### Author Topic: Using a diode to drop voltage  (Read 752 times)

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#### Mark K

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##### Using a diode to drop voltage
« on: January 17, 2019, 10:53:32 am »
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

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##### Re: Using a diode to drop voltage
« Reply #1 on: January 17, 2019, 11:07:44 am »

#### beanflying

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##### Re: Using a diode to drop voltage
« Reply #2 on: January 17, 2019, 11:09:57 am »
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

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#### Mark K

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##### Re: Using a diode to drop voltage
« Reply #3 on: January 17, 2019, 11:39:07 am »
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

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##### Re: Using a diode to drop voltage
« Reply #4 on: January 17, 2019, 11:41:38 am »
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.
« Last Edit: January 17, 2019, 11:45:42 am by Cliff Matthews »

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#### beanflying

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##### Re: Using a diode to drop voltage
« Reply #5 on: January 17, 2019, 11:46:10 am »
Diodes as a voltage drop are a really poor choice even though they sometimes get used. Forward voltage drops vary a lot with current.

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#### Mark K

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##### Re: Using a diode to drop voltage
« Reply #6 on: January 17, 2019, 11:52:23 am »
Thank you beanflying. That looks like just the unit for the job. I'll talk to the the customer about it tomorrow. At the low price, it's a no brainer. Thanks again guys for your help.

#### Mark K

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##### Re: Using a diode to drop voltage
« Reply #7 on: January 17, 2019, 11:57:56 am »
Thanks Cliff. I do have an old PC at home that I've been looking at scrapping out. I like the idea of the adjustable supply though so I can tune it for a happy medium between KOEO and KOER voltages. The 20A rating provides plenty of cushion for constant duty in hot weather. And it's under $10. #### ebclr • Super Contributor • Posts: 1704 • Country: ##### Re: Using a diode to drop voltage « Reply #8 on: January 17, 2019, 12:01:13 pm » #### beanflying • Super Contributor • Posts: 2515 • Country: • Toys so very many Toys. ##### Re: Using a diode to drop voltage « Reply #9 on: January 17, 2019, 12:09:44 pm » While it wasn't spelt out specifically by Mark in his original post you are powering either a DC motor which on startup draws a lot more than it's running current or a solenoid type which is a very peaky load. DC motor, 5A switchmode and Chinese totals in being to close to the limit IMO for the above unit. Saving a few$\$ in particular where you are getting paid to do it is likely to come back and bite you on the bum.
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#### fsr

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##### Re: Using a diode to drop voltage
« Reply #10 on: January 17, 2019, 12:11:34 pm »
I'm curious: why was the car converted to 8v? I've never heared about a car with an 8v battery.

#### Zero999

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##### Re: Using a diode to drop voltage
« Reply #11 on: January 17, 2019, 11:11:37 pm »
Have you looked into a 12V pump and a boost converter?

#### GeorgeOfTheJungle

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##### Re: Using a diode to drop voltage
« Reply #12 on: January 18, 2019, 12:13:31 am »
I'd wire it in series with the headlight bulbs, the only problem is it won't work with the headlight on.

int main (void) { while (1) fork(); }

#### wraper

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##### Re: Using a diode to drop voltage
« Reply #13 on: January 18, 2019, 12:29:29 am »

#### mariush

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##### Re: Using a diode to drop voltage
« Reply #14 on: January 18, 2019, 12:51:44 am »
A switching regulator would be indeed the best solution, followed by a linear regulator.
A linear regulator would be the easiest but you have problem with heat dissipation, as linear regulators dump the difference between input voltage and output voltage as heat.

Skip eBay because it's full of chip clones (or fake) or boards made with subpar chip. Distributors like Digikey have pre-made DC-DC converters that can do what you need, just go to the Power Supplies , dc-dc converters section.
Here's some examples:

4.5v - 14v to 0.6..6v up to 10A : https://www.digikey.com/product-detail/en/abb-embedded-power/NQR010A0X4Z/555-1161-ND/2270975
same up to 6A from other maker : https://www.digikey.com/product-detail/en/murata-power-solutions-inc/OKR-T-6-W12-C/811-2179-ND/2199629
or 8v..16v to 0.6v..8v 4A : https://www.digikey.com/product-detail/en/abb-embedded-power/APXK004A0X-SRZ/555-1141-1-ND/2170682

These are super simple, they just need a resistor or two to set the output voltage and maybe a couple of ceramic or electrolytic capacitors on input and ouput, and they turn on by either connecting a pin to ground or to input voltage, so you can add a switch or set it to always on.

If you want to go with linear regulators, keep in mind that a lot of them can only dissipate up to 15w or so, in the case of common packages like to-220 or dpak
You could configure your regulator to output 7v all the time (as the fuel pump is probably designed to work with 6v lead acid batteries, the max voltage should be more than 6v) and then you'd have (9.2v to 7v ) x 3.5A = 2.2v x 3.5a = ~8A.
You can try a regulator like LM1084 which has a maximum current of 5A : https://www.digikey.com/product-detail/en/texas-instruments/LM1084IT-ADJ-NOPB/LM1084IT-ADJ-NOPB-ND/363557
However, you'd probably have to screw it on a credit card sized heatsink (or a metal case that you would then fix to the metal frame), to keep the chip below 150 degrees Celsius ...
So such solution is cheaper, but it also wastes power and produces a lot of heat...

#### GeorgeOfTheJungle

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##### Re: Using a diode to drop voltage
« Reply #15 on: January 18, 2019, 02:25:14 am »
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.
« Last Edit: January 18, 2019, 02:27:20 am by GeorgeOfTheJungle »
int main (void) { while (1) fork(); }

#### spec

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##### Re: Using a diode to drop voltage
« Reply #16 on: January 18, 2019, 09:10:18 am »
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.

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.
« Last Edit: January 18, 2019, 09:27:04 am by spec »

#### Seekonk

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##### Re: Using a diode to drop voltage
« Reply #17 on: January 19, 2019, 12:58:33 am »
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.

Smf