Hello all,

I would like to make a very simple and preferably small transmitter on perfboard/prototype board (I do not have the materials to make PCB's).

Now while most of the circuits I find online are for FM transmitters, I am open to any other methods of tracking that you guys might have. For the FM transmitters, I found this to be the coolest/easiest one since it does not require a pcb: 1.5 Volt Tracking Transmitter . However, the LM3909 is discontinued. Can a 555 IC perform the same operation as that LM3909?

Any ideas will be helpful, but remember, I want simple, non-pcb, and somewhat small. Thank you.

Edit: Also, if there is a way to minimize the error in making a hand made inductor out of wire that would be great. By this I mean, I would prefer to have a variable inductor that I can just put a screwdriver to (like a potentiometer) and modify it's value to get the desired frequency

 

The very old LM3909 IC doubles the 1.5V to 3.0V for the transmitter to operate.
A 555 IC has a minimum supply voltage of 4.5V.
Try using a 6V battery and a 555 turning the transmitter on and off.

The transmitter uses a trimmer capacitor for tuning. It is adjusted with a plastic screwdriver.

__________________
Uncle $crooge

 

the 7555 can run off of 2VDC.

ICM7555, ICM7556 Low-Power, General-Purpose Timer

__________________
MechTronics

World of Warcraft is a disease

 

But its minimum output high current is only 2mA when its supply is 5V and its current is much less when its supply is less. The transmitter needs more current.

__________________
Uncle $crooge

 

What about using the TLC555? Here is the datasheet: http://focus.ti.com/lit/ds/symlink/tlc555.pdf

This IC is from Texas Instruments and can operate at as low as 2VDC. I cannot seem to find how much current it can source/sink at 2VDC from the datasheet. It says it can sink over 100mA and source over 10mA but those are probably not the numbers for a Vdd of 2V. If someone can tell me where on the datasheet this information is found (and what they call it) I would greatly appreciate it.

If the TLC555 still cannot provide the current needed, is there another IC that does what the LM3909 does (double the voltage)? I don't mind operating a 555 on 6V but if I don't have to then I'd prefer not to (to minimize size). Thank you guys for your responses.

 

I looked on the datasheet for the LMC555 that spec's only with a 5V and higher supply.

The datasheet for the TLC555 has spec's for a 2V supply and the output voltage drops to only 1.5V when its load current is only 300uA which is next to nothing.
The transmitter needs much more current.

I think the range of the transmitter depends on its supply voltage.
With a 2V supply its range might be across the street (30m).
With a 6V supply it might be 200m.
with a 5V supply it might be 400m.

__________________
Uncle $crooge

 

why does 5 V give more range than 6 V ? I think you got your figure switched

remember hugo that as distance doubles the power will be quartered when its three times the distance the power will be 1/9

__________________
I AM the exeption that disproves the rule in many ways but the rules still apply (unfortunately)

my site:www.simons-photography.com
http://rushdenrotaract.org.uk
see also
http://www.bigstockphoto.com/account...fid=m2URATYch5
http://www.redbubble.com/people/simonsphotography

 

Where did you see 300uA? Could you please let me know what page and what area of the page you are getting this from because I cannot seem to find it. Is it the Idd (supply current) that tells you how much current it outputs? If so... the data sheet says 250uA at 25degrees and 400uA for full range (these are max values).

 

Are you talking to me? I don't know what you mean with this. I would think that the higher the voltage the longer the range.

Yes, the power is inversely proportional to the distance squared. I graduated with a bachelors in electrical engineering so I have the book knowledge, now I just need the practical knowledge (which is only acquired by doing... which is why I am here )

 

Sorry. I punched 5V instead of 9V.

__________________
Uncle $crooge

 

There are several tables given on the data sheet. One table list test conditions @Vdd = 2 volts. the other tables 5v etc.

 

Yeah, I guess my question is which value on that table is the current that is present at the output of the IC (how much current does it source/sink)? Or does it always source 10mA and sink 100mA as long as the voltage source is between 2-15VDC?

 

On the page with the spec's for a 2V supply, it lists the Output Voltage with 300uA sourcing current (output high).

The output current from a Cmos IC depends on its supply voltage.
The output low current with a 15V supply is a minimum of 100ma with a max allowed current of 150mA.
The output high current with a 15V supply is at least 10mA with a max allowed current of 15mA.

The 2V page shows the minimum output high current of 300uA.
Didn't I repeat this on another website?

__________________
Uncle $crooge

 

I am really surprised that someone with a EE cannot read a simple spec from a data sheet.

 

Well, like I said before, book knowledge != practical knowledge. I wish my college would have made practical application of the learned material a requirement but unfortunately they didn't and so now I am trying to learn it on my own.

Yes, it is a simple data sheet, and it is easy to read numbers, I would just rather have someone tell me exactly which value on the table corresponds to the current source/sink of the IC instead of just going with what I think is the correct value and not being sure. If you solve a problem and you think the answer you got is correct when indeed it is not, then all you have done is learned the wrong way of solving that problem.

Besides, having a degree has nothing to do with how much you know or how smart you are. It only decides how much you are going to get paid