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Catalina/Capri 25/250 Sailor's Forums

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9 volts DC

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JimB517
Past Commodore

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USA
3285 Posts

Initially Posted - 03/10/2003 : 15:24:50

My 25 year old english made depthfinder uses a large 9 VDC battery. This looks like a 6V lantern battery, but it's 9. This depthfinder came as an option from Catalina in 1978. I have the original factory invoice for the boat, including all options the buyer chose back then. By the way - bare boat was a little over $10K in 1978 and this model with all options came to $15,100 (CA sales tax was 3.23% back then!)

Can anyone tell me how to create 9 volts from my onboard 12 V system? Obviously a resistor, but how big and what capacity?

Don't want to spend much money on this in case I plug it in and it still doesn't work.

I will post a picture of the ancient device when I get home.

Edited by - on

ClamBeach
Master Marine Consultant

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3072 Posts

Response Posted - 03/10/2003 : 16:14:27

Look on the device and find the rated current draw. That will determine resistor size.

E = I X R

Electromotive Force (volts) = Current (amps) X Resistance (ohms).

Plug your voltage drop needed in E, Rated Current of the device in I,
and solve for the resistor needed.

Currently maintaining two holes in the water...'77 Venture 23 and new to the family,
'78 Catalina 25

Edited by - on

dgabel
1st Mate

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USA
75 Posts

Response Posted - 03/10/2003 : 17:47:42

Also, make sure that you have a resistor that's big enough to handle the power that you will dissipate in the resistor.

For example, (to use simple numbers) suppose your device requires a 1 amp current. You need to drop 3 volts across the resistor. The power in the resistor then is the voltage drop times the current, or 3V x 1A, for 3W. You would need a 3-W (or larger) resistor in this case. All resistors have a power rating.

Your device probably needs much less than 1 amp, but check anyway to make sure you won't burn up your dropping resistor when current starts to flow. Most small resistors come in 1/4, 1/2 and 1-W power ratings.

Dave Gabel
Walkure
Sail # 1484

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eric.werkowitz
Captain

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USA
283 Posts

Response Posted - 03/10/2003 : 19:01:47

Jim,

A better way to generate 9 volts is with the LM7809 voltage regulator. Get info at:

http://www.fulcrum.ru/Read/CDROMs/NS/pdf/nsc07756.pdf

These regulators come in many voltages, you want the 7809 (09 means 9 volts). Look at the bottom left of page 9 of the pdf document. You can get all the parts you need at Radio Shack.

Eric Werkowitz
C25 #4969

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Arlyn Stewart
Master Marine Consultant

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USA
2984 Posts

Response Posted - 03/10/2003 : 20:38:14

A simple load resister does not drop the voltage but rather reduces the current. This will likely work if you limit the circuits current to match very closely to the normal current draw of the device. This is a bit precarious compared to Eric's suggestion to use a 7809 regulator which is far safer.

However, the 7809 is limited to 1 amp. Use a heat sink and if your depthfinder by some chance uses more than 1 amp... then a pass transistor can be used with the 7809.

Arlyn C-250 W/B #224
<img src="http://www.stewartfam.net/arlyn/rr.jpg" border=0>
N/E Texas and Great Lakes

[url="http://www.cox-internet.com/arlynstewart/"]Arlyn's C250 Mods n Cruisin Stories[/url]

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JimB517
Past Commodore

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USA
3285 Posts

Response Posted - 03/10/2003 : 23:46:15

here is a picture of the strange old depthfinder

<img src="http://www.indiscipline.org/cat25/pictures/depth.jpg" border=0>

If anyone knows anything about this please let me know.

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dgabel
1st Mate

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USA
75 Posts

Response Posted - 03/12/2003 : 17:49:47

A resistor will drop the voltage. If the device draws 50 mA, for example, then pick a resistor that will drop 3 volts at 50 mA. That would be a 60-ohm resistor. Problem is, of course, that you can't buy a 60-ohm resistor: They come in standard values, so you have to experiment to get a voltage drop that is appropriate. And it is true that when you insert more resistance into the circuit, overall current will be reduced, so, again, you have to experiment.

It WOULD be better to use a voltage regulator. The regulator will give you a lot less power dissipation than the resistor, which will contribute to longer battery life. But if you don't know anything about electronics, using an LM 7809 regulator with a pass transistor could be a bit daunting.

There are devices called dc-to-dc converters. But they are generally pretty expensive, and it may not be worth the purchase. But you could check at an electronics store to see if they have any reasonably priced.

Dave Gabel
Walkure
Sail 1484

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ClamBeach
Master Marine Consultant

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3072 Posts

Response Posted - 03/12/2003 : 20:56:07

Agree that some sort of regulated power supply would be better... Depends on how much time, effort and $ is justifiable putting into the unit.

Since you can buy a simple digital depth sounder for about $100.. you aren't talking a lot of headroom here.

As I recall, the early flashers are pretty current hungry compared to modern solid state devices.

Currently maintaining two holes in the water...'77 Venture 23 and new to the family,
'78 Catalina 25

Edited by - on

JimB517
Past Commodore

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USA
3285 Posts

Response Posted - 03/12/2003 : 21:01:51

I have wired 6, 9 volt batteries in parallel. I will try it this weekend and see if it lights up the unit.

Indiscipline 1978 FK #398

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