Help designing analog filter designs for an audio activated 8x8x8 LED cube

I am working on designing a music visualization system using LEDs. I would like to create an analog design that uses op amps to filter, amplify, and add DC bias to an audio signal so it can be read into the A/D of a 5V microprocessor. The input signal is a +/- 1.5V volt audio signal coming from an MP3 player. From this signal I need to produce three output signals. This means that I need to design three filters (a lowpass, a bandpass, and a highpass). These filters also need to amplify and bias the input signal so it will be a 0 to 5V signal that can be read by the processor's A/D at the maximum resolution. So basically what I have pictured is to first bias the signal, then amplify it, then filter it:

MP3 Player Audio signal --> Bias --> Amplify --> Filter --> uP

I am familiar with building filters with amplification, but I am not sure about the correct way to go about adding a DC bias to the filters with amplification.

Lowpass (bass range): cutoff f @ ~500 Hz


Bandpass (mid range): center f @ ~2250 Hz

Highpass (high range): cutoff f @ ~4kHz


The parts I have available for use are: LT1632 dual op-amps, resistors, 100nF capacitors
Measuring the output signal from the iPhone: V(peak-to-peak): ~2.85V centered around 0V
Required Gain: ~1.75
Available Power supply units: +5V and +12V DC

I've tried designing these filters by hand, and using TI's FilterPro software, however I have not been able to achieve the DC bias affect. I have been simulating the filters using LTSpice. I have the design files for the three filters though if it would help in describing my current problem. Any help in this area would be greatly appreciated.

I am not sure what you mean by bias because the term is used to mean different things.

Here is what I think you mean. Lets say we have an AC signal of arbitrary frequency ad amplitude. That signal can be riding on a DC signal as well that is something other than zero volts. This ,"Offset," some might call bias, or some may call it a level shift.

Given that you would have one input and three band pass filters, these could all be centered at zero volts to make it easy. You then would use a summing amplifier for summing all three signals but again this summer would be centered about zero volts as well. Now lets say you need the final signal ,"Level shifted," to 2,5V for what ever reason, that would produce a combined weighted signal (contains predetermined amounts of each of the three bands as in an EQ circuit) with a 2.5V offset.

A simple way is to use cap couple, since you aren't trying to pass DC but then at HF, more signal is amplified so that another filter would be required to correct that. So not a good choice.

A preferred way would be to use symmetrical differential amplifier with let's say a gain of one. In the inverting resistor network to ground, you would put a DC offset of what ever you choose say in this case 2.5V. So now when there is no AC signal the diff amp output is not zero volts but 2.5V.

This is one stage of offset, the final stage. On the other hand you mentioned having each stage with a bias (offset?) which you could do as well, you use same diff amp circuit and same single DC bias (offset) to each, which would work.

If this is what you are looking for and need the simple circuit diagram. I can send it.

Thanks for your response, yes by biasing I did mean DC offset. I have added biasing to my differential amplifier circuit before my low-pass filter and it seems to be working but off-phase with the input signal and not correctly amplified. I am going to try you're suggestion of adding a symmetrical differential amplifier to the input and output of the low-pass and get back to you, but the circuit diagram would be greatly appreciated. As of now this is how I have my low-pass setup:

"Your text to link here...(Lowpass Schematic)":http://i.imgur.com/LZoxi.jpg
"Your text to link here...(AC Analysis)":http://i.imgur.com/sptRy.jpg
"Your text to link here...(DC Transient Response @ Input(freq) = 300Hz)":http://i.imgur.com/DVIp4.jpg

The filters if they have lead/lag will have no phase shift over all a decade away from CF. The HF filter lets's say 2K - 20K will have phase lead and given you have some overlap between the sections should cancel more or less depending on the crossover symmetry.

You shouldn't have 180 deg phase inversion unless you purposely use inversion, which you can, just have to account for it.

Enclosed is a real simple approach to a single stage. I have left out any values, etc, but functionally it can provide some frequency response and DC offset at the output with a fixed gain of what you choose. It's best to fix the gain before the DC offset section, because you don't want to amplify the DC, presumably.

OK so I looked at your sims, and as you can see there is phase lag which is consistent with a LP filter -20dB/decade response if you would change input frequency, you would notice that the higher the input frequency the more the shift and the greater the attenuation.

Would it be possible to set up the display to where the x scale is log, just used to log scales myself. Also you should be able to do a Bode plot which is more telling in that you can show both gain and phase on the same chart.

If not, no big deal.

OK so some more looking at the response. The LP looks like it's DC coupled, is that why you need a DC offset circuit to null that out? A much simpler way would be to cap couple the LP stage from the input and that takes care of any offset.

Also the R/C components you have and op amps will work just fine for filter circuits. I don't actually see any real circuit schematics? Did you mean to include them?

My prior circuit has an error in that the feedback is connected to plus (+) terminal which is not correct of course. That should have been just a non-invering gain amplifier with appropriate filter time constants. in any event I have included a new circuit with values that is a LF band pass 20Hz - 200Hz with gain correction and o DC offset if you need.

Opps, 12db gain correction A3 should be before A2, otherwise it multiplies the DC offset. Sorry, for that.

Thank you so much for providing this circuit diagram. I was able to change the values and adjust the circuit to provide the appropriate gain and DC offset by changing the value for R3, however I still do not fully understand the relation for calculating the correct gain, cutoff freq, and offset. If I understand correctly R5 and R6 are used with R3 to create the gain, while R4 and C3 are used for impedance matching. (Please correct me if I am wrong) Not sure how cutoff freq would be calculated from here.

I also saw that the design for the LP filter is based off of the opamps having a positive and negative power supply. I set the negative supply terminal of the opamps to ground in LTSpice and it pretty much converted Vout to a DC value. How would I be able to account for my lack of a negative power supply?

Again thank you so much. You have been very helpful.

R3=R4=R5=R6, are simply easy convent assignments, The ratios R3/R4 & R5/R6 are what's important. The diffamp is set to have a differential gain of 1/2.

R3/C3 are not impedance matching. Imagine at low frequency C3 is open, then R3/R4 divide the signal in half. At high frequency C3 is a short and the signal into A2 is zero. The time constant for the cutoff frequency is R3 in parallel with R4 & C3. You can do the math or I can send you the equation.

Ok so have zero voltage as a reference is not magical. what you can do is use a source that is 1/2 of your op amp voltage, (-) goes to gnd and (+) is where you would reference all of your circuit, EXCEPT your op amp gnds, they go to gnd as usual. The positive (+) of this voltage source behaves like a new ground reference, sort of.

That way the input or output signals never swing below gnd because you have an offset. Although a more simple way is to use simple voltage controlled voltage source as an op amp. It would demonstrate the filter and does not have the limitation of the op amp component you are trying too use.

Included is the filter math for the diff amp cutoff filter.

Also I notice my polarity on my offset voltage is reverse which would give inversion so please change the polarity on the offset voltage supply. Reason if zero input (+) terminal of A2 is zero which means as is the (-) terminal is positive (as shown) which makes the output negative and opposite of what you want.

I should have run a sim on this myself and just sent the sim schematic free of the schematic bugs, SORRY about that.

The gain of the diff amp with all resistors the same is: V0 = (V(+) - V(-) )/2, which is text book. In this case it really isn't a true diff amp in that the signal isn't connected to V(-) so it really has a gain of one if all resistors are the same value. Therefore A3 gain correction needs to be just two, or 6dB.

I'll send you as sim tomorrow with values are results.

You are awesome. It's very much appreciated that you are willing to help a stranger this much. I have gotten the schematic you provided to simulate properly using the +/- 5V supply rails. I am still a bit unclear as to how to do this with my single +5V supply. Your response makes sense when it comes to how to reference the op-amps, but I still have the problem of Voffset needing a negative supply to prevent inversion. I am not sure how to work around this.

I feel like I am almost there. Once I have this circuit working, I should be able to adjust the time constants to make the filter into the three filters that I need. Is that correct?

Ok so here is the latest greatest simplified version. It actually puts an offset in the other leg but it too is just like the other components in terms of reference.

See if this helps, it should not having to use a negative Voffset. The (-) terminal of Voffset goes to the same reference as the resistors in this case.

C1/R1 same as before, R2-R5 &C2; same as before as well.

This is not my area (I deal mainly in digital and MCU programming), put I can speak to the signal mode at the input of the A/D input of the microprocessor:

What Pat wants is a signal that varies between 0 and +5 volts that conveys the "envelope" information of a range of frequencies (i.e. the amplitude [or intensity] of a band of frequencies). I recommend not feeding the actual audio signal to the A/D input, but instead, a rectified and low-pass filtered signal -- unless, Pat, you _want _ the high frequency content because you intent to digitally process that information -- but, if that was your goal, then wouldn't you be digitally filtering for bandpass?

So, these are the stages I foresee: Bandpass filter, ideal rectifier (which will double as voltage translation), low-pass filter with gain to achieve the 0-5V dynamic range.

But, I'm no where near as good at analog design as Earl is, so I will step back and see if he picks up the thread

OK so heree are my sims. The schematic. the transient response & the frequency response. The input has an arbitrary DC offset, which is filtered out with the cap coupling. The next stage adds back a DC offset if for some reason you want that, not sure why but if not then remove Voffset.

Unfortunately the text info on the graphs didn't come out like the original. I am running the simulator in VMware and it isn't quite WYSIWYG, unlike a Mac. PCis usually WYSIWYDW ( don't want).

Sorry for that. The transient resonse just shows that there is zero offset in the middle stage, and offsets in the input and outputs which are different and arbitrary.

Earl, we were able to get the project done successfully, much do to your help. Now we are graduating college! I am very appreciative to you for your kindness. Please keep being an awesome person and have a great holiday.

Yes I will always stay near a phone booth and keep my cape clean in the ever daunting crusade!