How Can I Increase Current from a CD4017 IC to Fire a Relay?

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To increase current from a CD4017 IC for firing a relay, using a transistor like the 2N3904 is recommended, as it can handle a couple of hundred mA. When switching high currents, ensure the transistor is rated appropriately for the load. A reverse biased diode, often referred to as a flyback diode, should be added across the relay coil to protect the transistor from back EMF. It's important to check the relay's datasheet to see if it includes a built-in diode. Properly implementing these components will enable the circuit to handle larger loads effectively.
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Hi guys I'm new here and need map some help. I have built a simple circuit with a cd4017 IC. The circuit works perfectly and has led connected to it. However I want to connect much larger load to it. A relay is the obvious answer but I'm not getting enough current from the IC to fire the relay. I tried using a triac but the triac latches and I'm led to believe that this is because it's a dc load, the relay. I'm stuck now any suggestions welcome please.
 
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a transistor would be a better/easier way to go unless you were switching more that say 10A

attachment.php?attachmentid=72084&stc=1&d=1407801780.gif


OK I have drawn this one ... still switching a LED
The 2N3904 transistor will switch a couple of 100 mA

choose a transistor rated for a current and load you want to switch

for high currents, you could replace the LED and 1k resistor with a relay ... don't forget to add a reverse biased diode (1N4001) across the relay coil, else you will kill the transistor

Dave
 

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More often than not you will find that you have to turn things on in stages because the control IC doesn't provide enough current to drive the switching device. Davenn's suggestion is good. I just wanted to add that the reverse biased diode he mentioned is often called a "flyback diode" and it is sometimes included inside the relay itself. Check your datasheet for your relay to see if it has a diode.
 
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