![]() For this reason each coil needs at least three plugs. The dips are extremely DEEP A resistor from the 9 volt battery is connected to the coil center. Hand-effects are nonexistent if you tune using the edge of the large pointer disk. The tuning capacitor is a single gang shafted-trimmer mounted on a perspex plate. The circuit is basically a multivibrator with a center-tapped coil in lieu of the normal collector resistors. the output level falls slightly on some ranges towards one end but is 100% useable. My design has no unwanted peaks or troughs. Yes :) A lot of dip meters don't work very well as they tend to stop oscillating at one end of their range, or worse, fluctuate over the range. This ensures that capacitors C8 and C9 are connected to R5 for equal lengths of time. The buffer amplifier has a gain of 2 which is set by the ratio of resistors R6/R8. IC4 acts as a buffer-amplifier between the 3 volt peak-to-peak signal from the synthesis and the 5 volt logic for IC2. The voltage gain of IC1 is set by the ratio of R4/R3 which equates to 1000/50 = 20 (26dB) giving an overall gain approaching 72dB which is suitable for head-phone listening. To obtain this gain R5 has been connected to the low impedance output of the RF (radio frequency) amplifier IC1. The audio gain of the op-amp IC5 is set by the ratio of R7/R5 which equates to a voltage gain of 10000/50 = 200 (46dB). The purpose of this network is to further attenuate high frequency signals and noise. XC13=R7 where XC13 is the capacitive reactance 1/(2*pi*cutoff-freq*C13). The reason for the 2*multiplier is that the input signal is only presented to each network for half the time which effectively doubles the time constant. The values of 50 ohms and 0.47uF produce a cutoff frequency of 3000Hz XC8=2R5 where XC8 is the capacitive reactance 1/(2*pi*cutoff-freq*C8) The differential amplifier IC5 sums the positive and negative outputs from the two networks and passes the audio signal through C15 to the "audio output" terminal of J2. the second network comprises R5, the switch 2B3, and C9. The first network comprises R5, the switch 2B2, and C8. The reason for two networks is that all waveforms have a positive-voltage waveshape and a negative-voltage waveshape. The above circuit has two switched RC (resistor - capacitor) networks. Amplification is required as the input signals are very weak (microvolts).The RC combination determines the highest audio frequency that can be heard.The switching frequency determines the receive frequency.The amplitude of this tone will drop off rapidly once the difference frequency exceeds the cutoff frequency (3000Hz) of the RC network. If the the difference frequency is, say, 1000Hz then we will hear a tone of 1000Hz across the capacitor. Should the incoming signal differ slightly from the switching frequency then the capacitor will start to charge and discharge as it encounters different shaped segments of the incoming signal. Even though the incoming signal is well above the cutoff frequency of 3000Hz, the capacitor is always being presented with the same uni-polar DC waveshape and will charge to the average value of that waveshape. Let's now pass a high frequency signal through a switch that is opening and closing such that the same portion of the incoming signal is presented to the RC network described above. Should we open the switch before the capacitor has fully charged then the voltage across C will stay constant until the switch is again closed. If we close the switch and apply a DC voltage to the input, the capacitor will start charging to that value. The cutoff frequency for my circuit has been set to 3000Hz which means that there is no AC output for broadcast frequencies and above. Frequencies above the cutoff frequency are attenuated at a rate of 6dB/octave. This frequency, known as the "cutoff frequency", occurs when the reactance Xc of the capacitor is equal to the resistance R. Of particular interest to us is the frequency at which the AC voltage across the capacitor falls to 70% of the input. If we close the switch and apply an AC signal to the input, an AC voltage will appear across the capacitor, the amplitude of which will decrease with increasing frequency due to voltage divider action. The above circuit shows a switch, resistor, and capacitor connected in series. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |