Battery saving LED connection, good idea?

[Clint Eastwood]

Hi folks, I would love to hear your comments on this:

In the circuit below, in stead of using just a source resistor, I added a LED plus a (smaller) resistor. The idea is to use the current that is flowing there anyway and so save a bit of current. Is there a reason why this would not be a good idea?

[GibsonGM]

Some tube circuits (and JFETs are considered to be an 'offshoot' of them) use diodes for biasing, in Hi-Fi, mainly. Can't see any problem with doing this except that the LED will always be on, rather than as a bypass indicator.  You'd want a bypass cap in || with it to shunt switching noise.

If the current in the source is such that the LED only lights when actively playing, then it could 'indicate signal is present'.   

[Clint Eastwood]

Thanks for the reply,

You'd want a bypass cap in || with it to shunt switching noise.

You mean the LED would introduce noise to the signal when just playing? and what value cap?

If the current in the source is such that the LED only lights when actively playing, then it could 'indicate signal is present'. 

Its function would be just a simple 'power on' indicator.

[antonis]

Its function would be just a simple 'power on' indicator.

Then J1 should always exhibit a V_GS of at least LED forward voltage drop plus voltage drop across R1 due to LED working current..
(you should definately replace R3/R4 with a trimpot..)

[GibsonGM]

Ok if it's a power indicator, if that's what you want.  I'd try 1n to 100n bypass cap, see what you get - that's what they use in tube biasing, but of course the JFET is going to be slightly different. Choose it 'to taste' as they say.

[antonis]

I'd try 1n to 100n bypass cap

Bypassing what, Sir..??

[GibsonGM]

If you insert an LED into the cathode of a tube gain stage to create a steady bias voltage, you need to bypass it since it may turn on and off with the signal (when clipping) and introduce switching noise.  I don't know if JFETs exhibit the same property when clipping, to be honest.  Last I read about it, it was considered good practice.
But the OP's circuit ALSO uses a resistor - biasing tubes with diodes does not... 

Rob Strand knows about this - there was a post from 2009 (similar but not enough to bother to link) where he suggests it might cause some trouble, and also suggested the bypass.

[Clint Eastwood]

As you can see it is a unity gain stage, and it is not supposed to clip. So then the LED would never switch off. I ran it in LTspice, and it won't clip until 3 volts peak. With an electric guitar input, that would be enough margin I would think.

[antonis]

I ran it in LTspice, and it won't clip until 3 volts peak.

That calls for about 200μA current..
(dunno if it should be enough for LED glowing..)

[R.G.]

in stead of using just a source resistor, I added a LED plus a (smaller) resistor. The idea is to use the current that is flowing there anyway and so save a bit of current. Is there a reason why this would not be a good idea?

LEDs have some difference in forward voltage  with changes in current through them. Not a lot, but some. The only way a source follower makes output voltage is to vary the current it lets through into the source resistor, and in this circuit, the LED. So the LED's voltage will change a little with changes in signal.
The LED replaces part of the the linear variation in voltage with current in a source resistor by a non-linear change in voltage with current. Exactly how much this affects the signal depends on how big the signal voltage is and how big the variation in the LED's forward voltage with forward current is. At higher currents, the  change in LED current gets more linear, and at low forward currents it's much bigger. I suspect that the output voltage will have some signal-level dependent distortion depending on the exact LED, as well as the exact JFET.

Whether this is a good idea or not depends on what you started out to do. It makes getting just buffering of a signal without much distortion harder, but then you might like the distortion.
In terms of biasing, the LED adds a (relatively) constant voltage to the source "off voltage". This might let you use JFETs with bigger Vgsoff ratings if you want to, and also have the power supply volts to spare.  The LED voltage probably pushes the JFET towards more off condition for a given current.

I guess my off-the-top-of-my-head response is that it adds something else to be controlled and makes getting some specific result out of the circuit more complicated. Whether that's worth saving some amount of LED current depends on how you like the resulting distortion of that JFET and that LED.  I'd fall back on a circuit simulator for more detail. 

Who knows? You might like the sound.
Edit: just as I hit the post button it occured to me that maybe just getting a higher efficiency LED might be almost as good as reusing current in the JFET source.

[Clint Eastwood]

I'd fall back on a circuit simulator for more detail. 

I have put the simulator to work, and it says

100mV peak input: 0.016% THD with LED  0.009% with resistor only, almost all 2nd and 3rd harmonics

1000mV peak input: 0.16% vs 0.09%, same harmonics spectrum.

So although the distortion almost doubles, it remains quite low. I will breadboard the thing and listen to it, I doubt that I will be able to tell the difference.

[antonis]

So although the distortion almost doubles,

Tenfolds..
(followes signal amplitude..)

[amptramp]

You should be able to put the LED in series with the drain, where it will not do much to the signal.  In circuits I have built before, a red LED dropped 1.88 volts and a green LED dropped 2.05 volts, so this is the same as operating the LED with the source at a lower voltage but the drain at a lower voltage as well.  This may require a change in the bias resistors but the non-linearity of the LED will be less in the drain than in the source.

[PRR]

Tenfolds..
(followes signal amplitude..)

I think he means with- and without- LED in the amplifier. Which I think is what he/we were wondering.

Yes, 10X for 100mV to 1000mV is unsurprising.