Your condensed version wasn't helpful. I bit the bullet and read your first post.

The second paragraph described your project.

If you want it to be something you'd be proud of, using that dual USB cigarette is going to look like a hack. If I was to do something like this, I'd make a faceplate that fit well in the opening. I'd find a way to disassemble the USB thing you have to make them look like they were designed for the faceplate.

You're not going to be able to measure the battery voltage of anything you're charging. And the USB charging devices don't have a charging indicator.

Displaying the car battery voltage (when the vehicle isn't running) or the regulator voltage is more realistic.

Color makes a difference because forward voltage is different for each color. We also need to know the maximum DC current they're designed for.

The most common way to power LEDs is to use a current limiting resistor. If you use multiple LEDs and you want them to appear to be the same brightness, you need to use a current source; because the forward voltages of the same color LED from the same batch isn't tightly controlled.

Hey... You...

That was useful information! Thank you. And FYI...

I bought a piece of 1/16" abs plastic. One side is baby butt smooth and the other looks like pretty much any other panel. I think they called it "skin cell" or something oddly.

But yes my reason in coming and asking is because I want it to look as natural as the original piece that was there. Which was a small digital clock.

Certainly trying to avoid the hack job as you call it. I refuse to do this any way but the proper and best looking way. If I don't get answers here I will get it somewhere.



Note the white space!

 

To monitor USB charging , a simple circuit that detects a V drop across a diode could be used. Diode is inserted between supply & buck converter + input.View attachment 156150

Bernard, this is quality info! Man where were you guys when I posted this? I'm still in the info gathering stage so your information will certainly be useful no matter what.

I just love how my welcoming party to this forum was a bunch of b.s.! Lol

Hey I can deal with it otherwise I wouldn't return right!? Like I mentioned in the other reply, if you can dish it you best be able to take it when you're served!

This isn't in reference to you, I just can't get over the wonderful welcome I got! I wish helpful guys like you had been around. I feel so at home here now...

 

What about a shunt and op-amp and comparator? You could also add one comparator to detect current and another to prevent a short, and more to get an LED bar indicator of current. The issue with a diode is that is wastes power and if you have a 5V converter it's now a 4.4V converter (for silicon).

Hey now not everybody all at once!!

Thanks again livewire! So back to what you had mentioned. By taking one, or two even USB ports out of a charger like you mentioned isn't necessarily cheating myself of a useful project that might teach me something I can use in the future or is it basically that all I'm going to learn from this is how to do this so I might as well just go the easy route electrically anyway and use the premade circuit?

If that sounded snide or something it in no way was. I know I've thrown a s***ton of sarcasm around here already but now I'm honestly trying to take all of this in seriously.

I did come here for help right!? Thanks again! Your ideas and opinions all count. I'm actually fairly intelligent. When it comes to electricity though I've never taken it in. Never to late though right!? Right. I think I say right to much.... Am I right

 

Your condensed version wasn't helpful. I bit the bullet and read your first post.

The second paragraph described your project.

If you want it to be something you'd be proud of, using that dual USB cigarette is going to look like a hack. If I was to do something like this, I'd make a faceplate that fit well in the opening. I'd find a way to disassemble the USB thing you have to make them look like they were designed for the faceplate.

You're not going to be able to measure the battery voltage of anything you're charging. And the USB charging devices don't have a charging indicator.

Displaying the car battery voltage (when the vehicle isn't running) or the regulator voltage is more realistic.

Color makes a difference because forward voltage is different for each color. We also need to know the maximum DC current they're designed for.

The most common way to power LEDs is to use a current limiting resistor. If you use multiple LEDs and you want them to appear to be the same brightness, you need to use a current source; because the forward voltages of the same color LED from the same batch isn't tightly controlled.

Also I should mention that I'm only trying to monitor the voltage of the battery under the hood. True it is likely going to be upwards of 14.4 volts when I'm driving but it's something i'd like to monitor.

In my really long.. really really long post where I summarized the project in the second paragraph I mentioned, possibly, that I'm adding some stereo equipment and though not a huge amount it's something I'd like to keep tabs on.

Knowing the voltage would help me in the event it was dropping below a desirable level so I could adjust whatever I needed to.

 

Re post #18: A shunt R will dissipate about 100 mW vs 1.4W with diode. The 5 V output is not changed by a diode on input to buck converter.

 

The 5 V output is not changed by a diode on input to buck converter.

Absolutely right! As long as it's ok to monitor on 14V side instead of 5V side, any voltage drop due to the sensing circuit doesn't really matter.

Re post #18: A shunt R will dissipate about 100 mW vs 1.4W with diode.

I don't quite follow this. Do we know how much current will be running through this diode or shunt? It's a function of charging current, the ratio of input/output voltages, and losses due to imperfect converter efficiency, right?

So, let's say charging is 500mA at 5V. The buck circuit only needs 179mA at 14V to deliver that same power, plus some amount for losses... say 10%, so round up to 200mA, maybe?

In this scenario diode dissipation would be 0.2A x 0.7V = 0.14W, or 140mW.

But even after all that, we can't compare shunt dissipation without knowing what value shunt resistor would be used, right?

Anyway, I'm not trying to be difficult, I swear! It did strike me as a little odd to be comparing dissipation figures when it seems like there are a lot of assumptions to be made before calculating them. As for the calculations, I ran through that for practice and a sanity check more than anything. I'm still learning and sometimes I'll work through these little exercises just for practice.

I would be interested in hearing your thought process with the dissipation figures though? Would it really be that much higher with a diode? And, what's the benefit of choosing the option with significantly higher power dissipation?

 

Then there is this: https://www.ebay.com/itm/2-in-1-Car...ter-Voltage-Meter-12V-24V-BC552-/191863706289

There plenty of these round gizmo things on ebay for separates if you will.

Then there is USB. Sometimes it works and sometimes it doesn't. Many times, it doesn't.

I've bought a number of USB chargers, but many won't charge a Motorola Backflip. One of the latest chargers I purchased was dual power in 120 and 12 V cigarette with 2 unlabled USB ports. The seller just sells so he didnt know anything and even the package doesnlt know anything.

there were two ports. I identified them as 0.5A and 2.1 A. The 0.5A port could charge the Backflip phone and a Garmin GPS. The Garmin GPS would actually charge using either port. They could be charged together providing the right ports are selected.

This https://www.ebay.com/itm/Flush-Moun...l-USB-Port-Charger-Adapter-Cover/292626944451 might have the BEST CHANCE of one of the ports working. Note one is labeled 0.5 and the other 2.1 A

Another issue I had with an MP3/Handsfree./2 port USB charger is that it died when the car was jump started,

 

@ ebeowulf17, minor screw up on my part; used 2 A output for figuring input data. Input current is about
900 mA counting efficiency for 2 A output. A 50mV @ 1 A meter shunt, 50 m ohm , should be available but I see that 75 mA
shunts are more common. Shunt power = 1 A X .0 5 V = 50 mW. Diode power about, 1 A X .7 V = 700 mW.
With max. charging power of about 10 W, either methods seem reasonable.