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Electronic Stethoscope


t_ang4
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Hi Subodhthok,
Welcome to our forum. ;D
It is good that it works perfectly.

The circuit has AC signals. Digital counters use DC pulses, not AC as their input. In order to count heartbeats then the signal must be rectified into DC pulses but the polarity of the pulses might be negative instead of positive. If the pulses are negative then they need to be inverted to be used as a clock input to a Cmos counter IC.

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Hi Subodhthok,
You must design the interface circuit yourself or ask your teacher to help you.

The output pulses of U5 are only plus and minus about 3V if the batteries are new, and only 2V if the batteries are old. A rectifier would reduce it to be too low to drive logic circuits.

The output pulses of U4 are plus and minus about 7.5V when the batteries are new. A rectifier would reduce it to 6.8V. A voltage divider made with two resistors would reduce it to 5V for logic circuits.

My stethoscope made heatbeats that were sometimes made of 16Hz vibrations for each beat. A filter would need to be made to integrate the rectified pulses into single beats.
The rectifier should drive an Schmitt-trigger inverter or Schmitt-trigger buffer that can be used if pulses of the opposite polarity are best.

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  • 2 weeks later...

Hi Steve,
It is good to hear that it works well. ;D

Hum is mains radiation picked up by the high gain input. Unshielded input cable usually causes it. If the wiring of your circuit is long like on a breadboard then it picks up mains hum.

Use shielded audio cable from the microphone and keep all unshielded wires very short.
I show the shield on the microphone cable on my schematic.

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  • 4 weeks later...

I'm a newbie here. How can i send the output of the electronic stethoscope to computer through serial port ? I use MATLAB to receive the signal.

I need a ADC and a Microcontroller as well as the RS-323 to do that ?

The level of the signals must be shifted to the positive range before i can digitize it rite ?

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Hi Cs,
Welcome to our forum. ;D
The output of the Electronic Stethoscope 2 project feeds an AC without any DC bias voltage to headphones.
The output pin 5 of the LM386 power amplifier IC is also AC but it has a DC bias voltage of half the positive battery voltage. It swings from about +1.2V to about +7.8V if its load resistance is high. It would need to be level shifted to be from 0V to no more than +5V to feed an ADC.

I think the heartbeat signal goes in a negative direction since the first opamp inverts the output from the microphone. If this is a problem then either the first opamp can be made non-inverting or the LN386 power amplifier can be connected as inverting.

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Hi AudioGuru,

Thanks about the valuable informations. I was jz tring to calculate each op-amp's gain (necessary for my project report).

Currently learning to use ADC and ATMEL8051 for the interfacing purpose. Will post my results here for comments if there's a breakthrough.

I will use a standard electronic stethoscope which is sold on the market and compare the signals obtained from it with the electronic stethoscope 2 posted here. Try to calculate the percent root-mean difference between these 2 and analysis the quality of the circuit.

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  • 2 weeks later...

I'm new to this site, and am analyzing this circuit for a project.  Im not exactly clear on what the function of the first amp is for

The first opamp amplifies the very low level from the microphone. The second opamp is an equal component Sallen and Key second-order Butterworth lowpass filter that should not have more gain than 1.6.

why you use a dual op-amp?

A dual opamp is inexpensive because they are used in stereos. They are smaller than two separate opamps.

i have seen on other circuits a buffer before the gain stage-- is this necessary?

The electronic Stethoscope-2 circuit has enough gain for its job.

what is the difference between using the 741 and 386 amps?

The 741 is an opamp that has a minimum load resistance rating of about 2k ohms. It doesn't have enough output current to drive low impedance headphones.
The LM386 is a small power amplifier that drives low impedance headphones perfectly.
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I am having issues with using PSPICE to simulate this circuit.  I keep getting errors for floating nodes at the LED and the headphone jack.  Any advice? 
The LM386 was invented before PSPICE so there is not a model for it. If you made your own model then connect a 16 ohm load resistor to the headphones jack.
Disconnect the LED, all it does is blink.
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This is worth a try. I have not tested it

*--------------------------------------------
* NO-FRILLS LM386 MODEL
* Dave Dilatush 5/30/95

* PSPICE analysis statements:

.probe
.ac dec 20 1 1e7
.tran 1u 3m 0 5u

* circuit to test the lm386 model:

vsupply vcc 0 dc 9
vsignal input 0 ac 1 sin 0 .05 1k
csnub output snub .05uf
rsnub snub 0 10
ccoupling output speaker 1000uf
rspeaker speaker 0 8
xamp input nc1 nc2 nc3 nc4 output vcc 0 lm386
*cgain nc3 nc4 10uf ;gain boost capacitor
*cbypass nc2 0 50uf ;bypass cap for PSRR

* lm386 subcircuit model follows:

* IC pins:    2  3  7  1  8  5  6  4
*              |  |  |  |  |  |  |  |
.subckt lm386 inn inp byp  g1  g8 out  vs gnd

* input emitter-follower buffers:

q1 gnd inn 10011 ddpnp
r1 inn gnd 50k
q2 gnd inp 10012 ddpnp
r2 inp gnd 50k

* differential input stage, gain-setting
* resistors, and internal feedback resistor:

q3 10013 10011 10008 ddpnp
q4 10014 10012 g1 ddpnp
r3 vs byp 15k
r4 byp 10008 15k
r5 10008 g8 150
r6 g8 g1 1.35k
r7 g1 out 15k

* input stage current mirror:

q5 10013 10013 gnd ddnpn
q6 10014 10013 gnd ddnpn

* voltage gain stage & rolloff cap:

q7 10017 10014 gnd ddnpn
c1 10014 10017 15pf

* current mirror source for gain stage:

i1 10002 vs dc 5m
q8 10004 10002 vs ddpnp
q9 10002 10002 vs ddpnp

* Sziklai-connected push-pull output stage:

q10 10018 10017 out ddpnp
q11 10004 10004 10009 ddnpn 100
q12 10009 10009 10017 ddnpn 100
q13 vs 10004 out ddnpn 100
q14 out 10018 gnd ddnpn 100

* generic transistor models generated
* with MicroSim's PARTs utility, using
* default parameters except Bf:

.model ddnpn NPN(Is=10f Xti=3 Eg=1.11 Vaf=100
+ Bf=400 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100
+ Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333
+ Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n
+ Tf=1n Itf=1 Xtf=0 Vtf=10)

.model ddpnp PNP(Is=10f Xti=3 Eg=1.11 Vaf=100
+ Bf=200 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100
+ Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333
+ Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n
+ Tf=1n Itf=1 Xtf=0 Vtf=10)

.ends
*----------end of subcircuit model-----------
.end

*--------------------------------------------
* NO-FRILLS LM386 MODEL
* Dave Dilatush 5/30/95

* PSPICE analysis statements:

.probe
.ac dec 20 1 1e7
.tran 1u 3m 0 5u

* circuit to test the lm386 model:

vsupply vcc 0 dc 9
vsignal input 0 ac 1 sin 0 .05 1k
csnub output snub .05uf
rsnub snub 0 10
ccoupling output speaker 1000uf
rspeaker speaker 0 8
xamp input nc1 nc2 nc3 nc4 output vcc 0 lm386
*cgain nc3 nc4 10uf ;gain boost capacitor
*cbypass nc2 0 50uf ;bypass cap for PSRR

* lm386 subcircuit model follows:

* IC pins:    2  3  7  1  8  5  6  4
*              |  |  |  |  |  |  |  |
.subckt lm386 inn inp byp  g1  g8 out  vs gnd

* input emitter-follower buffers:

q1 gnd inn 10011 ddpnp
r1 inn gnd 50k
q2 gnd inp 10012 ddpnp
r2 inp gnd 50k

* differential input stage, gain-setting
* resistors, and internal feedback resistor:

q3 10013 10011 10008 ddpnp
q4 10014 10012 g1 ddpnp
r3 vs byp 15k
r4 byp 10008 15k
r5 10008 g8 150
r6 g8 g1 1.35k
r7 g1 out 15k

* input stage current mirror:

q5 10013 10013 gnd ddnpn
q6 10014 10013 gnd ddnpn

* voltage gain stage & rolloff cap:

q7 10017 10014 gnd ddnpn
c1 10014 10017 15pf

* current mirror source for gain stage:

i1 10002 vs dc 5m
q8 10004 10002 vs ddpnp
q9 10002 10002 vs ddpnp

* Sziklai-connected push-pull output stage:

q10 10018 10017 out ddpnp
q11 10004 10004 10009 ddnpn 100
q12 10009 10009 10017 ddnpn 100
q13 vs 10004 out ddnpn 100
q14 out 10018 gnd ddnpn 100

* generic transistor models generated
* with MicroSim's PARTs utility, using
* default parameters except Bf:

.model ddnpn NPN(Is=10f Xti=3 Eg=1.11 Vaf=100
+ Bf=400 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100
+ Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333
+ Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n
+ Tf=1n Itf=1 Xtf=0 Vtf=10)

.model ddpnp PNP(Is=10f Xti=3 Eg=1.11 Vaf=100
+ Bf=200 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100
+ Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333
+ Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n
+ Tf=1n Itf=1 Xtf=0 Vtf=10)

.ends
*----------end of subcircuit model-----------
.end

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for the input, the mic has an output resistance of 5k, which is in parallel with 10k, and in series with 2.2k. combined with the 4.7u leads to frequency of 6.1Hz.

Yes. the frequency response is down to about 6.1Hz.

The 470u cap keeps a steady supply voltage for the mic and preamp, due to fluctuations in the internal resistance of the battery.

No. The signal causes fluctuations in the supply voltage (when the internal resistance of the battery increases as it runs down) if the 470uF capacitor is not there to smooth the fluctuations.

It is unclear is what frequency we are trying to amplify (what is the typical frequency of a heartbeat?)

A heartbeat is a DC pressure wave. A microphone cannot produce DC so the circuit responds to very low frequencies near DC.

unclear is what the role of the 1k is

The 1k resistor and the 470uF capacitor are a filter.

what voltage levels are expected at the input of the preamp.

Maybe the input to the circuit is 10mV peak.

For the preamp, .... The input resistance is 10E12

The input resistance of the preamp is 2.2k ohms (R2) which is too low but it works.

the gain is -8.5

Yes, it inverts with a gain of 8.5.

unclear how to figure what frequency is being amplified

U1A amplifies frequencies from 6.1Hz to 100kHz.

and what the output resistance is

The output resistance of an opamp is "zero ohms". It is actually the physical 75 ohms divided by the open loop gain of about 200,000, times the gain. So it is actually 0.003 ohms at very low frequencies.

For the low pass filter,  also a TL072.  function is to filter higher frequency noise.  unclear where the noise comes from-- only lower frequencies have been amplified.

The microphone picks up sounds of road traffic outside and people talking nearby. The first opamp amplifies it.

cutoff frequency is 103Hz (what is in this frequency range?).  it is a 2nd order butterworth filter, meaning it has a flat amplitude response.  its gain at 103Hz is .707, leading to an overall gain of about 1.6.

Yes, the response of the Butterworth filter cutsoff at 103Hz.

unclear what the 33k and 56k are for

The gain of the opamp is (33k/56k) + 1= 1.59. If one filter capacitor is twice the value of the other then the gain can be 1 and it will still be a Butterworth filter.

what the output resistance is.

The output resistance of an opamp is "zero ohms". It is actually the physical 75 ohms divided by the open loop gain of about 200,000, times the gain. So it is actually 0.0006 ohms at very low frequencies. 

The volume control is a variable resistor, which when increased, increases the input bias which results in higher output voltage

No. The bias voltage is not changed by the volume control. It changes the signal amplitude due to its adjustable voltage divider action.

unclear what the limit is for this resistor value.

The value of the volume control can be from 2k (the lowest value an opamp can drive) to about 50k which is the input resistance of the LM386 amplifier.

The output amp is a LM386.  its input resistance is 50k in series with the pot.  it has a very low load resistance and is able to power low impedance headphones.  although this amp is better, i will be using a 741 amp because i have to do a simulation is PSPICE, and i dont know how to create a subcircuit.  The 741 opamp has an input resistance of 2Meg, unclear of the output resistance.

A 741 is an opamp. An LM386 is a power amp which is completely different.
A 741 can be loaded with no less than about 2k ohms. An LM386 is made to have a load down to 8 ohms.
A 741 needs to be biased correctly and needs to have two resistors to set its gain.
An LM386 is internally biased and has its gain internally set at 10.
A 741 doesn't work if its inputs are at ground (0V).
An LM386 is designed for its inputs to be at ground.

unclear what the role of 3.9 and .1u are

They are the load on the LM386 at very high frequencies where the speaker's impedance becomes very high due to its inductance. The LM386 will oscillate at a very high frequency without them.

about to put circuit in PSPICE wish me luck.

Good luck. ;D
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I answered your Personal Message. I didn't know you posted the same questions here.
U1a and U1b are a TL072 which is a dual opamp. It has two low noise opamps inside. Look at its datasheet.

U1a is a preamplifier. U1b is a Sallen and Key 2nd-order Butterworth lowpass filter that cuts frequencies above 103Hz.

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Hi, Im not so new to electronics, but new to building from scratch.  I repair mobile phone circuits for a living, and Im not sure if its a confidense issue, but have never really built a decent project, and this is 1 that I want to build.

My girlfriend is pregnant and thought it would be good if I could pick up the babys heartbeat.  But Im not sure if this would work so well for that, dont they use US to detect heartbeats if you go to the hospital.

I would love to make this as a present for her.  Any ideas if it would work, if it needs further amplification Im sure I can figure it out.

Thanks, Philip

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