Hi all,

With reference to the schematic below, why is the output of the instrumentation amplifier connected to an AC coupled inverting amplifier before the signal goes to the output?

The instrumentation amplifier inputs are connected to ECG electrodes and apart from amplifying the signal, it is also offsetting the signal by 2.5V (hence the 2.5V at the reference pin). Why isn't the output connected directly to the output of the instrumentation amplifier if no gain is applied on the inverting amplifier?

Thanks in advance.

 

Let's rephrase the question: Why would the designer want an inverted wave shape with a high pass function that loses 1/2 of its voltage at 72 Hz and even more of its voltage below 72 Hz?

I don't know, but this form of the question includes the function of the high pass filter.

 

Let's rephrase the question: Why would the designer want an inverted wave shape with a high pass function that loses 1/2 of its voltage at 72 Hz and even more of its voltage below 72 Hz?

I don't know, but this form of the question includes the function of the high pass filter.

What kind of caffeine do you take - 72 Hz? Even a hamster has trouble hitting 10 Hz.

 

My calculator says 220 nf has a capacitive reactance of 10K at 72 Hz. Did I slip a digit? Did I fumble the function?

 

My calculator says 220 nf has a capacitive reactance of 10K at 72 Hz. Did I slip a digit?

No, you were dead on. I just was thinking of 72 beats per minute is a normal human pulse, so I assumed you slipped Hz instead of beats per minute. Then I calculated the cutoff frequency. I got 72.38Hz, like you.

 

Thanks for the replies. So the inverting amplifier with the capacitor are acting as as HPF? and how did you calculate the 72Hz please? is that the cutoff frequency?

Thanks.

 

I assumed you slipped Hz instead of beats per minute.

This is where the, "I don't know" comes in. I can guess that if a normal pulse rate is 72 BPM, then all the frequency components that happen within a beat must be above 72 Hz. Personally, I used to have a resting pulse rate of 55 BPM, so I would design the filter at a lower frequency.

 

how did you calculate the 72Hz

Xc = 1/(2PiFR)
That means the apparent resistance of the capacitor is equal to the 10K resistor at 72 Hz.
The apparent resistance of the capacitor is higher at lower frequencies, so frequencies lower than 72 Hz are attenuated more than 50%.

 

"connected to ECG electrodes" and being used for a traditional ECG test are two different things. We don't know the actual use for the circuit. Heart valve issues can appear in the hundreds of Hz. Still, that does seem like a high corner frequency. Maybe it is intended as a (poor) 60 Hz interference filter.

ak