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#1 |
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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I have a circuit that uses an LED to set a tube's bias. The amp, with a basic schematic, can be seen at http://www.ecp.cc/semha.html
The current tube is I am using is a 7788, and I have it biased with a single LED at about 2.35V. With a B+ of 160V, this is leading to a current draw ~30mA ... maybe a tad more. I would like to reduce this slightly by upping the bias to closer to 3V. So, the question is, should I pick up some 3V LEDs, or will a Schottky diode in series with the currently used LED work just as well? Anyone with any reason one solution is better than the other?
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#2 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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A pair of red LEDs in series will give about 3.2V and a nice low AC resistance. A yellow LED in series with an infra-red LED will give about 2.9V but not such a low AC resistance.
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#3 |
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diyAudio Member
Join Date: Dec 2003
Location: San Diego
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You can do as 8010 suggests, but no reason not to try a Schottky or other diode and see how it sounds to you. The dynamic impedance of the Schottky will not be as low as the LED and will cause some distortion, but don't know how audible it will be. Another option, if you can give up a little gain, to try just a small series resistor, say 100r or so. A little degeneration will reduce distortion, might sound better, might not.
Sheldon |
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#4 | |
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diyAudio Moderator
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Quote:
I've found certain red leds with a 1N4148 in series to be a reasonable compromise. (Larger diodes tend to sound distorted.)
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#5 | |
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diyAudio Member
Join Date: Nov 2005
Location: SoCal
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Quote:
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#6 |
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diyAudio Moderator
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There have been quite a few measurements published. Sometimes it's even a specification. In any case, it's quite easy to measure and reasonably consistent.
There's a nice table of typical AC impedances for different diodes in Morgan Jones's "Valve Amplifiers" 3rd edition.
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#7 | |
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diyAudio Member
Join Date: Sep 2004
Location: Brisvegas
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Quote:
BTW, very few of the commonly available red diodes produce a 1.6.- 1.7 forward voltage - most are now in the region of 2.1V. If you want a 1.7V drop red diode with a guaranteed low dynamic impedance pay a bit extra and get HLMP6000.
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#8 |
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diyAudio Member
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Might be of interest
National Semiconductor Online Seminar LED Application and Driving Techniques September 14, 2007 (Friday) 1130 - 1300 (India) 1400 - 1530 (Singapore/Malaysia/Philippines/Hong Kong) http://www.eeplace.com/eeplace/event...L&eventid=1447 |
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#9 |
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diyAudio Member
Join Date: Dec 2003
Location: San Diego
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May be of interest. Looking at the voltage/current measurements, it looks like most of these are lower current devices:
http://www.diyaudio.com/forums/showt...008#post417008 |
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#10 | |
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diyAudio Moderator
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Quote:
I can confirm this, all of the red leds I have are hovering around 1.95V - 2.1V at 10mA. (I even have a bunch of jumbo red leds from about 1982 which fall into this range.) You can also measure two or more points on the curve directly with your chosen LED type. Simply measure the difference in mA between two "operating" points and simultaneously measure the voltage across the LED at each of those operating points, then just take the difference in voltage and R just becomes E/I. (More correctly delta of E over delta of I) I would recommend doing the measurements using the expected range of current in the amplifier stage. You can take several sets of points (Just 3 points is actually enough) to make sure that R is not too dynamic over the current range you intend to operate. (R varying with current is a distortion generating mechanism.) Using this technique should allow you to choose diodes with the lowest dynamic resistance variation of the lot(s) you have in addition to determining the VF.
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