what is Self Adjusting Threshold

[DJSUOL]

In addition, The graph shows the signals from the integrator and the final gain stage of my surf PI circuit. It can be seen that some small spikes occur and why the output is not constant when there is no metal object?

[kt315]

is the SURF PI MINE DESIGN??? NOT. i do especcial trassing WITH CAPS on power lines in all my boards, i also add tantalums on power lines,
i do especial trassing of the power lines. DO YOU HAVE THAT ON YOUR BOARD? so why you ask me about the spikes?
thats problem of YOUR board and the lack of knowledge of whom did your board.

BTW original (WHITES) board design is PERFECT in terms of perfection of power lines trassing. SIC!!!

[DJSUOL]

Could you please show me the output graphs of these two stages?

[kt315]

White's board

[Carl-NC]

Sorry I am a little confused now. For example, in the surf PI design, the DC voltage from the integrator is directly fed into a high-pass filter, why the DC signal is not blocked by it? Is it because that the output of integrator is still an AC signal and the time constant for the high-pass filter is larger than the integrator? In addition, when a target is present, the output of integrator will increase so the change could be fed to the final gain stage. If the metal detector keeps detecting the same target, the output from the integrator will remain steady so the voltage will become the value before. Am I right? Thanks for your explanation.

For a target, the integrator produces a fairly fast-changing signal that goes right through the SAT stage (high-pass filter). But very slow-changing signals -- like circuit drift -- are removed by the SAT circuit. If you hold the coil over a target for too long, the SAT stage will tune it out.

[DJSUOL]

Why the integrator produces a fast-changing signal rather than a DC voltage? For example, if the time constant for integrator is 47ms, during this time the output keeps increasing. But what happens when after 47ms?

[Carl-NC]

I think you are confusing the time constant with the sample time. During the sample time (input switch is closed) the integrator is driven by the preamp output. If you are approaching a target, then the integrator output will increase. When the sample switch is turned off, the integrator voltage is held at a DC value, except for some small droop due to the feedback R.

[DJSUOL]

Thanks. Could you please tell me if my measured signal is correct? I am not sure why negative spikes occur.

[Carl-NC]

Probably due to noise on the supplies. PI designs tend to have a lot of clock noise running around. As long as it stays in place, it should be OK.

[DJSUOL]

So what should be the output of the integrator and SAT? a DC voltage?

[Ferric Toes]

So what should be the output of the integrator and SAT? a DC voltage?

Typically, the SAT stage has 0.47uF as a d.c. blocking capacitor and a resistor to ground of between 1Megohm and 220K. Ideally the integrator output should be zero in the absence of any signal, but it rarely is. The reason is that there is usually some d.c. offset either from the preamp or from some other small imbalances in the circuit components. One dominant source of imbalance is the fact that during the recovery of the preamp, following the cessation of the TX pulse, there is often a small ramp in the flatness which can last for a few hundred uS. There can be numerous causes of this but it is not something to be overly concerned about and it is hard to see on a scope, but when sampled and integrated, appears as an offset at the integrator output and which can be seen on a scope. The SAT stage also usually has some gain like 50 or even 100 times in some of my circuits, so that any noise, i.e. preamp noise or EMI noise picked up by the coil, will be superimposed on the d.c. level from the integrator. This will be passed by the relatively long time constant of the RC circuit and appear as a 'flutter' on the output of the following gain stage.

When a metal object passes the coil, the output of the integrator rises rapidly and the SAT capacitor transmits virtually the same voltage which then appears on the opamp input, and amplification follows. If there was no resistor and the input resistance of the opamp infinite, plus no circuit board or capacitor leakage, the result would be as if there were no capacitor in the signal path. However, that would be no good for our purposes as what is needed is a relatively slow reset and this is what the resistor does; it leaks the charge on the amplifier side of the capacitor back to zero when the integrator output ceases to change. That is why a stationary object over the coil will 'tune' out. 'Tuning' is a misnomer as no tuning in the accepted sense occurs. Zeroing out or auto-zero would be more accurate.

Any superimposed noise on the integrator output will appear at the output of the SAT amplifier, which only blocks the dc component. Clock spikes, inverter noise, EMI etc. all have to be dealt with in other ways. It's a big subject.

Eric

[DJSUOL]

When the sampling switch is on, the signals enter the inputs of the integrator will be integrated and averaged to a certain DC value and at this time the changing signal could feed into SAT to produce a output. If the switch is off, the output of integrator will become nearly 0. Am I right? But according to the content in the ITMD, I think the integrator is used to average a period of samples to eliminate the random noise which I am confused about the real working principle of the integrator now. So why we could not detect the fast changing output from the integrator using an oscilloscope? I really appreciate your help and your time.

[waltr]

I have stated to you several times to Search and READ the threads in the forum.

What you are looking at is NOT Simple even though the circuits seem simple.
There is a lot of dynamic action in the Integrator and SAT stages. And since this is a Sampling
Integrator do google searches and read about 'box car' integrator and Lock-in amps. The simple op-amp integrator is barely applicable here.

A few Forum threads (from Search I did) to get you started:
https://www.geotech1.com/forums/show...ght=integrator

Carl has posted a good explanation of how the integrator in a PI works. Can't find the post so I am re-posting:

The purpose of the integrator is to constructively add successive positive samples (resulting from a target) while
averaging out everything else. That is, random noise (which is both positive and negative) will average to zero. So the
integrator needs to have a charge rate that is fast enough (compared to the sample rate) to build up a signal, and a
discharge rate that will get rid of the signal fairly quickly once the target is gone, so you don't get an audio response
long past the target.

So let's make up some numbers. Sample rate is 1kHz, so sample period is 1ms. Sample width is 10us or 1%. Let's say a
weak target signal from the preamp is 10mV. A single sweep of the coil takes 1 second.

We get 1000 pulses per swing, so as we go over a target we might get 10-20 positive samples. Let's say exactly 10.
Without an integrator, this would result in 10 samples that are 10mV high with a 1% duty cycle.
This is a time-averaged signal of 100uV and probably will never trigger the audio.

Let's say we use a single-ended integrator shown below; R1=1k, R2=100k, C1=0.1uF. When the switch is closed vin (=10mV)
is converted to a 10uA current (i1) by R1 and charges the cap. dv/dt = i/C1 which gives 10uA/0.1uF = 100V/s, so in 10us the
cap charges up by a whopping 1mV. Wow, a whole millivolt? Yup.

When the switch opens the integrator is left with R2 discharging C1, at a time constant t = R2*C1 = 10ms. We have 990us
of discharge time (sample switch open) so the signal drops to e^(-990us/10ms) = 90% of whatever the sampled value is. So
when the switch is closed, the cap charges by 1mV, and when the switch is open the cap discharges by 10%.

We assumed a 10mV signal for 10 samples, so at the end of each of the 10 samples we have:

1: 1.00mV
2: 1.90mV
3: 2.71mV
4: 3.44mV
5: 4.10mV
6: 4.68mV
7: 5.22mV
8: 5.69mV
9: 6.12mV
10: 6.51mV

6.5mV doesn't sound like much, but it's 65 times better than the average 100uV without the integrator. And the integrator
is usually followed by a gain stage so if it has a gain of 100 (e.g., HH) then the final voltage applied to the audio is 650mV.
That's much better.

Finally, when there is no more target signal, the integrator cap will discharge in about 3t (95%) so a time constant of
10ms means the integrator output dies off in about 30ms.

- Carl

[Carl-NC]

When the sampling switch is on, the signals enter the inputs of the integrator will be integrated and averaged to a certain DC value and at this time the changing signal could feed into SAT to produce a output. If the switch is off, the output of integrator will become nearly 0. Am I right?

No, when the switch turns off the voltage across the cap remains, then slowly discharges through the feedback R. If you remove the feedback R, the voltage would just stay there, as a DC value. The R is needed for drift stability.

[DJSUOL]

So if there is no target, the output of integrator also varies due to the charging and discharging of the circuit? However, if this is ture, why the measured output at integrator is almost a DC signal?