# Using unsuitable relay in transistor switching circuit?

I'm trying to make use of this relay which operates on 5 volts, in a circuit powered by +9 volts. I'm kind of new to electronics, so bare with me. The schematic can be seen on the image below:

According to page 3 of the datasheet the maximum voltage the relay can handle is about 6.5 volts, if I understood correctly.

I have 3 ideas on how to solve this:

1. Using a voltage divider to divide V. How would I go about choosing the resistor values (kOhms, MOhms..)? Do I need to take care of how big the current through the relay's coil is?

2. Adding a few basic crystal diodes antiparallel to the diode D11 to drop the voltage, but this solution seems a bit clunky. (edit, meant to say: in series with the relay's coil, but differently oriented than the diode D11)

3. Using a 7805 regulator, by simply connecting its pins as follows: $V_{in} = V$, $V_{out}$ to the upper side of the coil, and GND to GND. Do I need to take care of anything when using this solution, or am I safe to just plop i into my circuit mindlessly?

If you have any better or more elegant solutions, please share them with me. And why are my Tex edits not showing up correctly?

• Have you considered using a 63ohm resistor? – Ignacio Vazquez-Abrams Jul 7 '16 at 9:43
• Using it where exactly? – HarryHey Jul 7 '16 at 9:45
• @HarryHey In series with relay. – Jakub Rakus Jul 7 '16 at 12:45

The best solution is to select the correct relay. The data sheet shows that the part is available with a 9V coil. The 9V relay coil only draws 40mA from your supply whilst the 5V coil will draw 72mA according to the data sheet.

The easiest solution if you have to stay with the 5V relay is to simply put a resistor in series with the coil of the relay that drops the voltage from 9V to 5V when the transistor turns on to energize the relay coil.

Note this data regarding the relay coil:

Knowing that the relay coil resistance of 70 ohms drops 5V we can compute the coil current as:

Icoil = 5V / 70 ohms = ~71mA

The suitable resistor for the drop becomes:

Rseries = (9V - 5V) / 71mA = ~56ohms

This resistor will need a power rating of at least:

Rpower = (71mA * 71mA) * 56ohms = 0.28W

So simply place a standard sized 56 ohm 1/2 Watt resistor from the 9V supply to top of the relay coil. (Leave the diode directly across the coil).

• Don't know how I didn't think of a resistor. The fact that it requires a certain amount of voltage AND current must have confused me. Thank you very much. – HarryHey Jul 7 '16 at 10:14

Option 1 is inefficient as the relay needs a large current, but the simples solution is a variant of this option as it uses the resistance of the coil's relay as one of the voltage divider resistors ! (more on this below).

Option 2 is plain wrong, do not limit the voltage like that as you will be overloading the supply. Instead, add some diodes in series with the relay's coil.

Option 3 is a possibility to convert 9V into 5 V, look at how others use 7805 regulators, do the same, it will work.

But the simplest option is to determine the current which the relay needs at 5 V, see the datasheet it is about 72 mA. So you need to drop 9 - 5 = 4 V at 72 mA that's a resistor of 56 ohm and it will dissipate about 0.29 Watt. So if you only have 0.25 W resistors, use two 27 ohm resistors in series or two 120 ohm resistors in parallel. Place those resistors in series with the coil of the relay. Keep D1 as is, directly connected to the coil. Basically the 56 ohm resistor will be in series with the collector of Q15.

• Thank you, I phrased myself incorrectly describing option 2. Don't know how I didn't think of a resistor. The fact that it requires a certain amount of voltage AND current must have confused me. – HarryHey Jul 7 '16 at 10:13