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

How old is your 6 volt car?

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

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

Here is the unit I ordered:

It's the first one, the #48-150.

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

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:

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

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

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)

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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.

?Ref=1141&Site=US&Cat=33817837

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.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

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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:

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.

?Ref=1141&Site=US&Cat=33817837

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

--
Baron:

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

snipped-for-privacy@yahoo.com wrote:

?Ref=1141&Site=US&Cat=33817837

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

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

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.

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?

snipped-for-privacy@yahoo.com wrote in news: snipped-for-privacy@b28g2000cwb.googlegroups.com:

*snip*

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

Puckdropper

--
Wise is the man who attempts to answer his question before asking it.

To email me directly, send a message to puckdropper (at) fastmail.fm

And the answer is . . . (drum roll please) . . . nothing--at least not for a normal, healthy person with no pacemaker, etc.

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.