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Please help me understand this 12VDC - USB circuit. Answered

This ties into my earlier question about the large resistor (Thank you, Quercus austrina et al).

I'm an electronics newbie working on hacking a boombox into an under-cabinet kitchen radio.  One of the things I want to do is provide USB power to charge my BlackBerry while I listen.  I want to repurpose this car power adapter to do that.  It has a cigarette lighter plug, two cig lighter jacks, and one USB port.

I have attempted to draw a schematic of the circuit as I understand it.  This is my first schematic, so be gentle.  Sorry about the small size.  You may need to click on it and choose "Original Size" to see it more clearly.

Here are my questions:

1) There's a surface-mount resistor (R2) between the big scary "sandstone" one and the LED.  The numbers on that resistor could be read as "102" or "201".  If I understand correctly, it's either 1000ohms or 200ohms.  That's a significant difference. Which is it likely to be?

2) Likewise, the ceramic disc capacitor (C2) appears to be labeled 104, but is very hard to read, since it's hidden behind the other two.  There may be a dot after 104.  Would it be .1µF, then?

3) Did I correctly decode the numbers on the group of four resistors that provide current to the USB data pins?

4) Here's the Big Question:  If I remove the 7805 and everything to the left of it, and hook in 5VDC from a computer power supply, will it function?

Thank you for your consideration.


1.) 200 Ohm, NOT 1K
2.) 104 = 10000 pF, or 10nF or 0.01uF
3.) Probably.
4.) Yes.

In an un-negotiated USB connection, you can only pull 100mA from the supply.

Thank you!

You say "un-negotiated".  It's my understanding that the purpose of the resistors on the far right are to supply a small amount of current to the data pins in order to fool the USB device into thinking that it's connected to a computer.  Is that not the case?

I'm looking at instructables like this one here that describe this.

An official USB connection has to ask for, and be allocated a supply current. If it doesn't get the authority, it can't draw the load.

You aren't going to break it by trying.


In the 1.0 spec wasn't it just said that the data pins should be shorted? It works for me. Ok it is not as safe when it comes to a pro charger but for something nocked together it should work for all devices. I've never come accros a device that needs 2.0

The current just isn't provided by the HUB if its compliant. Steve


7 years ago

steveastrouk suggests that 104 refers to 0.01uF.

is it not more likely:

10 x 10**4 pF = 10 x 10,000pF = 100,000pF = 100nF = 0.1uF ?

I know its off center but a couple years ago j & r stereo catalog had a car stereo that was strictly a SD car player stereo in other words no fm radio no cassette no cd or dvd just SD car and a mp3 player with no knobs just push buttom flat to surface controld not a touch screen as the i pod controls are but that would go right in a standard dc rom opening in the pc case hook righ up to a power supply and could be easily wired to a say 1000 watt amp in the plsace of mother board and out the back with two female pluga like used to plug in a guitar heavy duty these to skeakers of capacity to handel the wattage. people think your crazy but with little riding mower batter in case it be enough to go hours. maybe a solar rig to recharge it and handle on to to make it easy to tote. people are so used to computers everywhere now they never think it wasnt one and it totally will have them thinking you got portable desktop or something making your mp3 like superior in sound compared to their cheesey ear buds

If the 12v dc supply is already filtered and the 7805 if close to the power supply you can use just the 7805. Othewise you will at least want to have a filter cap on the input of the 7805.

in electronics it is common to label components using 3 or 4 digit method where last digit is "number of zeroes". "222" = 2200 Ohm or pF "471" = 470 Ohm or pF "472" = 4700 Ohm or pF "103" = 10000 Ohm or pF on small devices like resistor, colors are often used instead of characrters: 0 black 1 brown/tan 2 red 3 orange 4 yellow 5 green 6 blue 7 violet 8 gray 9 white on resistor there may be (nowdays usually is) additional mark indicating tolerance. "102" = 1000 Ohm or 1kOhm "201" = 200 Ohm since this is car adapter and it receives power from 12-14V outlet, this is not 200 Ohm, more like 1 or 1.2 kOhm. also please note that the LED is shown in wrong orientation. C2 as 0.1uF is very common ceramic capacitor, you will see few dozen of them on old motherboards or any digital board. not sure what the values are (27k or 2.7k?). you can remove 7805 and everything to the left from it and use any other 5V source to provide power to USB devices. if you decide to keep LED and R2, reduce value of R2 to 220Ohm for example (only 5V from USB).

I think you can just borrow some power from the supply of your boombox.

I am guessing that in this boombox you will be able to find a source of approximately 12 - 16V DC, capable of sparing ?maybe an ampere or so? of current, above and beyond the demands it already faces to supply current to the circuits of the radio. 

Anyway if you can find where in the radio to connect it, and use a voltmeter to confirm that you really do have 12 V or so at this node, you can just connect your power adapter circuit into this, and the 7805 will take care of the problem of regulating the voltage at its output to 5V. Also use a voltmeter to confirm that your adapter really is regulating the output.

So I think the next step is to take apart the radio, and see what's inside.

BTW, I think this idea of bolting a spare computer power supply onto this thing is inelegant.

I've already taken apart the boombox.  This particular Frankenstein build requires me to replace the CD player, and since I have a spare CD-ROM drive with a play/skip button on the front, I'll be using that as a replacement, a la this instructable.

The boombox was powered by wall current transformed to 12VDC.  Since I need 12VDC and 5VDC for the CD-ROM, my plan is to take a power supply from a dead DVR.  It's not a full-on PC PSU, but has the same standard power connector.

I'll plug the standard PSU connector to the CD-ROM and splice in connections for 12VDC to the boombox innards and 5VDC to the USB port.

Then the fun part will be figuring out how to take sound from the CD-ROM and a mini-plug for aux input and send them to the boombox.  I'm looking at the circuitry from the dead boombox CD player and cassette for that, based on this other instructable.

As I said, I'm a noob, and fully prepared for the whole thing to melt down at some point if I screw it up, but I'm trying to learn as much as I can in the process.

If you do fry that one i have a boom box with 1/4 aux input it runs on twelve volts and is hugh it old school boom box. i was using it to run my mp5 on the porch if you like 20 bucks plus postage.

Thanks. I'll keep that in mind if this build goes to hell. At this point, though, it's looking pretty good. I've got the power connections all worked out for the boom box mainboard and the CD-ROM, and even an extra computer case fan I'll be adding in. I've desoldered the extra components from this USB power circuit, and am waiting on some time to spend getting it reconfigured.

My real worry now is the audio connections. If you have any insight into that process, take a look at this question. :-)

Yeah, I think that'll work, this idea of removing the 7805 and connecting where its output used to go to your +5V supply.

You might be wondering about the circuits to the right of the 7805, and what its all there for. 

A true USB port has 4 wires.  These are +5V, ground, and D+, and D-. See:

In a computer, the D+ and D- wires are, of course, used for sending data, and negotiating whose turn it is to send data, and even negotiating how much current (in multiples of 100 mA) the computer is willing to let the device use.

However for a USB car-charger, or wall charger, there's no computer to attach to D+ and D-, so the designer is left wondering what to do with those D+ and D- wires.  And the answer that some (er most, all?) of these designers came up with is some kind of analog signaling.  The D+ and D- lines are either tied to each other, or tied to some network of resistors, like maybe a voltage divider between +5 and ground,or something...

I myself have not figured out what the "correct" answer is for what you do with the D+ and D- lines in a dumb, not-a-computer, USB charger.

It may turn out that your Blackberry, or whatever, might not care what it sees on D+ and D-, and just charge itself anyway without incident.

Others have just left D+ and D- unconnected and flapping in the breeze, yet the device still figures out how to charge itself, um here:

I think it depends on the device. 

Anyway, if you've used your Blackberry with this car-charger before, and it worked, then I think you can still expect it to work with its +5V comming from your other supply.

this might help pin one signal vcc wire color red comment cable power pin two - data wire color white camment data transfeer pin three + data wire color green comment data trans feer pin four ground wire color black comment cable ground shell signal name sheild no wire color comment drain wire

That was quite useful, despite wanting to strangle the guy halfway through.  I find his speaking style quite off-putting.  Reminds me of John Parkin from infomercials.  :-)

The 7805 is a pretty standard 3 pin 5V DC regulator, you will get tons of datasheets and application circuits for this on the net.

Everything on the left of the USB port in your diagram is the standard way in which the 7805 is connected. I don't remember it at the moment but you will find the maximum DC input voltage for the 7805 in the datasheets.

You could also consider buying a 12V transformer and 4 diodes to supply 12v DC to the above circuit, which will give you a standalone USB charger to operate directly from the mains outlet.

Be careful when working with mains voltages and also cross check the output voltage on the USB port with a multimeter just to make sure you are getting 5V DC.

Good advice!

I thought about using 12V on the circuit as-is, but it's not currently functional.  The big "sandstone" resistor (R1) was cold-soldered on one side.  It's worked loose, and there's discoloration around that joint.  That's why I'm wanting to bypass all of that.  Plus, as noted in my reply to Jack above, I will have a readily available 5V line.

If you have the 5 volt line available it is the way to go. the 7805 is a very powerful tool for creating a constant 5v however it is not without loss. It is a fairly simple linear regulator and not a switching regulator as found in the power supply which is much more efficient.

not only can you drop everything left of the 7805 but also the2 capacitors. Their job in this circuit is to attenuate any ripple due to changes in the load and the power supply should have some built in. The resistors shown to the right of the connector are the only critical parts if you are supplying your own 5v the rest is for converting 12v to 5v