Light Dependant Resistor Current Control

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Fixed voltage controlled LDR with circuit values of 5v /100 ohms is 50ma = .25w .. and you said only 3.1ma I am now really concerned !!! with that type of design 😱

Thankfully I chose a better and safer way of doing LDR circuit design.🙂

Cheers / Chris
This is totally incorrect. The 100 ohm is a fixed resistor only for over current protection. There is a log pot in front of that.
 
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I am coming to a conclusion.

Chris is bringing disrepute to current controlled LED/LDR.

Can we have some other examples of good current control that are likely to work, rather than have this one, allegedly flawed scheme, shoved down our throats?
 
No, he started the thread. It seems to me that he is without knowledge about electronics but intentionally trying to discredit other designs. If you look at all the questions and answers going on, he randomly changes his calculations to make them seem like he is right.

Can the moderators handle this before someone gets hurt?
 
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Why no?

In the earlier thread where it was suggested that a dedicated current controlled thread be started, it was the intention to show the various different ways of using current control as different from voltage control to see what can be done LED/LDR attenuators.
 
Because of the thread starter. If some person with more knowledge started this thread, then it is fine. But Chris seems to want to twist others to believe his solution is better. Nobody in his right mind will enter this discussion. We are wasting time here. I only stayed till I asked questions to make sure what was really going on. Personally, I think this thread should be killed to avoid problems. If someone more well know in the forum started this thread, it would be fine.
 
Where is the Nelson Pass schematic? I did a search but could not find it. BFNY, thank you.
been away from the 'puter for a few days, hmmmm intersting developments 🙄

The NP schematic was originally posted in the Lightspeed thread years ago, final version here

http://www.diyaudio.com/forums/anal...uator-new-passive-preamp-137.html#post1521700

but the sum of Nelson's contributions on distortion, type (SR2 vs. SR3) etc start here, and if you missed it, are worth re-reading.

http://www.diyaudio.com/forums/anal...uator-new-passive-preamp-133.html#post1520215

As per my previous comments, he chose to use the SR3.
Bob
 
This is totally incorrect. The 100 ohm is a fixed resistor only for over current protection. There is a log pot in front of that.

capable at the end of its travel to deliver 5v+ 😱

You have also misunderstood the inability of current to flow in virtual ground and ground in op amp circuits, have a look at figure 1.6(b) in the link http://users.ece.gatech.edu/mleach/ece3050/sp04/OpAmps01.pdf

" shows the current-to-voltage converter with a current source connected to its input.
Because RS connects from a virtual ground to ground, the current through RS is zero. "

As a result not only is there no current flow across the pot itself, but the circuit also offers ability for digital pots such as AD5228, or DS1869 to be utilised.


Cheers / Chris 🙂
 
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So measuring actual circuit we find from Adj the source of current and connecting to inverting input at its end travel .. to op amp gnd resistance above 90meg ohm, and Adj wiper to inverting input to actual ground higher still.... just what is needed, because we want the LDR to be controlled, The control circuitry is contained in very high impedance - the very wonder of op amps has finally come to the rescue.

Inferring in my circuit quite clearly there is no opposite potential for current to flow at all across the potentiometer, ie 4.24/90,000,000.00 .. or more, and Ohms law tells us V/R =I

The LDR devices are inputs to opamps, which is quite implicit provided in the schematic 🙂

I look forward to ideas from forum members to develop these ideas of current control of Light Dependant resistors, further still.

Cheers / Chris
 
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I asked this:
You are not going to solve the balance problem with this circuit. But I am kind of interested. Two questions:
1. What is the maximum total current that will go through the pot, and at what position is it maximum?

2. Why do you not use a log pot?

You said this:
Max current through the pot is 1.25/26 = 48ma to be divided to 2x LDR,

...
Cheers / Chris

Then I asked this:
If the 48mA goes through the pot, the pot would have to handle how many Watts? and the voltage drop across the pot at both ends if we put it in the middle? I am just looking at what load the pot is taking first.


Then you calculate using 1/1000 of the current value going through the pot.
Hi Soongsc

4.24v( is across Adj &Wiper ) x .048ma = 0.20w when pot is in middle and most of range
3.20v at bottom x .048ma = 0.15w
3.66v at top x .048ma = 0.175w

and voltage drop = at bottom 1.04v
at top .58v

The pot I am using is part no Bourns 91A1A-B28-B18L which actually is a single and is Conductive Plastic element,linear taper at .5w

Cheers / Chris 🙂

Now you say this:
...
As a result not only is there no current flow across the pot itself, but the circuit also offers ability for digital pots such as AD5228, or DS1869 to be utilised.


Cheers / Chris 🙂
Either you are wrong somewhere, or you are still playing with words. What is it? This places subtantial difference in requirements on the pot. I am just checking your words agains your own right now, because I noticed you have played this a few times.

The pot in that kind of circuit dissipates wattage based on the actual resistance current flows through, at the end, it acts like a normal conductor and dissipates near 0 watts.

How about some pics of your working circuit?

I am going to stay out until I see something that makes sense.
 
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Hi Soongsc

It is easy to see the characteristic op amp virtual ground used in my circuit offers advantage. I answered your questions providing answer that current to that potential, potentially has that ability. However in my circuit due to high impedance, there is no opposite potential of any concern, particularly for that current to be absorbed by the potentiometer itself.

Sorry I would really like to assist but, Pictures, particularly containing images or partial images that are advertised elsewhere, is, or could be construed as, advertising on forum...🙁.

As my schematic shows two TL072 op amps a Bourns potentiometer, 2x LM317, 7 NPN transistors, which are BC547 2x 13 ohm, and optional 22k resistor, and 4x NSL32SR2 are connected in unique original way. Outside of this circuit there is voltage stabilization, to upper and lower half, including cap multipliers,🙂

Cheers / Chris
 
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The attached image shows schematic available at Silonex and described as ' It is better to drive the coupler LED from a constant current source, to minimize the effects of variations in LED forward voltage from device to device and temperature. A simple circuit that gives 1mA per Volt input is shown in Figure 5

Has anyone had success trying this ? To work as a stereo attenuator though I see would require fairly precise voltage ranging, well below 1 volt, and some method of running in the opposite direction at the same time.... suggesting 2 of these working in tandem opposite in voltage change.

Cheers / Chris
 

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Channel balance I am finding is a good objective to get as good as possible. Throughout this thread I have made available figures that the schematic I provided at post 108 offers aggregate difference of L vs R no signal cables attached midway 2210 ohms, ( and end of travel max volume 2400 ohms.)

How do other LDR circuits ( without any politics please ) compare. at midway and end travel ? Although subjectively very good already It would be nice to know... to progress where we are with this channel balance objective, if aggregate midway 2210 ohms is way off, or acceptable, compared to what other members have built in LDR preamps.

If you want to help:
You do not have to mention by advertising Brand name, just indicate how your model performed at midway setting..... like this

Chris I measured at midway volume difference between L and R = ............... ohms

To measure this with DC power connected to your LDR preamp and no signal cables attached, move your LDR design volume control to approx midway volume With multimeter reading k ohms insert a meter test lead into R input and to Gnd, note the ohms figure then Without changing volume insert meter test lead from L input to Gnd and note the ohms figure. Now write down the difference in ohms, one channel vs the other.

Post it back here, if you want to contribute. 🙂

Cheers / Chris
 
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I'm puzzled. Why are you asking people to measure the least important characteristic of an attenuator - total resistance? What has this got to do with channel balance?

An analysis of Series and Shunt writing those individually down adding them up L v R will provide the same overall result as.
Series + Shunt = L
Series + Shunt = R

Such a measurement aims to find mid volume differences between L and R, ... however if you can suggest a better way..

The series + shunt L difference series +shunt R is
just an easier method, however i am not rejecting that people do it the longer way, that shows up individual performance of each LDR, if they want to.....doing each LDR will show up straight away, Why the aggregate comes about relative to each LDR.

And 10 posts or so should be enough... to work out what LDR's in the DiY community are doing.
 
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That method will tell you precisely nothing about mid volume channel balance, as anyone who knows Ohm's law can tell you. It is this type of thing from you which makes people doubt that you know what you are doing. Do you know how a volume control (i.e. potential divider) actually works?

Lost your multimeter ? I'm all ears, if you can suggest a better way ?

Cheers / Chris
 
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