[SOLVED] best practice method of lowering a voltage to meet a component's max?

[maark6000]

Just curious about something, I'm working with a circuit that has a single sided supply, high rail is 8 volts, low rail is ground. If I have a component (in this case it's a relay) whose max voltage is 5 volts, what is the best way to lower my rail voltage for this component? Obviously I could use a voltage regulator, but that seems a bit much. What about a 4732 or 4733 Zener diode? A voltage divider (via resistors) also gets me there but in the case of the relay it's really more about the current, right? Of course the datasheet on the relay doesn't specify what the minimum coil current needed to close the switch is... I'm going to monkey around with it right now to see if I can figure that out... but I just felt like there must be some "this is the best way" method that I should know about.

Thanks as always for your help.

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what I've come up with so far is this; the relay's coil has a resistance of 128Ω. So in effect it could be the bottom part of a voltage divider. The datasheet says the relay needs at least 3.75 volts to close. So with a 8 volt high rail, if I put a 130Ω resistor in series with the coil, this will get me to around 4 volts, and in this case about 30mA. That's obviously probably the simplest solution, but I still would like opinions as to what's the "best" way to go about this.

 

[BradtheRad]

A zener is the simplest method. Also led or plain diode (or a combination).

There is nothing wrong with a resistive divider, although it helps if you start from a regulated voltage.

The above are shunt regulators. They waste a certain amount of power. However when you speak of a relay, then you need enough current to energize it. If possible you might prefer a series regulator rather than a shunt type. A series regulator can be a transistor, or a regulator IC.

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So with a 8 volt high rail, if I put a 130Ω resistor in series with the coil, this will get me to around 4 volts, and in this case about 30mA.

This acts as a series regulator. It prevents overcurrent through the relay coil. It applies proper voltage.

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In effect, it acts as a resistive divider.

 

[FlapJack]

5 VDC relay has a 128 ohm coil.

I = E / R, 5 VDC / 128 ohm = .040 A

You have 8 VDC to run this from.

8 VDC power supply - 5 VDC relay voltage = 3 volts you have to drop.

R = E / I, 3 VDC / .040 A = 75 ohms.

P = I x E, .04A x 3 VDC on resistor = .12 W = 120 mW

De-rate the resistor by at least 50% for long life = 1/4 Watt = 250 mW minimum wattage rating.

The resistor is the simplest way to go. Relays are fairly forgiving there is no need for a regulated power supply to run them.

So bottom line is you need a 75 ohm 1/4 or even better 1/2 watt resistor to drop 3 VDC to power your 5 VDC relay.

 

[maark6000]

okay, yeah, sometimes the easy solution is the best one. thanks all!!!