -3 dB points are of particular interest as 3 dB's is a doubling in the power level while -3 dB's is a reduction in the output to input power by 1/2 (i.e. the half power level).

It is the -3 dB points on a filter which are used to define the bandwidth of the filter. If the -3 dB points on either side of a band-pass filter occur at 100 kHz and 105 kHz it is said to have a bandwidth of 5 kHz.

Why use decibels? Well as it turns out filters tend to have a linear drop-off in decibels when frequency is plotted logarithmically. In other words, if the x-axis is frequency with each tick defining a 10-fold increase in frequency then the gain or attenuation will often be a straight line. This is in fact what a Bode plot is for. On a Bode plot the ticks of the x-axis will be say 10Hz, 100Hz, 1KHz, 10KHz, etc resulting in a gain plot which is a line with a slope of say -20 dB's per decade of frequency. A slope of minus 20 dB's/decade of course means the voltage gain drops to 1/10 of its value every 10-fold increase (i.e. each x-axis tick) in frequency.

Last edited: Jun 17, 2017  