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Topic: Super Easy Noob Questions about Circuit (see pics inside) [Updated] (Read 733 times) previous topic - next topic

Synbro

Complete electronics noob here.  I have the following LED setup I pulled out of a USB hub that is super annoying.
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Images: http://imgur.com/gallery/8GSSS




So the simple questions:
1. Red = power, black = ground?
2. Why does something need to be grounded?
3. Is there anyway to tell, just from the picture the LED voltage and current requirements?
4. Why does the wiring that is closer to the LED look fatter?
5. What determines how wide/thick the wiring and the board is? (I assume it is an IV requirement but how would I go about calculating this?
6. How do I model this circuit on paper?
7. Since the power goes to both LEDs, are these in series or parallel and why does it matter?

Due_unto

Your pics don't show and your links come up as dangerous website - blocked.
1. Common practice, de facto standard. You won't be arrested if you use different colors.
2. Ground is the common reference and also the current source for circuits, hence Vss.
3. Usually not unless you can see the entire circuit.
4. Can't see the picture
5. Current determines wire size. How much you put on the board determines board size.
6. Start with one component, add connections and other components as observed on the board.
7. Unable to see circuit
Do not look into laser with remaining good eye.

Synbro

Thanks for the reply,

did you see the Imgur link - I updated the pictures.

CrossRoads

1. Red = power, black = ground?
Probably. If not, switch them around. Got a 5V source? Connect a 270 to 1K resistor to +5, to red wire, then black wre to Gnd.

2. Why does something need to be grounded?
Electricity  needs to make a loop from positive to negative - otherwise there is no current flow.

3. Is there anyway to tell, just from the picture the LED voltage and current requirements?
No. But you can figure about 40mA max should be allowed to flow, and there will be ~2.5 to 3.5V across the LEDs.
Call it 3V. Then (5V - 3V)/.04A = 50 ohm resistor as the lowest value. With 270 to 1K you can measure the Vf of the LEDs and adjust your resistor from there for max brightness if needed.

4. Why does the wiring that is closer to the LED look fatter?
Thicker insulation? More durable wire? Really, 30 AWG wirewrap wire is all that is needed for 40mA, but it's not very durable in a vibration environment.

5. What determines how wide/thick the wiring and the board is? (I assume it is an IV requirement but how would I go about calculating this?
Wire thickness determined by current and voltage drop needs.
Board thickness determined by engineering/manufacturing. 1.6mm thick is common.

6. How do I model this circuit on paper?
Draw two LEDs in parallel.

7. Since the power goes to both LEDs, are these in series or parallel and why does it matter?
Two in parallel will share same source voltage, if their Vf is different enough then 1 may not turn on. Current needed is double.
Two in series need a higher source voltage to turn both on, but will share the same current.
If both have a Vf of 3V and the source is only 5V, then there is no chance of both turning on.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Synbro

Thank you, follow up questions:

1. Red = power, black = ground?
Probably. If not, switch them around. Got a 5V source? Connect a 270 to 1K resistor to +5, to red wire, then black wre to Gnd.

2. Why does something need to be grounded?
Electricity  needs to make a loop from positive to negative - otherwise there is no current flow.

Follow Up Question: this is a bit strange for me: If I stick a wire into my outlet and attach something to the end it will receive charge without completing a loop.  Can you explain further?

3. Is there anyway to tell, just from the picture the LED voltage and current requirements?
No. But you can figure about 40mA max should be allowed to flow, and there will be ~2.5 to 3.5V across the LEDs.

Why 40mA max?

Call it 3V. Then (5V - 3V)/.04A = 50 ohm resistor as the lowest value. With 270 to 1K you can measure the Vf of the LEDs and adjust your resistor from there for max brightness if needed.

Why do I need a resistor?

4. Why does the wiring that is closer to the LED look fatter?
Thicker insulation? More durable wire? Really, 30 AWG wirewrap wire is all that is needed for 40mA, but it's not very durable in a vibration environment.

5. What determines how wide/thick the wiring and the board is? (I assume it is an IV requirement but how would I go about calculating this?
Wire thickness determined by current and voltage drop needs.
Board thickness determined by engineering/manufacturing. 1.6mm thick is common.

6. How do I model this circuit on paper?
Draw two LEDs in parallel.

7. Since the power goes to both LEDs, are these in series or parallel and why does it matter?
Two in parallel will share same source voltage, if their Vf is different enough then 1 may not turn on. Current needed is double.
Two in series need a higher source voltage to turn both on, but will share the same current.
If both have a Vf of 3V and the source is only 5V, then there is no chance of both turning on.

tinman13kup

Follow Up Question: this is a bit strange for me: If I stick a wire into my outlet and attach something to the end it will receive charge without completing a loop.  Can you explain further?
You are talking AC mains voltage now. Without getting too deep, you are completing a circuit. The ground/neutral are connected to earth ground, which you are standing on. People are capacitive, so you also feel the effects of AC more readily than DC voltages. Don't jam stuff into outlets. It can kill you


Why 40mA max?
Leds are somewhat an oddity in that they require current limiting, by resistor or current limiting circuits. Without something to limit the current, the led acts as a short and will burn itself out faster than you can see the flash. Leds are not all the same. Some have very small voltage drops and can only handle small current loads, while others require large amounts of current and voltage. The little led on your computer uses little, while the led driving lights on a car take a lot. You cannot just pick an led up and know what the Vf and I requirements are without a datasheet.


Why do I need a resistor?
The resistor is used for small leds to limit the current. They are easy to get and easy to use on small leds, like the ones on your computer. High power leds require a different approach, such as using a current limiting power supply.
Tom
It's not a hobby if you're not having fun doing it. Step back and breathe

Jiggy-Ninja

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Follow Up Question: this is a bit strange for me: If I stick a wire into my outlet and attach something to the end it will receive charge without completing a loop.  Can you explain further?
I certainly hope you are not sticking wires into your outlets! You should be using properly rated cables. That last word is key; cables have more that one wire in them. Mains power cables usually have 2 or 3. For AC mains those are called (in the US at least) hot and neutral, with ground being the optional 3rd one.

See how there's two metal prongs on one end and two holes on the other? Those are connected to two separate wires in the cable. Just like there's two wires going to the LED board, every electrical appliance in your house has two wires going from the plug to it. They're just bound into one cable for convenience.
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Why 40mA max?
Experience. Most indicator LEDs are pretty interchangeable since they are used for the same applications. Maybe if you had the datasheet for it you'd see that it can handle 50 mA or something, but 40 is a pretty safe bet.
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Why do I need a resistor?
The short answer is because that's just how LEDs work. Every component acts differently (that's why there are so many different kinds anyway), and LEDs (and diodes in general) are just not the kind of component you can apply a voltage to and get good results.

The long answer is because the relationship of current to voltage in a diode is not linear like a resistor, but exponential. I don't just mean "it increases really quickly" like the slang meaning of exponential, it is literally an exponential function like y = ex.

As you can see, above a certain voltage the current shoots up very rapidly. This makes it extremely difficult to control a diode using just a voltage source, because small changes in voltage can cause huge changes in current. The graph's values are also sensitive to the diode's temperature. As the junction heats up, the current will actually increase. Depending on how the diode is used in the circuit, this can cause a situation called thermal runaway, where the increasing temperature causes increased current, which causes an even higher temperature in a positive reinforcement loop until the diode is destroyed from overheating or overcurrent.

To combat this, an external circuit is usually used to tame the LED's current. For low power indicators, a series resistor is usually enough. For higher wattage LEDs, the power dissipated in the resistor becomes substantial, so they tend to be used with more complicated active circuitry (transistors and amplifiers) that can be more precise about regulating the current.

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