Author Topic: What if 10nF is cut in this circuit?  (Read 4573 times)

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Online coppercone2

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Re: What if 10nF is cut in this circuit?
« Reply #50 on: August 16, 2023, 04:50:57 am »
lol man get some soldering tweezers or whatever so you don't worry about destroying parts. It looks like a EMI precaution.

But I find it hilarious that people are up in arms over like a 50 cent gamble/addon.. the sky is falling. the thread sounds like a broke ass company is doing a major product release

but I guarantee you, the most important thing you can do is learn to solder that board without worrying about destroying parts, rather then trying to figure out what you can destroy...

you can try heat sinks, heat shields, different tools and techniques in this circuit to actually get better at something
« Last Edit: August 16, 2023, 05:01:59 am by coppercone2 »
 

Offline ataradov

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Re: What if 10nF is cut in this circuit?
« Reply #51 on: August 16, 2023, 04:58:55 am »
I suspect this is the same OP that was obsessing over ADC in some 40 year old MCU. So, obsessing over things that don't matter is a theme here.
Alex
 

Online coppercone2

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Re: What if 10nF is cut in this circuit?
« Reply #52 on: August 16, 2023, 05:02:38 am »
the rant about potentiometers being the end all for MCU is comedy gold.

hi, sensors calling.
 

Offline bonyzTopic starter

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Re: What if 10nF is cut in this circuit?
« Reply #53 on: August 16, 2023, 05:26:22 am »

I'm reading a lot about the Driven Right Leg  (DRL) circuit and electrode. It is not just a simple ground. Somehow the common mode in the body can be inverted and feed back to the body via the DRL electrode. Which of the circuit in the diagram exactly does the Inverting and feeding back? Or maybe it uses some kind of short cut and nothing is inverted?

Reference about the DRL I read at Quora

"The reason this is needed is because the body acts like a radio antenna, producing signal variations much greater than the tiny impulses produced by the cardiac tissue. Unless these larger signals are filtered out, the ECG signal is lost in the noise.

To counter this, the monitor circuit compares the raw signals from the other leads to determine what to filter out and what to pass on. The “noise” signal is then inverted and fed back to the skin via the right leg lead, canceling out the noise component. This turns the riot of noise from external sources into a stable, zero-voltage baseline."

Also I read https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7219345/

"According to the AAMI EC 11 standard, the maximum system noise allowed is 30 µVp-p for an ECG".

Well ECG or EMG has milliVolt level waveforms only much like microphones (they are not EEG which uses 1 to 10 microVolt). So how can 30 microVolt noises affect 1mV to 10 mV waveforms?
 

Offline bonyzTopic starter

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Re: What if 10nF is cut in this circuit?
« Reply #54 on: August 16, 2023, 08:19:29 am »

The following is when the right leg electrode was placed. The waveform noises at the end are due to my clicking the mouse to take the screen shot



The following is when the right leg electrode was removed.



So the inverting circuit and feedback is fully working, or can simple grounding do the trick?

Ok here is what I planned to do. I want to change the bandwidth of the module from 40 Hz to 20,000 Hz.

Then I'll try to create capacitive coupling electric field that vary from 1 to 20,000 Hz that can couple with body (no contact) and see the waveform of the modified EMG conform to the capacitive coupling frequency.

Then final test is by rapidly changing the capacitive coupling frequency, I can transmit music into the body and the circuit display it at the screen and maybe even playing back the music. Is this possible?


Also how do you create capacitive coupling transmitter circuit that vary from 0 to 20,000 Hz and couple it to the body non-contact? Has this been done before?
 

Online wraper

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Re: What if 10nF is cut in this circuit?
« Reply #55 on: August 16, 2023, 10:51:41 am »

I'm reading a lot about the Driven Right Leg  (DRL) circuit and electrode. It is not just a simple ground. Somehow the common mode in the body can be inverted and feed back to the body via the DRL electrode. Which of the circuit in the diagram exactly does the Inverting and feeding back? Or maybe it uses some kind of short cut and nothing is inverted?
Nope, in this particular circuit it just feeds constant 1.5V.
Simplified ECG circuit example in INA321 datasheet (page 14) https://www.ti.com/lit/ds/sbos168d/sbos168d.pdf?ts=1692145737378 and this seems to be heavily bastardized version of that.
« Last Edit: August 16, 2023, 10:57:37 am by wraper »
 

Online wraper

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Re: What if 10nF is cut in this circuit?
« Reply #56 on: August 16, 2023, 11:12:16 am »
I see that V_REF actually a virtual ground, not voltage reference. Still it's a poor design which can only be explained by a poor knowledge of whoever designed that.

How so?  It's buffered (op-amp output), not virtual ground (op-amp input held at GND by action of feedback through a network).
It's buffered 1.5V used thorough of the circuit the same way as if it was GND in a split supply. IMHO it has all signs of the virtual ground.
Quote
If the op-amp buffer were handled properly (damping resistor feeding bypass caps), and the capacitance increased a bit (I'd use one or two 1uF+ ceramic placed very near D1 and D2).

Also if the input signal is very low, D1 and D2 could just be back-to-back tied with VREF.  Instead of clamping positive to VCCA (which needs a local 1uF+ for the same reason).
There is 47uF to GND + a bunch of small caps to GND and VCCA. I meant more like it's not an ADC reference. Virtual ground is still a reference point.
 

Offline bonyzTopic starter

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Re: What if 10nF is cut in this circuit?
« Reply #57 on: August 16, 2023, 04:22:10 pm »

Here is what is so puzzling.

The main purpose of right left drive is to improve the CMRR by filtering mainly powerline 60 Hz noise. The Olimex doesn't have any right leg drive but why does it eliminate powerline 60 completely when you put the right leg electrode (as shown in the output graphs a few message back?

I read this part in physics stack exchange:

https://electronics.stackexchange.com/questions/329476/what-does-this-op-amp-circuit-do-part-of-an-ecg

"What the Driven Leg circuit does is use feedback techniques to measure the common mode voltage, and feed it back through the reference electrode. This effectively reduces the impedance of the connection at the reference electrode by a factor of the gain of the feedback.".

Could the Olimex somehow able to pull the same trick, by maybe also  effectively reduces the impedance of the connection at the reference electrode without using any feedback? Maybe the designer is really genius after all?
 


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