Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I wanted to make a modification to the clap clap switch with a schematic diagram below but instead of a LED lighting I want to make it turn on a small fan..enter image description here

share|improve this question
    
Are you aiming to power a fan from the same 9V battery, or run a fan from some other power source? Have you a link to the sort of fan you plan to use? A BC547 transistor is only rated to handle 100mA, so it is not likely to run much of a fan directly. –  gbulmer Sep 1 at 15:14

3 Answers 3

up vote 4 down vote accepted

Probably the easiest way is to change the T1 to an NMOSFET to increase current switching capability. Since a lot of fans are designed for PCs and rated at 12V I would also recommend upping the supply to suit the fan. No other changes would be needed.

enter image description here

EDIT NOTE:

Have removed connection of pin 12 (carry out - output) from ground. (See EM Fields answer - nice catch sir!)

share|improve this answer
    
Q1: Max working voltage of the given schematics? Can I place a 15 Volt Battery? Q2: Can I put a switch to prevent the fan from turning "on" if someone suddenly clap? If possible, where should I put it? Q3: Can I make the mic more sensitive to clap volume? Q4: Can you help me do the math ^^, Not that I am forcing you to help me but It would be a great help for my part. –  Dexter Sep 2 at 14:07
    
Comments are not really the place for answering subsidiary questions 15V supply - yes as long as fan will work. Electronics should have no problem with that voltage. Switch for fan on/off - yes. Put switch (SPST type) in series with fan. I'd probably put it between the source of Q1 and ground (0v). Mic more sensitive. Lots of ways but as Dave points out this is not a particularly good circuit to start with (+1 from me Dave). What I suggest is you ask another question in this exchange quoting Dave's analysis and asking for an improved circuit. . math help? no need, ask the question –  JIm Dearden Sep 2 at 16:39

The best way to be able to control arbitrary loads would be to connect the input side of a solid-state relay in parallel with the LED, across points A and B in your schematic. With the appropriate selection of SSR, this can be used to switch AC or DC loads at any voltage or current.

Note that this is a poor circuit to begin with. The opamp is being used as a comparator on the audio signal directly (rather than on the envelope of the audio), so it will be prone to mis-counting the claps if there are multiple cycles per clap, like you might get in a reverberant room.

share|improve this answer
    
How can I make the op-amp be used on the envelope of the audio. Let's say I clap twice simultaneously in the given schematic will the lamp turn on? –  Dexter Sep 2 at 14:10
    
If it were me, I'd put an additional opamp stage between the microphone and the existing comparator, configured as a precision rectifier. I would filter the output of this with a time constant of about 100-150 ms, which should be longer than the reverberation, but shorter than the period between two distinct claps. (It's hard to clap twice in less than about 200-250 ms.) –  Dave Tweed Sep 2 at 14:23
    
How should I place it could you show me a schematic diagram? –  Dexter Sep 2 at 15:48
    
Sorry, but no. If you can't figure it out from the pieces you've been given so far -- and given your questions about the other answers -- then this project may still be a bit beyond your capabilities, at least for now. –  Dave Tweed Sep 2 at 16:20

Pin 12 is an output which is high whenever the count is less than 5, and should be left floating instead of being shorted to ground, as it is now.

share|improve this answer
    
Thanks for the edit. And the up-vote? –  EM Fields Sep 1 at 19:08
    
Nice catch sir - wasn't really looking at the rest of the circuit to check if it was correct so +1 from me. –  JIm Dearden Sep 2 at 19:40
    
Thanks, Jim! :-) –  EM Fields Sep 2 at 19:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.