Shunting LED

[UnlikeKurt]

Hi all.

I have multiple small circuits running off of the same 9V supply.
One of which is an LED. Due to the wiring and limitations of the switching arrangement i'm not able to turn the led on/off by engaging/disengaging ground to its cathode. But what i could do is wire the anode to the switch and shunt it to ground. 1. Would this work? 2. Would it screw up the whole 9v supply to the other circuits?

thanks
james

[The Scum]

You'll have a resistor in series with said LED?

Then you can put the switch on either side.

If no resistor, expect very short LED lifetime.

(scratch resistor math: 9V minus a 2V forward drop leaves 7V...using V=IR, a 700 Ohm resistor will give you 10 mA through the diode. 1K is probably easier to find, and should work about as well.)

[Scodiddly]

You don't want to put a short to ground on the + side of the LED. That will drag down the power supply, maybe causing damage.

You want to put the switch in series between the power supply + and the LED +. Or between power supply - and the LED -... the power supply - is also ground, which fits your description of "shunting to ground".

[The Scum]

Good catch, Sco,

You can indeed switch either side, but the LED orientation has to be such that the current still flows anode-to-cathode (following the arrow contained in the schematic symbol).

One other possible arrangement is with the LED and switch in parallel. When the switch is closed, it presents a lower impedance path, and will carry all of the current - when it opens, the LED will. You still need the series resistor to limit the current in the LED and switch.

This technique is used in mixers, because it means that the current being used by the circuit doesn't change when the switch is operated, avoiding a possible source of pops. But it also presents a constant load on the supply - a bit wasteful, since you're drawing the current even when nothing is lit, particularly if the supply is a battery.