Understanding of how HH circuit works

[EEE]

Hi everyone,

I am trying to understand how HH circuit works. I have looked at the manual and would like to discuss what I understand. Please correct me if I am wrong.

The 555 timer generates the pulse control signal to switch the MOSFET IRF740 on and off. The circuitry connected to the gate of the MOSFET provides the function of gate driver.

The coil is connected in parallel with the damping resistor which its value should be between 200-800 Ohm and needs to be experimentally adjusted. Should not be over or underdamped.

The clamping diodes will only conduct after the back emf voltage of the coil (after it is switched off) drops below 0.7 V

The preamp will amplify the decaying signal (from 0.7V to 0) by a gain of 1000. Hence, the output of the preamp would be clamped to the power supplies! This part I don't understand because the decaying signal will not be properly amplified as it would be amplified to 700V to 0 and the power supplies of the preamp is just 5/-5V.

The output of the preamp is connected to two 4066 switches. Both are controlled by 74HC221 IC. There is a main delay sample and 2nd delay. The main delay is set to lets say 75us (After the 555 timer pulse) and the second delay is set to 150us.

When the main sample switch of 4066 is on, the 2nd sample switch is off and vica versa. The differential integrator will only integrate 1 signal at a time and not both signals of the switches. How does this part of the circuit work? I don't understand how the integrator will eliminate the noise and detect the induced voltage in the metal target.

What is the circuit trying to sample, the back emf of the coil or the induced voltage of the metal target? What I understand from PI, that the coil will induce voltage in any nearby metal targets. This will cause eddy currents to flow in the target which will decay with time hence the metal target will generate a magentic field. This generated magentic field is picked up by the coil.

I would like to build this circuit but first i am interested in exactly how it works.

Regards,

[Carl-NC]

Hi EEE,

The clamping diodes will only conduct after the back emf voltage of the coil (after it is switched off) drops below 0.7 V

Actually, they conduct until the back emf drops below 0.7V.

The preamp will amplify the decaying signal (from 0.7V to 0) by a gain of 1000. Hence, the output of the preamp would be clamped to the power supplies! This part I don't understand because the decaying signal will not be properly amplified as it would be amplified to 700V to 0 and the power supplies of the preamp is just 5/-5V.

The preamp output will be railed out until the input reaches the last few millivolts.

The output of the preamp is connected to two 4066 switches. Both are controlled by 74HC221 IC. There is a main delay sample and 2nd delay. The main delay is set to lets say 75us (After the 555 timer pulse) and the second delay is set to 150us.

Typically the main delay is around 15us.

When the main sample switch of 4066 is on, the 2nd sample switch is off and vica versa. The differential integrator will only integrate 1 signal at a time and not both signals of the switches. How does this part of the circuit work? I don't understand how the integrator will eliminate the noise and detect the induced voltage in the metal target.

A slow signal (such as from Earth field) will have significant content at both sample points, while a fast signal (metal target) will only show up at the first sample. So when you subtract the sample points, the slow signal cancels and the fast signal does not. Integration is slowly cumulative (relatively) so the samples don't have to be simultaneous.

What is the circuit trying to sample, the back emf of the coil or the induced voltage of the metal target? What I understand from PI, that the coil will induce voltage in any nearby metal targets. This will cause eddy currents to flow in the target which will decay with time hence the metal target will generate a magentic field. This generated magentic field is picked up by the coil.

The induced voltage from the target is superimposed on the normal back EMF. This delta voltage is what we're trying to pick out.

- Carl

[simonbaker]

I appreciate this back-to-basics explanation too. Thanks.

-SB

[EEE]

Thanks Carl for the info.

[EEE]

Hi EEE,

The preamp will amplify the decaying signal (from 0.7V to 0) by a gain of 1000. Hence, the output of the preamp would be clamped to the power supplies! This part I don't understand because the decaying signal will not be properly amplified as it would be amplified to 700V to 0 and the power supplies of the preamp is just 5/-5V.

The preamp output will be railed out until the input reaches the last few millivolts.

What is the circuit trying to sample, the back emf of the coil or the induced voltage of the metal target? What I understand from PI, that the coil will induce voltage in any nearby metal targets. This will cause eddy currents to flow in the target which will decay with time hence the metal target will generate a magentic field. This generated magentic field is picked up by the coil.

The induced voltage from the target is superimposed on the normal back EMF. This delta voltage is what we're trying to pick out.

- Carl

Hi Carl,

Thanks for the help. How is the output of the preamp is railed out? is that to do with the clamping diodes as they are off when the back emf drops below 0.7 V or because the output is sampled after 15us?

is the delta voltage affected by the mutual inductance between the coil and metal target? How can I detect the eddy-current decay signal of the target?

[DOOLEY]

op amp "railed out" exp.

dont want to type in the far end of a fart about an op amp, so simple responce below,

and lets say op amp is on a 0v / 10v single supply ,non inverting, for simplicity.

if op amp feedback is set to say x10 then a 0.7v signal differance in =7v out
if set x100 then 0.7v doesnt =700v , it will go to top rail (about 9v), + rail or "rail out" as carl said.
op amps (most as far as i know , only go to within 1v of each rail)

and as the 0.7v drops to say 0.09v output will be 9v etc....etc...
all the way to 0.001v = 1v output , bottom rail + 1v.

so on a scope your nice curve from 0.7v down to 0v now looks like a straight line clinging to top rail , followed by a slope to 1v above bottom rail.

hope this makes sense , hope it helps , somewhat.

[bklein]

This has me bringing up the question, maybe what you are saying,...
As many are interested in obtaining the fast response of gold return signals, it seems we should design the preamp so that we can begin sampling the signal as soon as possible. But as there is so much gain in the preamp it rails out like you say even though the input signal is below the diode protection levels. So sampling this soon is pointless. But if the gain were reduced you could...?
Another thing I wonder about is the offset biasing of the preamp. I usually read that this is supposed to be set for VCC/2 (ground). But I wonder if setting this at ground is correct as you may be able to get sooner samples if offset is intentionally added? I did pick up some depth with my SurfPI by tweaking this off from ground - didn't have a scope to see what I was doing but had one for the original setting.

Barry

[Carl-NC]

How is the output of the preamp is railed out? is that to do with the clamping diodes as they are off when the back emf drops below 0.7 V or because the output is sampled after 15us?

What Dooley said.

If you have a preamp gain of 1000 and +/-5V supply rails, then any input beyond +/-5mV is guaranteed to "rail" the output.

is the delta voltage affected by the mutual inductance between the coil and metal target?

Yes, targets will cause a small deflection in the decay.

How can I detect the eddy-current decay signal of the target?

That's the function of the sampling integrator, too look for deviations in the decay. You could also sample the decay and do it in software.

[Carl-NC]

As many are interested in obtaining the fast response of gold return signals, it seems we should design the preamp so that we can begin sampling the signal as soon as possible. But as there is so much gain in the preamp it rails out like you say even though the input signal is below the diode protection levels. So sampling this soon is pointless. But if the gain were reduced you could...?

HH was designed for overall simplicity, so I chose a single-stage high-gain (~1000) preamp. If you split it into a 2-stage amp, each stage with ~32 gain, then the overall preamp will be faster. Or, as you say, you can just go with a lower single gain, and boost the gain in the post-integrator section.

Another thing I wonder about is the offset biasing of the preamp. I usually read that this is supposed to be set for VCC/2 (ground). But I wonder if setting this at ground is correct as you may be able to get sooner samples if offset is intentionally added? I did pick up some depth with my SurfPI by tweaking this off from ground - didn't have a scope to see what I was doing but had one for the original setting.

Yes, you could assume the decay will always be well-behaved and on one side of "ground," and skew the preamp bias to give more effective headroom. However, you could plug in a coil that is slightly underdamped and suddenly the response of interest is on the other side of "ground."

[taliesin]

op amp "railed out" exp.

dont want to type in the far end of a fart about an op amp, so simple responce below,

and lets say op amp is on a 0v / 10v single supply ,non inverting, for simplicity.

if op amp feedback is set to say x10 then a 0.7v signal differance in =7v out
if set x100 then 0.7v doesnt =700v , it will go to top rail (about 9v), + rail or "rail out" as carl said.
op amps (most as far as i know , only go to within 1v of each rail)

and as the 0.7v drops to say 0.09v output will be 9v etc....etc...
all the way to 0.001v = 1v output , bottom rail + 1v.

so on a scope your nice curve from 0.7v down to 0v now looks like a straight line clinging to top rail , followed by a slope to 1v above bottom rail.

hope this makes sense , hope it helps , somewhat.

"dont want to type in the far end of a fart about an op amp" .i haven't heard that expression before nice one

[DOOLEY]

"far end of a fart" exp : long wind'ed explination / to much information

it's a british thing me thinks.

didn't want to fill a page full about op-amp's

that and i didn't want to offend anyone by telling many folks on here "how to suck egg's"

note : open minded , i'm learning every day , know quite a bit , been tinkering with electronics for 30 years ish , only set my mind to metal detectors a few months ago , like a dog with a bone , my mind won't let the puzzle go.

thought i'd throw in my 2 peneth's worth.

[Qiaozhi]

"far end of a fart" exp : long wind'ed explination / to much information

it's a british thing me thinks.

I think it originated in the North East of England. It's certainly not a southern expression.

[DOOLEY]

also , great respect to carl , the designer ,
so didn't want to post to much as this increces the risk of me typing something inacurate , and then being pulled up and corrected.

there is poss only about 6 varieties of pi circuits on the net the study and compare,
and think about , and one of them is his , so i didnt want to "mess with the master"
and butt in and offend.

ps : is it just me , or does everyone that keeps trying new idea's and to better / perfect something go round in circles , loose the way , and end up right back where you started , and not knowing where youve been or how you got there ?

frustrating field of electronics this , theory , simulation and in practice , not always end up with the same result.

[simonbaker]

also , great respect to carl , the designer ,
so didn't want to post to much as this increces the risk of me typing something inacurate , and then being pulled up and corrected.

there is poss only about 6 varieties of pi circuits on the net the study and compare,
and think about , and one of them is his , so i didnt want to "mess with the master"
and butt in and offend.

ps : is it just me , or does everyone that keeps trying new idea's and to better / perfect something go round in circles , loose the way , and end up right back where you started , and not knowing where youve been or how you got there ?

frustrating field of electronics this , theory , simulation and in practice , not always end up with the same result.

not just you...

[taliesin]

it's not just you dooley

i think there is always a better way .it's just finding it