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The following circuit is the transmit coil of an IB metal detector. I recognize most of this: an op amp astable with a CE push-pull output stage driving a resonant circuit. But, I have questions.

enter image description here

The frequency of the tank is about 27KHz, which I'm guessing is the fundamental frequency of the astable. I understand the general astable frequency formula, but I'm finding it difficult to calculate the positive feedback fraction in this instance. How to calculate the feedback fraction?

My other question relates to capacitors C1 and C2. They seem to act like a charge pump, pumping the centre-biased input signal up (and down) to either rail. My feeling is that it makes the push-pull pair more efficient by turning off one or other of the transistors, but I'm not sure. Can anyone shed any light?

Incidentally, here's the original project document (for context) Buccaneer Metal Detector.

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The frequency of the tank is about 27KHz, which I'm guessing is the fundamental frequency of the astable.

No, the op-amp isn't acting as an astable multivibrator; the 100 nF capacitor fed from R5 (100 kohm) is there to self bias the op-amp so that it produces a pretty decent 50:50 square wave. The 100 kohm and 100 nF provide a cut-off at about 16 Hz so are way, way below the operating frequency of the tank.

The transistor output stage is inverting (collectors at the output point feeding the tank) and this means the sinewave signal from the tank gets turned into a square wave by the op-amp and the transistors force the tank in the opposite direction. This causes sustained oscillation.

Your feeling about the way the output transistors are driven appears correct - it's just a way of saving a few mA and extending battery life.

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  • \$\begingroup\$ OK, so if I understand you correctly (and it's shaky at best), the tank is driving the op amp. I calculate the resonant frequency of the tank to be ~213kHz and interestingly, if I divide that by 10 (22n/220n), I get the answer. But, that's just guesswork. \$\endgroup\$
    – Buck8pe
    Sep 23, 2016 at 17:21
  • \$\begingroup\$ Dividing frequency by ten just because the C ratio is ten makes no sense. What makes you think the tank runs at 213 kHz - do you know the inductance of the buccaneer coils? \$\endgroup\$
    – Andy aka
    Sep 23, 2016 at 17:24
  • \$\begingroup\$ @Andyaka - What do you think L1 is? \$\endgroup\$
    – WhatRoughBeast
    Sep 23, 2016 at 18:48
  • \$\begingroup\$ The 27khz tank f comes from the spice simulation and, although a little high, is ballpark the 20khz quoted in the project doc. The doc doesn't give a coil inductance, so I stuck 100uH in there to figure things out. I'd like to work out how the 27khz comes about and thanks to your answer, I'm not barking up the wrong tree. \$\endgroup\$
    – Buck8pe
    Sep 23, 2016 at 18:54
  • \$\begingroup\$ It almost certainly be closer to the mH range for a metal detector of this type. @WhatRoughBeast - it won't be 100 uH - that was just an estimate by the OP - that's why I asked. \$\endgroup\$
    – Andy aka
    Sep 23, 2016 at 19:42
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Feedback fraction is not a particularly useful concept in this case. Since the impedance of C6 (0.1uF) will be quite low at kHz frequencies (160 ohms at 10 kHz), the gain of the op amp will be very high at AC, while essentially 1 at DC. This produces the (roughly) square wave output which drives the tank. Whenever you see oscillators putting out square waves, you know that the amplifier is not operating as a classic linear amplifier.

The circuit diagram is also confusing in showing the tank coil as having 100 uH inductance. For the coil spec in the link, the inductance will be in vicinity of 2300 uH. Combined with C3 and C4 gives a nominal tank frequency in the vicinity of 22 kHz. Eh, close enough for government work.

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  • \$\begingroup\$ Hmmm. I'm having some difficulty calculating the 22khz figure you've arrived at using 1/sqrt(LC)? Also, what's really odd (considering Andy's answer) is that changing L1 from 100uH to 2300uH produces 27khz and 22khz respectively. I'd expect a greater f difference if the resonant circuit was controlling the frequency?? What am I missing? \$\endgroup\$
    – Buck8pe
    Sep 23, 2016 at 19:52
  • \$\begingroup\$ @Buck8pe - You have to take into account BOTH C3 and C4. And the 2300 uH figure is for an unloaded, ideal coil. Try using 1500 uH instead. You can then use C4 as a trimmer, if frequency is important - and in this case it's not. \$\endgroup\$
    – WhatRoughBeast
    Sep 23, 2016 at 20:49
  • \$\begingroup\$ Dah, computation error! 23khz it is, thankyou. \$\endgroup\$
    – Buck8pe
    Sep 24, 2016 at 8:17

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