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Opamp Circuit for metal detector

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pal114525

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Hi,

I have seen a circuit as attached for metal detector.
Can anyone please explain me what is basic functionality of the opamp in that circuit arrangement. Pin2 is going to the input (-)of another opamp.
Please explain me the functionality of this type of circuit.

Thanks.
 

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Audioguru

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Why is your schematic a negative image in a Word document instead of a normal positive (black lines on a white background) picture saved as a PNG file type??
The opamp integrates and rectifies a little the signal. The input signal to this opamp is not shown and might feed its pin #12.
The output of this opamp goes through the series diode and resistor on the left side to the input pin #2 of another opamp.
Here is your schematic as a positive picture:
 

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pal114525

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Hi,
Thanks for the schematic update.
The other input which is connected to Pin12 (Non-inverting input) of the opamp is from the sensitivity switch.
Can anyone please explain me how is the signal getting processed in this circuit?
What is the functionality of the two capacitors C9 and C118 in the circuit?

Thanks.

- - - Updated - - -

Hi,

Thanks for the schematic update.
The pin12 (non-inverting terminal) of the opamp is connected to the sensitivity switch.
Can you please explain me how is the signal getting processed in this circuit?
What is the functionality of the two capacitors C9 and C118 in this circuit?

Thanks.
 

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pal114525

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Integrator using opamp

Hi,

I have seen this type of integrator in a metal detector circuit.
In the attached schematic, Pin13 is connected to nowhere. The specialty of this type of circuit is that, C10 and C11 are connected in an interesting manner( The negative leads are connected together).
1. Can any one please tell me what is the functionality of this type of capacitor configuration?

Thanks.
 

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dahuangpeng

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Re: Integrator using opamp

maybe it is for layout matching
 

FvM

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Re: Integrator using opamp

It's a poor man's way to make a bipolar electrolytic capacitor, not exactly suggested by capacitor manufacturers, but it should usually work. See below an Epcos application note about back-to-back connection of tantal capcitors.

4.8 Series back-to-back connection

For applications where higher polarity reversal voltages occur, two capacitors with identical rated voltage and identical rated capacitance can be connected back-to-back in series (e.g. cathode to cathode). In this way, blocking in each polarization direction is achieved. To avoid damage to the reversed polarity capacitors during charging, it is necessary to connect diodes in parallel to the capacitors with the center of the diodes and the capacitors connected together.

This non-polar or bi-polar version (which only has half the capacitance value as a result) can be operated at voltages of up to the rated direct voltage of any polarity or with double the superimposed alternating voltage of the value permitted for the individual capacitor. The capacitors connected back-to-back in this way can also be operated at a pure ac voltage. At 20 °C ambient temperature, the surface temperature of the capacitor is not allowed to increase by more than max. 10 °C, while the upper temperature limit should not be exceeded.
 

Borber

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Give a complete circuit or link to it. Your fragments are useless.
 

pal114525

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Re: Integrator using opamp

Hi,

Would you please explain me the point?

Thanks.

- - - Updated - - -

Hi,

In this circuit, I am dealing with millivolt (mV), not even a volt. In this type of circuit, what is the application of C10 and C11 arrangement?

Thanks.
 

Audioguru

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C10 and C11 are high value electrolytic capacitors that are polarized (look up the definition).
Since they are exposed to AC that has both polarities then they are connected together in series with their polarities opposite so that the AC will not damage them.
The inverting opamp with a capacitor as negative feedback (look up the definition) is called an integrator (look up the definition).
 

pal114525

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Hi,

I have simulated the circuit with C10 and C11 as per above arrangement. I have used Proteus 8 for that simulation. But this is giving me simulation error saying timestep too small, timestep = 1.25e-019.
Would you pleaes explain me how to solve it out?

Thanks.
 

BradtheRad

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simulation error saying timestep too small, timestep = 1.25e-019.
Would you pleaes explain me how to solve it out?

Thanks.
I believe you can make the timestep longer.

Suppose your highest frequency in your circuit is 1 MHz.
Calculate a timestep which breaks a cycle into segments, between 50 and 500 of them.

This gives you a range of 2 to 20 nSec for your timestep. Your scope trace should reveal a few cycles of your operating frequency.

I have no idea whether Proteus will then say your timestep is too large. It may help to play with timestep values, in case it needs to be smaller so as not to skip very brief events.
 

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