[SOLVED] Basic electric circuit questions (amp, volts, resistance, oh my!)

[jimj]

17FEB2022 my original question takes up only a tiny amount of this thread.  Lots of good electrical circuit info is found here, so I updated the subject (old subject was "How to safely use just the 5V wire from a SATA power cable?").  My original question is below:

*****************

I know using PC power for bulbs has been discussed a lot, it seems easy enough, but I'm still not sure of the best way to just utilize the 5V power wire.

To power some 5V LEDs I'm thinking it'd be easiest to:
1.  Buy a SATA power extension cable like this
   https://smile.amazon.com/Cable-Matters-3-Pack-Extension-Inches/dp/B00V6QA4AS
2.  Plug one end into the power supply SATA connector
3.  Clip off the other end and wire the red 5V and black ground wires to my bulb

However, what abut the unused yellow 12V (and is there a 3V wire too?)?  Do I just clip the unused power wire(s) as close to the connector as possible?  What's the best/safest way to keep the unused power wire(s) from touching anything?

And if I'm way off base and missed something totally easy/obvious, please point that out too. 

One last bonus question, I could swear I've seen links to "PCI power brackets" that simply extend a molex or SATA power connector to a female port on a PCI bracket (instead of just having a wire snake out of the PC case), but I can't seem to find one of these on Amazon.  Does anyone have a link to a SATA PCI power bracket they should share?

[Zebidee]

What you propose seems sensible enough on all counts.

Some sugestions:

I use standard 4-pin molex connectors (like they use for older drives) for accessing power from PCs. You can easily buy a sata-molex power adapter cable. Then snake the female molex head of that cable out of the PC. Wire your LEDs to a male molex (just the 5v and one GND, ignore the other 2 pins), then you just plug it in.

In general, I use Molex connectors because they are easier to wire by hand and I always keep some headers/pins around.

Another tip is that some sata connectors (not all) can be easily pulled apart and moved further down the wire, then they snap back in (cutting the wires at right places to make connections). Then you can use the extra wire to put on a molex connector female end.

You could also consider using USB for power, as the USB ports will be already externally mounted on your PC. There are four USB wires. You want to use the red (5v) and black (GND), and ignore the green/white data wires. Remember that USB 2.0 ports are theoretically rated for only 500mA, but that should be enough for around 20 standard LEDs. Make sure the PC doesn't supply standby power to the USB ports you want to use (otherwise the LEDs might stay on after you shutdown the PC).

[jimj]

Some suggestions:
I use standard 4-pin molex connectors...
You could also consider using USB for power...

Thanks for the advice, what you say about molex makes sense.  However, your USB idea does sound even easier.  I actually happen to have a USB cable I had just tossed in my garbage I can rescue for this project (I'm pretty sure my USB powers off when my PC is off).  If that doesn't work then I'll go the molex route.

[jimj]

If I may take my post off topic, I'm trying to determine if I need resistors when running two coin door 5V LEDs off of USB power.  If I'm reading this thread correctly:
http://forum.arcadecontrols.com/index.php?topic=10586.0
I definitely do or do not need resistors. 

If I assume I do need resistors, then according to reply 45 I need to know the current the LEDs can handle:

Personally the easiest way for most people to do this is to wire a resistor inline with each led to limit its current. 
First, find out what the maximum current is that the led can handle.  (example 20ma [same as .02A])
Second, how much voltage is available.  (example 5V)
Third, figure out how much resistance is needed to keep the led from getting too much current (example   Resistance =  Voltage / Current)
Resistance = 5 / .02
Resistance = 250 ohms
This simply gives is the smallest size resistor (or series of resistors) that you want to use inline with the led.  Larger resistors than the value found will result in a dimmer led.

However, when I look at two coin door LEDs I've seen referenced in the forum, neither of them list a current (just volts):
https://www.cometpinball.com/products/2smd-bulbs?variant=29411784654950
https://www.pinballbulbs.com/products/frosted-555-cool-white-pinball-led

Does anyone have a final answer if resistors are needed or will 5V LEDs just magically work?

[Zebidee]

Does anyone have a final answer if resistors are needed or will 5V LEDs just magically work?

Ask the vendors. If they can't tell you, then it tells you something about them.

[PL1]

However, when I look at two coin door LEDs I've seen referenced in the forum, neither of them list a current (just volts):
https://www.cometpinball.com/products/2smd-bulbs?variant=29411784654950
https://www.pinballbulbs.com/products/frosted-555-cool-white-pinball-led

Does anyone have a final answer if resistors are needed or will 5V LEDs just magically work?

The bulbs that you linked to are direct replacements for incandescent 555 bulbs that draw 250mA@6.3v IIRC.
- They have a built-in current limiting resistor for 6.3v.  It's slightly larger than the resistor for 5v, but the LED should still light up just fine.
- As you mentioned those LED bulbs don't list the current draw, but even with a couple of super-bright SMD LEDs it will be considerably less than an incandescent.
- You can measure the precise current draw of a single bulb using your multimeter to be sure your setup won't draw more current than your USB can safely supply.

You should not need an additional current limiting resistor to run them if you're using 5v from a USB powered hub or ports that are directly connected to the motherboard. (500mA)
- If there's an un-powered USB hub in the path, the USB spec only requires it to provide 100mA per port, but many un-powered hubs exceed that minimum spec.

You definitely do not need an additional current limiting resistor to run them if you're using 5v from Molex.


Scott

[Zebidee]

If you're running 6.3v 555 LEDs from molex then you may as well grab the 12v for that as well, while you're there.

[jimj]

...
You definitely do not need an additional current limiting resistor to run them if you're using 5v from Molex.

If you're running 6.3v 555 LEDs from molex then you may as well grab the 12v for that as well, while you're there.

Thanks for the info PL1 and Zebidee.  I think a light bulb just went off in my head!   Based on what you two have written do I correctly infer that the resistor is NOT to protect the bulb from getting too much current, but instead to protect the power source from from the bulbs drawing too much current?  This is why molex 12V works as well as 5V and neither 12V or 5V from molex needs a resistor?

[PL1]

do I correctly infer that the resistor is NOT to protect the bulb from getting too much current, but instead to protect the power source from from the bulbs drawing too much current?

The molex or powered USB hub port can supply far more current than a couple of 555-replacement LED bulbs will draw.

The one way you might draw too much current is if you're powering two of those 2-LED bulbs through a single unpowered USB hub port.
- If the LEDs draw 30mA and there are two LEDs in parallel per bulb, that would be 60mA per bulb.  2 bulbs wired in parallel would draw 120mA which would exceed the 100mA USB spec for unpowered hub ports but most can handle it anyways.
- If the LEDs are in series like the example below, it would be only 30mA per bulb.  2 bulbs wired in parallel would only draw 60mA.  No problem.

--------------

The current limiting resistor is there to protect the LED in two ways:

1. The LED is only rated for a certain number of miliAmps -- most are 20mA, but some of the large or SMD ones are higher.

If you allow more current to flow through the LED than it is rated for, you'll see the same result as running 5A through a 1A fuse.  You'll let out the magic smoke that makes it work.   

2. The LED is only rated for a certain voltage -- usually between 1.8v and 3.2v depending on the chemical composition of the PN junction.

If you try to drop more voltage across the LED than it is rated for you'll let out the magic smoke that makes it work.   
-----------------
To put it in plumbing terms:
  * Voltage is comparable to water pressure
  * Amperage is comparable to the amount of water flowing through per second
  * The LEDs and resistor are like water wheels
  * If the water pressure is too high, too much water will flow through causing the water wheel to spin too fast and fail
  * After the water flows over one water wheel, it will flow through the next one in series

For this example, let's use two red LEDs (rated for up to 2v) and a current limiting resistor.
- The first two wheels (the LEDs) drop 1.5v each.
- The third wheel (the current limiting resistor) drops 2v.

If you tried to run one LED (fourth wheel) on 5v without the resistor, it would have to drop all 5v and it would draw more current because there would be less resistance between 5v and ground.
- Ohm's law states that voltage = current * resistance so if the voltage is the same and the resistance decreases, the current increases.

By changing the value of the resistor, you control how much voltage it drops and the amount of current flowing through the circuit.




Scott

[jimj]

...
The current limiting resistor is there to protect the LED in two ways:
1. The LED is only rated for a certain number of miliAmps -- most are 20mA, but some of the large or SMD ones are higher.
If you allow more current to flow through the LED than it is rated for, you'll see the same result as running 5A through a 1A fuse.  You'll let out the magic smoke that makes it work.   
...

Thank's Scott for taking the time to explain electric circuits 101 to me (and if you never reply on this thread again because I'm a lost cause I would totally understand .

I think the main takeaway is that I'll be fine running a couple 5V LEDs off my motherboard's USB 5V line.

However, I still don't fully understand the first reason for releasing magic smoke.    If I wired up 10 20mA 5V LEDs in parallel they'd be trying to pull 200mA.  If I plugged these LEDs into an unpowered hub that was truly limited to 100ma then why would magic smoke appear?  Wouldn't their volts just be cut in half causing them to either not work or be dim?  My (almost certainly incorrect) logic is below:
I=V/R
If I could supply 200mA then for 5 volts R = V/I = 5/0.2 = 25 ohms (i.e. under normal/full power working conditions each LED is supplying 25 ohms of resistance).
If I'm limited to 100mA I assume the resistance can't change, so therefore my voltage must drop in half?  V = R*I = 25*0.1 = 2.5 volts.

Or can't voltage drop and I need to dissipate 2.5V with a resistor?  Is it this extra 2.5 V that causes the magic smoke?

I feel like I'm close to understanding this, but I still don't really have a good feel for what happens with the extra 2.5V in my example above.

[Zebidee]

Supply voltage is the same and you need a resistor for each LED. This is because you'd wire them in parallel, which means each LED gets its own little circuit including 5v supply and a resistor to ground

You "release the magic smoke" when you try to run (say) a typical 20mA 2v rated LED from 5v with no current limiting resistor. It will burn very brightly for a short time. It doesn't matter how many LEDs you have, you will need a resistor for every LED.

The most important thing to remember is that series circuits divide voltage, parallel circuits divide current. You could wire them in series with a single resistor, but you'd run out of volts after just 2 LEDs with a typical USB 5v/500mA supply (e.g. from a powered hub). However, wired in parallel you divide current, and can easily have over 20 x 20mA LEDs with the same USB power supply.

I= V/R

As you add more LEDs in parallel, total resistance (R) must go down, so current (I) must go up each time you add a new parallel branch. It is simple addition, just add them up. Supply voltage stays the same for each parallel branch, so long as the USB can supply the total current demanded.

[PL1]

I think the main takeaway is that I'll be fine running a couple 5V LEDs off my motherboard's USB 5V line.

You got it.

The current limiting resistor is there to protect the LED in two ways:
1. The LED is only rated for a certain number of miliAmps -- most are 20mA, but some of the large or SMD ones are higher.
If you allow more current to flow through the LED than it is rated for, you'll see the same result as running 5A through a 1A fuse.  You'll let out the magic smoke that makes it work.   

However, I still don't fully understand the first reason for releasing magic smoke.    If I wired up 10 20mA 5V LEDs in parallel they'd be trying to pull 200mA.  If I plugged these LEDs into an unpowered hub that was truly limited to 100ma then why would magic smoke appear?  Wouldn't their volts just be cut in half causing them to either not work or be dim?

The "If you allow more current to flow through the LED than it is rated for" part would be the result of using a too small value for the current limiting resistor.
- It won't happen with 5v and the 6.3v bulbs you linked to since they have a large enough resistor in the base of the bulb.
- It could happen with 12v and the 6.3v bulbs you linked to if the resistor you add drops less than 5.7v which forces the bulb to drop more than 6.3v.

In your example, the unpowered hub is where the magic smoke might escape from.
- The USB specification says that each port on the hub must be able to provide 100mA.
- Trying to draw 200mA through one port is far enough above the spec that it could damage the hub unless there are current-limiting circuits built into the hub.  Those circuits might be in some hubs, but AFAIK there are quite a few unpowered hubs that are not current-limited to 100mA per port. 
- It's not a good habit to exceed published specs by 100%.  Better safe than sorry.   

The best approach when you don't know one spec for a particular LED is to test it before wiring everything up in the final form.

To test the current draw on the LED bulbs you linked to, set your multimeter to "mA" and use this test setup.
- No separate current limiting resistor needed because there's one inside the bulb.
- If you get the LED anode (+) and cathode (-) backwards, the LED will be reverse biased.  No damage done, it just won't light until you swap the connections so it is forward biased.

    USB 5v ---- red lead--meter--black lead ---- +LED bulb- ---- USB ground

Once you confirm how much current one bulb draws, you'll know for sure that it is safe to wire the two bulbs in parallel like this.   

    USB 5v  --------|---------|
                            |             |
                         bulb 1    bulb 2
                            |             |
    USB ground ----|---------|


Scott

[jimj]

Thank you so much for the detailed explanations Zebidee and Scott.  I actually think I'm starting to understand the basics of electricity now (first few days pounding my head against this were frustrtating) .  Since it's my bedtime I ran out of time to think of any questions in reply to Scott's post .  I do have a couple clarifying questions form Zebidee's post though.

Supply voltage is the same and you need a resistor for each LED. This is because you'd wire them in parallel, which means each LED gets its own little circuit including 5v supply and a resistor to ground

You "release the magic smoke" when you try to run (say) a typical 20mA 2v rated LED from 5v with no current limiting resistor. It will burn very brightly for a short time. It doesn't matter how many LEDs you have, you will need a resistor for every LED.

So 5V suppply minus 2V LED means we need to drop 3V with a 150 ohm resistor?
   R=V/I
   R=3/0.02
   R=150 ohms
Regardless of how many lights you wire in parallel they'll each get their own 150 ohm resistor?

Could I wire a 125 ohm resistor R in series with a set of parallel 20mA 2v lights behind it like this*:
    USB 5v  ---R----|---------|
                           |            |
                         bulb 1    bulb 2
                           |            |
    USB ground ----|---------|
   R=V/I
   R=5/0.04   //Two 20mA bulbs add to 0.04A
   R=125 ohm

Is it better to use multiple resistors in parallel or a single resistor in series?

Actually, as I'm writing this post I just realized I think I know the answer.  You should not put a resistor in series before a parallel set of lights because if any of the lights were to fail the resistance of the circuit would change.  With my two light example the single 125 ohm resistor would work until one of the lights fails.  With only one working light (i.e. path/circuit) I would now need a a 150 ohm resistor, not 125 ohm.  Can anyone confirm if my logic/understanding is sound?

*ASCII art shamelessly stolen from Scott without permission. 

The most important thing to remember is that series circuits divide voltage, parallel circuits divide current. You could wire them in series with a single resistor, but you'd run out of volts after just 2 LEDs with a typical USB 5v/500mA supply (e.g. from a powered hub). However, wired in parallel you divide current, and can easily have over 20 x 20mA LEDs with the same USB power supply.

I= V/R

As you add more LEDs in parallel, total resistance (R) must go down, so current (I) must go up each time you add a new parallel branch. It is simple addition, just add them up. Supply voltage stays the same for each parallel branch, so long as the USB can supply the total current demanded.

Ahh, another mental breakthrough for me.  Each parallel light adds another path for the electricity to flow, so with more paths to flow there is less resistance.  Just like if bucket of water has more holes in the bottom of the bucket each hole you add means there's less resistance to the water flowing out (each hole is like a light wired in parallel).  Writing Ohm's law as V=I*R it's clear to me that with V being constant as I goes up R must go down.

[PL1]

Is it better to use multiple resistors in parallel or a single resistor in series?

Actually, as I'm writing this post I just realized I think I know the answer.  You should not put a resistor in series before a parallel set of lights because if any of the lights were to fail the resistance of the circuit would change.

Once again, you got it.   

If you wire it like this* and bulb 1 burns out, current stops flowing through that branch of the circuit and the voltage drops (3v resistor + 2v LED) and curent flowing through the bulb 2 branch remain the same.

    USB 5v  --------|---------|
                            R            R
                            |             |
                         bulb 1     bulb 2
                            |             |
    USB ground ----|---------|

*ASCII art shamelessly stolen back without permission.   


Scott

[Zebidee]

Ahh, another mental breakthrough for me.  Each parallel light adds another path for the electricity to flow, so with more paths to flow there is less resistance.  Just like if bucket of water has more holes in the bottom of the bucket each hole you add means there's less resistance to the water flowing out (each hole is like a light wired in parallel).  Writing Ohm's law as V=I*R it's clear to me that with V being constant as I goes up R must go down.

I love these "light-bulb" moments (please excuse the pun )

 

Ohms law is the bomb - whenever I'm confused about my electrical circuits, I go back to basics and apply Ohm's law, and then things become clear.

[jimj]

Yes, light bulb moments are MUCH more enjoyable than the moments I usually have.    However, I thought of one more thing that's not clear to me even after staring at Ohm's Law. 

Is it possible to know what happens to the current when the voltage is decreased?  I verified that the 2SMD Bulb LEB draws 25mA at 6.3V.  Using Ohm's Law R = V / I then
   R = 6.3/0.025
   R = 252 ohms
At this point life is good, I know all three parts of Ohm's Law.  However I'm going to run these bulbs at 5V.  Since V=I*R I know that if V goes down then I and/or R must go down.  Is it safe to assume the resistance is constant so the amps used by the LED go down like this:
   I = V/R
   I = 5/252
   I = 0.020A
I.e. since we know that at 6.3V the LED uses 25mA then at 5.0V the LED uses 20mA?

Of course this is all academic at this point.  Once I have the LED in hand I know I can use Scott's instructions in reply #11 above to measure how many mA it really draws.

[seawolfgoo]

Since most PCs components use the 12v rails, perhaps go with 12v bulbs. 

I use pc fan header splitters.  You can harvest some old fan connectors to make a quick disconnect.  This might address your safety concern.
https://www.amazon.com/gp/product/B07TRV6XZX/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1

For existing incandescent bulb holders, like the one under a happ trackball, I use 12v T10 led bulbs, they fit perfectly and come with a remote.
https://www.ebay.com/itm/125022337915?hash=item1d1be97b7b:g:o9UAAOSwqWFhoZBn

The 12v rail is also nice because it powers my legacy 14v incandescent bulbs inside the coin door. 
https://www.amazon.com/gp/product/B00JNFB31I/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1

If you use the motherboard controlled fan headers, you may want to force them to max in the bios, otherwise the brightness will auto adjust based on the cpu load. 

[jimj]

Since most PCs components use the 12v rails, perhaps go with 12v bulbs. 

I use pc fan header splitters...

I'm going to 5V USB for my coin door lights, but thanks for the suggestions.  I wasn't aware of fan header splitters, may use that in the future.

[jimj]

...To test the current draw on the LED bulbs you linked to, set your multimeter to "mA" and use this test setup...

One last post on this.  Comet Pinball actually told me that the mA used by their 2SMD Bulb LEB depends on the batch and color, but it shouldn't be much more than 25mA at 6.3V.

I ordered an assortment of colors and measured their current as:

• 17mA cool white
• 19mA warm white
• 30mA yellow
• 31mA red

This is with these LEDs powered by my motherboard's onboard USB ports which provide 4.85V.

I liked the look of the yellow LEDs best, so I'm using two of those for a total current draw of 60mA.

Thanks again Scott and Zebidee for the crash course in basic electric circuits.

[PL1]

Glad to assist and thanks for posting your test results.   


Scott