One should not continue the experiment with a radio transmitter with >2km range as it is risky and polluting the radio dial of community.. if your frequency or ghost signals is somewhere over a particular local FM radio station then you are at high risk!
One should not continue the experiment with a radio transmitter with >2km range as it is risky and polluting the radio dial of community.. if your frequency or ghost signals is somewhere over a particular local FM radio station then you are at high risk!
Are you talking about the TL499 IC that was used in the Australian project?
Maybe it is obsolete because the biggest American parts distributor (Digikey) doesn't have any in stock but they say they can get thousands of them.
Some manufacturers say if a part is obsolete but not Texas Instruments.
Here is a very simple boost using transistor and 555 timer. If you want 9V out, just change to a 9V zener on the output. With 3V input, you can get 9V out at about 12 mA.
9v SUPPLY FROM 3V You can replace a 9v battery with this circuit.
The output is about 10.4v on no load and 9.6v @30mA .
The advantage is the voltage stays over 9v for the life of the cells.
A normal 9v battery drops to 7v very quickly.
The output voltage is set to 9-10v by the 6k8 and 390R resistors. The 470R gives the circuit an idling current of about 20mA and the spikes are about 75mV.
By increasing the 470R, the quiescent current decreases but the voltage drops more when the current is 30mA.
ppl,
i need a 3V to 9v converter circuit for my new project, so i want know and learn about them, can anyone post a schematic that is easy to build? I got a 3.7V battery with 750mA current and want 9v or 12V out of it
9v SUPPLY FROM 3V You can replace a 9v battery with this circuit.
The output is about 10.4v on no load and 9.6v @30mA .
The advantage is the voltage stays over 9v for the life of the cells.
A normal 9v battery drops to 7v very quickly.
The output voltage is set to 9-10v by the 6k8 and 390R resistors. The 470R gives the circuit an idling current of about 20mA and the spikes are about 75mV.
By increasing the 470R, the quiescent current decreases but the voltage drops more when the current is 30mA.
Adding a 5K from Q4 base to ground would speed up the turn off. Thanks BH.
Collin, I don't think 30 ma is enough to run his circuit, but like you said, not sure which one he is using.
heya all...
Welcome back again to this thread!. I built the circuit that colin posted and made following changes in the circuit.
Replaced bc547 with bc548, replaced bc338 with s8050 and built inductors on a ferrite core which i got from a tiny transformer and used copper wires (don't know abt its thickness but it's small). Powered the circuit using a 3.7v battery (not 3V),. Result: i got 3.65V in output and i realized that the s8050 transistor is getting very "hot"!!. What should i do?