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my explaination

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The inductor in the receiver's circuit is the most sensitive part for the signal.
so if i increase the inductance 10 times (100mH), and reduce the capacitance for 10 times (0.0022uF), will i get better sensitivity?

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.
i will try both and see the result.

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.
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 ;)
 
bananasiong said:
so if i increase the inductance 10 times (100mH), and reduce the capacitance for 10 times (0.0022uF), will i get better sensitivity?
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.
 
It might peak at 9600Hz but not be very sensitive at 10kHz
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.
 
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.
 
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.
 

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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 ;)
 
about the 555 and the 393

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 :D ;)
 
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bananasiong said:
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?
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.

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?
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.

2. The internal transistor in LM393 is npn or pnp type? Does it need 0.7V to turn the transistor on?
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.

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.
Yes, up to a few megohms.
 
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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.
i mean.. just the 555 alone, not including the detector. Why is the ln2 included?

If +in is more positive than -in then the output is high. If -in is more positive than +in then the output is low.
For any comparator? I thought the larger one minus the lower to get the output?

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.. it is npn type. it is Q16, am i right?. from https://www.electro-tech-online.com/custompdfs/2006/05/LM393-DPDF.pdf
The gain can turn on the bjt with very small voltage?? how?

Yes, up to a few megohms
thanks.. i will try it..
 
bananasiong said:
Why is the ln2 included?
What is "ln2"?

For any comparator? I thought the larger one minus the lower to get the output?
Most comparators have balanced inputs and a voltage gain of typically 200,000. Therefore if the + input has a voltage slightly higher than the - input then the output voltage is high. Also if the + input has a voltage slightly less than the - input then the output voltage is low.

The gain can turn on the bjt with very small voltage?? how?
Since the voltage gain is 200,000, to make the output change 0.6V then the input voltage changes only 3uV.
 
What is "ln2"?
The formula used for calculating the frequency of 555 timer.
T = (Ra+2R2)C*ln2
f = 1/T
i found it from reference book.. why is it needed?

Most comparators have balanced inputs and a voltage gain of typically 200,000. Therefore if the + input has a voltage slightly higher than the - input then the output voltage is high. Also if the + input has a voltage slightly less than the - input then the output voltage is low.



Since the voltage gain is 200,000, to make the output change 0.6V then the input voltage changes only 3uV.
thanks.. can i know what's the voltage value to the -ve when the tank circuit detects the electromagnetic field?

about the 555 again, if i don't connect the capacitor from valtage control (pin 5) to ground, what will happen?
 
bananasiong said:
The formula used for calculating the frequency of 555 timer.
T = (Ra+2R2)C*ln2
f = 1/T
i found it from reference book.. why is it needed?
because it has charging of a capacitor is involved , try some 555 tutorial/book

bananasiong said:
can i know what's the voltage value to the -ve when the tank circuit detects the electromagnetic field?
as explained earlier , since the gain is in the order of 1E5 ,i/p in uV will swing o/p .
 
The datasheet for an LM555 shows a formula and a graph to determine its frequency.
It also shows that having a capacitor at pin 5 keeps the frequency constant when the supply voltage has ripple.
The comparator's voltage gain is very high so it works when its input voltage is very low.
 
The comparator's voltage gain is very high so it works when its input voltage is very low.
as explained earlier , since the gain is in the order of 1E5 ,i/p in uV will swing o/p .

nono... i mean, the voltage of the tank circuit to the -ve when it detects the electromagnetic field, before it goes into the comparator. The input offset voltage is a few miliV due to the 100K right? then how much is the voltage from the tank when it detects? I know it must be greater than i/p offset voltage, but how much is it?

can it be calculated? or measure using multimeter of CRO?


555
https://www.electro-tech-online.com/custompdfs/2006/05/NE_SA_SE555_C_2.pdf
from the datasheet, the ln2 is not included... but the reference book yes.. which one to be followed??

thanks..
 
bananasiong said:
The input offset voltage is a few miliV due to the 100K right? then how much is the voltage from the tank when it detects? I know it must be greater than i/p offset voltage, but how much is it?

can it be calculated? or measure using multimeter of CRO?
You can measure it with a sensitive multimeter, or measure it with a 'scope.
You can also calculate it but it has a wide range. The datasheet shows a wide range for the input bias current for the LM393 comparator. It also has a wide range for its built-in offset voltage.

from the datasheet, the ln2 is not included... but the reference book yes.. which one to be followed??
The 2 datasheets are nearly the same. National uses 1.44 in its formula, Philips uses 1.49 and the text book uses the log which might be the same.
 
yup..
1/ln2 almost equal to 1.44.
but why is it needed? as i know, the RC time constant is calculated by R*C only. for the 555, charging time is (R1+R2)C, discharging time is R2C. so the time should be (R1+2R2)C, why is the ln 2 or the 1.44 needed??
 
The current in the capacitor charges and discharges exponentially.
 
exponentially.. sorry.. suddently feel that, i don't really know about the 555.
when charging, the output of the comparator is high, the output of the internal FF, Q is high right? then how it turns on the internal transistor?
 
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