What is this filter?

Thread Starter

mike_canada

Joined Feb 21, 2020
239
I'm trying to tune the lights to the incoming music from my amplifier (yes I'm making a VU meter). My circuit is in grey and the output of the class AB push-pull amplifier (transistor emitters connected together) are connected to C3 and I noticed by changing any combination of any of the 5 components shown that the waveform can be modified. So I have a hunch I somehow have made a special active filter from 2. I think the first filter (from R13 and C3) is a high-pass filter and the second filter is a band-something filter.

I would use an electrolytic capacitor for C3 but I'm wondering if replacing it with a small ceramic capacitor and a high value resistor would be more beneficial (as opposed to low value resistor and a high value capacitor value) because there's been too many bad stories about electrolytics, one namely being accuracy loss with respect to temperature.

The yellow circuit came from http://www.talkingelectronics.com/projects/StereoVUmeter/StereoVU.html I know I didn't add a resistor in series with the input. I did get better results when I replaced the diode (in the talking electronics version) with a resistor because at all input audio volumes, the lights were following the sound more naturally whereas with the diode in place, if the sound volume is too low, then the lights would always stay on.

My LED stages are the same as whats in talking elecronics stereo VU meter except the diodes are 1N914 and transistors are 2N3904.


filter.png
 

MrChips

Joined Oct 2, 2009
27,686
The circuits shown use PNP transistors.
If you wish to use 2N3904 which is an NPN transistor you will have to make some changes to the circuit.
 

Audioguru again

Joined Oct 21, 2019
5,418
The very high output current of your amplifier (peaks of 0.5A or more) has probably destroyed the base-emitter junction of the 2N3905 transistor with nothing limiting the base current.

Your sketch does not show the speaker that is probably causing an extremely high DC current to ground (1A or more) at the base of the 2N3905. The high current lasted for only the instant that the base burned out.
 

Thread Starter

mike_canada

Joined Feb 21, 2020
239
Ok this is the circuit I'm playing with to make part of my meter. I have the capacitors separating the VU part and the speaker from the main amplifier. It would be something if that "jolt" from a capacitor can destroy that transistor. I tried this circuit running at 7.2V for a good say 20 minutes. The lights danced wonderful to the music. Only things that was warm at first was the TIP42. then TIP41 was a bit warm too. I guess I can't be surprised considering that I deliver excellent output to the speakers. Heck, I managed to make the system audible for up to 15-20 feet away.

Anyways, does anyone have an answer to my original question?
For R13, C3, C6 and R4, don't they make up 2 filters? I think R13 and C3 are high-pass but I don't know the name of the C6 and R4 filter. And if my transistor is shot then I'll have to replace it.

x.png
 

Audioguru again

Joined Oct 21, 2019
5,418
With 470uF for C3 and its R13 load is 160k ohms then it is a highpass filter that passes frequencies down to 0.002Hz (each cycle is 470 seconds). If you use 0.47uF then it will pass frequencies down to 2Hz. But the base-emitter resistance of the transistor increases the cutoff frequency.

With 2.1uF for C6 and its R4 load is 2.2k ohms then it is a lowpass filter that cuts frequencies above 34.6Hz. But R4 is parallel with many other resistors in the VU meter outputs which increases the cutoff frequency.

If your RC filters have their frequencies corrected then they are so simple that they barely filter frequencies and are almost useless.

You did not notice that the speaker in the circuit you copied is used as a microphone that has a very low signal level. It uses the two transistors as a preamp that can feed the low level input of a speaker amplifier. The output signal from the two preamp transistors feeds a 470 ohm resistor feeding a 100k ohms volume control to attenuate the signal.

Your circuit feeds the 2V peak output of your high current power amplifier through C3 which will blow up the base-emitter junction of Q4 since the signal voltage and current fed to Q4 are not attenuated.
 
Top