plzz..help..electronics eavesdropping device detector

[opampopamp]

i hv a circuit wid me..its an electronic eaves dropping device detector..i'm doing it as project..nd want 2 include the design of resistors and capacitors in my project report...so plzz..help..i want the design...its urgent..

R1 = 39 K; C1 = 10nF ceramic
R2 = 470Ω; C2 = 47µf/16V electrolytic
R3 = 22Ω; C3 = 10nF ceramic
R4 = 100Ω; C4 = 1nF ceramic
R5 = 560Ω; C5 = 10nF ceramic
R6 = 1 MΩ; C6 = 10nF ceramic
R7 = 220 Ω; C7 = 10nF ceramic
R8 = 100 KΩ; C8 = 10nF ceramic
R9 = 4.7 KΩ; C9 = .1µF ceramic
R10 = 1.5 KΩ; C10 = 10 µ/16V electrolytic
R11 = 10 KΩ; Chip :
R12 = 220 Ω; IC1 = LM324 1C
R13 = 10 K; Transistors :
R14 = 100 K; TR1 = BFR 90A
R15 = 470 Ω; TR2 = 2N2222
R16 = 47 K; TR3 = 2N2222
R17 = 47 K; Diode :
R18 = 1 K; D1 = NTE 112
R19 = 10 K; Potentiometer :
R20 = 100 K; P1 = 4.7K
R21 = 47 K; Other :
R22 = 10 K; ΑNT = telescopic antenna. Radio type.
R23 = 47 K; LS = Loudspeaker 8Ω

 

[opampopamp]

this is hw d ckt works...plz..help...
The detector detects an RF signal ranging from 1 to 1000 MHz. Figure 1 shows the schematic diagram of the detector. When the electronic bug operates, it emits an RF signal. This signal is received by the antenna (point 1 of figure) and is applied to the base of the transistor TR1 through a high-pass filter. The high-pass filter consists of the capacitors C4, C5 and the resistor R7.
When low frequency signals are applied to the input of the filter, the capacitors C4 and C5 operate as open switches and the filter rejects the low frequency signals. The cut-off frequency of the high-pass filter is defined to be 60 Hz rejecting any interference which come from the mains. The transistor TR1 is in common collector conjunction and it is used to amplify the signal. The gain that is provided is 10dB for a signal that varies in the frequency range 1 - 1000 MHz. The resistors R1, R2, R8 form the biasing network for the transistor TR1. The amplified signal is then applied, via the capacitor C6, to the anode of the diode D1. The diode is particularly manufactured to operate in high frequencies, since a common diode can not go from the conduction condition to the cut-off condition very fast preventing the appearance of reverse currents.
However the operation of the diode in the high frequencies range results in the appearance of conduction at the beginning of the negative half period. The potentiometer P1 adjusts the proportion of the signal at the terminals of the diode D1 and applies it to the inverting input (pin 2) of the operational amplifier A1. The operational amplifier A1 is configured as a very high gain amplifier. The value of the gain is defined by the capacitor C8 and the resistors R6 and R12. With no signal input from the antenna, the output of A1 at pin 1 is near ground potential. When the antenna detects a signal in the range 1 - 1000 MHz, it applies it to the base of the transistor TR1 producing a negative-going voltage at the cathode of diode D1.

That voltage is applied to the inverting input of A1 which amplifies and inverts the signal, producing a positive-going output at pin 1. The next stage after the opamp A1 is a voltage controlled oscillator (VCO). The oscillation frequency of VCO is controlled by the output voltage of the operational amplifier A1. The operational amplifiers A2, A4, along with the resistors R11, R13, R14, R16, R17, R18, R19, R20, R21, R23, the capacitor C9 and the transistor TR2 are arranged to form a voltage-controlled oscillator (VCO) that operates out of the audio-frequency range. As the output of A1 increases, the frequency of the VCO increases. The VCO output, at pin 8 of the operational amplifier A4, is fed to the input of A3, which is configured as a non-inverting unity gain (buffer) amplifier. The output of A3 is used to drive the transistor TR3, which, in turn, drives the output speaker

 

[mneary]

You should try to write whole words and assemble them into your own sentences instead of copying a whole page straight from here. You might accidentally learn something.

 

[audioguru]

The circuit doesn't work. People complained about it at Electronics-Lab.

The VCO's old quad opamp has such a high input offset voltage that it amplifies it and causes most of the opamps to be saturated or cutoff all the time.
It does not have a null adjustment for its input offset voltage.

Its input will pickup and be overloaded by nearly every radio and TV station in town.

I corrected errors in its schematic but it still doesn't work.