How this circuit works?
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Posted by on June 25, 2008, 9:20 am
Please log in for more thread options I have the following questions, can anyone explain to me how the following circuit works, how does the first op-amp oscillate, and how does the buzzer sound? The first op-amp is configured with negative feedback, which normally means that the negative input will try to equal the positive input, which is always at half vdd, therefore it is unclear why it would oscillate. Secondly, the 2nd op-amp is connected to a dc level, so how would we hear the piezo, which needs an ac signal to be heard? The circuit is here http://www.geocities.com/maxcotester/S5030995.jpg http://www.geocities.com/maxcotester/S5030996.jpg http://www.geocities.com/maxcotester/S5030997.jpg | ||||||||||||||||||||||||||||||||||
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Posted by GM on June 25, 2008, 9:59 am
Please log in for more thread options http://en.wikipedia.org/wiki/Relaxation_oscillator > Secondly, the 2nd op-amp is connected to a dc level, so how would we
> hear the piezo, which needs an ac signal to be heard? > The output of the first op-amp will oscillate between Vdd and Gnd so that the final signal that arrives at the 2nd op-amp input will oscillate too and will make the buzzer to oscillate as long as the final signal is something more than Vdd/2 > The circuit is here
> > http://www.geocities.com/maxcotester/S5030995.jpg > http://www.geocities.com/maxcotester/S5030996.jpg > http://www.geocities.com/maxcotester/S5030997.jpg | ||||||||||||||||||||||||||||||||||
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Posted by Rich Webb on June 25, 2008, 10:17 am
Please log in for more thread options On Wed, 25 Jun 2008 06:20:43 -0700 (PDT), use@slax-he.info wrote:
>Hello
>I have the following questions, can anyone explain to me how the >following circuit works, how does the first op-amp oscillate, and how >does the buzzer sound? > >The first op-amp is configured with negative feedback, But it's a comparitor, not an op-amp, so the output "wants" to be either all the way high or low (insofar as a given part can reach either rail). > which normally
>means that the negative input will try to equal the positive input, >which is always at half vdd, therefore it is unclear why it would >oscillate. If the (+) terminal at Vcc/2 is more positive than the (-) terminal then the output will rise towards Vcc. The (-) terminal will then rise towards Vcc at a rate determined by the R3 and C2. At some point it will be greater than the (+) terminal, at which time the output switches to ground and the process reverses. >Secondly, the 2nd op-amp is connected to a dc level, so how would we
>hear the piezo, which needs an ac signal to be heard? It's AC coupled to its input. Imagine what would happen if R4 and C4 were connected together. -- Rich Webb Norfolk, VA | ||||||||||||||||||||||||||||||||||
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Posted by Tim Wescott on June 25, 2008, 11:04 am
Please log in for more thread options use@slax-he.info wrote:
> Hello
> I have the following questions, can anyone explain to me how the > following circuit works, how does the first op-amp oscillate, and how > does the buzzer sound? > > The first op-amp is configured with negative feedback, which normally > means that the negative input will try to equal the positive input, > which is always at half vdd, therefore it is unclear why it would > oscillate. > > Secondly, the 2nd op-amp is connected to a dc level, so how would we > hear the piezo, which needs an ac signal to be heard? > > The circuit is here > > http://www.geocities.com/maxcotester/S5030995.jpg > http://www.geocities.com/maxcotester/S5030996.jpg > http://www.geocities.com/maxcotester/S5030997.jpg Right there in the text "The left side of the circuit uses a low-power dual comparator (MAX9022) to form a relaxation oscillator..." So, (a) you've been _told_ what's going on, and (b) it's not an op-amp, it's a comparator. On the right side of the circuit, consider what happens when a bunch of AC shows up on the open end of C4. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html | ||||||||||||||||||||||||||||||||||
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Posted by on June 25, 2008, 1:42 pm
Please log in for more thread options
> Right there in the text "The left side of the circuit uses a low-power
> dual comparator (MAX9022) to form a relaxation oscillator..." > > So, (a) you've been _told_ what's going on, and (b) it's not an op-amp, > it's a comparator. OK, this is clear now, shame engineers use the same symbols for different things... >
> On the right side of the circuit, consider what happens when a bunch of > AC shows up on the open end of C4. This is still not understood. The text talks about 155KHz, which is more than the human ear can hear, so which AC are we talking about? And even if we could hear 155KHz, D3 and C5 for a rectifier which converts any ac to dc, so how can a dc signal vibrate the piezo? >
> -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" gives you just what it says. > See details at http://www.wescottdesign.com/actfes/actfes.html | ||||||||||||||||||||||||||||||||||
> I have the following questions, can anyone explain to me how the
> following circuit works, how does the first op-amp oscillate, and how
> does the buzzer sound?
>
> The first op-amp is configured with negative feedback, which normally
> means that the negative input will try to equal the positive input,
> which is always at half vdd, therefore it is unclear why it would
> oscillate.
>