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).
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?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.
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?Looking at older posts about heart rate monitors and ecg circuits i have stumbled on this diagram
i am failing to see where the output on this is though compared to the first attachment i posted.
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.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?
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.do modern day exercise machines that work out heart rate use ECG techniques?
The lousy old LM358 opamp has a lot of noise and the circuit amplifies it.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 input 220k resistor feeds the inverting (-) input of the opamp so guess what the opamp does??does the 0.47µF capacitor and 220kO resistor configure it to become an inverting op amp?
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.Why is it of importance to have get a gain of 500? How did was this figure calculated?
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