• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

EMF detector

Status
Not open for further replies.

satch922

New Member
I am building an electromagnetic field detector for a school project, and I need some help getting started with the analysis of the circuit.



1) How do I figure out the gain of the op-amp? Without an input resistor, I am lost.

2) Why does the meter not work when the headphones are plugged in? Is it because that branch is the path of least resistance?
 

colin55

Well-Known Member
When the headphones are plugged in, you have taken away the 0.65v bias on the base of the transistor.

You need to feed the headphones via a separate 220u from the output of the op-amp.
 
Last edited:

dougy83

Well-Known Member
Gain is frequency dependant. You can use laplace transform for the impedance of the inductors and capacitors: Z(L) = sL = 2j*pi*f*L, Z(C) = 1/(sC) = 1/(2j*pi*f*C) = -j/(2*pi*f*C). Using these impedances you can work out the gain for any frequency.
 
Last edited:

MikeMl

Well-Known Member
Most Helpful Member
The input circuit to your detector is a parallel resonant tuned circuit which resonates at 15.9Khz. The current circulating in the tuned circuit is determined primarily by the Q of tuned circuit. That current also flows in the feedback network of the LF351 because of the "virtual ground" at the inverting input. That means that the output voltage is the circulating current times the reactance of the 150pF capacitor.

btw- what is this supposed to detect? As shown, it resonates near the horizontal freq in an old analog TV set (15575Hz). Also, there is no RF detector, just linear amplification at the "carrier" frequency, so are you expecting to hear the 15kHz in the earphone? Your hearing must be much better than mine:D
 

Sceadwian

Banned
Acoustical downmixer? Works along the same lines as an RF mixer but in the AF range. I've played around with this a little for fun, using a 15khz tone to encode a 3khz audio sub carrier. Basically frequency shifting the whole thing up. If you produce a tone close to the main carrier in free air you'll start hearing the beat frequencies from the sub carrier. It's really creepy to hear. Invisible audio. But I digress.
 
Last edited:

MikeMl

Well-Known Member
Most Helpful Member
Acoustical downmixer? Works along the same lines as an RF mixer but in the AF range. ...
But doesn't this need a non-linear device somewhere in the detection chain? The op-amp is within its linear region, so where does the detection take place? In your eardrum?:)
 

Sceadwian

Banned
In my example yes =) In his, no idea sorry =) Analog is a bit new to me, I have no idea what kind of feedback or possible mixing could be going on in there.
 

satch922

New Member
Thanks for the replies. By the way, I didn't design this circuit, I'm just using it for my project. How do I determine the Q of the tuned circuit, and then the current circulating in the circuit?
 
Last edited:

colin55

Well-Known Member
The "Q" of the inductor will depend on how it is made and the core material.
It has to be provided by the manufacturer.
 

Mikebits

Well-Known Member
Isn't that op amp being used as a comparator? 2.2M is for hysteresis.
 

Roff

Well-Known Member
Isn't that op amp being used as a comparator? 2.2M is for hysteresis.
No, it has negative feedback. Hysteresis requires positive feedback.
I look at the pickup coil as a transducer which converts EMF to a current. That current is multiplied by the impedance of the feedback loop, giving a voltage out. I have no idea how you would get an audio signal for EMF frequencies above the audio range. Perhaps it is designed to pick up mains fundamental and harmonics, lightning(?), and maybe ghosts. Anyone know the frequency range preferred by ghosts?:rolleyes:
 

Mikebits

Well-Known Member
No, it has negative feedback. Hysteresis requires positive feedback.
I look at the pickup coil as a transducer which converts EMF to a current. That current is multiplied by the impedance of the feedback loop, giving a voltage out. I have no idea how you would get an audio signal for EMF frequencies above the audio range. Perhaps it is designed to pick up mains fundamental and harmonics, lightning(?), and maybe ghosts. Anyone know the frequency range preferred by ghosts?:rolleyes:
Thanks for the corection, looks like I am batting 0 for 4, back into the dugout with me...:eek:
 

satch922

New Member
I look at the pickup coil as a transducer which converts EMF to a current. That current is multiplied by the impedance of the feedback loop, giving a voltage out.
I'm going to give an example using the info you gave me, so if I make a mistake, tell me what I did wrong. Let me know if I'm right also.

Let's say the frequency is 1kHz and the inductor produces a current of 0.1uA.

The impedance of the feedback loop would be: ω=2(pi)(1kHz)=6.28kHz; Xc=1/6.28kHz(150pF)=1.06MΩ; 2.2MΩ in parallel with 1.06MΩ=715.3kΩ; so the voltage out=0.1uA*715.3kΩ=71.5mV
 
Last edited:

Roff

Well-Known Member
I'm going to give an example using the info you gave me, so if I make a mistake, tell me what I did wrong. Let me know if I'm right also.

Let's say the frequency is 1kHz and the inductor produces a current of 0.1uA.

The impedance of the feedback loop would be: ω=2(pi)(1kHz)=6.28kHz; Xc=1/6.28kHz(150pF)=1.06MΩ; 2.2MΩ in parallel with 1.06MΩ=715.3kΩ; so the voltage out=0.1uA*715.3kΩ=71.5mV
It's not quite that simple.:(
The current through the capacitor leads the current through the resistor by 90 deg, so you can't just use the parallel resistor formula. There are several ways to get to the impedance. For a resistor in parallel with a capacitor, you have to do vector summation of the admittances, then take the reciprocal of the result to get the impedance.

Yrc=√[(1/Xc)² + (1/R)²]

Zrc=1/Yrc

In your example, Zrc=955.7kΩ

Multiply that by 100nAand you get 95.57mV.
 
Last edited:

satch922

New Member
I'm now working on the transistor gain. It looks like a common emitter amp with collector-feedback bias, and the meter is the load resistor. Is this right?

Also, somebody told me that full-wave rectification takes place via the diodes, meter, and capacitor. Why is this?
 

MikeMl

Well-Known Member
Most Helpful Member
I'm now working on the transistor gain. It looks like a common emitter amp with collector-feedback bias, and the meter is the load resistor. Is this right?
Yes. The load is complex, so it is hard to determine the gain. However, on positive excursions of the collector voltage, current from the 10K pull-up resistor flows through the meter into the 220uF capacitor. On negative excursions, the current flows out of the 220uF capacitor, through the other diode and into the collector of the transistor and thence to the emitter.

Also, somebody told me that full-wave rectification takes place via the diodes, meter, and capacitor. Why is this?
I would not call it full-wave rectification. Only half-waves flow through the meter. The meter "ballistically averages" the half-wave current pulses to produce a meter reading. The other diode is there simply to reset the 220uF capacitor for the next cycle.
 

satch922

New Member
Thanks for the replies. I have a couple more questions. Since the -V of the op-amp is supplied by the negative terminal of the 9vDC, the negative cycle out of the op-amp will not truly be negative, it will be zero, is this right?

Also, I am wondering what the capacitors that haven't been explained already are used for.
 

Roff

Well-Known Member
Thanks for the replies. I have a couple more questions. Since the -V of the op-amp is supplied by the negative terminal of the 9vDC, the negative cycle out of the op-amp will not truly be negative, it will be zero, is this right?

Also, I am wondering what the capacitors that haven't been explained already are used for.
Let's call the negative terminal of the battery zero volts, or ground. Most schematics will have a ground symbol attached to that line. The 10k/10k voltage divider on the op amp +in terminal biases that pin at 4.5V nominal. Due to feedback through the 2.2M resistor, the output will also be at 4.5V, allowing maximum signal swing out of the op amp. Any signal on the output will swing plus and minus around 4.5V.

The cap from +in to ground makes that pin low impedance at the frequencies of interest, so the bottom end of the transducer is effectively at AC ground. The 100nF cap in series with the transducer prevents the inputs from being shorted together by the transducer.
The 220uF cap on the output of the op amp blocks the op amp's 4.5V output voltage from appearing on the 2.2k resistor, which would screw up the transistor biasing.
The 100uF cap across the power rails lowers the AC impedance of the power supply, eliminating the impedance of the battery and the inductance of the wiring of the power supply.
 
Status
Not open for further replies.

EE World Online Articles

Loading

 
Top