I would use the LDRs for low value resistors. Say 20k or less. They are heat sensitive. Below a few thousand ohms they are rock solid but over that and they get touchy with changes in heat. Over 20k and there can be major changes in value with a degree of heat.
I think its a great project.
Uriah
I think its a great project.
Uriah
I had a thought about why a dual pot is needed. Instead, why can't a single pot be utilized, with the 5V coming into the wiper, and the series LDRs connecting to one end of the pot, and the shunts on the other side? You'd have, in effect, the same type of control with less expense.
Or am I missing something?
Or am I missing something?
Actually Uriah does that and changes the load impedance on the source to "fine tune" the sound.
According to him it makes profound differences.
Any idea of the page or post number? there are 325 pages!
I am at least glad to know that it works, and I'm not crazy. It seems like an easy way to save expense and complexity. You can also easily control L/R levels with a single pot for each LDR pair.
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The reason I used dual pots is I had a couple of failures with the single pots, and the reason is the pots didn't like dc voltage on them at these currents, so the best solution was to share it over two pots.
Cheers George
Yeah, I went back to 100k dual logs. Something funky was happening. I was getting max resistance on series when in minimum volume. I mean we are talking megohms. Also AndrewT and George both have stated that the wipers can die. Spose you would just replace the pot, but it would be annoying.
Uriah
Uriah
You could parallel a 400K trimpot set in the middle (or two 200K resistors -wiper to pin 1&3) with a 200K single log potentiometer. This will also give you some flexibility to tweak if your signal and shunt pairs don't match.
A Beefy Log Pot
How much current are we talking about at the wiper for a typical single end stereo volume control? I have a 2.2 watt single audio taper 100K pot. At 5V that should give me a margin of 440 mA.
How much current are we talking about at the wiper for a typical single end stereo volume control? I have a 2.2 watt single audio taper 100K pot. At 5V that should give me a margin of 440 mA.
Not really but again dont sweat it. What if you only used 1volt? Then how much current could you put into a 2.2watt pot? Dont tell me 2.2Amps 
So go to Rod Elliots site and read the short article on beginners potentiometers. It gives the math and shows that pots are sensitive devices to current and can burn out the wiper fast.
Uriah
So go to Rod Elliots site and read the short article on beginners potentiometers. It gives the math and shows that pots are sensitive devices to current and can burn out the wiper fast.Uriah
Sharing the Load
Okay, if I'm getting this I take two 200K resistors in series and connect the center of this pair to the wiper of a now 200K single log pot. Then each end of the series pair is connected to the resistive track connections of the pot, 1 & 3 I think. That would divide the current and lessen the load on the pot. But would signal and shunt then add up to 100K at every setting?
At volume full up or full down it's 100K, but how about a moderate level of 140K signal and 60K shunt: 140K parallel with 200K is 82K Ohms net for the signal side of this bridge. Shunt, 60K parallel with 200K is 46K. 82K + 46K = 128K Ohms. Right at half it's 133K. Sounds like overlaying an upward-opening parabola onto the log volume curve. How would that affect the LDRs and the input impedance of the control if the 5 volts supplied isn't divided as evenly?
Okay, if I'm getting this I take two 200K resistors in series and connect the center of this pair to the wiper of a now 200K single log pot. Then each end of the series pair is connected to the resistive track connections of the pot, 1 & 3 I think. That would divide the current and lessen the load on the pot. But would signal and shunt then add up to 100K at every setting?
At volume full up or full down it's 100K, but how about a moderate level of 140K signal and 60K shunt: 140K parallel with 200K is 82K Ohms net for the signal side of this bridge. Shunt, 60K parallel with 200K is 46K. 82K + 46K = 128K Ohms. Right at half it's 133K. Sounds like overlaying an upward-opening parabola onto the log volume curve. How would that affect the LDRs and the input impedance of the control if the 5 volts supplied isn't divided as evenly?
Bud Barclay Answers Tom Swift
Thanks for the reference! If the maximum dissipation is 2.2W and the resistance is 100k, then if P=I^2R, I=(P/R)^(1/2), I=(2.2/100,000)^(1/2)=4.7mA.
Not really but again dont sweat it. What if you only used 1volt? Then how much current could you put into a 2.2watt pot? Dont tell me 2.2Amps
Uriah
Thanks for the reference! If the maximum dissipation is 2.2W and the resistance is 100k, then if P=I^2R, I=(P/R)^(1/2), I=(2.2/100,000)^(1/2)=4.7mA.
Now, that said... I have been using a dual 100k pot from Radio Shack for a few years with no problem. I dont leave it in minimum volume position when I am done listening so it does not sustain full current for longer than an hour a day or less. Never had any trouble and never heard of anyone having trouble with the dual gang of 100k or less. I just think single gang is not a great idea. Dual works much better. I think mine has been a 1 watt. When you use the Lightspeed at any volume other than minimum there is so little current draw that it will not harm any pot. Maximum volume has the 1k trimmers to help out so you can hardly over do it with max volume current.
I also like the Alpha pots available at Mouser. Very good, very cheap, nice feel to them. Have not checked for 100k dual log but betting they have them at Mouser.
Uriah
I also like the Alpha pots available at Mouser. Very good, very cheap, nice feel to them. Have not checked for 100k dual log but betting they have them at Mouser.
Uriah
Okay, if I'm getting this I take two 200K resistors in series and connect the center of this pair to the wiper of a now 200K single log pot. Then each end of the series pair is connected to the resistive track connections of the pot, 1 & 3 I think. That would divide the current and lessen the load on the pot. But would signal and shunt then add up to 100K at every setting?
At volume full up or full down it's 100K, but how about a moderate level of 140K signal and 60K shunt: 140K parallel with 200K is 82K Ohms net for the signal side of this bridge. Shunt, 60K parallel with 200K is 46K. 82K + 46K = 128K Ohms. Right at half it's 133K. Sounds like overlaying an upward-opening parabola onto the log volume curve. How would that affect the LDRs and the input impedance of the control if the 5 volts supplied isn't divided as evenly?
Hi Paulfx,
You're right, it will cause a variance of the impedance. Perhaps not a great idea after all. As usual, George is right again to KISS
Hey Alazira,
Per Uriah I checked out Rod Elliot's Potentiometers for Beginners, but I couldn't help noticing one of the advanced diagrams had a 22K parallel resistor between wiper and terminal 2 of a 100K pot for a desirable hybrid S-shape audio taper. Your idea would have another resistor between wiper and terminal 3, and 5VDC to the wiper, of course. Played with getting a 2 watt 500K pot and using 110K resistors for the current bypass so it would have the same ratio as Elliot, but the wiper would still have most of the current at minimum volume. The impedance would sum up to around 150 ohms at middle volume. And after all that, with my luck it would work for a month instead of 2 weeks.
Per Uriah I checked out Rod Elliot's Potentiometers for Beginners, but I couldn't help noticing one of the advanced diagrams had a 22K parallel resistor between wiper and terminal 2 of a 100K pot for a desirable hybrid S-shape audio taper. Your idea would have another resistor between wiper and terminal 3, and 5VDC to the wiper, of course. Played with getting a 2 watt 500K pot and using 110K resistors for the current bypass so it would have the same ratio as Elliot, but the wiper would still have most of the current at minimum volume. The impedance would sum up to around 150 ohms at middle volume. And after all that, with my luck it would work for a month instead of 2 weeks.
a switched pot to send control voltages to the LEDs would be OK.
If you adopt the wiper as input from the 5Vdc supply and the two ends of the switched resistors as the LED feeds, then you only need a single wafer switch.
If you happen to have a dual wafer switch then one wafer controls Left LEDs and the other wafer controls the right LEDS. Now you can trim the individual resistors for exact channel balance that will stay in balance for the life of the two way switched attenuator.
If you adopt the wiper as input from the 5Vdc supply and the two ends of the switched resistors as the LED feeds, then you only need a single wafer switch.
If you happen to have a dual wafer switch then one wafer controls Left LEDs and the other wafer controls the right LEDS. Now you can trim the individual resistors for exact channel balance that will stay in balance for the life of the two way switched attenuator.
Stepped Potentiometer
Andrew,
Check this out:
Stepped Attenuator Audio Level Controls 12, 24 and 46 positions IR remote control
A single is $300US. Dual is 500. You could put beefy resistors in for low volume positions. But what if THAT wiper burned out?
Come to think of it, it's not a wiper, it's a switch contact. Lot of money, and the precision is way more than needed.
I'd like to power 11 channels eventually, and this beast might just handle the current. I hope someone else tries it! Meanwhile, I'll find a way to gang my 2 pots...
Andrew,
Check this out:
Stepped Attenuator Audio Level Controls 12, 24 and 46 positions IR remote control
A single is $300US. Dual is 500. You could put beefy resistors in for low volume positions. But what if THAT wiper burned out?
Come to think of it, it's not a wiper, it's a switch contact. Lot of money, and the precision is way more than needed.I'd like to power 11 channels eventually, and this beast might just handle the current. I hope someone else tries it! Meanwhile, I'll find a way to gang my 2 pots...