MPP Sampling Integrator Simulation

[Qiaozhi]

For qurashimoin, and anyone else who is interested.
This is an LTSpice simulation of the sampling integrator used in the MPP.

Note to qurashimoin:
This is all I have in the way of LTSpice simulation files for this design. However, as a final year engineering student, you should be quite capable of simulating the rest of the design if necessary. Please don't ask for further information, as I'm not going to do your final year project for you. Everything is clearly described in the Build Document.

[qurashimoin]

Thank you sir

[6666]

Thanks George, just found this sim, very handy.

[green]

For qurashimoin, and anyone else who is interested.
This is an LTSpice simulation of the sampling integrator used in the MPP.

Note to qurashimoin:
This is all I have in the way of LTSpice simulation files for this design. However, as a final year engineering student, you should be quite capable of simulating the rest of the design if necessary. Please don't ask for further information, as I'm not going to do your final year project for you. Everything is clearly described in the Build Document.

https://www.geotech1.com/forums/atta...0&d=1484661372

For a long time I thought increasing sample time to GB wouldn't work with the MPP. Members were posting they had success and I convinced myself I was wrong. Was playing with Qiaozhi"s simulation and didn't get the gain to double when I doubled sample time. What am I doing wrong?

[waltr]

The integrators are NOT Linear. They also have a Maximum V out regardless of intgration time periods.

You are looking at this wrong.

Instead of looking at the 'final' output V look to the output Voltage at say 20msec after a change of the input Voltage.
Your simulations show a huge increase of output Voltage with the longer integration time at 20ms, ~.25V for 116us verse -2.75V for 58us.

Also remember how long a Target is under the coil when sweeping the coil across the ground. For a 1 meter/second sweep rate and a coil sensitive size of 2cm, the detector has 20ms max to respond to the target.

However, you also stated "increasing sample time to GB".
What do you mean?

I and others have done GB with a simple two input integrator. This is done by increasing ONLY the second Sample then adding a third sample just before the TX pulse the 'even' out the difference and adding EFE.
But this is not within the title of this thread.

[Qiaozhi]

I think you're incorrectly expecting the output voltage of the sampling integrator is going to increase in amplitude with increasing sample pulse width. This is not the case. Run the attached simulation, and note that the sampling integrator has a faster response with increasing sample pulse width. This allows it to respond to weaker targets. Remember also that the sampling stage is followed by a self-adjusting threshold and a further integrator stage.
The simulation has been modified to use 10us, 20us, 60us and 120us sample pulse widths for comparison, and the input signal has been reduced to 1mV.
The simulation results also shows VOUT has a faster rise time with longer sample pulses. Note that the apparent increase in gain (sensitivity) is not linear.

[waltr]

Good response Qiaozhi.
This is what I was getting at.

Also the original simulation is really only showing the 'start up' settling of the integrator.

Another way to run the simulation is add a 'pulse' to the input that is delay by a 500 msec. Make the pulse 1mV and 10ms to see how the integrator responds to a target sweep under the coil.
Now change the integration periods as in the Q3 simulation.

[green]

Good response Qiaozhi.
This is what I was getting at.

Also the original simulation is really only showing the 'start up' settling of the integrator.

Another way to run the simulation is add a 'pulse' to the input that is delay by a 500 msec. Make the pulse 1mV and 10ms to see how the integrator responds to a target sweep under the coil.
Now change the integration periods as in the Q3 simulation.

I think the original simulation shows the response and gain of the integrator. Reason I changed the input to include zero volts. Gain is output/input. Step response changes with sample time, gain doesn't. Not seeing how the integrator can cancel a ground signal by having a different target sample time and ground sample time. Think I remember making a MPP simulation where changing second sample time would GB but don't remember how.

Changed the simulation to 1C integrator. GB has a hole so I adjusted target TC and second sample time to find the hole. Is there a way to simulate the MPP integrator to have a hole?

[6666]

I am following this with great interest thanks.

[green]

Found the simulation that I thought GB. Replaced the target and ground signal with the decaying target to see if there was a hole. Low signal but when I reduced the input TC in in half still a low signal compared to the 1C integrator. Don't know if the real circuit can GB but so far I can't get the simulation to GB. Including the simulation with target, ground, target EF and ground EF samples. Maybe someone can see where I went wrong with the simulations. When I first tried I thought adding the EF samples solved the problem, now I don't think so.

[green]

Still trying to understand the MPP integrator. Made a 1C integrator almost same schematic for comparison. 1C first stage gain=Rfdbk/Rin*sample time*sample rate, second stage gain=Rfdbk/Rin, total gain=5.6. Thought MPP gain was Rfdbk/Rin second stage, 5.6, appears to be 2.6. Anyone know formula for MPP integrator gain? Simulations with ground sample and ground EF sample off, target and target EF only. Anyone see any errors in the simulations?

[waltr]

There is NO "Gain" as a simple equation. This is an Integrator which changes the Cap through the input series resistor and the Cap is discharged through the feed-back resistor. The dominate components are the input series R and the feed-back Cap.

Carl posted a explanation some years ago on how an integrator in PI circuits work. I can not find a copy but have the details.

Assue:
R1 = input series R
R2 = Feed-back R
C1 = feed-back C
Vi = input Voltage
t = sample time
p = pulse period

Then When sample switch is ON:
R1 passes a current: i = Vi/R1
This charges C1 as dv/dt = i/C1 (V/sec)

When the sampling switch is OFF:
C1 is discharged through R2. The discharge Tau = R2*C1
The discharge time = pulse period - sample time
The Vdrop per period = e^(-discharge time/discharge Tau)

Also, the sampling Duty cycle = sample time/pulse period


Now we calculate the integrator output Voltage per pulse period. We must do it this way since the Cap's discharge rate is dependent on Voltage stored.

[Altra]

Carl posted a explanation some years ago on how an integrator in PI circuits work. I can not find a copy but have the details.

I think this is what you are looking for

https://www.geotech1.com/forums/show...lculating+gain

[green]

Thanks for the replies, I was asking for MPP integrator gain. reply #11, doesn't appear to be the same as 1C or 2C integrator

1C first stage gain=Rfdbk/Rin*sample time*sample rate, second stage gain=Rfdbk/Rin, total gain=5.6.

1C integrator gain same as Carl's formula? Calculates same as simulation for 1C integrator. reply#11

[green]

Thanks for the replies, I was asking for MPP integrator gain. reply #11, doesn't appear to be the same as 1C or 2C integrator

1C first stage gain=Rfdbk/Rin*sample time*sample rate, second stage gain=Rfdbk/Rin, total gain=5.6.

1C integrator gain same as Carl's formula? Calculates same as simulation for 1C integrator. reply#11

Think MPP integrator gain formula is (R3/R1)*((R5/(R4+R5)) https://www.geotech1.com/forums/atta...8&d=1574962175 MPP is left side, 1C integrator is right side