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Using a diode to drop voltage
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GeorgeOfTheJungle:
If the pump is to fill the receptacle of the floater of the carburetor, would only have to be run before cranking, with the engine off, so it'd only see 7.2V.

You've got four big ~50W resistors in the car in the head lamps, just use them.
spec:

--- Quote from: Mark K on January 16, 2019, 11:53:32 pm ---I have a question that will certainly seem newb to many of you. I tried to work out the answer on my own but I'm just not sure and I don't want to purchase unneeded components. I need to reduce 9.2V to no more than 7.5V in a circuit that will draw 2.5 - 3.5 amps. The circuit will mostly see short duty cycles, but it should be capable of constant duty. The case in point is a 1955 Studebaker(originally a 6 volt positive ground system) that has been converted to 8 volts. The generator is regulated to 9.6V and with nominal voltage loss through wiring and switches, the system runs at 9.2V with the engine running, 7.2V key on engine off.

The customer has requested an electric fuel pump be installed for both priming the fuel system after lengthy periods of sitting and also to prevent vapor lock in hot weather. A small, low pressure armature fuel pump will be perfect for this. The problem then becomes, these pumps are available in either 12V or 6V. The 12V pump will operate poorly at 9.2V and likely not at all at 7.2V. I've spoken to the tech line at one manufacturer and been assured that their 6V pump will not last long at 9.4V.

This brings me to the question of how to best reduce 7.2-9.2V to 6.0-7.5V. Quick math puts power rating at about 25 watts. For constant duty, the circuit should be capable of a least 50 watts. This seems to make resistor(s) a poor choice. So I'm looking at using a large diode. Something like a zener diode perhaps. I would appreciate any quideance I can get from people with more basic understanding of electronics.

Thank you in advance for any thoughts on the solution to the problem.

--- End quote ---

Hi Mark K

I have been tying to figure your calculations. Your task seems to be to lose 2V at a maximum of 3A6. That works out to be a loss of 2V * 3A6 = 7W2 which is quite a way down from 25W. The solution is quite simple: just put a 2V/3A6 =0.56 Ohm resistor of 15W of greater in series with the pump and all will be well. Motors like a bit of series resistance, but not too much. But the motor in a fuel pump has a nice life and is not that heavily loaded, and also the load is fairly constant.

But I would be inclined to try a 1 Ohm resistor first and if the pump works OK stay with 1 Ohm.

The best resistor for the job is one of the bolt down wire-wound types. Your resistor should be out of the way of any heat and bolted to a heatsink which, could be the metalwork of the car. Typical resistors are shown in the link below:

https://www.amazon.co.uk/Gold-Axial-Aluminum-Housed-Resistor/dp/B00LURTUKS/ref=sr_1_12?s=diy&ie=UTF8&qid=1547762676&sr=1-12&keywords=.5+OHM+resistor

Alternatively, three 10A to 20A rectifier diodes (not schottky types) in series will also be fine and would be my choice. The diodes would also need to be bolted to some form of heatsink.
Seekonk:
Why didn't you just make the car 12V. I have vintage boats and have made them all 12V.  But, hands down the best solution is diodes. Use two 25A bridge diodes with spade connectors and screw mount so it will be easy for someone later to replace or test.  Under voltage is perfectly fine. Nothing will get a customer pissed more than getting stuck somewhere because you used some Chinese converter. Go for dumb and rugged. I ran a fuel injected engine with a headlight in series at low voltage because I didn't want to have it go into a lot of bypass. I had a 5 gallon bucket of gas in the back seat. Any bypass would go to the tank and I wouldn't get home.
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