High Voltage LED

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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?
 
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.
 
.........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 ?
 
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:

21576d1076789180-ultrasimple-mm-mc-riaa-preamp-2-mad_k-pacific1.jpg
 
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
 
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.
 
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.
 
You can connect the cathode of the LED to ground in the lower half of the adj. pin divider circuit. You can bypass the LED with various capacitance and even resistance if you don't want all the current going through the LED. I'm not sure how light you can go on the divider current, it may depend on exactly what you expect from the regulator, but the application circuits suggests a minimum of 1mA program current for 20V out.
 
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?

Just tap off one of the batteries a 9volt rail and run the led off that through a resistor.
 
You can build a crude converter like this one:

It will drive the LED with a constant ~20mA, while using less than 3mA @ 24V.
It is usable from 12 to 40V and oscillates at ~60KHz, well in the ultrasonic range.

The efficiency is pretty awful, but it is nevertheless a big improvement compared to a plain resistor.
 

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just to clarify, i read it that the circuit has only 1, 24 V positive suppy. and no negavive supply and you are going to use a LM317 pos regulator, is this correct?

if so why do you need 8X9Volt batteries?

either way i would use LED in B+ supply, no power loss (well no more than without the LED) and no resistor.
 
just to clarify, i read it that the circuit has only 1, 24 V positive suppy. and no negavive supply and you are going to use a LM317 pos regulator, is this correct?

if so why do you need 8X9Volt batteries?

Just going for long battery life, it will also run off just 4; also hard to describe, but 8 sit better in the box than 4, I'll post a pic of the finished product in a few days. I've just run the LED off the main supply at less than 1mA for now, I want easy access to the two preamp circuits to tweek, the two channels are completly independent including their own regulator circuit I'd like to keep them that way.
 
still do not understand,
do you mean that each channel will run from 4 X 9=36V regulated to 24V?

or 8 X 9=72, regulated to 24V each channel.

bearing in mind the regulator has maximum input voltage of 37V or there abouts.

9V batteries are considered exausted at 6-7V at low currents.
 
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