Circuit question

On Wed, 15 Jun 2011 20:42:12 +0800, Man-wai Chang
<toylet.toylet@gmail.com> wrote:

Is this a regulated supply. You cannot guess, you must

Could I use a multimeter to find out?

Do you know how to use a multimeter?

Tell me what to do!!! :)
Click to expand...
---
Answer the question.

--
JF
 
---
Answer the question.
Click to expand...
Yes. Basic operations.

--
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On Wed, 15 Jun 2011 22:00:13 +0800, Man-wai Chang
<toylet.toylet@gmail.com> wrote:

---
Answer the question.

Yes. Basic operations.
Click to expand...
---
View using a fixed-pitch font:

1. Set your multimeter to read > 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:

.. +-----+
..MAINS>---|~ +|----------------+
.. | | |
.. | | [VOLTMETER]
.. | | |
..MAINS>---|~ -|----------------+
.. +-----+
CHARGER


2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.


.. +-----+
..MAINS>---|~ +|-----+----------+
.. | | | |
.. | | [8R2] [VOLTMETER]
.. | | | |
..MAINS>---|~ -|-----+----------+
.. +-----+

3. Compare the multimeter readings.

If they're both close to 6.2V, then the charger is regulating its
output at no load and at full load into a resistive load.

Keep in mind that this is a battery charger designed to feed a
battery load, and not a general purpose power supply, so you may be
in for a nasty surprise unless you can fully characterize its
outputs into whatever load you have planned for it.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a
resistor adequately rated; 5 watts or greater if you plan to leave it
connected for a long time.

--
JF
 
Op 6/15/2011 5:45 PM, John Fields schreef:
On Wed, 15 Jun 2011 22:00:13 +0800, Man-wai Chang
toylet.toylet@gmail.com> wrote:

---
Answer the question.

Yes. Basic operations.

---
View using a fixed-pitch font:

1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:

. +-----+
.MAINS>---|~ +|----------------+
. | | |
. | | [VOLTMETER]
. | | |
.MAINS>---|~ -|----------------+
. +-----+
CHARGER


2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.


. +-----+
.MAINS>---|~ +|-----+----------+
. | | | |
. | | [8R2] [VOLTMETER]
. | | | |
.MAINS>---|~ -|-----+----------+
. +-----+

3. Compare the multimeter readings.

If they're both close to 6.2V, then the charger is regulating its
output at no load and at full load into a resistive load.

Keep in mind that this is a battery charger designed to feed a
battery load, and not a general purpose power supply, so you may be
in for a nasty surprise unless you can fully characterize its
outputs into whatever load you have planned for it.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a
resistor adequately rated; 5 watts or greater if you plan to leave it
connected for a long time.
Click to expand...
8.2 ohm will draw 0.75 A at 6.2 V.
The adapter is rated for 0.72 A
You are overloading the adapter.

--
pim.
 
On Wed, 15 Jun 2011 22:54:38 +0200, tuinkabouter
<dachthetniet@net.invalid> wrote:

Op 6/15/2011 5:45 PM, John Fields schreef:
On Wed, 15 Jun 2011 22:00:13 +0800, Man-wai Chang
toylet.toylet@gmail.com> wrote:

---
Answer the question.

Yes. Basic operations.

---
View using a fixed-pitch font:

1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:

. +-----+
.MAINS>---|~ +|----------------+
. | | |
. | | [VOLTMETER]
. | | |
.MAINS>---|~ -|----------------+
. +-----+
CHARGER


2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.


. +-----+
.MAINS>---|~ +|-----+----------+
. | | | |
. | | [8R2] [VOLTMETER]
. | | | |
.MAINS>---|~ -|-----+----------+
. +-----+

3. Compare the multimeter readings.

If they're both close to 6.2V, then the charger is regulating its
output at no load and at full load into a resistive load.

Keep in mind that this is a battery charger designed to feed a
battery load, and not a general purpose power supply, so you may be
in for a nasty surprise unless you can fully characterize its
outputs into whatever load you have planned for it.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a
resistor adequately rated; 5 watts or greater if you plan to leave it
connected for a long time.

8.2 ohm will draw 0.75 A at 6.2 V.
The adapter is rated for 0.72 A
You are overloading the adapter.
Click to expand...
---
Right you are!

I misread the 720mA as 750, so what value of resistor (resistance and
wattage) should he use?

--
JF
 
"John Fields" wrote in message
news:10aiv61m1rgtjtipom8bnotm4fi9c0iu7g@4ax.com...

On Wed, 15 Jun 2011 22:54:38 +0200, tuinkabouter
dachthetniet@net.invalid> wrote:

8.2 ohm will draw 0.75 A at 6.2 V.
The adapter is rated for 0.72 A
You are overloading the adapter.

Right you are!

I misread the 720mA as 750, so what value of resistor (resistance
and wattage) should he use?
Click to expand...
The 30 mA additional current is negligible unless it will be applied for
long-term continuous use, and at the extreme ambient temperature and heat
dissipation conditions. Also the 8.2 ohm (standard value) resistor you
suggest is probably 5%, so it may actually draw 720 mA.

As I'm sure you are aware, overloads on transformers and simple power
supplies are well-tolerated, and it is generally a matter of holding a duty
cycle in order to keep the temperature from exceeding a safe level. An
overload current of 41% will usually be allowed with a 50% duty cycle, and a
2x overload current for 25%. Depending on thermal mass, the actual ON time
may vary from several seconds to several minutes.

This entire discussion seems to be futile in terms of helping the OP. You
can get a 5V 4A regulated wall-wart for about $6:
http://www.mpja.com/prodinfo.asp?number=18520+PS

And a 4 port USB hub for about $6:
http://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=7248125&CatId=392
Or even $4:
http://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=7248141&CatId=4753
And a powered hub for $15:
http://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=1187799&CatId=4753
or $10:
http://www.tigerdirect.com/applications/SearchTools/item-details.asp?EdpNo=7248128&CatId=4753

I know some people who will spend hours trying to make some sort of
ill-fitted surplus or freebie device work, often without success, than just
buying what is needed and have done with it. As a learning experience, maybe
it's valid, but otherwise, not so much.

Paul
 
tuinkabouter wrote:
Op 6/15/2011 5:45 PM, John Fields schreef:
On Wed, 15 Jun 2011 22:00:13 +0800, Man-wai Chang
toylet.toylet@gmail.com> wrote:

---
Answer the question.

Yes. Basic operations.

---
View using a fixed-pitch font:

1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:

. +-----+
.MAINS>---|~ +|----------------+
. | | |
. | | [VOLTMETER]
. | | |
.MAINS>---|~ -|----------------+
. +-----+
CHARGER

2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.

. +-----+
.MAINS>---|~ +|-----+----------+
. | | | |
. | | [8R2] [VOLTMETER]
. | | | |
.MAINS>---|~ -|-----+----------+
. +-----+

3. Compare the multimeter readings.

If they're both close to 6.2V, then the charger is regulating its
output at no load and at full load into a resistive load.

Keep in mind that this is a battery charger designed to feed a
battery load, and not a general purpose power supply, so you may be
in for a nasty surprise unless you can fully characterize its
outputs into whatever load you have planned for it.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a
resistor adequately rated; 5 watts or greater if you plan to leave it
connected for a long time.

8.2 ohm will draw 0.75 A at 6.2 V.
The adapter is rated for 0.72 A
You are overloading the adapter.

It doesn't have to be that precise just to determine whether the supply
Click to expand...
is regulated or not. I was going to recommend 10 ohms at 5 watts.

Good Luck!
Rich
 
1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:
2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.
Click to expand...
OK, to look for significant voltage change after giving the AC adaptor a
small load.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a
Click to expand...
V=IR
P=VI=V*V/R

I don't have any 5W resistor. Could I use 2 1/4W 100ohm instead?

--
@~@ You have the right to remain silence.
/ v \ Simplicity is Beauty! May the Force and farces be with you!
/( _ )\ (x86_64 Ubuntu 9.10) Linux 2.6.39.1
^ ^ 18:57:01 up 5 days 3:06 0 users load average: 1.12 1.14 1.08
不借貸! 不詐騙! 不援交! 不打交! 不打劫! 不自殺! 請考慮綜援 (CSSA):
http://www.swd.gov.hk/tc/index/site_pubsvc/page_socsecu/sub_addressesa
 
On Thu, 16 Jun 2011 19:02:58 +0800, Man-wai Chang
<toylet.toylet@gmail.com> wrote:

1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:
2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.

OK, to look for significant voltage change after giving the AC adaptor a
small load.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a

V=IR
P=VI=V*V/R

I don't have any 5W resistor. Could I use 2 1/4W 100ohm instead?
Click to expand...
---
12 would be better.

--
JF
 
Man-wai Chang wrote:

1. Set your multimeter to read> 6.2VDC, hook up your equipment like
this, and make a note of the voltmeter reading:
2. Disconnect the charger from the mains, connect an 8.2 ohm resistor
in parallel with the multimeter leads, plug the charger back into
the mains and make a note of the multimeter reading.

OK, to look for significant voltage change after giving the AC adaptor a
small load.

Just as an aside, be aware that 6.2 volts into 8.2 ohms will cause
the resistor to dissipate about 5 watts, so make sure you use a

V=IR
P=VI=V*V/R

I don't have any 5W resistor. Could I use 2 1/4W 100ohm instead?

Anything that will draw more than voltmeter draws will give you some
Click to expand...
idea of whether or not the thing is regulated. But I'd put maybe 5 or
10 of those things in parallel - you could also use a 6V lantern bulb.

Good Luck!
Rich
 
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