I have included an attachment that shows the circuit. It should be easy to hook up.

And the advantage with this circuit is that the output is actually going someplace useful!

All that the current output is doing (without making any modifications) is giving itself to the resistor. and what I like to know is, other than generating heat, Why would anyone (with experience) want to just connect a grounded resistor to the output, and nothing else?

anyways, the tank circuit in my diagram is the inductor (the bottom-left-most component) and the capacitor (two T's with their heads almost touching each other) next to it. The coupling capacitor is the remaining component.

I suggest using a large value capacitor so that a large portion of the signal can be passed on. I suggest a 0.1uF capacitor. Capacitors do block DC, so there is NO chance of a short circuit.

Now on to the tank. The goal is to pick a low inductor, and a low capacitor value. and use the following equation to determine the resonant frequency (the frequency you will will be transmitting on).

f = 1/(2*pi*sqr(l*c))

pi is equal to whatever appears on your calculator after you press the "pi" button. sqr means "square root of".

Attached Images

ckt_326.gif (1.5 KB, 517 views)

__________________
-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

oh i see! thanks a lot! so just ignore the word mouser... just get a same value inductor, am i right? from the transmitting part, the sentence "use a 10k pot to adjust frequency".. is it variable resistor? there is no variable resistor.. and.. for the receiving part, how do i know if i get the signal? don't we need to tune at the output??

thanks for answering my question..

 

u mean.. a large value capacitor, which one? both or only one? the one parallel with the inductor? or the one straight from output?? then how about the value of the inductor?? thanks a lot...

 

Mstechca, you are an interesting character. It apparently never crossed your mind that Russ has been around the block a few more times than you, and might know something you don't. I don't want to imply that you know nothing, but you might want to consider that Russ is a retired engineer, while you are a hobbyist still struggling to understand the fundamentals of electronics. It would behoove you to ask how his circuit works, instead of proclaiming that it doesn't.
Russ's circuit is setting up a alternating current in the loop. This causes the loop to radiate an electromagnetic field. The receiver will pick up that field when it gets close to the loop. More current will be launched into that loop (which is an inductor, BTW) his way than you will ever get by trying drive one end of it off a tank circuit.
One way to improve the efficiency might be to resonate the loop (possibly a multi-turn loop) with a capacitor, and drive it with a transistor operated in class C mode.

__________________
Ron

 

A tank circuit is a high impedance parallel resonant circuit. It should be driven from the very high impedance of the collector of a transistor, not the very low impedance output of a 555 where it won't do anything.

__________________
Uncle $crooge

 

seems to me that u have not understood the requirement .. :roll:

 

the resistor labelled 5600 is actualy a variable resistor .
u have an LC tuned ckt at the receiver . so u need to adjust the transmitting freq to that of the tuned ckt to get the proper o/p

The LC is the tuned ckt . the o/p is marked in the dia. when the ckt receives a tuned freq , it will give a sq wave o/p .

 

sorry for disturbing again, the transmitting part. u said that the 5600 is a variable resistor, but is is a VR should have 3 pins? and why there is a sentence "use 10k pot"??

for the receiving part, i only need to get a 27mH inductor? or a variable inductor?

thanks

 

yes , it may be a VR with a val set to 5.6K. yes a VR has 3 legs , tie the center with a side leg ..u get a VR with 2 legs

no need of variable inductor.
if ur planning to deploy the above system in a no/very low EMI region ,
then u can construct the receiver ckt as follows.
a receiver coil (not an LC) connected to a high gain amplifier(refer op-amp based amplifiers) + band pass filter (band freq=osc freq of 555).

 

the vcc is connected to the +ve terminal of the battery and the ground is connected to the -ve terminal of the battery right? but it doesn't work! i used a cro to measure the output (after the 0.1k resistor), it shows a constant -12v (supply is 12V dc).

how about the receiving part, the pin 7 of the opamp is connected to nothing?? where is the antenna of the receiver??

thanks for helping me..

 

the ckt is a normal 55 psc , and it will work .
u have wrongly selected the test point , measure the freq&voltage at the pin3 of 555. the 100R is placed for current limiting at the o/p section of 555.

the lm393 contains 2 opamps with open collector transistor , we are not using the second half of the ic , and it is left free.
probably u wont require an antand it should work in ur case) , the coil will pick the freq. and , if u need connect it to pin 2

 

a simple radio can be made by using LC components im attaching one simple circuit chech out

Attached Images

	mge106.gif (13.1 KB, 408 views)
	xmitter_schematic.gif (9.5 KB, 409 views)

 

so.. what i need to do is.. just follow the figure shown by Russlk, right? anything need to be modified? i couldn't get 1100resistor, can i use 1K? or 1K+0.1K??

 

go ahead..use 1k

 

yes!! thanks.. another question, is.. i couldn't find the 27mH inductor.. i only get the 10mH.. do i need to change anything to the transmitting circuit?? can i know the range of this circuit?? thanks for answering..