Light Dependant Resistor Current Control

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#101
soongsc said:
This is my normal discussion mode. I think of course there are things that can be improved in any design. But if you want to do it in an open forum, make claims of superiority, but publish incomplete schematics, and hope to generate massive massive discussion to get people to do more work than you have done, this is really the wrong place.

Generally people first post complete schematics if they want other people to try and comment. Then based on various experience, the discussion evolves. Lots of threads here evolve very nicely. At least the schematics should allow people to compare your explanation to fully understand what you are trying to say so that effective communication can take place.
Click to expand...

Hi soongsc
I respect the interest people are taking and the time that takes as well, and your approach is very considerate.( I really had a smile when you asked about the music ) so a few days now to produce some meaningful figures, and perfect the schematic with sequential transistor numbering

Cheers / Chris 🙂
 
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#102
Hi Tom
Re voltage/ current relationship, In my circuit there are 2 sources of voltage, that delivered by current regulation by the wiper to the LED, ....and the voltage inherent that the opamp gives up, as a result of being single ended connected, the latter needs to be assessed as to how much current it is capable of. 1.4v is start point of the NSL32SR2 LED as your graph shows and 1.45 is the inherent voltage The inherent voltage is just there and as I see independent of resistance from potentiometer .... which starts to explain the small voltage swing of .350 vs the larger current swing.... exciting.🙂.

Cheers
 
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#103
Schematic attached, showing simplified, away from op amp voltage supply components.

As I see it Vref of the LED ( in parallel in each half ) becomes target for op amp inverting input to adjust.

The transistors Q2,3,4 in top half and Q5,6,7 bottom half are current steering, and setting by their diode drop, voltage start point of LED.assisted by inherent voltage of single supply op amp available at TL072.negative input.

Q1 also contributes to this start point, as it is diode drop above cathode of LED, which is ground. Regulated voltage set by LM317 1.25 x 520/220 +1 and current via LM317 1.25/26 provides to 50k pot then steers current provided to LM317 voltage regulator by no more than the inherent voltage available at TL072 output, onwards via the transistors to each half to function as an attenuator. There is also buffer by each channel of op amp back to same feedback path .🙂

Circuit performs as I observe between 12v and 24v V+, and functions due to voltage regulation within feedback path with little change from variation within that range.


Cheers / Chris
 

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#106
soongsc said:
Oregano can do simulation. If it can do that, it would also be able to check drawing errors. Why do you not use it's simulation capabilities?
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Hi soongsc
Oregano, despite having a LM317 in its parts provides error back saying LM317 is absent as a directory and "none analysis found" Probably easiest just to correct the dot. which I will attend to. See attached.

Also if a more pronounced log curve ie less volume at lowest volume, is desired this is a matter of placing resistor 18k- to 22k between negative in of upper most opamp and the collector base ground lift Q1. Circuit responds to make volume less apparent, at low volume see attached as 22k and labelled as R5 22k Option.

Noticed too the dot at wiper leading to op amp inputs is very naughty, I place it and it disappears when file converts to pdf to upload. I think it has been well described already as a buffer leading from wiper to other channel of upper and lower op amps.

Cheers / Chris
 

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#107
The gain curve issue is a complicated one. Wireless World Oct.~Nov. 1980 has a pretty good discussion on the various topologies by Peter Baxandall.

The LM317 models are not uncommon. It should be possible to add it to the simulation library.
 
#111
And channel balance measured unloaded volume midway on this same schematic model (+ voltage stabilization components ) is:

Shunt R 13.82 L 14.60k

Series R 4.26k L 5.69k


aggregate R 18.08k L 20.29k

difference 2210 Ohms

Cheers / Chris
 
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#112
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?
 
#114
soongsc said:
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?
Click to expand...

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

Current is at maximum at each end of travel, relative to the loads in theory, in practice
approx 40ma gets delivered to top and 30ma to bottom. Inferring top has conduction
going on available to 10ma when pot is at bottom.... need to measure this tomorrow to
see


And Andrews question;
at maximum volume, unloaded

series L 124 ohms series R 119.2
shunt L 53.7k shunt R 56.1k
aggregate 53.8k 56.2k

difference 2400 ohms

Cheers / Chris
 
#115
Chris Daly said:
Max current through the pot is 1.25/26 = 48ma to be divided to 2x LDR,

Current is at maximum at each end of travel, relative to the loads in theory, in practice
approx 40ma gets delivered to top and 30ma to bottom. Inferring top has conduction
going on available to 10ma when pot is at bottom.... need to measure this tomorrow to
see
...
Cheers / Chris
Click to expand...
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.
 
#116
soongsc said:
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.
Click to expand...

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 🙂
 
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#117
Thanks for those numbers.

What is the maximum current that can pass through a 50k half watt pot?

Imax = sqrt(P/R) = sqrt(0.5/50k) = 0.0031A = 3.1mA

You should never pass more than 3.1mA through any terminal of an adjustable 50k 500mW pot.
 
#118
AndrewT said:
Thanks for those numbers.

What is the maximum current that can pass through a 50k half watt pot?

Imax = sqrt(P/R) = sqrt(0.5/50k) = 0.0031A = 3.1mA

You should never pass more than 3.1mA through any terminal of an adjustable 50k 500mW pot.
Click to expand...

Agreed, but way too simplistic, assumes the pot is the power absorbing component and suggests that it is directly grounded one end, is the only component and it is absorbing all of the load...not the case here. In fact just checked for you, if we measure resistance at wiper, and end of pot travel to gnd there is resistance of 90 Meg Ohms due to use of high impedance op amps, my handheld DMM ranges out above 10meg, . .so there is in fact extremely small, basically un- measurable current through potentiometer... despite it being the source.

Therefore there is no power issue at all with this implementation.

If instead we examine a fixed voltage controlled potentiometer LDR, the power power at end travel of pot is of real concern.

Cheers / Chris
 
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#119
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
 
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#120
You said this:
Chris Daly said:
Max current through the pot is 1.25/26 = 48ma to be divided to 2x LDR,

...
Cheers / Chris
Click to expand...

Then I asked this:

soongsc said:
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.
Click to expand...

Then you calculate using 1/1000 of the current value going through the pot.
Chris Daly said:
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 🙂
Click to expand...

So you are either hustling us with a fake diagram or you are telling a lie about having a circuit working. If I am wrong, you'd better have a good explanation or I will request the moderators to lock you out of the forum.😡
 
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