Electronics > Beginners
Using a diode to drop voltage
Mark K:
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.
ebclr:
I believe this is what you need
https://www.ebay.com/itm/LM2596-Buck-Step-down-Power-Converter-Module-DC-4-0-40-to-1-3-37V-LED-Voltmeter/400802470941?epid=9020791273&hash=item5d51b05c1d:g:jWEAAOSwXSJbFl3q:rk:2:pf:1&frcectupt=true
beanflying:
It can be done with a Linear Regulator which in simple terms burns the power up as heat. Using your worst case your voltage drop required will be 3.2V @ 3.5A or about 11W dissipated. This can be done using an Linear Regulator with a bypass Transistor such as this http://www.bristolwatch.com/ele/lm317.htm
A simple way to do it is to use a switchmode power supply as they are more efficient they generate less heat. Something along these lines, yes it is overrated but NONE of these converters should be trusted to full rating. https://www.ebay.com.au/itm/Step-Up-DC-DC-Converter-Step-Down-20A-300W-Buck-Boost-Power-Adjustable-150KHZ/254011625330?hash=item3b2445cf72:g:RE0AAOSwdhdb-r1J:rk:33:pf:0
or if you can find one a 6V version of this Automotive one https://www.ebay.com.au/itm/DC-DC-Buck-Converter-Step-Down-Power-Supply-Regulator-12V-24V-to5V-50W-10A-AU/142929816552?hash=item214747ebe8:g:NWEAAOSwHWJbkb-b:rk:2:pf:1&frcectupt=true
Mark K:
Thanks ebclr. I looked at that unit earlier. I was concerned about the 2A rating if the owner uses the pump for extended periods. It is a good solution though.
Cliff Matthews:
Simple so keep it cheap. Rip-n-test 6 amp bridge(s) from barfed-up PC supply, clip-off center AC pins, hook positive source to negative pin and Presto! 1.8 to 2.2v less on rectifiers positive pin. If it gets hot, bolt it to something to sink away the heat.
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