Digital signal, transistor switch, ring the bell

Discussion in 'General Electronics Chat' started by kgstewar, May 10, 2012.

1. kgstewar Thread Starter Active Member

Apr 5, 2012
152
6
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

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2. R!f@@ AAC Fanatic!

Apr 2, 2009
9,647
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Any value from 1K to 10K would suffice

3. kgstewar Thread Starter Active Member

Apr 5, 2012
152
6
Thanks, and the rest of the circuit looks ok? 2n2222 should suffice?

Kevin

4. R!f@@ AAC Fanatic!

Apr 2, 2009
9,647
1,110
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

5. #12 Expert

Nov 30, 2010
18,076
9,680
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.

6. R!f@@ AAC Fanatic!

Apr 2, 2009
9,647
1,110
or OP could use a buffer in at the 40106 output

7. kgstewar Thread Starter Active Member

Apr 5, 2012
152
6
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