volume attenuator LDR

Status
Not open for further replies.
A

arivel

Member
HI .
I thought I'd ask for an opinion on something.
it is a volume attenuator that uses 2 pairs of LDRs as you can see in the photo.
the two opamp chains for each LDR pair are used to ensure that as the input signal varies, when the value of a photoresistor increases, the other connected one decreases in synchrony. same thing for the other branch being 2 channels.
the curves of the LDRs are not equal therefore the task of U7 is to vary the output voltage of the opamp chain of the group below to vary the voltage that controls the led of the photocouplers and obtain the same attenuation value of the 2 channels.
it is true that the u7 opamp inputs are on the audio path but having very high input impedance it is irrelevant.
it is a draft that I tried in the simulator but if I don't connect the U7 output to create the feedback of the other opamps it works but not anymore if I connect .
of course everything that comes before is missing but it is not necessary to know it.
what do you think about it ? Can it work if done right?
 

Attachments

  • ldr+fda.jpg
    75.4 KB · Views: 378
Sort by date Sort by votes
There is no need to arrange U7 in the signal path. A well designed variable LDR L pad is capable of 60 ohms to well over 40 mega ohms by controlling the anode and cathode, which your diagram is not showing. Begin by having separate circuitry for series and shunt, don't use op amps as you are using them , Set your initial current for each pair to be maximum of 5ma per device, then work on ripple reduction to be 2mv or lower. At all points of design learn from what the circuit provides with audio resolution , which will or should begin to suggest the proper parts to use, which are not in common integrated circuits rather are in discrete semiconductors such as jfets , mosfets with some use of voltage references.
 
Upvote 0 Downvote
This won't work because the signal is AC and gain is DC so a precision rectifier is needed then delays require compensation for attack and decay control.
 
Upvote 0 Downvote
perhaps there is an easy way to do this.
I thought of a SPDT relay switch to be applied to the LDR terminal. then one end is connected to DC and the other end is connected to audio signal.
at each volume change the contact passes from the audio signal to DC, a circuit detects the voltage difference resulting from the pairs of LDRs then we need a simple circuit that stores this difference and then transforms it into feedback and equalizes the two LDR outputs. made the switch returns from CC to the audio signal .
but what to use to memorize the tension ?
I hope I explained myself well enough.
 
Upvote 0 Downvote
No ,do not make any DC connection to signal paths , as others have already said. your creativity needs to be directed to making the LDR anode and cathode , achieve the result you require from the signal side. it appears you need to match firstly the LDR to the next, rather than short cut with techniques that invite and leave DC residues .

Begin by making the signal side a simple L pad - no relays no DC LDR's typified by the NSL32SR3 are current devices - begin trying setting current maximums at 8ma for the anode and cathode and then reduce ripple to be below 2mv - if you can. Listen at all times to what you have constructed.
 
Upvote 0 Downvote
arivel said:
if I use an MDAC driven by two CD40193s in cascade, do I get a linear or logarithmic output?.
I ask because it could be used to drive LDR pairs.
Click to expand...
Unlikely to be successful, as there is no actual change to current seen by the LDR pairs , sufficient to be a attenuator for audio purpose. Instead I would suggest studying the electrical characteristics , particularly how current interacts with the LDR device. https://www.advancedphotonix.com/wp-content/uploads/2022/03/DS-NSL-32SR3.pdf

As can be seen 8ma or less provides sufficient current for each LDR. Viewing the graph on page 2 should indicate the main focus of a circuit is controlling extremely small currents with precision.
 

Attachments

  • Screenshot from 2023-09-02 23-53-01.png
    17.6 KB · Views: 165
Last edited:
Upvote 0 Downvote
we didn't understand each other well.
the chain would look like this:
two cd40193 in cascade that drive an MDAC whose outputs connect to the circuit that I have already posted. i can set the current flowing in the LED via the gain of the first opamp with RF (btw i made a mistake with the inputs).
it remains to understand if from a MDAC I can get logarithmic output
 
Upvote 0 Downvote
Referring to your post on this same page where you show a diagram, the series pair cathode has no lower potential thus has no way of changing resistance. The shunt pair fails to benefit from a current sink to its cathode, and appears will only offer fixed resistance rather than affording attenuation as a volume control. I am ignoring the signal side alterations as they have no benefit and only serve to introduce DC potentials.

i would suggest you work out where actual attenuation is going to occur, by introducing lower potentials that admit current from the LDR cathodes. At all times do not have signal ground and cathode ground the same. Begin by testing a NSL32 SR3 pair to see if you can obtain 60-80 ohms R on - through to at least 20 mega ohms for the series pair , and 60- 80 ohms - 100k for the shunt pair, if so you then have viable resistance range to suit a volume control.
 
Upvote 0 Downvote
Status
Not open for further replies.

Similar threads

A
  • Question
volume attenuator and LDR
Replies
1
Views
1K
danadak
  • Question
How to modify cheap USB electret microphone for higher volume
Replies
4
Views
688
aliarifat794
A
T
  • Question
Rotary switch as volume controller
Replies
6
Views
3K
crutschow
A
  • Question
LDR Equivalent from an SMD (phototransistor)
Replies
5
Views
2K
ZipZapOuch
S
  • Question
KRK 10s sub cutting out at a certain volume
Replies
5
Views
3K
aliarifat794
A
Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…