# Very basic question about placing an LED in a circuit

Discussion in 'Electronic Basics' started by Thot, Apr 29, 2005.

1. ### ThotGuest

Hello all.
I have been loorking at this group for a while.

I am an absolute beginner just now experimenting with hobby electronics on
my own.

I am building a simple ADC from a project that I found on a magazine
(Popular Electronics).

My question is:
I would like to improve it by adding an LED so that I know when the ADC is
powered.
The circuit is powered at 5V (9V battery through a 5V stabilizer IC).
Do I just connect the leads of the LED to any + and ground points, in series
with a proper resistor?
Or will this affect the behaviour of the rest of the circuit?
If so, is there any better strategic point to connect the LED?
Perhaps to the 9V source, before the 5V IC?

Sorry if the question appears trivial.

Any hint will be appreciated

Luca

2. ### John PopelishGuest

You have several choices. LEDs run on current (their light output is
roughly proportional to their current) and drop a certain voltage
while this current passed through them. For instance, red LEDs
typically drop about 2 volts when they are operating somewhere near
their rated current. Yellow and green LEDs drop 2.5 to 3 volts and
blue ones may need 3.5 (shorter wavelengths need more voltage across
the junction to produce the higher energy photons).

You could operate the LED from the regulated 5 volt output. For
instance, a red LED would need a resistor in series to control its
current that wastes the about 3 volts extra. If you decide that 10
milliamps provides the needed light, that resistor would be about
3V/0.01A=300 ohms.
But this additional load will make the regulator hotter and it will
drop out of regulating well with a bit higher battery voltage. But
the LED will shine steadily till the regulator sags.

If you run the LED directly from the 9 volt battery, you will need a
higher value of resistance to limit the current to the same peak
value. In this case, the 9 volt battery has 7 volts more than the LED
needs, so you will need a 7V/.01A=700 ohm resistor. I would probably
pick 680 ohms because it is the nearest 5% standard value. Now, the
LED will dim smoothly as the battery sags, but the regulator will
operate to a slightly lower battery voltage because it doesn't have to
pass the LED current.

A stranger possibility would be to operate the LED in series with the
regulator input. This would use no extra power from the battery, but
would transfer power that would have been burnt up in the regulator to
the LED. This case is practical only if the 5 volt load current
happened to be about right for the LED to operate upon. It also
raises by a couple volts, the battery voltage that allows the
regulator to begin to sag.

I recently built a fourth version based on the LP2950CZ5.0 low drop
out regulator. Since this regulator works with an input voltage below
5.5 volts, I connected a PNP transistor, emitter to the battery,
resistor from base to the regulated output (keeping the value high
enough that it didn't pull the regulated voltage above 5 volts) and
put an LED and current limiting resistor between the collector and
ground. As long as the regulator has more than .6 volts across it,
the transistor is on and the LED lit. This makes the LED go out just
just before the sagging battery drops low enough to sag the regulator
output. A lit LED indicates not only power on, but regulator
regulating. It draws the same power from the battery that connecting
the LED directly across the battery does. But I was able to achieve
acceptable brightness with only a milliamp or 2.

3. ### John SmithGuest

Put the LED (with a suitable resistor in series) across the +5V and GND.
That way you know that the ADC has power applied to it.

Putting it (the LED and resistor) across the 9V only tells you that you have
9V: it doesn't tell you if the 5V regulator is operational.

4. ### Andrew HolmeGuest

Yes, you may connect them between any convenient points.
No, it won't if you choose points with a low resistance path to +5V and GND.
The worst you can do is cause a voltage drop by drawing the LED current
through the distributed resistance of your power supply wiring. This will
be a VERY small and constant drop so it can't do much harm.

5. ### ThotGuest

Thank you for the replies, they were very helpful!

Luca