Another lightspeed lives, it is connected to my DC-B! and I am using the Salas "baby" shunt powersupply. Sounds swell, even though I wired the pot backwards. I am using George's schematic exactly as posted with the exception of only using one trimmer. 100R resistors in from of each LDR and a dual gang 100k pot.
The issue I am having is that the range of control is very small only about the 1st 20% of the pot controls anything beyond is full volume. Could this be the 100k pot should I try the 500k pot?
The issue I am having is that the range of control is very small only about the 1st 20% of the pot controls anything beyond is full volume. Could this be the 100k pot should I try the 500k pot?
A 500k pot may work but it also may get to dark. Its worth a try but a lot of LSA's have only a small working range. One thing to try first is a 10, 20, or 30R resistor in series with the 5VDC going to the pot. This will extend range a little bit and will also increase impedance which may or may not be what you want. Again its to easy to not try.
Uriah
Uriah
Well if that's the case then I am good to go. I'll try adding a resistor in the range you mentioned. What is a good way to measure the impedance? Can I do it in circuit?
As you use only a small fraction of the pot, with 100r in series your pot won't go far.
Leds are consuming too much current for it to stay safe.
Uriah suggestion or i think 4x470r instead of 100r will do the trick. Even small resistors before the pot can do "strange things"...
Anyway, you will always raise the overall impedance.
But if you now have a 5/10k total impedance it will not be a problem if you end within 50k with a normal amp (not my_ref or any other ultra-feedback amp).
Regards,
Gianni
Leds are consuming too much current for it to stay safe.
Uriah suggestion or i think 4x470r instead of 100r will do the trick. Even small resistors before the pot can do "strange things"...
Anyway, you will always raise the overall impedance.
But if you now have a 5/10k total impedance it will not be a problem if you end within 50k with a normal amp (not my_ref or any other ultra-feedback amp).
Regards,
Gianni
Thanks Gianni,
My goal for impedance is 22k although anywhere between 20-25k should be ok. I need to do some figuring here, not quite sure what I will do. Most likely try Uriah's suggestion of adding a small bit, 10-30R, between the +5v and the pot.
My goal for impedance is 22k although anywhere between 20-25k should be ok. I need to do some figuring here, not quite sure what I will do. Most likely try Uriah's suggestion of adding a small bit, 10-30R, between the +5v and the pot.
Which makes me wonder
Where would half attenuation fore lowest wear on LDR be in this case
I have been puzzling till my puzzler is sore to quote Dr Seuss. It has to do with my basic understanding of how this works.
1) The LDR is a photo resistor and LED in a little tube. the resistance of the photo resistor varies with the intensity of the light shed upon it by the LED.
2) Matching would entail insuring that for a given quantity of power the LEDs in the LDR would shed a similar amount of light or provide the appropriate amount of light in each so that the resistance of the photo resistors in each would be the same (or close enough) in both devices at any given point.
3) Control is provided by providing full intensity to zero intensity regarding the light inside the LDR thereby changing the resistance from all to none or anyplace in between based on the intensity of the light.
4) The only thing in the circuit acting as the pot are the photo resistors.
How then does changing the circuitry that provides the power to the LEDs change the impedance of the photo resistors? I mean if all I am doing is to vary the intensity of the light, how does the "dimmer" affect the properties of the light to enact a change in the impedance or resistance of the photo resistor?
1) The LDR is a photo resistor and LED in a little tube. the resistance of the photo resistor varies with the intensity of the light shed upon it by the LED.
2) Matching would entail insuring that for a given quantity of power the LEDs in the LDR would shed a similar amount of light or provide the appropriate amount of light in each so that the resistance of the photo resistors in each would be the same (or close enough) in both devices at any given point.
3) Control is provided by providing full intensity to zero intensity regarding the light inside the LDR thereby changing the resistance from all to none or anyplace in between based on the intensity of the light.
4) The only thing in the circuit acting as the pot are the photo resistors.
How then does changing the circuitry that provides the power to the LEDs change the impedance of the photo resistors? I mean if all I am doing is to vary the intensity of the light, how does the "dimmer" affect the properties of the light to enact a change in the impedance or resistance of the photo resistor?
LDR = Light Dependant Resistor.
LED = Light Emitting Diode.
Changing the current in the LED, changes the light emitted by the diode and this in turn changes the resistor value.
LED = Light Emitting Diode.
Changing the current in the LED, changes the light emitted by the diode and this in turn changes the resistor value.
Pretty much what I said in distilled form. I does not address why folks refer to changing the LED (The Light) circuit would change the impedance of the photo resistor (The Dependant) circuit. It is still in my mind - X amount of current = X amount of light and therefore a given impedance at that setting. At one end of the control pot it's on at the other it's off.
of course it does.
Light Dependant Resistor tells all.
Change the current and you change the resistance.
There are two LDRs in each channel.
They are connected in series. The top is connected to the Source.
The middle (series junction) is connected to the receiver.
The bottom is connected to the Signal Ground to both the Source side and the Receiver side.
The input impedance of the pair of LDRs is the sum of each of the pair. But we know that the LDRs are arranged that as one increases in impedance as the other falls in impedance. It is this inverse resistance change that gives rise to the attenuator action.
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That's obvious!
I guess I am not asking the right question.
Lightspeed Attenuator as in the schematic 5VDC, 100k pot, and 100R resistors in front of each LED section. For the sake of conversation, would give the equivalent of a 22k pot for use as an attenuator in my system. Why would changing the pot to 500k or changing the 100R to 470R change the overall impedance of the attenuator? Say for the sake of conversation again, to a 50k pot; it still goes from fully lit LEDs to dark LEDs, yes?
I guess I am not asking the right question.
Lightspeed Attenuator as in the schematic 5VDC, 100k pot, and 100R resistors in front of each LED section. For the sake of conversation, would give the equivalent of a 22k pot for use as an attenuator in my system. Why would changing the pot to 500k or changing the 100R to 470R change the overall impedance of the attenuator? Say for the sake of conversation again, to a 50k pot; it still goes from fully lit LEDs to dark LEDs, yes?
no!!!!!
the LED circuit has absolutely nothing to do with the audio side.
The audio side is connected solely to the LDRs.
the LED circuit has absolutely nothing to do with the audio side.
The audio side is connected solely to the LDRs.
EXACTLY!
However, that is not what I am getting from other posts. for example
This kind of thing... very confusing.
Bill,
If you are at the 500k end of a 500k pot you have a lot less light than at the 100k end of a 100k pot.
Uriah
If you are at the 500k end of a 500k pot you have a lot less light than at the 100k end of a 100k pot.
Uriah
So then the LDRs are not making use of the full range from dark to light?
Also if I understand the operation, when the one LDR is dark the other is lit and they invert as the pot is turned from 0R to 100k, how does that fit? How can one get "darker"?
Also if I understand the operation, when the one LDR is dark the other is lit and they invert as the pot is turned from 0R to 100k, how does that fit? How can one get "darker"?
The LDRs can be controlled into their own megohms. We are not making anywhere near full use of their resistance potential. We split the pot, remember? So one side of the pot is letting through more current and the other side is letting through less. One side controls both series the other side controls both shunt.
Uriah
Uriah
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