What does this OpAmp circuit do?

[mhsnrah]

Hi,

I have a board to measure resistance of a dental canal.
I want to know what's the exact job of elements in this circuit.

The PCB:
**broken link removed**

Schematic from PCB:
**broken link removed**

My idea about the above SCH:
I think U2A(upper section) is generating a non-DC signal and feeding it to the JL(Lead/Probe) and the lower part is converting the AC to DC with suitable amplification.

So I rearranged the SCH as follow:
**broken link removed**

Now what do you think about this circuit?
Please share your ideas with me, I know the analog men will solve this.

Very thanks in advance.

 

[ericgibbs]

hi,
The circuits will not work in the same way.

Original: U2C is used to change the polarity of U2B output when Jl signal reaches a certain level.

Copy: U1C will not give the same action on U1B.

EDIT:
I will recheck the two circuits,,, a little doubt about my above statement.

I have downloaded your circuits so that I can read them correctly,,, so what do you want to know.???

 

[mhsnrah]

Dear ericgibbs,

I want to know how does it measure the impedance or resistance of the canal to make a more accurate one.

Oh, I forgot to say that pin 14 (output of last op-amp) is the outputted signal to the rest of the circuit (although it's obvious).

 

[Warpspeed]

Dear ericgibbs,

I want to know how does it measure the impedance or resistance of the canal to make a more accurate one.

Oh, I forgot to say that pin 14 (output of last op-amp) is the outputted signal to the rest of the circuit (although it's obvious).

U2A is a square wave oscillator to generate a continuous voltage waveform train.
C1 gives equal positive and negative voltage swings, and drives a small ac current (set by R4 and P1 into what you wish to measure the resistance of.
U2B and U2C form a precision rectifier which only passes the positive half cycles.
C1 charges up to about half of the pulsing dc voltage coming from U2B
U2D is a buffer and amplifier to isolate C1 from the output.

This is just an ohm meter that operates on alternating current instead of dc current.
This must be done, because feeding a dc current into many complex chemical things, (including living tissue) can create all kinds polarization and electro chemical changes that can influence what you are measuring. So ac is a necessity.

 

[ericgibbs]

U2A is a square wave oscillator to generate a continuous voltage waveform train.
C1 gives equal positive and negative voltage swings, and drives a small ac current (set by R4 and P1 into what you wish to measure the resistance of.
U2B and U2C form a precision rectifier which only passes the positive half cycles.
C1 charges up to about half of the pulsing dc voltage coming from U2B
U2D is a buffer and amplifier to isolate C1 from the output.

This is just an ohm meter that operates on alternating current instead of dc current.
This must be done, because feeding a dc current into many complex chemical things, (including living tissue) can create all kinds polarization and electro chemical changes that can influence what you are measuring. So ac is a necessity.

Hi warpspeed,
I appreciate your post, but I know exactly how that circuit works, I posted an almost identical one a few weeks ago for another OP

Perhaps it would be better if you posted your explanation to the OP.

My initial mistake was to misread the OP's version of the circuit.

E.

 

[Warpspeed]

Perhaps it would be better if you posted your explanation to the OP.

I though that is what I did ???

 

[mhsnrah]

Thank you Warpspeed.

I think your response is totally true, especially about "a small ac current".
And what's your advise for increasing the accuracy and lowering the noise and ripples of final voltage?

I appreciate any suggestion.

Hi warpspeed,
I appreciate your post, but I know exactly how that circuit works, I posted an almost identical one a few weeks ago for another OP

Perhaps it would be better if you posted your explanation to the OP.

My initial mistake was to misread the OP's version of the circuit.

E.

Would you please put the link of that post here?

 

[Warpspeed]

Thank you Warpspeed.

I think your response is totally true, especially about "a small ac current".
And what's your advise for increasing the accuracy and lowering the noise and ripples of final voltage?

It may not be possible or desirable to increase the ac energizing current, especially with direct biomedical measurements.

The main problem with this circuit is as you say, there will be extraneous noise voltages which will simply be added to the measured ac output voltage, introducing fluctuations in the output.
This is a fairly common problem with many low level biomedical measurements.

There are several defenses against noise with this type of circuit.

Probably the single worst enemy will be noise voltages related to radiated mains frequency power, and it's harmonics.

So what could help might be some frequency discrimination or reasonably narrow band filtering, and to make sure your ac energizing frequency is well removed from the mains frequency.

Another more complex scheme might be to sample the signal voltage through a very narrow time window coincident with the peak of each cycle, and use sample and hold techniques to measure right at the voltage peaks. With a fairly long integration time, any residual random noise should average out to zero, while the signal frequency is very strongly reinforced, because the sampling points will be exactly synchronous with the peaks of the desired signal.

There are quite a few ways to improve the circuit, but they all involve adding some complexity.

 

[ericgibbs]

.
Would you please put the link of that post here?

hi,
I can only go back to the last 200 posts, [early Oct 2011] the Thread I referred too I posted 16 May2011.

This image is what that OP was asking about.

 

[ericgibbs]

Dear ericgibbs,

I want to know how does it measure the impedance or resistance of the canal to make a more accurate one.

Oh, I forgot to say that pin 14 (output of last op-amp) is the outputted signal to the rest of the circuit (although it's obvious).

hi,
This is an LTspice simulation of your circuit.

Do you have any details of the probe and how the output voltage is displayed.??

 

[mhsnrah]

Thank dear friends,

I got similar simulation result for oscillator(rectangular pulse with about 1kHz frequency) but the output result is different from yours, it contains a DC part around 800mV and a non-desired ~250mVp-p ripple. That's because you disregard the cap of last stage
Power supply voltage is +5V but it seems in your simulation it is +10V.

In reality when I short circuit the probe, I get about 3V and with 10k+ or open loop, I get 0V.
In simulation I used 10k for unknown impedance.

The whole device contains another board with 2 7-segments and some LEDs.
It connected to the analog board with 10 wires(the connection is recognizable at the bottom of attached PCB image).
The analog board has 3 LM324 and the display board has two CA3161 BCD to 7-seg driver.
So I think the OpAmps of analog board converts the unknown impedance to BCD.

The probe is simple as wire.
I don't have a dental maquette so I connect the probe to a trimmer.

The device is built to help dentist to determine the apical area of tooth.

Ok,
Replacing the 1uF with 47uF in simulation has a very nice effect on removing ripples, but in action it doesn't have so much effect.
So I want to use an active LPF to remove undesired ripple and noise.
**broken link removed**
With R1=47kΩ, R2=100kΩ, and C2=100nF the cut-off frequency is 1/{(2π)(R2)(C1)}≈0.16Hz so I think it'll remove noise and ripples very well.
Am I right?

 

[ericgibbs]

hi,
Your drawing did not show the Vcc value, so I used 14V,
I missed the 1uF smoothing cap, adding it I get the same as your Sim.

Why don't you add that integrator circuit onto the output of your circuit and simulate it.??? also do an AC Analysis, that will show the cut off frequency

I would also choose a different OPA, with a lower noise figure.

EDIT:
I guess you know that you need to use a non inverting version of your posted integrator.??

 

[Warpspeed]

Integrating the output will only stop the display figures from jumping around, it will not reduce measurement errors introduced by a noise voltage.

In other words, if the noise voltage is the same amplitude as the wanted signal, the output will read maybe twice the amplitude it should, even after smoothing with an output integrator.

I cannot help feeling we really need to filter out the signal from the noise before rectification and integration.

A low pass filter at the signal input may not be that effective, because induced 50/60Hz mains noise is probably the main problem. Both the patient and this magic box are both fully floating, the impedances are quite high, and the whole thing ripe for noise pickup.

A bandpass filter centered around the oscillator frequency or sampling techniques may be more effective.