the designer said:

{{{Q1 and Q2 are wired as a Darlington pair to obtain the highest possible output from a Hartley type oscillator running at about 135KHz frequency. The 230Vac mains is reduced to 30Vdc without the use of a transformer by means of C1 reactance, a two diode rectifier cell D1 & D2 and Zener diode D3.}}}

now :
1- what is the use of the R and C in parallel
2- does this look like Hartley Osc. (nodes X ,Y should be connected to the input and the output of the amplifier)


to explain more about the circuit :

Pressing the pushbutton of the transmitter, a sound and/or light alert is activated in the receiver. The system uses no wiring or radio frequencies: the transmitted signal is conveyed into the mains supply line. It can be used at home, in any room from attic to cellar, simply plugging transmitter and receiver in the wall mains sockets. Transmission range can be very good, provided both units are connected to the mains supply within the control of the same light-meter.

Attached Images

	rx.png (6.8 KB, 307 views)
	tx.png (8.0 KB, 307 views)

 

The resistor is there to discharge the capacitor, as it could be left charged to a high voltage.

Looks fine - an oscillator requires positive feedback, and enough gain to over come any losses. From base to emitter of a transistor there's no voltage gain (only current gain), but the tap on the transformer gives the required voltage gain for it to oscillate.

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PIC programmer software, and PIC Tutorials at:
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[quote="Nigel Goodwin"]The resistor is there to discharge the capacitor, as it could be left charged to a high voltage.


but combination of the resistor and the cap ,, what does it do??

i reduces the mains voltage i guess but how ?

 

it does not reduce the mains voltage.

The resistor is in parallel with the capacitor so it must sit at the same potential as the capacitor.

Now that cap will charge upto some voltage. When power is removed that cap will still hold that potential and will (very) slowly discharge. Now if you go and play around with the cct when you have just turned it off then you will get zapped

Now what that resistor does is when the cct is in operation it will dissipate some power (hence why burn-resistors like these are in 10's and 100's of kOhms), when power is removed the capacitor now has a path to discharge its stored charge, alot quicker then via its leakage.

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Nothing is impossible.
Once a problem is realised, the rest is just details

 

I think we're assuming he knows more than he does?.

Simple answer:

The capacitor drops the mains voltage, by acting as a capactive dropper.

The resistor discharges the capacitor when it's unplugged for safety reasons.

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

 

tanks for the simple answer