Circuit question

[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]

You didn't say if the output of this adapter is DC? I can only

Sorry. It's DC. I should have labelled "AC adaptor output wires"!

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[Jamie]

Man-wai Chang wrote:

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?

You didn't say if the output of this adapter is DC? I can only

assume it is AC ? If so, shouldn't you be using a rectifier, cap
and a 7805 Post reg? Or something close to that?

Jamie

 

[Jamie]

Man-wai Chang wrote:

You didn't say if the output of this adapter is DC? I can only


Sorry. It's DC. I should have labelled "AC adaptor output wires"!

the you should be passing that into a REG...

use a 7805 or some LDO type

http://www.madvapes.com/LDO-5-volt-15-amp-regulator_p_2100.html

Something like those.

Jamie

 

[Man-wai Chang]

the you should be passing that into a REG...
use a 7805 or some LDO type

Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the
USB hub.

--
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[Jamie]

Man-wai Chang wrote:

the you should be passing that into a REG...
use a 7805 or some LDO type


Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the
USB hub.

Ripple removal, stability.. etc..

suit yourself.. It's your hub..

Jamie

 

[John Fields]

On Sun, 12 Jun 2011 22:06:10 +0800, Man-wai Chang
<toylet.toylet@gmail.com> wrote:

the you should be passing that into a REG...
use a 7805 or some LDO type

Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the
USB hub.

---
Here's your circuit: (View using a fixed-pitch font)


.. 6.2V E1
.. |
.. [10k] R1
.. |
.. +----> E2
.. |
.. [40k] R2
.. |
.. GND

With the resistors you've shown:

E1 * R2 2.48e5
E2 = --------- = -------- = 4.96 volts
R1 + R2 5.0e4

and the current in the string will be:

E1 6.2V
I = --------- = ------ = 124 microamperes
R1 + R2 50kR

Now, let's say you have something connected to E2 that wants to take
1mA.

Then you'll have this:


.. 6.2V E1
.. |
.. [10k] R1
.. |
.. +---->>----+
.. | |
.. [40k] R2 [5k] R3
.. | |
.. GND GND

But, since R3 is in parallel with R2 the combination will look like:


R2 * R3 40kR * 5kR
Rt = --------- = ------------ ~ 4400 ohms
R2 + R3 40kR + 5kR

Your circuit will look like:


.. 6.2V E1
.. |
.. [10k] R1
.. |
.. +----> E2
.. |
.. [4k4] R2
.. |
.. GND

and the voltage at E2 will fall to about 1.89 volts.

For that reason you need to interpose a regulator between the 6.2V
source and the load so that 5V will be supplied to the load regardless
of the current it draws, but within the limits of the source and the
regulator, of course.

Unfortunately, since it has a dropout voltage spec of 2 volts, the
common 7805 won't work, so you'll need to find, most likely, an LDO
regulator which can do the job.

--
JF

 

[IanM]

Jamie wrote:

Man-wai Chang wrote:

the you should be passing that into a REG...
use a 7805 or some LDO type


Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the
USB hub.

Ripple removal, stability.. etc..

To expand this slightly, you (Man-wai) proposed: "I am going to use a
10k and a 40K resistor to make a potential divider to reduce 6.2V to
about 5V."
That gives you 80% of the input voltage and is fine when nothing is
connected to the common point but what happens when the hub chip draws
some current and you have four high current devices connected each
drawing up to 500mA so the hub tries to draw about 2A from the junction
of the resistors?

Also what actual voltage does the power supply output? You said it is a
6.2V one, but did you measure that or just read the label? If you
measured it, did you also check what the voltage is under load? Can it
even supply enough current? For a hub you need 500mA per port + a small
margin for the controller chip so a 4 port hub need a 2.1A PSU and a 7
port hub needs a 3.6A PSU.

You don't even know if your hub needs 5V or somewhat more. If it has an
internal regulator it may need between 7.5 and 9V. It is *MUCH* easier
to design a circuit to switch properly between Self powered (external
supply) and Bus powered (from the 5V on the USB cable)if the external
supply is a little more than 5V. It is absolutely forbidden by the USB
spec to 'back-feed' power to the port that goes to the host and simply
connecting another supply to the USB bus WILL cause problems and may
damage the host so it MUST have some sort of switching circuit.

Your idea isn't going to work the way you expect. It may do nothing, it
may be worse than useless i.e. it prevents the hub working from bus
power, but with those resistor values you are unlikely to cause damage
if you get the polarity right.

suit yourself.. It's your hub..

Jamie

and your PC, hub, USB devices, adaptor etc. As long as you can afford
to replace them and dont leave them unattended till you know they
aren't going to catch fire by all means experiment, but if you don't
have lots of money, use a load that doesn't matter like a collection of
6V torch bulbs wired in parallel to test your power supply modification
and be certain it works not expensive computer peripherals possibly with
valuable data on. . . .

p.s. If you are trolling, you are good at it and have got me!

--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
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[Man-wai Chang]

p.s. If you are trolling, you are good at it and have got me!

Because I was not a trained professional in electronics, so I asked. I
thought it's just a simple dc circuit.

--
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[John Larkin]

On Sun, 12 Jun 2011 19:27:44 +0800, Man-wai Chang
<toylet.toylet@gmail.com> wrote:

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.

Was the hub originally unpowered?

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?

That won't work. The resistors will fatally reduce the ability to
supply current to the loads.

Can you buy a powered USB hub? They are fairly cheap.

John

 

[whit3rd]

On Sunday, June 12, 2011 7:06:10 AM UTC-7, Man-wai Chang wrote:

the you should be passing that into a REG...
use a 7805 or some LDO type

Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the
USB hub.

USB supplies power to the devices plugged into the hub.
If you connect four devices, it might change the current
requirement by 4 x 500 mA, and a regulator
would keep the hub voltage nearly constant when the
current requirement changes by up to two amps.

Some hubs have internal regulators. Some expect that
function of their power supply module.

 

[IanM]

Man-wai Chang wrote:

p.s. If you are trolling, you are good at it and have got me!

Because I was not a trained professional in electronics, so I asked. I
thought it's just a simple dc circuit.

If I had thought there was more than a 20% chance you were trolling, I

wouldn't have bothered giving a detailed reply. We all started with the
basics and a genuine question from anyone willing to learn (at least a
little) is always welcome.

It *IS* a simple DC circuit, but unfortunately it seems you initially
misunderstood the likely requirements of the hub. That stuff I
mentioned about a switching circuit would all be inside the hub - unless
it is a dirt cheap one from the dollar store that ignores the USB
standard.

IF the power supply is rated for a couple of amps output current and is
REGULATED (i.e. outputs the same 6.2V no matter whether you measure it
unloaded or with a 1A load), and you can confirm the hub needs 5V, it
may be worth taking this project further, otherwise 5V regulated SMPSUs
with a suitable current rating are readily available from many
suppliers, although buying a powered hub *WITH* a PSU may be simpler and
cheaper!

--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
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[tuinkabouter]

Op 6/12/2011 3:10 PM, Man-wai Chang schreef:

You didn't say if the output of this adapter is DC? I can only

Sorry. It's DC. I should have labelled "AC adaptor output wires"!

You can put two diodes in series, it will drop 2 x 0.6 volt.

--
pim.

 

[Nobody]

On Sun, 12 Jun 2011 19:27:44 +0800, Man-wai Chang wrote:

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

A potential divider produces a specific voltage *if* the current
drawn is negligible compared to the no-load current. Otherwise, it
produces a voltage which decreases linearly with the current drawn.

In practical terms, a potential divider is okay if you need a few
milliamps or less. Above that, use a linear regulator; when you start
measuring in amps rather than milliamps, consider a switching regulator.

 

[IanM]

tuinkabouter wrote:

Op 6/12/2011 3:10 PM, Man-wai Chang schreef:
You didn't say if the output of this adapter is DC? I can only

Sorry. It's DC. I should have labelled "AC adaptor output wires"!



You can put two diodes in series, it will drop 2 x 0.6 volt.

If you are lucky and you use standard 3A silicon diodes, it *MAY* get

the output voltage in spec for USB. You probably need a small minimum
load to keep the voltage down and prevent it from damaging the hub when
no devices are connected - a 100 ohm 1/2W resistor across the output
after the diodes would be a good idea.

--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
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[Man-wai Chang]

You can put two diodes in series, it will drop 2 x 0.6 volt.

What diode? Name?

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[Rich Grise]

Man-wai Chang wrote:

You can put two diodes in series, it will drop 2 x 0.6 volt.

What diode? Name?

1N4148, 1N914, etc. - any general-purpose silicon diode.

Have Fun!
Rich

 

[Man-wai Chang]

1N4148, 1N914, etc. - any general-purpose silicon diode.

Quite expensive a piece. Thanks.

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[Man-wai Chang]

If you are lucky and you use standard 3A silicon diodes, it *MAY* get
the output voltage in spec for USB. You probably need a small minimum
load to keep the voltage down and prevent it from damaging the hub when
no devices are connected - a 100 ohm 1/2W resistor across the output
after the diodes would be a good idea.

I won't connect more than 2 USB devices to the usb. SO I am NOT trying
to overdraw from the AC adaptor.

--
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[IanM]

Man-wai Chang wrote:

1N4148, 1N914, etc. - any general-purpose silicon diode.

Quite expensive a piece. Thanks.

I would probably grab 1N540x series or similar 3A general purpose

silicon rectifier diodes. As you only want to connect two devices,
unless they both draw the maximum 500mA you could use 1N400x series 1A
diodes. If you use too weak a diode it will probably fail short-circuit
and blow up the hub and all connected USB devices and maybe your USB port.

If cost is a problem, you could salvage suitable diodes from the mains
power input bridge rectifier circuit of just about any scrap equipment
or appliance that has a switched mode power supply with its mains input
fused at 3A or over - e.g. a scrap PC power supply. If you find the
bridge rectifier is a single component, short its positive and negative
output terminals and you will have a two diode drop if you put it in
series via its input terminals with your 6.2V PSU and load. DON'T
FORGET the 100 ohm half watt load resistor I mentioned, and CHECK the
maximum output is less than 5.25V BEFORE you connect it to the hub.
--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
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