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?
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.
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.
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