Dir Dobromir Dobrev

Welcome!

I would like to greet the respected colleagues Lutz and Nicolay.

I have some comments on the the circuit above!

What is the probe internal resistance?

In biomedical probes the internal impedance is very large may be much larger than the potential divider in the input. The divider in the input limits the input impedance of the instrumentation amplifiers.

The instrumentation amplifiers have the capability to cancel the common mode voltages. So, one can directly connect the electrodes to the input of the instrumental amplifiers. The active electrode and the reference electrode.

If frequency filtration is required it can be accomplished as the last step.

This circuit has components that do not add to the functions of the circuit:R3 and R4 together with the UA1,A.

One do not need the differential amplifier configuration. Single ended amplifier with noninverting input is sufficient after the input divider. The input is single ended after the divider.

Best wishes

Hello Dobromir D., interesting question. However, I must admit that at the moment I am a bit short in time (and perhaps also a bit too lazy?) for analyzing the whole circuit.

But one comment: You are asking for pros and cons. Don`t you think that a detailed answer (pros, cons,) can be given in comparison to another circuitry only?

Hello Lutz,

The other circuit is an usual AC coupling, made from two simple high-pass RC filters. Suppose in this schematic that R3, R4 and U1A are omitted, and the midpoint of R1 and R2 is connected to the signal ground AGND.

The circuit reminds classical ECG 'right leg' or 'common mode' holder active conditioning circuit ( above, as taken from https://www.egr.msu.edu/classes/ece480/capstone/spring13/group03/documents/ElectrocardiographyCircuitDesign.pdf ).

However, I don't understand what benefits it may give in the configuration given.

Hello Nikolay,

It is AC coupled amplifier without reference electrode.

Advantages: none

Disadvantages: ??

One joker more:

Assuming no input signals, what is the bias point (DC potential) of U1A output?

Dir Dobromir Dobrev

Welcome!

I would like to greet the respected colleagues Lutz and Nicolay.

I have some comments on the the circuit above!

What is the probe internal resistance?

In biomedical probes the internal impedance is very large may be much larger than the potential divider in the input. The divider in the input limits the input impedance of the instrumentation amplifiers.

The instrumentation amplifiers have the capability to cancel the common mode voltages. So, one can directly connect the electrodes to the input of the instrumental amplifiers. The active electrode and the reference electrode.

If frequency filtration is required it can be accomplished as the last step.

This circuit has components that do not add to the functions of the circuit:R3 and R4 together with the UA1,A.

One do not need the differential amplifier configuration. Single ended amplifier with noninverting input is sufficient after the input divider. The input is single ended after the divider.

Best wishes

Dear Abdelhalim abdelnaby Zekry,

Thank you for your answer.

I like your statement:

"This circuit has components that do not add to the functions of the circuit: R3 and R4 together with the U1A."

I would add that these components make the circuit unable to work.

Let's consider the static behavior (DC operating point) of the stage with R1, R2, R3, R4 and U1A, assuming that the U1A input offset voltage is 1mV.

What will be the bias voltage (DC potential) at the U1A output?

Greetings

Dear Emad Alnasser,

Thanks for your reply.

Simply we can analyze the circuit as follows.

The resistors R1, R2, R3, R4 can be replaced by one resistor of 3.8Meg connected between U1A output and its noninverting input.

Now the opamp input offset voltage appears as a differential input overvoltage.

The opamp amplifies this overvoltage and depending on its sign, the output is saturated to one of the supply rails.

Now, what is the dynamic behavior in case of power-line interference?

Dear Emad Alnasser,

Your sim circuit is not the real case.

The opamp inputs do not have a physical connection to the signal ground.

The opamp inputs do not know which potential is signal ground.

The opamp inputs are connected to the input differential pair/pairs and to the input ESD protection.

The ESD protection generates leakage current towards one of the supply rails.

In your sim, connect both resistors R7 and R5 to VCC, after that to VEE, next remove them, and say the results.

Rgds

Dear Emad,

The circuit has only two stable points, and these points are the supply rails!

Let's discuss dynamic behavior. It's very simple.

We have two 2Meg resistors connected to the opamp output.

If Vio is 1mV. The power-line interference current greater than 1mV/2Meg=0.5nA per input, or 1nA per both inputs, will cause the opamp output to oscillate between the supply rails. It will alternate the supply rails with a power-line frequency.

This switching will result in a disastrous CMRR, distort the useful signal, and compromise the performance of the inamp stage.

Cheers

Dear Emad,

Can you tell one advantage of this circuit?

Rgds

Dear Emad,

As I said before, the circuit has only two stable points, and these points are the supply rails. Due to implemented positive feedback, it tends to be latched to one of them. Depending on circuit parasitics, e.g. Vio, ESD leakage currents, input bias current of diff. pair (JFET), etc., the opamp output could be pulled somewhere in the supply range, but if the temperature changes, it again will go towards one supply rail, because only these points have stable DC potential.

I never said that the circuit is a real oscillator. It is not an oscillator. It will oscillate due to external interference, in our case power-line interference.

This link could be also interesting to you:

https://www.researchgate.net/post/This_circuit_claims_to_be_bootstrapped_with_Rd500M_Rcm135k_Is_it_true

Regards

Dear Dobromir, I think there is a drawing error. Capacitors C1 and C2 must be moved horizontally to be placed between the bars R1-R2 and R3-R4. Anyway, I don't understand the purpose of the circuit...

Regards,

Enrique.