Cheapest and "Smallest" way to cut 6.5v to 5v

I have been designing a very simple circuit for a project which works with 6.5v, it can work with any voltage from 4 to 12v but I'm using 6.5v just because I had a 220v to 6.5v adapter laying around.
In the circuit, there is a relay, which works with the circuit main voltage (6.5v now) but I couldn't find a 6.5v relay so I'm gonna use a 5v relay instead, but since this relay will be turned on for long periods of time, I don't want to over-drive it and shorten its life span.
now I'm looking for a simple way to cut down the 6.5v to 5v for that relay, any advice?

p.s: I don't want to use a buck converter module or a 5v regulator like 7805, first because I should keep the PCB as small as possible and second because I should keep it as simple as possible.

p.s2: I don't want to make the 5v for the whole circuit, it will be used only for the relay

If it's just a relay, which will draw constant current, put a resistor of appropriate value in series with it.

The total resistance is (6.5 / 5) * the resistance of the relay.
Subtract the resistance of the relay from the result above to get the resistor value.

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Yes, or two silicon diodes in series... that drops about 1.4V.


so if the relay is 420 ohm (measuring a 12v relay just for an example),
a 546 ohm (the nearest one obviously) in series with the coil will do the job?

That's also a great idea! thanks
1n4007 is fine right?

Err, no.
6.5 / 5 = 1.3
1.3 * 420 = 546
546 - 420 = 126

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oh yes sorry I missed the subtraction part, Thanks a lot! I'll try it out.

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It's a 1A diode IIRC, should be enough for most relays.

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If you make your project for 5V, then you will benefit from it later.
Suppose you need to replace that adapter some day, then a 5V adapter is easier to find.
Did you know that a USB "charger" is just power supply of 5V ?


Not a meaningful concept. :upside_down_face:

The only consequence of the excess voltage, is increased operating temperature.

The standard "Songle" relay operating at 90 mA has a coil resistance of 55 Ohms. At 5 V it will dissipate 450 mW; at 6.5 V it will dissipate 770 mW. It will get a little warm if actuated for long periods.

To drop 1.5 V at 90 mA, you need a resistor of about 17 Ohms. The nearest common value is - well either 15 or 18 Ohms would be just perfect! :sunglasses:

Not a good example though. :grin:

470 or 560 would be the nearest common values.

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It should be connected directly to a plug,
I can make it 5v I just need to buy a 220-5v adapter but I just have a 6.5v so you know, doesn't really matter that much.
Also, it may be a school project for an A.H :poop: professor and I really don't feel like ordering anything else for him lmao :joy:

I know yeah it's not a big deal, that's why I said I don't wanna use a regulator or a buck converter. but if just a single resistor can fix it, then why not?

Which is the most reliable way to shorten a component's life-span. Runs hotter, expands and contracts more, stresses the solder joints with the PCB fatiguing them. Also the contacts oxidize more and become less reliable.

Ha ha. How many times have I seen the area underneath a through hole resistor on the PCB, brown or black from the heat? :slight_smile: Because someone only looked at the optimistic numbers.

Well the usual problem is not knowing that a resistor run close to its max power should not be mounted flush to the PCB, but spaced up above it for increased air-flow and to avoid scorching the board - power resistors can run at pretty high temperatures.