Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

LED Audio Spectrum Analyzer

Status
Not open for further replies.
Your filters are Sallen Key highpass and lowpass but the filters in the article are Multiple Feedback bandpass filters which are completely different.
Why do you have an opamp at the input with very high gain of 21.3? It drives a volume control that messes up the filter fed from it. Also, the input opamp with very high gain will easily be overloaded. The volume control should be at its input, then it directly drives all 3 filters without coupling capacitors. Then it biases the lowpass filter opamps.

Why does the original circuit use an opamp for its "virtual ground" when it bias the other opamps with a very low current? The voltage divider that feeds the virtual ground opamp can bias hundreds of opamps.

The input opamp in the original article is not an audio amplifier, it is a modern opamp that can drive a load as low as 600 ohms that is not in the original circuit. Your opamp from an LM324 is fine to drive 5hree 20k volume controls.

I corrected your tiny schematic that is covered with Multisim chicken pox dots.
 

Attachments

  • spectrum analyser.png
    spectrum analyser.png
    231 KB · Views: 216
Your filters are Sallen Key highpass and lowpass but the filters in the article are Multiple Feedback bandpass filters which are completely different.
Why do you have an opamp at the input with very high gain of 21.3? It drives a volume control that messes up the filter fed from it. Also, the input opamp with very high gain will easily be overloaded. The volume control should be at its input, then it directly drives all 3 filters without coupling capacitors. Then it biases the lowpass filter opamps.

Why does the original circuit use an opamp for its "virtual ground" when it bias the other opamps with a very low current? The voltage divider that feeds the virtual ground opamp can bias hundreds of opamps.

The input opamp in the original article is not an audio amplifier, it is a modern opamp that can drive a load as low as 600 ohms that is not in the original circuit. Your opamp from an LM324 is fine to drive 5hree 20k volume controls.

I corrected your tiny schematic that is covered with Multisim chicken pox dots.
OK i fixed my schematic according to your picture, I'm a bit confuse on the whole power supply thing. Is Vcc suppose to be 12 or 6? Also since Vee=0v can i just ground Vee?
 

Attachments

  • LEDSPECTRUMCIRCUIT.PNG
    LEDSPECTRUMCIRCUIT.PNG
    80.3 KB · Views: 205
The output of an audio opamp needs to swing up and down. Then the (+) input is biased at half the supply voltage. Then the output swings more positive and less positive.Ground is usually 0V. There is no VEE.
If you have a positive supply (VCC) and a negative supply (VEE) then ground is the 0V between them which biases the (+) input of the opamps and their outputs can swing up (positive) and down (negative).

Thank you for making your schematic much clearer. Because I had trouble seeing your resistor values then some capacitor values are wrong: C12, C13, C19, C20, c31 and C32. The capacitor values should be calculated for the frequencies they must pass.

Your filter at the bottom has its highpass and lowpass frequencies too close to each other.
You have the lowpass filters first so the levels are cut to 1/50th the input level.
Also you have the gain of each filter at 2 causing a peak when the R and C values are the same.
I show the 1/50th with peaks and I show my corrections:
 

Attachments

  • bandpass filter with lowpass first.png
    bandpass filter with lowpass first.png
    52.1 KB · Views: 199
  • bandpass filter corrected..png
    bandpass filter corrected..png
    37.7 KB · Views: 205
The output of an audio opamp needs to swing up and down. Then the (+) input is biased at half the supply voltage. Then the output swings more positive and less positive.Ground is usually 0V. There is no VEE.
If you have a positive supply (VCC) and a negative supply (VEE) then ground is the 0V between them which biases the (+) input of the opamps and their outputs can swing up (positive) and down (negative).

Thank you for making your schematic much clearer. Because I had trouble seeing your resistor values then some capacitor values are wrong: C12, C13, C19, C20, c31 and C32. The capacitor values should be calculated for the frequencies they must pass.

Your filter at the bottom has its highpass and lowpass frequencies too close to each other.
You have the lowpass filters first so the levels are cut to 1/50th the input level.
Also you have the gain of each filter at 2 causing a peak when the R and C values are the same.
I show the 1/50th with peaks and I show my corrections:
So you pretty much change both filters, lowpass became highpass and vice versa and took away the gain? Would i take away the gain of all filters or just the one in your corrections? Also C12, C13, C19, C20, c31 and C32, are they coupling capacitors? If so would my coupling capacitor for the 256hz also be 62nF? And Thanks for all your help so far.
 
Make the gain of all the filters 1 by removing the capacitors C12, C13, C19, C20, C31 and C32 and the two resistors associated with each one and make the frequency setting resistors double in parallel and the capacitors double in parallel as I show.
 
Make the gain of all the filters 1 by removing the capacitors C12, C13, C19, C20, C31 and C32 and the two resistors associated with each one and make the frequency setting resistors double in parallel and the capacitors double in parallel as I show.
OK how does this look? Also would i connect c12 c13 and c19 to the virtual ground? or how would i connect the sallen key lowpass filters to the virtual ground?
 

Attachments

  • AUDIOSPECTRUM.PNG
    AUDIOSPECTRUM.PNG
    79.7 KB · Views: 198
Make the gain of all the filters 1 by removing the capacitors C12, C13, C19, C20, C31 and C32 and the two resistors associated with each one and make the frequency setting resistors double in parallel and the capacitors double in parallel as I show.
One more question, we're only using one power source correct? +6V
 
The circuit uses a 6V supply and the input opamp and filters are DC biased at half the supply voltage.
Filter capacitors connect to 0V without affecting the bias voltage.
I cropped your schematic so that it is not with a lot of wasted space and not as big as my neighbourhood.
I made some more corrections:
 

Attachments

  • audio spectrum analyser again.png
    audio spectrum analyser again.png
    106.4 KB · Views: 196
The circuit uses a 6V supply and the input opamp and filters are DC biased at half the supply voltage.
Filter capacitors connect to 0V without affecting the bias voltage.
I cropped your schematic so that it is not with a lot of wasted space and not as big as my neighbourhood.
I made some more corrections:
Alright, i made those corrections and added the lm3916 circuit too it. How does it look? I'm thinking of testing the circuit this week.
 

Attachments

  • SPECTRUMANALYZER.PNG
    SPECTRUMANALYZER.PNG
    94.6 KB · Views: 201
You have pins 7 and 8 of the LM3916's set for the 10th LED to light when the input at pin 5 is +10V when they have a +11.5V supply. In your circuit only the 1st LED will light when it should and none of the other LEDs will light.
The inputs of the LM324 work properly as high as only +4.5V with a +6V supply but then their peak output is 4.5V- 3V= 1.5V. So the pin 7 reference voltage (and pin 6 voltage) on each LM3916 must be no higher than +1.5V.
Simply connect pin 8 to 0V and connect pin 6 to pin 7 and a 1k resistor to 0V, then the reference will be about +1.28V and an input signal of +1.28V peak will cause the 10th LED to light and all other LEDs will light at lower input levels as shown on the datasheet.
 
You have pins 7 and 8 of the LM3916's set for the 10th LED to light when the input at pin 5 is +10V when they have a +11.5V supply. In your circuit only the 1st LED will light when it should and none of the other LEDs will light.
The inputs of the LM324 work properly as high as only +4.5V with a +6V supply but then their peak output is 4.5V- 3V= 1.5V. So the pin 7 reference voltage (and pin 6 voltage) on each LM3916 must be no higher than +1.5V.
Simply connect pin 8 to 0V and connect pin 6 to pin 7 and a 1k resistor to 0V, then the reference will be about +1.28V and an input signal of +1.28V peak will cause the 10th LED to light and all other LEDs will light at lower input levels as shown on the datasheet.
ok althought im not sure how you got Vref=1.28v. If the formula is Vref=1.25(1+R2/R1)+R2(80uA) and R2=0, then shouldn't Vref=1.25?
 

Attachments

  • lm3916.PNG
    lm3916.PNG
    8.9 KB · Views: 199
The datasheet for the LM3916 says that Vref is typically 1.28V but is roughly 1.25V. It can be as low as 1.20V or as high as 1.34V, their average is 1.27V.

National Semi invented the LM series of ICs so I look at their datasheets. Who copied the IC and made their own datasheet you are looking at?
 
The datasheet for the LM3916 says that Vref is typically 1.28V but is roughly 1.25V. It can be as low as 1.20V or as high as 1.34V, their average is 1.27V.

National Semi invented the LM series of ICs so I look at their datasheets. Who copied the IC and made their own datasheet you are looking at?
woops my mistake then i just took a look again and you're right. Anyways I have most of the parts at the moment but the MC33171p and a few capacitors which would arrive sometime this week. While waiting for the parts to arrive im trying to fully understand how the circuit works.
From left to right, so C10 is a coupling capacitor that feeds the 20k volume control and cuts frequencies below 36 Hz. What purpose do c20, r16, and r17 served? I know that since I lack an additional negative supply then UC1 must be biased at half the supply voltage which is what the virtual ground is for. I believe i have a good enough understanding on how the sallen key filters that make up the bandpass filter work as well as the display circuit. Now, I’m a bit confuse on how your voltage peak detector works. I think c5 and c8 are bypass capacitors but im a bit curious as to how you choose these values, as well as the resistor in parallel with these capacitors. And what's the purpose of C3,C4,R9,R8 and R12?
 

Attachments

  • p1.png
    p1.png
    40.2 KB · Views: 206
C20 passes the signal but blocks the 0VDC from the grounded volume control from changing the bias voltage at the (-) input of the opamp. The opamp would be saturated and do nothing if C20 was a piece of wire. See my attachment.
C5 in the peak detector grounds any AC interference picked up by the opamp input that has the fairly high value for R9. R11 has the same value as R9 so that the idling opamp input voltages are the same since they have the same input bias current.
C8 is charged very quickly to the peak voltage of the signal with fairly high current from the transistor and the low value of R123, then C8 discharges fairly slowly (330nF x 330k= about 0.1 second) by R13.

A Butterworth lowpass or highpass filter has an absolutely flat response to near the -3dB cutoff frequency then a sharp change to a straight slope. It can have equal value resistors and capacitors only if the gain of the opamp is about 1.6 times. If the opamp has a gain of 1 then the feedback capacitor value must be double the value of the capacitor to ground for a Butterworth response.
 

Attachments

  • preamp.png
    preamp.png
    36.6 KB · Views: 202
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top