Increase 'sensitivity' of a sound to light circuit

Discussion in 'The Projects Forum' started by Fenris, Nov 8, 2008.

  1. Fenris

    Thread Starter Active Member

    Oct 21, 2007
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    Hi all

    Just a quick question if I may. I have built a circuit which incorporates a sound to light feature. The amplifier which supply's the audio out and which is the source for the sound to light is capable of 2.6-4Watts. This is at the lower end of the sound to lights circuits range which is 2-60W input. Now everything works fine but what I would like to do if possible is tweak the circuit into making the sound to light allow more power to the lights despite the low end audio input. The area in the picture in the red square is the front end of the input which includes an opto isolator. I understand That I can't do anything this side. However on the output side of the opto isolator there may be. R20 is, if I have understood the data sheets, referred to as RL in the data sheets. I assume this governs the flow of power when the opto is active. Can I adjust this to allow more power to pass and therefore increase the final output of the circuit? or am I likely to fry the opto? The circuit is a velleman MK114 sound to light unit I believe. DC power is 12V i have a pair of 12V/24W bulbs in parallel on it. As I say it does work fine but if I can increase the final output it would be even better.

    Regards

    Fenris
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    You can reduce the resistor to increase the power but reads the datasheet of the optoisolator as not to exceed the maximum current it can supply. Also, note that your output power transistor is capable of dissipating this more power and not burn.
     
  3. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi there

    Thanks for your reply. I have read the data sheets but I lack the understanding
    to apply what I find in them. I am working on the theory that as the input can take up to 60W max then the output side of the circuit is up to a point geared
    to handle a larger flow of power. So up to a point, in theory, If I am putting
    in such a smaller level then changing R20 downwards should be safe so long as I don't go so low that it exceeds it's limits. I hope that made sense :D

    regards

    Fenris
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Looking at a datasheet for a 4N35 (Fairchild) absolute maximum power dissipation for the detector side (output) is 150mW. If you cut that in half (for safety) you get 75mW, or 0.075W.

    Since P = EI, and you know that your E is 12v, you can calculate backwards that I = P/E, or I = 0.075/12 = 6.25mA maximum safe current.

    Right now, the 100k resistor you've marked with a blue dot (R20) allows a maximum of I=E/R = 12/100,000 = 0.12mA (disregarding Vce(SAT)=0.3v). So you do indeed have some leeway ;)

    Since R = E/I, E=12 and I=6.25mA, the smallest value for R20 you should use is 1,920 Ohms. Let's just call it 2k.

    Try starting out by reducing it from 100k to 47k. If that's not enough, try 33k, then 20k.
     
  5. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Cheers for that SgtWookie I will
    give it a whirl. The maths of
    electronics is the bane of my life
    I can do the formula 'mostly' but
    I either can't implement it or I can't
    figure out where to pull the figures from
    and I have read and read and read to
    try and get something to go 'click' in my
    head :(

    thanks again

    Fenris
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    The input volume control makes the input more sensitive when it is turned up and less sensitive when it is turned down. The volume control or the opto will fry if the input is higher than 60W.

    To make the circuit more sensitive then R20's value must be increased. But then the current from it will be too low to turn on the BC547 so it must be changed to a BC547C that has more current gain.

    The sensitivity has nothing to do with the max output power. If you overload the output (its max is 50W at 12V) then it might smoke and burn. I couldn't find the complete schematic at Velleman.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    I don't understand whether you are trying to make the circuit respond to lower level sounds, or to drive higher wattage bulbs with the existing sensitivity. Apparently everyone else does.:(
     
  8. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi Chaps

    I have continued to think about this 'improvement'. Could it be that I have looked at this the wrong way round. The amplifier which supply's the audio
    can produce 2.6-4Watts well under the maximum the sound to light can handle. I think what I also meant was not to make it drive bulbs bigger than it can handle (that is fine as is) but they are not as bright as they could be for obvious reasons. What I meant was because I am using the lower end of the audio I need to make it more sensitive to this lower level input. To this end I think where I need to look at the circuit is as follows;

    R19 is there to protect the opto in the event that a full 60Watts was present
    due to the variable resistor being turned so that the maximum audio would be present. If R19 was absent under these circumstances then the opto is likely to fail on the emitter side due to overload. I hope I explained that well enough :) . So given the fact that the maximum possible audio input would be 4Watts any way, this would mean that R19 could be of a lower value without harm to the opto and thereby facilitate a brighter light at the output of the sound to light. I have just tried to measure the voltage of the audio but it looks like my meter has upped and died.......new battery and nothing :( So if you think I have looked at this better could you suggest a minimum value for R19. The current audio is of the order of 6.7% of what the existing circuit could take. It would be too simple to think in terms of a resistor that was 6.7% of the value of R19 I guess?
    Heres the full schematic :)

    I have been looking at the datasheets for my amp and the 4N35.
    4N35 has a DC average forward current of 60-100mA
    Input forward voltage 1.18V typical, 1.5V max

    TDA2003 At 28V has an output of 3.5A my circuit is 12VDC so I am assuming 1.75A output
    I cant tell anything from this info in respect of what value R19 could be I dont even know if
    I have selected or looked at the relevant information :/

    regards

    Fenris
     
    Last edited: Nov 9, 2008
  9. Audioguru

    New Member

    Dec 20, 2007
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    896
    Since your amp has a 28V supply, its max peak output voltage is about 12V. The IR emitter in the opto has a 1.2V drop so the voltage across the 330 ohm input resistor that somebody reduced is 10.8V and the peak input current is 33mA.

    The max allowed input current to the 4N35 is 60mA continuously and 3A peak for a moment.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    I GOT IT!
    The circuit lights the light bulbs only during the half-wave peak voltage of music and speech.
    The peak voltage seldomly occurs so the light bulbs look dim.
    You need a peak detector and slow release to indicate the peaks of the music or speech. It should light LEDs for about 50ms and light your light bulbs longer so that the have enough time to get bright.

    The datasheet for the LM3915 bar graph driver has a few peak detector circuit.
     
  11. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi there

    The circuit is already set in stone on a PCB. 20+ at the last count.
    My Amp is only powered at 12VDC the figures I gave for the chip were
    the Maximum figures the Data sheet provided. But I am right aren't I?
    If R19 is there to protect the 4N35 from the maximum input of 60Watts
    if the unit was so used then as I am only ever going to have 4Watts max
    driving it then the value of R19 could be lowered to improve the brightness
    and response of the sound to light to the lower level of audio input.

    regards

    Fenris
     
  12. bertus

    Administrator

    Apr 5, 2008
    15,645
    2,344
    Hello,

    If you change the formula P = (U*U) / R to U = √(P * R), then you can calculate the resistor R19.
    For 60 Watt at 8 Ohm you will end up with √(60 * 8) = 21.9 Volts.
    The led inside the 4 N 35 has a voltage drop of 1.2 Volts.
    The current at 60 Watt is now (21.9 -1.2) / 390 = 53 mA max. (the led can handle 60 mA).
    For 4 Watt at 8 Ohm you will end up with √(4 * 8) = 5.6 Volts.
    The current in the led will be (5.6 - 1.2) / 390 = 11 mA.
    If you want to have more current at 4 Watt you can lower the R19.
    For 50 mA , (5.6 - 1.2) / 50 = 88 Ohm , so take a 100 Ohm.
    That will give you a current of (5.6 - 1.2) /100 = 44 mA. (that is still on the safe side).

    Greetings,
    Bertus
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    Then the lights will be turned on only for the very short peak duration of sounds then look dim. The circuit needs to be fed from a peak detector that allows enough time for the lights to get bright.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Consider replacing the lamps with others rated for lower voltage. You may need a series current limiter, like a power resistor.

    "Fixing" an early PCB run is indeed a challenge, and a learning experience.
     
  15. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi all

    Thanks for your input all. Gratefully received as ever :)
    I will I think leave well alone (if it ain't broke.......)
    no one else has complained so I may well be being
    to fussy. The alternate bulb voltage etc may well
    be a simpler solution so I will tinker with that idea
    and see what I can do. Cheers all :)

    regards

    Fenris
     
  16. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi all

    Well I bit the bullet on his one eventually. I built a complete voice mod to try out the change to the sensitivity of the sound to light. The new voice mod now has a bridged TDA2003 amplifier circuit on it and it sounds much better so having a go at the lights was deemed a requirement as well.

    The answer in the end was simple. Working on the premise that there was no way the sound to light circuit is ever going to get 60W input I went for the 390Ω resistor at the input end of the circuit.

    My original theory that it was there to protect the opto chip from getting the potential full 60W should the variable resistor be turned to fully off IE. maximum audio signal gets through and as such could be reduced as the actual maximum input was so low from my amplifier.

    Well I now fit a 1Ω resistor and everything is all the brighter for it. I actually went down to 150Ω, 47Ω, 22Ω and down eventually to my final setting.

    regards

    Fenris
     
  17. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi all

    Just a heads up on the resistor change I performed. It has bitten me in the a***! Recent circuits have had a tendency for the 4N35 to fail. At first I didn't get it because my one keeps working.

    I twigged today. When I pull the ear piece out to plug in a speaker, without turning the unit off, the amplifiers output is straight across the sound to light input. Obviously without the load of the speaker the 4N35 gets a bit of a blast.

    Having searched out this thread I re read Bertus's calculations and 100Ω is the minimum with a safe margin. I missed re reading it when I went for the change. That will teach me!!

    Something I would like help with on the same circuit is. The transistors I usually get are BC547 x 2 and BC557 x1. This time they are discontinued, at rapid at least, so I went with the BC547B x 2 and the BC557B. Considering they are surrounded by components which were selected to suit their predecessors will they be working as well as the old ones or are changes to the surrounding resistors needed. The circuit seems to be working OK but I just want to be sure.

    regards

    Fenris
     
  18. bertus

    Administrator

    Apr 5, 2008
    15,645
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    Hello,

    The calculation of the 100 Ohms resistor I made then where based for 4 Watts in 8 Ohms.
    For other wattage and speaker impedance you can recalculate the resistor again.

    Greetings,
    Bertus
     
  19. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Thanks Bertus

    I ran through your figures and for a 4 ohm speaker I can go to a 56Ω resistor.
    Thats odd........... I was able to work the math!!! Thats a first :D

    regards

    Fenris
     
  20. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The circuit is a half-wave rectifier. When the input signal is continuously strong then the output is turned on for only half the time (for each cycle of the audio) so the light bulb will be dimmed and will never be at full brightness.

    The BC547B and BC557B are selected ones that have a narrower range of spec's and are better.
    Some of the ordinary BC547 and BC557 transistors have hardly any hFE or almost too much hFE.
     
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