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|6/2/1999 4:39 PM|
The dod envelope filter 440 which i am currently building needs led/ldr diodes, is this the same as a normal light emitting diode? if not is it critical to the curcuit to use led/ldr 's?
|6/2/1999 7:08 PM|
An LED/LDR is a Light-Emitting-Diode plus a Light-Dependent-Resistor (photoresistor) close enuf to each other that the LED controls the LDR and ambient light does not.
Usually in one package (when they can be found), but you can roll your own if you need to. Clairex makes some self-contained ones (website?). The ones you make for yourself will probably use a cadmium sulfide (CdS) photoresistor. Cruise over to Jack Orman's site - I think he has links and more info.
|6/2/1999 7:16 PM|
LED/LDR packages consist of two elements, a fairly ordinary LED optically coupled to a Light Dependent Resistor (LDR) which may also be called a photocell.
The classic LED/LDR optoisolator is the CLM6000, which is not available any more, or available erratically. You can get substitutes in the Silonex NSL32 or the Vactec VTL5C3.
Or you can use a standard LED and a separate LDR and enclose them face to face in a light tight enclosure.
Mouser electronics sells suitable LDR's. Newark electronics sells the VTL5C3.
In some circuits you can use the H11F1, F2 or F3 LED-photoFET optoisolator instead.
Yes, if a circuit goes to all the trouble to specify an LED/LDR pair, it is usually critical to that circuit's functioning correctly.
|6/2/1999 10:38 PM|
"Functioning correctly" has several implied meanings, which should probably be spelled out:
1) Although LED's work instantaneously, LDR's have a certain inherent rise and fall time. This can be extended a bit by adding a few components, but can't be shortened. So, if the circuit is designed around LED/LDR packages that take a certain amount of time to do what they do, then random subbing of different types may or may not produce the kind of sound that the designer intended (and that users have come to know and love). RG mentions the H11F1/F2/F3 units as possible substitutes. Given that one can easily stretch out the rise and fall time of a photosensitive element, but can't shorten it, using something which is virtually instantaneous, like an LED-photoFET, is a nice starting point as far as design goes; a clean slate, if you will.
2) Different makes/models of LDR vary in their range of resistance. Some can have the same ON resistance, but vastly different OFF resistances, in comparison to each other. Since the LED/LDR package functions like an automatic control, using an LED/LDR package with a vastly different LDR range will give very different sounds. For example, in the DOD 440, the LDR values set the frequency of the filter. If the LDR values are wrong, then it is a bit like having a wah that only travels along 1/3 of its distance; it "works" but not quite completely.
3) LED's vary in how brightly they shine, for any given current level. The brighter they shine, the lower the resistance of the LDR. For example, you will note the presence of a "mystery" resistor adjacent to the LED on the various Mutron III schematics floating around. The mystery resistor is to tweak the LED performance, since even with the correct model number of LED/LDR, one is not guaranteed flawless performance.
Finally, as the many postings to this site have demonstrated, you can sometimes accidentally get something BETTER than the original when subbing different parts. Not always, but sometimes. So, even though my comments suggest being anal retentive about which parts you use, there is much to gained by serendipity. It's DIY, my friend. The pedal is whatever you make it and want it to be.
|6/3/1999 4:07 AM|
in the schematic it says the led/ldr module is probably a vactec VTL module w/ LED to center tapped LDR, i'm not quite sure what to look for , can you help?
|6/3/1999 12:57 PM|
It's likely the Vactec VTL5C3/2, available from Newark Electronics for about $5 last time I got some. They may have a minimum order.
|6/3/1999 8:03 PM|
Can you point to a schematic?
If the LDR is center-tapped one side of it may be part of the control feedback loop. That might tend to help 'cause you could sub in two non-center-tapped parts. (of course you'd have to match 'em)
I should just shut up now.
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