Circuit question

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Man-wai Chang

I found a un-used 6.2V AC adaptor, and would like use it to power a USB hub.

We all knew that USB takes 5V as input.

I am going to solder the 2 output wires of the AC adaptor directly onto
the circuit board of the USB hub after removing its DC input socket.

The 3 "o" symtbols represent the orginal pins for the DC input socket.
The top & the bottom "o" should be the sleeve; middle "o", the pin.

I am going to use a 10k and a 40K resistor to make a potential divider
to reduce 6.2V to about 5V.

|
| USB hub circuit board
AC |
Adaptor | o
wires |
|
6.2V ----10K-------------o----+
| |
| |
GND ---------------o---40K---+
|
|

Is the wiring diagram correct?

Would the input current be affected?

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Man-wai Chang said:
I found a un-used 6.2V AC adaptor, and would like use it to power a USB hub.

We all knew that USB takes 5V as input.

I am going to solder the 2 output wires of the AC adaptor directly onto
the circuit board of the USB hub after removing its DC input socket.

The 3 "o" symtbols represent the orginal pins for the DC input socket.
The top & the bottom "o" should be the sleeve; middle "o", the pin.

I am going to use a 10k and a 40K resistor to make a potential divider
to reduce 6.2V to about 5V.

|
| USB hub circuit board
AC |
Adaptor | o
wires |
|
6.2V ----10K-------------o----+
| |
| |
GND ---------------o---40K---+
|
|

Is the wiring diagram correct?

Would the input current be affected?
Click to expand...

Well, how much of a current load is the A/C adapter rated for? Voltage
is just electrical pressure. You actually have to deliver the flow.
How many ports does the USB hub have? If 4, for example, then is the
adapter rated for 2A continuous (500 mA per port X 4 ports)? Since
you're dropping voltage and wasting current outside the USB hub, the
adapter will have to supply more than 2A.

What's the rest of the circuit look like in parallel with the 40Kohm
resistor? What'll be the heat (watts) for the resistors and how are you
going to dissipate it? Why wouldn't you use a voltage regulator? While
you can get regulator chips, a simple circuit just uses a resistor and
zener diode.

http://www.reuk.co.uk/Zener-Diode-Voltage-Regulator.htm

Put in 6.2 for Vin, 5 for Vout, and 2200 mA for Imax (500ma/port * 4 for
a 4-port hub plus their recommended 10% overage). Notice how many watts
you'll be dissipating. Notice the resistor is zero ohms. My guess is
the zener drops all the voltage. If you setup for a separate regulator
for each port, you enter 6.2 Vin, 5 for Vout, and 500 mA for Imax to get
a 2-ohm, 2.5W resistor. In either case, you'll need heatsinks and some
[convection] air flow.

So why not just find an A/C adapter whose secondary windings and voltage
regulator already put out 5V @ 2-2.5A if it's a 4-port hub? Rather than
putz around with the cost and time to build a voltage regulator for
which you won't build cheaper than what's mass produced, why not take
the easy route of buying the power adapter?

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=400175393836
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=130477221871

You think your resistor potential divider with the necessary heatsinks
will work while not knowing what the load is (and which will vary)
across the 40Kohm resistor? I doubt you get a voltage regulator with
caps and resistors or the zener diode setup along with the needed
heatsinks, thermal paste, PCB, and wiring for cheaper than the $9 cost
of an 120VAC 5VDC 2A power adapter.
 
Well, how much of a current load is the A/C adapter rated for?

720mA only. It's a Nokia ACP-9X. Made in Finland.
Voltage is just electrical pressure. You actually have to deliver the flow.
How many ports does the USB hub have? If 4, for example, then is the
adapter rated for 2A continuous (500 mA per port X 4 ports)? Since
you're dropping voltage and wasting current outside the USB hub, the
adapter will have to supply more than 2A.
Click to expand...

I am not going to charge more than 2 USB devices at any time.
What's the rest of the circuit look like in parallel with the 40Kohm
resistor?
Click to expand...

I don't have the circuit diagram for the Samsung Crystal USB Hub. It has
4 standard USB ports.
What'll be the heat (watts) for the resistors and how are you
going to dissipate it? Why wouldn't you use a voltage regulator? While
you can get regulator chips, a simple circuit just uses a resistor and
zener diode.
Click to expand...

If the AC adaptor's DC output is just 720mA, would it matter now?
So why not just find an A/C adapter whose secondary windings and voltage
regulator already put out 5V @ 2-2.5A if it's a 4-port hub?
Click to expand...

To make some use of the junk. The Nikia AC adaptor is no longer in use.
You think your resistor potential divider with the necessary heatsinks
will work while not knowing what the load is (and which will vary)
across the 40Kohm resistor? I doubt you get a voltage regulator with
caps and resistors or the zener diode setup along with the needed
heatsinks, thermal paste, PCB, and wiring for cheaper than the $9 cost
of an 120VAC 5VDC 2A power adapter.
Click to expand...

220VAC. It's Hong Kong.

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I found a un-used 6.2V AC adaptor, and would like use it to power a USB hub.

We all knew that USB takes 5V as input.

I am going to solder the 2 output wires of the AC adaptor directly onto
the circuit board of the USB hub after removing its DC input socket.

The 3 "o" symtbols represent the orginal pins for the DC input socket.
The top & the bottom "o" should be the sleeve; middle "o", the pin.

I am going to use a 10k and a 40K resistor to make a potential divider
to reduce 6.2V to about 5V.
Click to expand...

No, don't bother with resistors. Go look up LDO voltage regulators.
They don't cost much and will do a much better job.
 
Man-wai Chang said:
I found a un-used 6.2V AC adaptor, and would like use it to power a USB
hub.

We all knew that USB takes 5V as input.

I am going to solder the 2 output wires of the AC adaptor directly onto
the circuit board of the USB hub after removing its DC input socket.

The 3 "o" symtbols represent the orginal pins for the DC input socket.
The top & the bottom "o" should be the sleeve; middle "o", the pin.

I am going to use a 10k and a 40K resistor to make a potential divider
to reduce 6.2V to about 5V.

|
| USB hub circuit board
AC |
Adaptor | o
wires |
|
6.2V ----10K-------------o----+
| |
| |
GND ---------------o---40K---+
|
|

Is the wiring diagram correct?

Would the input current be affected?
Click to expand...

To start, we have to be *really* careful with the terminology.

There are three kinds of AC wall adapters.

1) AC output adapter. 115VAC input, 6.3VAC output.
The output is a sine wave, going both positive and negative
with respect to ground.

2) Unregulated DC adapter. 115VAC input, 6 to 9VDC output.
This is similar in construction to the previous adapter, only
it includes a diode to rectify the sine wave. Only the top of
the sine wave remains, so the output signal is always some
positive value. By including a 1000uF or bigger capacitor, they
can filter off some of the ripple. The voltage changes, because the
transformer has an output impedance. I have an adapter like this
on my Dymo label maker.

3) Regulated DC adapter. 115VAC input, 5VDC (exactly) output.
This builds on the adapter in (2), and includes a regulator
circuit. The regulator circuit removes the variation in
output voltage. Whereas the previous adapter would put
out 9VDC at zero amps loading, and 6VDC at 1 amp loading say,
the regulated adapter puts out 5VDC at anywhere from zero to
one ampere. A regulated DC adapter, is what you use with a
typical USB hub.

Now that we've identified the adapter types, can they be mixed ?

Generally, NO. That's why we have to be very careful.

If I were to connect my Dymo unregulated adapter to my USB hub,
it would blow up (too much voltage, and simple resistor circuit
cannot satisfactorily fix it).

Can I use a 12V regulated DC adapter with a 5V regulated DC input
USB hub ? Yes, if as DaveJ suggested, you used some kind of
three terminal regulator. For example, a 7805 three terminal
regulator chip, can be connected to the output of a 12V
adapter, to make a 5V (exactly) signal for a project.

But don't get an "AC output" wall adapter, anywhere near that hub!

In this diagram, the transformer alone, produces an AC output.
Adding the bridge rectifier and filter capacitors, makes an
unregulated DC voltage.

http://www.team-bhp.com/forum/attac...2656502-bennys-first-diy-leds-ikon-bridge.gif

This shows the inclusion of a three terminal regulator, to the basic
circuit. C1 and C2 filter ripple coming from the bridge rectifier.
C3 and C4 filter high frequency noise, and provide bulk capacitance
to improve the transient response of the supply as the load changes.
Diode D1 handles the situation where the AC input power fails, and the
capacitors are still fully charged. D1 provides a safe discharge
path, so discharge current doesn't flow through the
substrate of U1 regulator (7805 or equivalent). Diode D2 is
for reverse polarity protection (might be useful if tantalum
capacitors happened to be in the circuit). D2 might also clamp
a transient, if the load was suddenly unplugged from the supply.

http://www.aaroncake.net/circuits/supply6.gif

This is not really a job for simple resistors. Resistors are
fine, if the load isn't changing with time. But real circuits,
draw variable amounts of current, and then the resistor method
isn't precise. A three terminal regulator, can tame a lot of
ills in the circuit.

When I built my own stereo amplifier a few months back, when I
first turned it on, I got a horrible "hum" in the circuit.
Modifying the power supply, to use a three terminal regulator,
fixed it. Nice pure power comes from it. The regulator I got,
is rated for up to 3 amps, and the amplifier isn't very powerful.

Paul
 
Far better to use a 5 volt regulator like a LM7805, Up to 30V dc input
and out comes a regulated 5 volts.
[image: http://img62.imageshack.us/img62/2443/16721078.jpg]
Click to expand...

Can't see the image

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Two diodes is series with the AC adapter should work. I've used diodes
to drop DC voltages before, and they work.
Click to expand...

Could I have the name of this component?

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http://www.fairchildsemi.com/ds/LM/LM7805.pdf
Figure 7 on page 22, shows how to use it (a minimal
implementation). The two caps are for stability.
That datasheet covers the various members of the 78xx
family, and only one page of specification numbers
there, is specifically for the 7805 chip.
Click to expand...

Thanks.

So I could just connect 6.2V output from the Nokia AC adaptor to Vin and
then the input of the USB hub to Vout?

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Thanks.

So I could just connect 6.2V output from the Nokia AC
adaptor to Vin and then the input of the USB hub to Vout?
Click to expand...
It may not work as the minimum input voltage according to
the Fairchild specs is 7 VDC. The suggestion to us 2 diodes
in series is the best so far. Also the max current is 1
Ampere, which is well below the possible needs of 2.5 Amps.
 
It may not work as the minimum input voltage according to
the Fairchild specs is 7 VDC. The suggestion to us 2 diodes
in series is the best so far.
Click to expand...

What's the name for this diode?
Also the max current is 1
Ampere, which is well below the possible needs of 2.5 Amps.
Click to expand...

I will limit the usage by not plugging more than 2 devices.

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Pen said:
A diode is a rectifier. If you hold the current down then
one 1N4001 should work.
http://www.fairchildsemi.com/ds/1N/1N4001.pdf
or an equivalent.
Click to expand...

I've used the diode technique, to convert a 6V output into
a 5V output, but I wasn't really happy with the results.
I later changed to a three terminal regulator (when I could
find the correct part number for the job), and the voltage
was then "rock solid".

If the input was exactly 6.2VDC and well regulated to start
with, and the load needed 5VDC, then I would search for the
"LDO" or "Low DropOut" version of regulator, which operates
properly with a lower differential between input and output.
That was already mentioned in this thread, as solution.

Now, at my one and only "good" electronics store in town,
we don't have much selection in three terminal regulators, and
LDOs aren't available. If I needed one, I'd have to order from
the Internet.

For example, compare the "regular" 7805 7V input to make 5V
input, versus this LDO version. This device takes an input
as low as 5.5V and makes 5V from it, at up to one ampere of
loading. So that would easily work with a 6.2VDC 1A supply,
to make a 5VDC 1A supply.

http://www.national.com/JS/searchDocument.do?textfield=LM2940CT-5.0&x=0&y=0

When ordering the part, you have to be careful to order
the right IC package for it. The TO-220 with the three
legs on one end, is relatively convenient for hobby usage.
The legs can be stuffed through a perf board and soldered
as required. And small heatsinks that bolt to the TO-220
metal tab are available. In the 6V to 5V circuit, I used
a perf board to hold the components in place. And bare wire,
solder, and heat shrink tubing underneath, for the wiring
job. Bare wire, makes it easy to whip the wire around the
legs of the components and solder them into the circuit,
one leg at a time.

LM2940CT-5.0 TO-220 package

This is an example of a perf board. They come in different designs,
and you select the type of perf board, based on the style of
circuit you're building. The mounting holes on the four corners,
can be combined with a small plastic or metal box, to
hold the project when you're finished. In this case, since
a heatsink is likely to be connected to the regulator chip,
that will make it more difficult to neatly package the
project. The heatsink should be exposed to the air somehow.

http://www.thesource.ca/estore/prod...ine&category=PC+board/Sockets&product=2760150

Some of the higher power regulators (capable of 3 amps output),
come in a TO-3 package, and a more substantial heatsink can be
bolted to one of those. Years ago, I built a power supply,
that looked similar to the following approach, with about 70 sq.in.
of cooling fins. Linear regulators are less efficient than
switching regulators, and (VIN - VOUT) * IOUT watts of power
is dissipated in the package of the integrated circuit. At
max load, your (6.2 - 5) * 1 amp gives 1.2W of power dissipation,
so the heatsink won't get too warm.

http://ts1.mm.bing.net/images/thumbnail.aspx?q=1021985360836&id=c221b08d0cc95b2ceddcebdfcd31a7ac

Paul
 
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