ECG/Heart Rate Monitor

MrChips

Joined Oct 2, 2009
30,712
Why do you think they used so many extra components? Because recording ECG is not a simple task. The common-mode signal (noise) is going to kill you. (sorry, not literally).
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
Haha its not killing me yet...just making my head hurt haha!! :( is that portable ECG appilcation link i posted just a small part of the whole ECG project? Or are all the fundamental components all there?
 

Audioguru

Joined Dec 20, 2007
11,248
The complicated ECG circuit you posted recently is the FRONT END of a portable unit. Its analog output feeds the ADS8321 Analog to Digital converter IC that records the waveform digitally.
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
The complicated ECG circuit you posted recently is the FRONT END of a portable unit. Its analog output feeds the ADS8321 Analog to Digital converter IC that records the waveform digitally.
Therefore the BACK END (so to speak) is MY output that i have chosen? If i want to show this digital waveform in a ECG program i would needs the likes of a radio transmitter circuit connected somehow to the FRONT END circuit?
 

t06afre

Joined May 11, 2009
5,934
Looking at older posts about heart rate monitors and ecg circuits i have stumbled on this diagram

http://forum.allaboutcircuits.com/attachment.php?attachmentid=14566&d=1261267520

i am failing to see where the output on this is though compared to the first attachment i posted.
The components used in that design are somewhat exotic. They are Rail to Rail type opamps and also designed for use in single supply applications. The component ADS8321 is a AD converter. That you can omit in your design. Then measuring ECG. It is very important to use proper electrodes. What do you use for electrodes at this moment?
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
The components used in that design are somewhat exotic. They are Rail to Rail type opamps and also designed for use in single supply applications. The component ADS8321 is a AD converter. That you can omit in your design. Then measuring ECG. It is very important to use proper electrodes. What do you use for electrodes at this moment?
Well tbh at the moment i have not got any electrodes. I was trying to build the first circuit (1st attachment post #1) with the LED and LDR, that gave me a nice looking scope reading that resembled a heartbeat but that was just a one off because i unable to produce that reading again.

Therefore instead of doing heart rate monitor circuits i am more inclined to research ECG/EKG circuits a bit more and see if this is a more reliable method. I have seen home made electrodes that people have posted on the internet, mostly coins! What do you suggest as possible electrodes?
 

Attachments

Audioguru

Joined Dec 20, 2007
11,248
You are confusing the functions of the circuits:
1) A heart RATE monitor measures the RATE of blood flow pulses per minute.
It shows a number, not the electrical waveform of a heart that is shown by an ECG unit.
2) An ECG unit measures a waveform of the electrical signal from the heart and can display the waveforms on a screen and display the heart rate number of beats per minute.
3) An electronic stethoscope plays the sound of the motion of the heart in headphones.

The first circuit is lucky to produce a waveform on your 'scope (but the waveform looks wrong) because it does not show the electrical waveform from the heart. It was designed to pulse the meter in its output.

Electrodes can be purchased from a medical supplies company.
 

t06afre

Joined May 11, 2009
5,934
do modern day exercise machines that work out heart rate use ECG techniques?
They sample the ECG signal. And do all that in software using some sort of peak detection algorithm. It sounds fancy but it is not that hard. I have done this in several Labview applications.
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
also im looking at using IR emitters and receivers to see how effective these are compared to normal LEDs and LDRs...im looking on farnell for emitters and receivers and it is asking me to select certain parameters to narrow down the search, are there certain carrier frequencies i should be looking at for the receivers or certain forward current for the emitter in relation to heart rate monitors/ecgs?

sensors
http://uk.farnell.com/jsp/search/br...search&Ntt=IR+sensor&Ntx=mode+matchallpartial

emitters
http://uk.farnell.com/jsp/search/br...earch&Ntt=IR+Emitter&Ntx=mode+matchallpartial
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
you know the first attachment, the circuit diagram i posted in entry #1, its says "Pin 3 is biased by a high impedance voltage divider consisting of two 3.3MO resistors. The feedback resistors to pin 2 set the gain to 11 times. The output of IC1a is fed via a 0.47µF capacitor and 220kO resistor to IC1b. This is configured as an inverting op amp with a gain of 45 so that the total circuit gain is about 500. The output of IC1b is used to drive an analog meter which may be a multimeter set to the 10V DC range or any panel meter in series with a resistor to limit the current to less than its full-scale deflection. The prototype used an old VU meter with a 47kO resistor fitted in series"

but my understanding is that if your gain is high the noise will be visible at the output...is having a gain of 11 on PIN2, gain of 45 on PIN3 and an overall gain of 500 high?
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
What are feedback resistors? What do they do?

Am i correct in thinking The 1M resistor connected between PIN 2 and PIN 1 and the 10M that is connected between PIN 6 and PIN 7 both act as feedback resistors?
 

Thread Starter

mattyleeboy

Joined Jan 27, 2012
23
"The output of IC1a is fed via a 0.47µF capacitor and 220kO resistor to IC1b. This is configured as an inverting op amp with a gain of 45 so that the total circuit gain is about 500"

The LM358 is a dual op amp anyway so does the 0.47µF capacitor and 220kO resistor configure it to become an inverting op amp? Why is it of importance to have get a gain of 500? How did was this figure calculated?
 

Audioguru

Joined Dec 20, 2007
11,248
my understanding is that if your gain is high the noise will be visible at the output...is having a gain of 11 on PIN2, gain of 45 on PIN3 and an overall gain of 500 high?
The lousy old LM358 opamp has a lot of noise and the circuit amplifies it.
If a low noise modern opamp is used instead then there will be much less noise.

does the 0.47µF capacitor and 220kO resistor configure it to become an inverting op amp?
The input 220k resistor feeds the inverting (-) input of the opamp so guess what the opamp does??
If the input to the opamp is to its (+) input then it does not invert.

Why is it of importance to have get a gain of 500? How did was this figure calculated?
The LDR senses a small amount of light through your finger. Blood blocks some of the light so the LDR changes its resistance a small amount for each heartbeat. The LDR is in a voltage divider circuit so the output changes the voltage a small amount.
The opamps amplify the small signal.

The first opamp is non-inverting. The gain of a non-inverting opamp is 1+ (Rf/Rin) + 1 which is (1M/100K) +1= 11 in this circuit but the gain of a lousy old LM358 opamp will be less because its input resistance is fairly low.

The second opamp is inverting and its gain is 10M/220k= 45.5 but the gain of a lousy old LM358 opamp will be less because its input resistance is fairly low.

The gains will multiply to 455 if the opamps are much better. A TL072 dual opamp is much better than a lousy old LM358 dual opamp.
 
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