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Very basic LED circuit question

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Hi, I'm not generally into this stuff, but I'm having to build something and I'd like to use some LED's and I have a question about the circuit.

I've read that generally you use LED's that add up to 2/3rds of the voltage drop and use resistors to make up the rest and limit the current, but why couldn't I just use LED's in series until I get to the proper resistance and skip the resistor?

Thanks for your time;
Bobby.
 
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RCinFLA

Well-Known Member
LED are diodes. Their current rises exponentially once their conduction voltage is reached. You need to drive them with a current source.

If you measure their forward voltage at the current you want to run them at and match them all in series you can set up a smaller ballast resistor drop. Don't forget to take into account any supply voltage variance and temperature range.

I would allow at least a 18 to 20% voltage drop across ballast resistor.

If you want maximum efficiency then use a switching power supply that does a power conversion and operates as a current source.
 
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colin55

Well-Known Member
When a LED is placed in a circuit and voltage is applied, a voltage develops across the LED called the CHARACTERISTIC VOLTAGE DROP and this voltage depends on how the LED is manufactured as well as the colour of the LED.
This voltage cannot be altered and is a characteristic that comes with the LED.
If you supply this exact voltage to the LED, everything will be ok. This exact voltage occurs when the current though the LED is 17mA for most LEDs but can be 1mA or as high as 300mA, depending on how the LED is manufactured.
If you reduce the voltage slightly, the brightness will decrease but if you increase the voltage slightly, the LED may or will be damaged.
Rather than go through all the complexities of delivering an exact voltage to the LED, it is much easier to add a resistor and allow the current to drop slightly when the voltage drops.
Also, if the voltage increases slightly, the current though the LED will increase slightly, whereas if the LED is connected directly, the current will increase considerably.
This is the simple answer.
 
Thanks for your help, guys.

Understanding transistors in general is helping me too. Now I know why it's a "voltage drop", instead of resistance.

**broken link removed**

Here is a picture of a transistor. It runs on water but is a very good example of how a transistor operates. There are three openings labelled "B" (Base), "C" (Collector) and "E" (Emitter). We provide a reservoir of water "C" (the "power supply voltage") but it can't move because there's a black plunger in the way which is blocking the outlet to "E". The reservoir of water is called the "supply voltage". If we increase the amount of water sufficiently, it will burst our transistor just the same as if we increase the voltage to a real transistor. We don't want to do this, so we keep that "supply voltage" at a safe level. If we pour water current into reservoir "B" (the base “voltage pressure”) this current flows along the "Base" pipe and pushes the black plunger upwards, allowing quite a lot of water to flow from "C" to "E". Some of the water from "B" also joins it and flows away. If we pour even more water into "B", the black plunger moves up further and a great torrent of water current flows from "C" to "E".

This is exactly how a transistor works.


excerpted from The Beginners Guide to Electronics; M.T. Pickering.
 
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