The tank doesn't go on and off, it has an AC signal in it from the magnetic AC field it picks-up from the loop coil around the room. Since one end of the coil is connected to ground then the other end goes positive and negative a little.
Yes, the output of the comparator goes low when its inverting input has a higher voltage than its non-inverting input.

Because it is an inductor without a shield. The inductor acts like the secondary winding of a transformer with the loop coil around the room as the primary winding.

The inputs of the LM393 comparator are the bases of PNP transistors. Their input bias current is 25nA to 250nA and creates a small positive voltage drop across a 100k input resistor to ground.

__________________
Uncle $crooge

 

without a shield? u mean the coil is only a thick wire that turned for a couple of rounds? but i'm using a 10mH inductor (i think it is with shield) (i couldn't get 27mH) and i have increased the capacitance 2.7 times so that i get the resonant frequency almost 10kHz.

*what's the value of the coil which is without a shield? it can be any value? i don't know about that, i have just seen them before.
what's the different between with shield and without shield?

 

The inductor has many turns of wire.
The receiver's inductor won't pickup much magnetic signal if it has a shield. An aluminum or plastic cover won't block the magnetic signal but a steel one will.

Did you tune the frequency of the transmitter to be at the peak frequency of the receiver's tuned circuit?

__________________
Uncle $crooge

 

yes. both tx and rx frequency also almost 10kHz.
i think the 10mH inductor which i'm using is a ceramic type, it looks like a big big resistor with brown color. i just want to know whether there is any other way or not to increase the sensitivity and which part of the receiver is the most sensitive to the signal.


does this type of coil have any value?

 

You need to change the frequency of the transmitter and monitor the receiver for its highest sensitivity to a frequency. It might peak at 9600Hz but not be very sensitive at 10kHz.

The big inductor in the picture has only a few turns and doesn't have an iron or ferrite core, so its value is measured in uH. On the web are instructions for making them.

The inductor in the receiver's circuit is the most sensitive part for the signal.
I don't remember if it is the most sensitive when it is vertical or horizontal when compared to the transmitter's loop that is flat on the floor.

__________________
Uncle $crooge

 

so if i increase the inductance 10 times (100mH), and reduce the capacitance for 10 times (0.0022uF), will i get better sensitivity?

i will try both and see the result.

actually the rx frequency is around 10730Hz, i assume it as 10kHz. so i will change the tx frequency to 9600Hz by changing the value of the resistor and see the result. thanks for helping me a lot

 

You have a good point. The inductor is the secondary winding of a transformer. A higher inductance will have more turns and therefore will have a higher output level.

__________________
Uncle $crooge

 

is it because of from the circuit, the tank circuit is the parallel of a 27mH and a 0.01uF? the calculated frequency is 9685Hz. Now i'm using 100mH coupled with 2.7nF to increase the sensitivity. So the frequency of the 555 has to be around that. i have found the nearest, R1=1K, R2=7.4K, C=0.01uF which gives the frequency of 9130Hz. Do them need to be equalizer?


How if i use 100mH and 2.2nF? The rx frequency now is 10730Hz. And the 555 tx frequency is determined by R1=3.3K, R2=5.6K, C=0.01uF which is 9949Hz.

Which one is better??

 

You don't theoretically calculate the values (except roughly). The coil is fixed, and you have a fixed capacitor across it, however, it's VERY easy to adjust the 555 frequency, and you should have a preset resistor in circuit for that very purpose. Adjust the frequency of the 555 to give the maximum range - if you have a scope you can adjust it for maximum voltage across the tuned circuit.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Calculation the value of parts is only approximate since they have a tolerance. Then you need to TUNE the frequency of the transmitter to A PEAK in the response of the receiver. If you don't then the circuits won't be matched and will have poor sensitivity.

__________________
Uncle $crooge

 

oookay.. i didn't do that because of the VR canot fixed well on the breadboard, besides that it is also big and heavy.
but now i think i have to do that to get the maximum range

*once i get the max range, can i use a normal resistor to replace the VR which is same value as the tuned VR??
thanks for helping ^_^V

 

VR= Variable Resistor?
Don't use a big and heavy volume control, instead use a small and lightweight trimmer potentiometer. They are turned with a small screwdriver.

Attached Images

trimmer pots.PNG (46.5 KB, 5 views)

__________________
Uncle $crooge

 

hey... thanks for your helping... i get the best range i want just by changing the inductor to the larger value. Thank you so much

 

555:
1. I thought i only need to calculate the charging and discharging time by using RC time constant, why should i include the ln2?

393:
1. For the normal comparator, when input to +ve is larger then -ve, the output is the different between them or what? How if when -ve larger than +ve?
2. The internal transistor in LM393 is npn or pnp type? Does it need 0.7V to turn the transistor on?
3. Instead of using 100K, can i use other value (i think larger value) to increase the input offset voltage so that the output will only go low when larger voltage go into the -ve. This is because i don't want it to be affected by other sourse.

Thanks a lot

 

You cannot calculate parts values accurately because of their tolerance. You need to change the value of the 5600 ohm resistor to tune the 555's frequency to be the same as the peak of the receiver's tuned circuit.

The comparator has a gain of a few hundred thousand, so a very small voltage difference between its inputs causes its output to saturate to near ground, or to be open-circuit and be pulled up to the supply voltage by the load resistor.
If +in is more positive than -in then the output is high. If -in is more positive than +in then the output is low.

Look on the datasheet to see all the transistors inside. The very high gain turns on the output transistor with an extremely small input voltage.

Yes, up to a few megohms.

__________________
Uncle $crooge