Hi all,

I have a circuit that sends out a 12V pulse from a 40106 inverter. I would like this pulse to ring a mechanical bell. I believe the bell draws too much current to plug directly into the 40106 (and it's rated at 3-6V DC) so I was wondering if the attached circuit would work. If so, I'm not sure how to calculate the value of R1. Many thanks!

Kevin

bell is this one: http://www.homedepot.com/h_d1/N-25e...0053&langId=-1&keyword=doorbell&storeId=10051

 

Any value from 1K to 10K would suffice

 

Thanks, and the rest of the circuit looks ok? 2n2222 should suffice?

Kevin

 

Thanks, and the rest of the circuit looks ok? 2n2222 should suffice?

Kevin

The transistor is chosen according to the load it is driving.
{ed}
I believe 2N2222 is OK as long as u don't use a load more than 800mA

 

There is still a limitation not discussed. The transistor base needs a tenth of the current required for the bell. If the 40106 can't do that, you'll need 2 transistors or a Darlington transistor.

(12V -.6V)/10k ohms = 1.14milliamps. Not enough drive current!
11.4V/.03 amps = 380 ohms. Try 390 ohms. If that doesn't work, add more current amplification.

 

or OP could use a buffer in at the 40106 output

 

Thanks for all of the good info! I will try the 390 ohm resistor and I have a few ULN2003s lying around so may just go with one of those. I'm not exactly sure which pins to connect to where because I see the data sheet for the ULN2003 contains an inverter. So perhaps I need to bypass the 40106....


Kevin