Resistance in parallel with LED?

Started by sfr, February 12, 2007, 03:36:28 AM

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sfr

This is probably a simple basic electronics thing that I should be getting, but I'm not quite sure -

Looking at the Tremulus Lune schematic at Tonepad - the two LEDs (one the rate indicator, the other one half of the LED/LDR combo) you see each one has a 1k resistor to ground, (I'd assume as a current limiting resistor, but I don't think the voltages coming out of the LFO is that high) as well as another 1K resistor in parallel to the LED (across both legs of the LED)  What's the purpose of this set up?  Are the two resistors working as a voltage regulator, feeding a reduced voltage to the negative end of the LED, and keeping the LED from being as forward biased (and thus constantly lit) at all times?  Or is something else going on?
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R.G.

I don't know what that specific circuit does, but I do know what mounting an LED on a voltage divider does.

Assume a circuit with two 1K resistors in series, and an LED parallel with the 1K to ground.

As you increase the voltage across the two resistors from 0V what happens?

Initially, the LED does not have enough voltage to conduct, so it's effectively not there. The voltage at the middle of the two resistors goes up linearly at 1/2 the voltage applied to the top resistor.

Then the voltage at the divider gets to the LED turn-on voltage. Now the LED starts sucking current. But how much?

The LED forward voltage is roughly constant. Not exactly, but close. So with a constant voltage across the lower 1k, its voltage remains constant by the same "roughly". Now as you increase voltage, all of the increased voltage appears across the top 1K resistor, and the current is ((Vsupply-VLED)/1K) + (VLED/1K) The first term being the variable current in the top 1K resistor, the second being the constant current in the bottom resistor.

It's a turn-on delayer. The LED turns on at twice the voltage it otherwise might.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

sfr

Okay.  Thank you once again.  That makes sense to me.

What happens when the LED is in parallel with the "top" resistor of the voltage divider, however?  In this case, the LED is in parallel with 1K from the voltage source to the middle of the voltage to divider, rather than the 1K from the middle of the divider to ground?

Sorry for all the stupid questions.
sent from my orbital space station.

R.G.

Same thing. The LED makes no difference at all until the voltage across the resistor it's paralleling gets to the LED threshold. Then it clamps the voltage on its parallel resistor to the LED voltage and the LED eats any more current that comes through, limited by the un-paralleled resistor.

This is a great illustration of the series-order-doesn't-matter argument. We have two "components". One is a 1k resistor, the other is a 1k paralleled by an LED. The series-order-doesn't-matter principle says that if we don't care about the voltage in the middle, it doesn't matter which order we put the two components in. The voltages and currents at the outside pins will be the same for all orders. And it is true in this case.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

amz-fx

#4
sfr,

The parallel resistor acts as a brightness control for the LED.

In the example shown here, there is 7ma through the circuit... the 1k resistor at the top carries the full 7ma and because we know that the voltage is 2v at the junction of components, we can calculate the equivalent resistance of the LED (400ohms).   THEREFORE,  instead of 7ma flowing through the LED as is the case without the parallel resistor, the current is divided between the resistor and the LED, so the current in the LED is reduced to about 5ma.



Remove the parallel resistor and there will still be 2v at the junction so the equivalent resistance of the LED is now about 286ohms and the LED is carrying the full 7ma.

As you make the resistor smaller in value, it will carry more current and make the LED dimmer.  Make it larger and...  :icon_biggrin:

In the T-Lune, the LED is on all the time so the parallel resistor is a way of controlling the range...

regards, Jack

sfr

Thanks you guys.  I get it now.  I should have gotten it the first time, and I certainly should have gotten it after the first reply.

Sometimes it seems like I beat my head against a wall understanding things that should be simple enough.  I think I need to go back to reading the basics of electricity as it's obvious a lot of this stuff didn't sink in the first time through.   I appreciate you guys taking the time to explain things to me until I understand it.
sent from my orbital space station.