electronic stethoscope

[Cinek]

Hello,

My goal is to design electronic stethoscope (the mos simple one) and i have big problem. I was thinking about using about microphone connected to circuit with audio amplifier and on output i have headphones. It's gonna work? and could anyone give some hints how to design this circuit ?

Regards.

[kobunko]

Originally Posted by Cinek

My goal is to design electronic stethoscope (the mos simple one) and i have big problem.

Hi. I think this one is almost that you need. I hope will be good for you. Regards from Argentina.

P.S.: the web is in spanish, but I think is easy to understand the circuit.
http://www.hispavila.com/3ds/tutores/estetoscopio.htm

[Cinek]

ooo thanks so much btw. now i am on student's exchange in spain, so spanish is not a problem .. ;]

[Audioguru]

They copied my circuit but used noisy old opamps:

[Audioguru]

Originally Posted by Audioguru

They copied my circuit but used noisy old opamps:

EDIT:
Their date is before mine but I never saw theirs until now. So we both fixed the horrible original project a www.aaroncake.net .
They made a few changes that I should have made.

[Cinek]

hey, i have few questions concerning this design:
1. Why we use capacitor C1?
2. in amplifier U2 why we construct this feedback formed by C4 and R7?
3. what does C5 and R8 are doing ? (can i use as C5 two normal condensators - not eletrolytic?)
4.R11 and C7 form next filter?
5.what elements : C10, C9,R14, C11 are making ?

Thanks in advance for the answer

[Audioguru]

Originally Posted by Cinek

hey, i have few questions concerning this Spanish design:
1. Why we use capacitor C1?

It filters the +9V supply to prevent the +9V from jumping up and down when the LM386 power amplifier uses a fairly high current to feed a speaker or headphones.

2. in amplifier U2 why we construct this feedback formed by C4 and R7?

U2 is a 2nd-order inverting lowpass filter.

3. what does C5 and R8 are doing ? (can i use as C5 two normal condensators - not eletrolytic?)

C5 uses two 1.2uf polarized capacitors in series and back-to-back to make a non-polarized 0.6uf capacitor. The capacitors are not needed and can be removed.
U3, R8 and R10 do nothing and can also be removed. The volume control R11 should be 10k and C7 should be removed.

5.what elements : C10, C9,R14, C11 are making?

C11 blocks the DC at the output of the power amplifier U5 from burning out the speaker or headphones. The other parts prevent the LM386 power amplifier from oscillating at a high frequency as shown on its datasheet.

[Cinek]

so we can simplify this SPANISH design to this form ?

and the last question i hope what is the job of C6 and C8 ?

[Audioguru]

Originally Posted by Cinek

so we can simplify this SPANISH design to this form?

You used two very noisy old LM324 quad opamps wasting 6 opamps and needing 28 pins.
I used a single low noise dual opamp that has only 8 pins.

Your opamps have a total gain of only 2, then you add C6 to the LM386 power amp to boost its gain 10 times so it amplifies its own noise plus the noise from your old opamps.
I used gain in both opamps so that the LM386 power amp did not need to have its gain boosted then the noise level is low.

I think the second opamp in the original Spanish circuit is wrong. Yours is a wrong copy since theirs inverts and yours is non-inverting. I have only made non-inverting Sallen and Key second-order Butterworth lowpass filters. I have never seen inverting second-order filters like the Spanish circuit. Yours has the values of its capacitors wrong (C4 should have double the value of C3) so your circuit might oscillate.

what is the job of C6 and C8 ?

They are explained in the datasheet for the LM386.

[Cinek]

i read your comment, then traveled along internet to find some solutions and i changes my design to some kind of mix of both :

values in low-pass filter are calculated for cutoff frecuency of value 150 Hz , with this value my results will be satisfactory? or should i choose another value of frequency? And one more question is about total gain. Now it has 10 times higher value than in your design , isn't it a problem ?

in advance and as for now.. thank you very much for your help :-)