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Need to drop 6v to 5v for my DC video players at 2A

R

Roveer

Jan 1, 1970
0
I've got two different video toys, one being an archos AV 500 and the
other being a Vantec Avox jukebox. Both are basically hard drive based
video players. Both came with AC power bricks that put out around 5v
DC at 2A.

I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

I need to drop the 6 volts down to 5 volts to drive these toys that
I'll put in different cars.

How do I do this?

Thanks for your help.

Roveer
 
L

Lord Garth

Jan 1, 1970
0
Roveer said:
I've got two different video toys, one being an archos AV 500 and the
other being a Vantec Avox jukebox. Both are basically hard drive based
video players. Both came with AC power bricks that put out around 5v
DC at 2A.

I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

I need to drop the 6 volts down to 5 volts to drive these toys that
I'll put in different cars.

How do I do this?

Thanks for your help.

Roveer

How old is your 6 volt car?
 
M

mkaras

Jan 1, 1970
0
Roveer said:
I've got two different video toys, one being an archos AV 500 and the
other being a Vantec Avox jukebox. Both are basically hard drive based
video players. Both came with AC power bricks that put out around 5v
DC at 2A.

I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

I need to drop the 6 volts down to 5 volts to drive these toys that
I'll put in different cars.

How do I do this?

Thanks for your help.

Roveer

Before even trying to think of using the multi voltage cigarette
lighter adapter you should study some about how these work. The basic
cheap versions of these are just a slide switch that selects various
resistors to drop the auto socket voltage down to the output voltage.
In the most sleasy designs the output voltage at any particular setting
has virtually no regulation and will change a large amount depending
upon how much current is drawn by the load that you attach. Better
models may include a power transistor on a heatsink that can provide
some degree of regulation BUT it can get rather HOT at the lowest
voltage settings when the load draws current at the full label
amount...so use care.

The best designs use a DC to DC switcher circuit (similar to those
found in many modern cell phone chargers). These are generally much
better in that the output stays regulated at the set voltage through
the whole range from low load current to the full load. The switcher
design also converts the DC to DC voltages in an efficient manner that
stays much cooler than the power transitor linear styles mentioned
above. This is the type of unit you should be working with. (It may be
possible to change the output voltage of this type of unit simply by
changing a resistor value or two inside the unit). Some work would be
required to investigate this and the methods employed would depend
completely upon the type of device you are working with.

- mkaras
 
H

Homer J Simpson

Jan 1, 1970
0
I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

Will either handle 2 amps? (Unusual). If so select 4.5 volts.
 
R

Roveer

Jan 1, 1970
0
mkaras said:
Before even trying to think of using the multi voltage cigarette
lighter adapter you should study some about how these work. The basic
cheap versions of these are just a slide switch that selects various
resistors to drop the auto socket voltage down to the output voltage.
In the most sleasy designs the output voltage at any particular setting
has virtually no regulation and will change a large amount depending
upon how much current is drawn by the load that you attach. Better
models may include a power transistor on a heatsink that can provide
some degree of regulation BUT it can get rather HOT at the lowest
voltage settings when the load draws current at the full label
amount...so use care.

The best designs use a DC to DC switcher circuit (similar to those
found in many modern cell phone chargers). These are generally much
better in that the output stays regulated at the set voltage through
the whole range from low load current to the full load. The switcher
design also converts the DC to DC voltages in an efficient manner that
stays much cooler than the power transitor linear styles mentioned
above. This is the type of unit you should be working with. (It may be
possible to change the output voltage of this type of unit simply by
changing a resistor value or two inside the unit). Some work would be
required to investigate this and the methods employed would depend
completely upon the type of device you are working with.

- mkaras


Here is the unit I ordered:

http://www.minute-man.com/acatalog/Automotive_DC_to_DC_and_DC_to_AC_Power_Converters.html

It's the first one, the #48-150.
 
M

Michael Black

Jan 1, 1970
0
Lord Garth" ([email protected]) said:
How old is your 6 volt car?
You're misreading the post. He wants to use the 6volts out of
the cigarette lighter adaptor, but since it's 6v he wants something
to drop it to the needed 5v.

So either the 6v is too high, or it will work fine, or a silicon diode
of sufficient amperage will drop it by about .7v and that likely puts
the voltage close enough.

Or the 4.5v from the adaptor is sufficient, in which case nothing
more is needed.

Or of course, the cigarette lighter adaptor can't supply enough current.
Maybe it can, but the voltages will drop too much to be of use.

Michael
 
M

mkaras

Jan 1, 1970
0
Roveer said:

It looks like you ordered a switching regulator type unit. As indicated
before this is the best type for efficiency and keeping heat down. The
other types on that referenced page appear to be the power transistor
type and this is the reason their current rating is lower at the lesser
output voltages so that the unit does not have a heat problem and burn
up.

A closeup picture of the #48-150 can be seen at this link:
http://www.philmore-datak.com/images/48-150.jpg

It is unknown if it would be possible to modify resistor values inside
this unit to customize it for a direct 5V output but at the fairly
reasonable cost of one unit it may be interesting to open up a unit and
have a look.

You could also just use the unit set to 4.5V as suggested already and
just use it that way.

- mkaras
 
Roveer said:
I've got two different video toys, one being an archos AV 500 and the
other being a Vantec Avox jukebox. Both are basically hard drive based
video players. Both came with AC power bricks that put out around 5v
DC at 2A.

I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

I need to drop the 6 volts down to 5 volts to drive these toys that
I'll put in different cars.

How do I do this?

Thanks for your help.

Roveer

There's at least one off-the-shelf adapter available that plugs into a
cigarette-lighter and outputs 5V @ 2A. It's at:

http://www.alk.com/store/whats_included.asp?sku=CIGPOW5V2A

Another option would be to cut one of the wires coming from your
adapter and install a resistor and use the 6V output. You could use
shrink tubing to cover the wires after you install the resistor.
Depending on the actual current your device(s) use, you would need
resistors with one the following values:

2A - 0.50 ohms, 3 watts (23JR50E-ND) (Wirewound)
1.5A - 0.68 ohms, 3 watts (PN TWW3JR68E-ND) (Wirewound)
1.0A - 1.0 ohms, 3 watts (33J1R0E-ND) (Wirewound)

http://www.digikey.com/scripts/DkSearch/dksus.dll?Criteria?Ref=1141&Site=US&Cat=33817837

I would probably just go with the 0.7 ohm resistor or you could cut a
wire and measure the actual current draw (amps) and select a resistor
with the formula R=1/Amps. The wattage required, by the way, will be
the same as the measured current, e.g., if the current is 2 amps, the
resistor should be rated for at least 2 watts. It always a good idea to
go higher. That's assuming a conversion of 6V to 5V. The amount of
voltage drop (reduction) you get using the resistor can be calculated
by multiplying the current the video player(s) uses times the amount of
resistance. For example, if the player uses 1.5 Amps and you use a 0.7
ohm resistor, the amount of voltage reduction will be 1.05 Volts.

It's possible, by the way, that your video players will operate without
a resistor on the 4.5V setting. These types of devices usually aren't
all the fussy about the voltage being precise. It's also possible that
it might work on the 6V setting, but doing so might cause it to fail
prematurely.
 
R

Roveer

Jan 1, 1970
0
There's at least one off-the-shelf adapter available that plugs into a
cigarette-lighter and outputs 5V @ 2A. It's at:

http://www.alk.com/store/whats_included.asp?sku=CIGPOW5V2A

Another option would be to cut one of the wires coming from your
adapter and install a resistor and use the 6V output. You could use
shrink tubing to cover the wires after you install the resistor.
Depending on the actual current your device(s) use, you would need
resistors with one the following values:

2A - 0.50 ohms, 3 watts (23JR50E-ND) (Wirewound)
1.5A - 0.68 ohms, 3 watts (PN TWW3JR68E-ND) (Wirewound)
1.0A - 1.0 ohms, 3 watts (33J1R0E-ND) (Wirewound)

http://www.digikey.com/scripts/DkSearch/dksus.dll?Criteria?Ref=1141&Site=US&Cat=33817837

I would probably just go with the 0.7 ohm resistor or you could cut a
wire and measure the actual current draw (amps) and select a resistor
with the formula R=1/Amps. The wattage required, by the way, will be
the same as the measured current, e.g., if the current is 2 amps, the
resistor should be rated for at least 2 watts. It always a good idea to
go higher. That's assuming a conversion of 6V to 5V. The amount of
voltage drop (reduction) you get using the resistor can be calculated
by multiplying the current the video player(s) uses times the amount of
resistance. For example, if the player uses 1.5 Amps and you use a 0.7
ohm resistor, the amount of voltage reduction will be 1.05 Volts.

It's possible, by the way, that your video players will operate without
a resistor on the 4.5V setting. These types of devices usually aren't
all the fussy about the voltage being precise. It's also possible that
it might work on the 6V setting, but doing so might cause it to fail
prematurely.

Thanks for the response. I'm going to try the 4.5v setting and see
what happens. I'm sure the 2A rating is just to get the hard drive
spun up then it's probably much lower than that. How do I measure amp
draw when operating?

Roveer
 
J

Jamie

Jan 1, 1970
0
Roveer said:
Thanks for the response. I'm going to try the 4.5v setting and see
what happens. I'm sure the 2A rating is just to get the hard drive
spun up then it's probably much lower than that. How do I measure amp
draw when operating?

Roveer
put a Diode in series with the 6 volt setting.
this will drop it down to 5.4 volts.
by the time you load it, it will even out.

a nice 6 amp diode will do.
maybe even 2 in series. that will give you
4.8 volts , load the diodes and you may get
a little more.
just a thought.
diodes by nature will drop a dc volt down to aprox.
.6 volts.
 
Roveer said:
Thanks for the response. I'm going to try the 4.5v setting and see
what happens. I'm sure the 2A rating is just to get the hard drive
spun up then it's probably much lower than that. How do I measure amp
draw when operating?

Roveer

To measure current, you will have to cut one of the wires going from
the adapter to the player (either wire will work). Then you will have
to put a Digital Multimeter between the two cut wires.

If you don't have a Multimeter, one of my favorite places to buy things
like this is at a Pawn Shop. Cash America Pawn, is my favorite pawn
shop. Make sure you get the test leads that go with it. You can also
get inexpensive multimeters at Harbor Freight:
http://da.harborfreight.com/cpisearch/web/search.do?keyword=multimeter

Radio Shack has multimeters also, but they are expensive.

5V and 12V power isn't dangerous. So you don't have to worry about
that. It's a good idea to not be wearing a ring, though, and make sure
you don't short a hot wire to ground.
 
B

Baron

Jan 1, 1970
0
To measure current, you will have to cut one of the wires going from
the adapter to the player (either wire will work). Then you will have
to put a Digital Multimeter between the two cut wires.

If you don't have a Multimeter, one of my favorite places to buy
things like this is at a Pawn Shop. Cash America Pawn, is my favorite
pawn shop. Make sure you get the test leads that go with it. You can
also get inexpensive multimeters at Harbor Freight:
http://da.harborfreight.com/cpisearch/web/search.do?keyword=multimeter

Radio Shack has multimeters also, but they are expensive.

When I was in Florida earlier this year, I bought one in Walmart for
$2.95 Very good value !
 
R

Roveer

Jan 1, 1970
0
I've got a RS auto-ranging multimeter. I'm guessing I have to use the
DC-mA setting to measure amps in DC? Volts DC and AC I got, much
beyond that and I have to ask questions. Doh!!!

Roveer
 
M

Michael A. Terrell

Jan 1, 1970
0
I guess that you've never worked around a 5 VDC 1000 Amp power
supply, or a pir of 1.5 VAC, 1000 transmitting tube filament supplies?


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
T

Tim Auton

Jan 1, 1970
0
Roveer said:
I've got a RS auto-ranging multimeter. I'm guessing I have to use the
DC-mA setting to measure amps in DC? Volts DC and AC I got, much
beyond that and I have to ask questions. Doh!!!

Yup, DC-mA is the one. You sometimes have to use different sockets on
the meter for measuring current, or for different current ranges. Use
the right ones, or you might blow a fuse in the meter and then you've
got to go and buy a new fuse.

Oh, and it's generally preferred to quote just enough of the previous
message to give context, then reply underneath or intersperse your reply
with what you're replying to. People not using Google Groups (most
readers of this group) see this stuff more like emails - one message at
a time - so it helps to give a bit of context. Google Groups hides all
the quoted stuff and makes it look like a web forum, which it isn't.


Tim
 
W

waz

Jan 1, 1970
0
Roveer said:
I've got two different video toys, one being an archos AV 500 and the
other being a Vantec Avox jukebox. Both are basically hard drive based
video players. Both came with AC power bricks that put out around 5v
DC at 2A.

I've got 2 12 volt DC cigeratte lighter plugs power supplies that put
out 3, 4.5, 6, 9 or 12 volts. So here's what I need.

I need to drop the 6 volts down to 5 volts to drive these toys that
I'll put in different cars.

How do I do this?

Thanks for your help.

Roveer

A diode that can handle over 2A should do the trick. I'd go with a 4A
silicon diode. Voltage drop of about 0.7 volts if I remember right, should
put you in the 5.3 volt range.
 
Michael said:
I guess that you've never worked around a 5 VDC 1000 Amp power
supply, or a pir of 1.5 VAC, 1000 transmitting tube filament supplies?

When I was a teenager, I remember getting my finger burned while
working on a car with a ring on my finger. If you had a 5V, 1000 Amp
supply in that situation, I imagine it could burn your finger off.

On the otherhand, if you had a 5V, 10000000 Amp supply and put your
tongue on the positive and one of your hands on the negative (no
jewelry), what do you suppose would happen?
 
P

Puckdropper

Jan 1, 1970
0
[email protected] wrote in

*snip*
On the otherhand, if you had a 5V, 10000000 Amp supply and put your
tongue on the positive and one of your hands on the negative (no
jewelry), what do you suppose would happen?

You'll get real hot and start glowing? Any one got any bread?

Puckdropper
 
Jamie said:
put a Diode in series with the 6 volt setting.
this will drop it down to 5.4 volts.
by the time you load it, it will even out.

a nice 6 amp diode will do.
maybe even 2 in series. that will give you
4.8 volts , load the diodes and you may get
a little more.
just a thought.
diodes by nature will drop a dc volt down to aprox.
.6 volts.

You don't want to put the two diodes in series, in general. Ideally,
it works fine, but the real world is far from ideal. It should be
fine in this case, as the ammount of power is far below the rated
maximum of both diodes, but the load across the diodes will be
unbalanced. Adding a high value resistor (which carries about the leak
current of the diodes) in paralell to each diode will balance it
somewhat.

You could also use a simple voltage regulator. Minimal support
circutry and high tolerance to overvoltage.
 
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