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Old 3rd February 2010, 04:29 PM   #1
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Default High Voltage LED

This is probably a stupid question but googling is getting me nowhere; I'm building a Phono Pre-amp it's going to be battery powered by 8x 9v Batteries at 36v (regulated to 24v). as it's battery powered every milliwatt is precious and the current limiting resitor for the power LED is going to burn over a watt, is there a simple way of driving a regular LED from this relativly high supply voltage without burning-off a load of energy into a resistor?
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Old 3rd February 2010, 04:33 PM   #2
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There is no magic solution to this except by using a switch mode voltage converter. Even then the efficiency gain will be marginal and the complexity relatively great.

Are you sure you need a high current in the LED? Try it at 100uA - you will be surprised how bright they can be.

Incorporate the LED into one of the amplifier's biasing circuits so that its current is not extra to requirements.
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Old 3rd February 2010, 04:40 PM   #3
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Originally Posted by cliff View Post
.........Are you sure you need a high current in the LED? Try it at 100uA - you will be surprised how bright they can be.

Incorporate the LED into one of the amplifier's biasing circuits so that its current is not extra to requirements.
That's the way to do it. There could be several places that you could introduce the LED without affecting the performance much. As Cliff says it needn't be very bright if it's just to indicate 'power on'.
What's your basic circuit diagram like ?
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Old 3rd February 2010, 05:14 PM   #4
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Originally Posted by ashok View Post
That's the way to do it. There could be several places that you could introduce the LED without affecting the performance much. As Cliff says it needn't be very bright if it's just to indicate 'power on'.
What's your basic circuit diagram like ?
A bit like this:

Click the image to open in full size.
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Old 3rd February 2010, 05:53 PM   #5
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OK.

In that case there is no perfect bias point.

You could put it at the top end of the first 24K load, but then would probably need a second decoupling cap.

Or simply use a 100K series R across the 24volt rail. ~ 200 uA current lost - I bet it is bright enough!
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Old 3rd February 2010, 05:57 PM   #6
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Sounds like a plan!
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Old 3rd February 2010, 06:01 PM   #7
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how about in series with the B+, after the battery, this would require the current drawn from the battery to be any thing up to 30mA or so.
this way there is the same overall power dissipated/wasted in the regulator or regulator and led.
dissadvantage is there is a led drop to the positive reg so battery will get to its end point sooner.

well somthing to think about at least
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Old 3rd February 2010, 06:05 PM   #8
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how about in series with the B+, after the battery, this would require the current drawn from the battery to be any thing up to 30mA or so.
this way there is the same overall power dissipated/wasted in the regulator or regulator and led.
dissadvantage is there is a led drop to the positive reg so battery will get to its end point sooner.

well somthing to think about at least
Good one! Yes, I've done that in the past, but with a R across the LED to reduce the brightness.

It "loses" a couple of volts, though. But here that would be no problem at all.
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Old 3rd February 2010, 06:10 PM   #9
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You may use a LED as a reference voltage source for your voltage regulator you are going to implement.
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Old 3rd February 2010, 06:55 PM   #10
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I recently made a battery operated white noise generator. I stuffed a SLI-580UT3F in the collector circuit of the emitter-follower portion of the output stage which is biased at about 100uA. It's plenty bright. Some of the newer high efficiency red diodes glow impressively bright just with the .5mA diode drop test current of a DMM.
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