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Old Sat 21 July 2007, 15:09
Just call me: Aubrey
South Africa
Originally Posted by Richards View Post
The 4.7K resistors on the emitters of the 2N2222 transistors will limit the current to 1mA, probably not enough to turn on the coil of the relays. Normally, the coil has enough resistance to be used without a resistor. The emitter of the MCT2E can normally be connected directly to the base of the 2N2222 to form a Darlington circuit. You might have to 'play' with those components a little. The trick is to turn the 2N2222 fully on, but not have so much current flowing through the base that the 2N2222 latches on. Sometimes it's necessary to use a transistor with less 'beta' or gain when it is used in a Darlington circuit.
If I understand you correctly, the 4.7k may be too much resistance and a direct connection may be too little.

This may be a dumb thought but how about a 4.7k trimpot? That way one could "fine tune" that portion of the circuit to perform exactly as required without having to swop out resistors all the time.

Only question is which would work the best, analog or linear?

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Old Sat 21 July 2007, 21:59
Just call me: Mike
South Jordan, UT
United States of America
Trimpots are usually used in addition to resistors. If the trimpot is adjusted to 0 ohms, it will normally fry. Typically, the coil of the relay has enough resistance that no other resistor would be required.

Just use Ohm's Law to check things before applying power. For instance, if the coil is rated at 50mA and the transistor is rated to pass a maximum of 200mA, then pick a number higher than 50 and less than 200, for example 100mA. 100mA is 100 / 1000 or 0.1A, so 5V / 0.1A = 50 ohms and 0.1A X 5V = 0.5 watts.

That will get you started.

I've never had to use a resistor in series with the coil in any of my designs. And, since a 2N2222 transistor costs so little, I would build a test circuit without a resistor and then measure the current flowing through the transistor when things were turned on. If the transistor is hot, then it would be better to use a bigger transistor with a heat sink attached.

(Designing your own circuits will require you to own a digital multimeter and an oscilloscope, along with breadboards, jumper cables and assorted other test equipment. If you're careful, you can buy everything new that you'll need to test your circuits for $500 to $1000 U.S.)

The coil on a relay is more like the filament in a normal light bulb than a resistor. You've probably noticed that most light bulbs 'burn out' when the light is being turned on. That is because of the surge current that flows through the filament at that particular instant. A relay's coil does the same thing: it causes a brief surge. If that surge is greater than the transistor's maximum rating, the transistor will fail sooner than expected. In that case a resistor in series with the coil will limit the surge current.

Last edited by Richards; Sat 21 July 2007 at 22:11..
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