Is a resonant PI possible?

[Unregistered]

JC1

Hi Kev,

OK you made me look. 4148s should be easy to get but Kev is right the 4448 is half the cap 2pf. So if you're worried about stress maybe parallel about five of these.

10pf

[Kev.]

Some capacitance needed

Hi Tinkerer,
I sure do hope you've got something there, it'll be a while before I can investigate it for myself. 600 to 800 Ohms damping seems good to me, and I've heard it said that one needs a few extra tens of puffs capacitance to get the perfect knee....so you may have a good combination...I'm wondering about the coil shielding, is it complete?? like what you'd use in the field? I've seen a really fast frontend that turned out to be all due to poor shielding, soon as I touched a bit of grass off she'd sound.

I don't fully appreciate just what a resonant frontend would mean? My guess is that it can be more tuned to the target of interest i.e. small nuggets.

Is this the post you mentioned in your other post regarding preamps?
http://www.findmall.com/read.php?34,...370#msg-129370
You don't have to answer that if you don't want to of course.
But if it is, then one can easily see that if the amp isn't looking down the cable at the coil, then you've done away with a whole lot of crud that's going to mess up the reception. With surface mount stuff, like 0402., etc you should be able to embed it all in the lower shaft, or even the coil. Problems getting NE5534 SOICs though. Maybe I'm off in the weeds, still comprende Carl?

Thanks for sharing your findings Tinkerer.
Cheers
Kev.

[Ferric Toes]

This is the setup. More tomorrow Tinkerer

Hi Tinkerer,

What you have is a basic logarithmic amplifier, with some slight modifications. The V/I relationship of the transistor base/emitter junction is the controlling element which gives non-linear feedback. In this simple arrangement it is temperature sensitive, and in practical circuits, temperature compensation has to be added. Normally there is no feedback resistor, so with no signal you will either get a lot of noise, or due to offsets, the amplifier will saturate in one direction or another. Practically you can't have a log of 0 so you have to have some signal or a teeny bit of offset. However, you have a 2.2M feedback resistor which limits the maximum gain and makes offset adjustment easier. I presume you have an offset adjusting preset?


What shape do you get for the decay of a ferrous, and a non-ferrous target? Rings and coins should show a substantially linear decay, while ferrous should still show considerably curved.

Eric.

[Tinkerer]

preamp

Hi everybody, Attached two scope pix, the first one set at 1 uS division, the other one set at 20 uS div. As you see, the pulse repetition rate is about 10Khz. I can crank up the power by increasing the Tx to about 60uS, then I use a repetition rate of about 6K hz. The preamp as shown works fine anywhere between. The coil has about 310 uH inductance, used with a 10 Ohm resistor in series, still the same coil as used in earlier posts, but now the coil is fully shielded and potted in the form, with the one difference being, I pinched the coil into an oval of 6.5 x 13.5 inches. The coil is connected to the breadboard by a shielded 2 conductor cable of 7 ft. I don’t know how to calculate the actual gain of the preamp, but it is high. The results are repeatable. Thanks, JC1, for the good idea of putting 5 4448 in parallel. I am sure this will help a lot. Thanks also Kev for your input, but no, for the time being the setup is still as described above. Very good post, from Eric. Tinkerer

[Tinkerer]

1uS div. preamp

Hi Tinkerer,

What you have is a basic logarithmic amplifier, with some slight modifications. The V/I relationship of the transistor base/emitter junction is the controlling element which gives non-linear feedback. In this simple arrangement it is temperature sensitive, and in practical circuits, temperature compensation has to be added. Normally there is no feedback resistor, so with no signal you will either get a lot of noise, or due to offsets, the amplifier will saturate in one direction or another. Practically you can't have a log of 0 so you have to have some signal or a teeny bit of offset. However, you have a 2.2M feedback resistor which limits the maximum gain and makes offset adjustment easier. I presume you have an offset adjusting preset?


What shape do you get for the decay of a ferrous, and a non-ferrous target? Rings and coins should show a substantially linear decay, while ferrous should still show considerably curved.

Eric.

Thanks Eric, The idea of a log amplifier came from one of your posts. I have not observed the differences of different metals yet, but sweeping my test coin set across the coil I did note big differences in the decay curve, that were not linear with the size of coin. Unfortunately I have no information of what the coins are made of, an example being the US$ 1 cent that is copperclad zinc or something of the kind. Tinkerer

[Tinkerer]

Question to Eric

I will refine the preamp, adding a temperature compensation. Here is a question: Now that I have learned how to make a high gain fast settling front end, why shuld I not make a 5oo uH coil? I think I could bring it in with under 10uS? Tinkerer

[Unregistered]

JC1

Hi Tinker,

Logarithmic Amplifier? My bad, without knowing which way your signal was going, I thought that was some kind of clamp to prevent saturation. Where is the ground (0 volt) on your signal trace? (turn on the other channel and set at 0), this output must be negative.

MHO

[Unregistered]

JC1

Hi Eric,

It seems to me that without the feedback resistor the op amp is free to seek the point at which the transistor just comes on somewhere around -0.5 volts on the output. Once the feedback resistor is added (yes I understand the super high gain you get without it) the op amp will drop to roughly zero and the transistor is off. Now some form of offset is required to get back to -0.5 volts and that is fine, but how do you now manually find that sweet spot where the xsistor starts to come on? measure where the gain just begins to drop? Or just crank in a little more to make sure?

And of course the resistor introduces a bit of non-logarithimic (is that a word?) curve to the whole thing.

[Tinkerer]

Where is 0

Hi Eric,

It seems to me that without the feedback resistor the op amp is free to seek the point at which the transistor just comes on somewhere around -0.5 volts on the output. Once the feedback resistor is added (yes I understand the super high gain you get without it) the op amp will drop to roughly zero and the transistor is off. Now some form of offset is required to get back to -0.5 volts and that is fine, but how do you now manually find that sweet spot where the xsistor starts to come on? measure where the gain just begins to drop? Or just crank in a little more to make sure?

And of course the resistor introduces a bit of non-logarithimic (is that a word?) curve to the whole thing.

Thanks for the input, JC1, I adjust the offset with the adjustment potentiometer to reach 0 at the end of the decay curve just before the next cycle. The pot is not shown on the circuit because it is standard for offset adjustment like the power supply (+/-5V is not shown either. I have to point out that this amplifier is not the only merit of the frontend, it is the combination of fast coil with the amplifier that is the importand part. To get this high gain with a short first sample delay. Tinkerer

[Unregistered]

JC1

Hi Tinkerer,

er, whoops there one for last time.

You will not get any log action until the output is

-0.5 to -10.0 volts. The rest is at the gain I said before.

Here's your penny.

and my two cents.

http://en.wikipedia.org/wiki/Cent_(United_States_coin)

http://www.usmint.gov/about_the_mint...ion=fun_facts2

[Unregistered]

JC1

Hi Tinkerer,

Whoa! we are posting in real time, you got that one in above me. doesn't matter still applies.

0 volt at the end, then about that at the beginning.

or at least not -0.5 volts.

looking at your signals the presence of metal will produce a negative going signal, but unless very big very close will not produce a half volt signal.

Not seeing any log action yet?

[Unregistered]

JC1

Oh by the way, your damping looks just fine.

My bad, again!

I assume by resonant you mean the coil inductance and capacitance form a resonant circuit at what? 10KHz.

Interesting, remember a talk about that somewhere once.

[Ferric Toes]

Hi JC1,

I have slipped up here, as your post made me think a bit more about what is happening. The amp is wired as a non-inverting type, so if connected to a TX using an N channel mosfet, the amplified signal is opposite to what we normally see. i.e the TXon saturation is negative going and the recovery period and signal, positive going. The way the transistor is connected, it only conducts on the non-signal negative saturation. For the signal, the transistor is turned off and the opamp simply amplifies 2200 times, linearly. To be a log amplifier, the signal input would have to be applied to the inverting input, or the transistor changed for a pnp type.

Eric.

[Unregistered]

JC1

Hi Eric,

I don't know about you, but I am getting

2old2think

I've been screwing up alot lately.

Is this supposed to happen?

Have a good day!

Tinkerer is pretty smart, he'll get it working.

[Unregistered]

JC1

Word

From Maxim,

The Classic DC Logarithmic Amplifier
In the classic pn-junction-based implementation of the DC log amp, a bipolar transistor is used to generate the logarithmic I-to-V relationship. As shown in Figure 1, bipolar junction transistors (BJTs) are placed in the feedback path of an operational amplifier. Depending on the type of transistor chosen, npn or pnp, the log amp is either a current-sinking or current-sourcing circuit, respectively (Figures 1a and 1b). Through negative feedback, the op amp places enough output voltage on the base-emitter junction of the BJT to ensure that all available input current is drawn through the collector of the device. Note that a floating-diode implementation causes the op-amp output voltage to include input-referred offset; the grounded-base implementation does not possess this problem.

With the addition of an input series resistor, the DC log amp can also function as a voltage-input device. Input voltages are converted to a proportional current though the resistor, using the op amp's virtual ground as the reference. Clearly, op-amp input-referred offset must be minimized so that accurate voltage-to-current conversion can be achieved. The bipolar-transistor approach is prone to temperature variations but, as will be discussed, this sensitivity is drastically reduced by using a reference current and on-chip temperature compensation.

Single-supply operation is a new improvement appearing in some modern-day DC log amps, making them desirable for use with single-supply ADCs/systems. The MAX4206 can operate from either a single +2.7V to +11V supply or a dual ±2.7 to ±5.5V supply. A consequence of single-supply operation is that these log amps generally hold a typical 0.5V common-mode voltage at their input terminals in order to maintain proper biasing on the logging BJTs. Because these log amps are current-input devices, this internally generated common-mode voltage is generally not a problem in most current-measurement applications.

The presence of an on-chip current reference has become quite popular in most contemporary DC log amps. This reference can be connected to the reference input of the log amp, thereby permitting an absolute, rather than ratiometric, measurement of the current presented to the log amp's main current input. In the case of the MAX4206, a reference current is obtained by means of a 0.5VDC voltage source, a voltage-to-current converter, and a 10:1 current mirror. An external resistor is required to program the desired reference current.

{uh,,,, read it all}

http://www.maxim-ic.com/appnotes.cfm/an_pk/3611

and old national

http://www.national.com/an/AN/AN-311.pdf