Pulse meter circuit

[MikeK]

I saw this online.  Yeah, I know it's not a pulse oximeter, since it's not measuring a red/IR differential.  It just reads a person's pulse.

But what is the input cap for?  The phototransistor at the bottom is what feeds the opamp.  But the output from the phototransistor is always between GND and Vcc.  So why the need for an input capacitor?



Here's the full size version, if you can't read that: http://www.swharden.com/blog/images/2013/04/lm324-opamp-pulse-oximeter-ecg-schematic.jpg

[4to20Milliamps]

It must have to do with isolating the input from the virtual ground:

http://www-k.ext.ti.com/SRVS/Data/ti/KnowledgeBases/analog/document/faqs/ssexpert.htm

[w2aew]

It's most likely there because the photo transistor is not being switching fully on and off. It is likely giving a very small "wiggle" in response to the small change in transmitted light during a heartbeat. The first op amp stage is configured for a lot of gain with that large feedback resistor. The AC coupling allows the feedback to establish a linear operation point for the first stage to amplify the small blips from the phototransistor.

[codeboy2k]

while I kinda agree that it's probably needed to isolate it from the virtual ground, that's not the only reason.  For example, if you removed the virtual ground and powered the circuit with a bipolar supply, you would still need the AC coupling capacitor... for the reasons stated by w2aew. 

The ambient light is going to bias the transistor on in a small linear region.  w2aew described it just about perfectly, it will be like little wiggles on the DC level, probably just a few microvolts, and the AC input capacitor will remove this BIAS caused by the LED and allow the signal to be amplified.

[MikeK]

Great, thanks.  I had trouble simulating it.  The output of the first stage kept coming out a square wave.  Time to get some parts and do a real demo.  I guess the simulators didn't like the 1.8M feedback resistor.

[MikeK]

I think I understand now.  So, since the phototransistor has a small input its output operates close to the positive supply rail.  The cap is used to pull that tiny output signal down.  And since the non-inverting input is set to Vpos/2, that signal will wiggle around that point.

Thanks.

[w2aew]

Great, thanks.  I had trouble simulating it.  The output of the first stage kept coming out a square wave.  Time to get some parts and do a real demo.  I guess the simulators didn't like the 1.8M feedback resistor.

It sounds like your simulation was making the assumption that the phototransistor was being switched on and off.  The reality is that, as described above, it is going to operate in the linear region, with only a very small variation due to the blood flow.

[MikeK]

Thanks, Alan.

Yeah, that was it.  The sine input I was using was too large.