Radio Circuit Description and Calculations

Discussion in 'Homework Help' started by anonymous11, Aug 19, 2010.

  1. anonymous11

    Thread Starter New Member

    Aug 15, 2010

    I have these two circuit (transmitter and a receiver) i have to understand the circuit and pick values for it. The operating frequency is meant to be 12 MHz and the transmitter circuit is meant to have 50mW MPT.

    I have attached the circuits and i will write up what i understand so far, please feel free to add to my description so i can get a clear picture of the circuits which should enable me to analyse the circuit by hand roughly then in LTSpice.

    Analogue Transmitter (1):
    The input is a 0-9V square wave that is manchester encoded (aka modulating signal).
    R1 is a filter, both R1 and R2 are potential dividers to make sure we don't blow up Q2
    Since the waveform is a square it's made up of many sine waves (Fourier series) so C1 is meant to filter out those high order harmonics ie it controls the width of bandwidth (anything else?)
    Q2 is a transistor switch which is meant to turn the crystal oscillator Y1 one and off.
    When Q2 is on R4 is shorted and oscillator is off and vice versa

    The term that was coined around was it's a OOK modulation (which is also known as ASK not sure if it's same or OOK is a form of ASK)
    So modulation is the act of adding information to an unmodulated signal (carrier), having 100% modulation for this circuit. With amplitude modulation there will be side bands (AM is for sine or square waves so for square we are doing ASK i am assuming). This will probably make the output like bursts of OFF (absence of carrier) with bunched up waves ON.
    Not really sure on this modulation, ASK,OOK thing

    Y1 is a crystal oscillator, it provides a stable frequency it has a very high Q factor (measure of sharpness or selectivity of frequency response of circuit). An oscillator is an amplifier with positive feedback from output to input. It has series and a parallel resonance frequencies (i can use the 12MHz specification but not sure which one to work out first what do it mean by series/parallel resonant frequency?). So over this very narrow band the crystal is basically inductive, but i do know i will be using the series one as it's a good approximation to the operating frequency. Which transistor is providing the feedback?

    R3 and R4 are biasing resistors which are meant to drive Q2 to active region (thus an amplifier), why do we need VDD? from specs max current through a BF199 is 25 mA and Beta is 40. They form a potential divider as well.

    C2 C3 and Y1 form the colpitts oscillator
    All i get is, in normal operation amplifier shifts the phase by 180, the tapped capacitance provides another 180 shift thus 0 in total (ie positive feedback?). Is there an RFC choke in this circuit?

    Q1 and R5 plus R3,R4 makes an emitter follower configuration, according to the reference sheet i have at A i am supposed to get about 4V V p-p with 5V DC.

    C4 blocks DC

    B C is meant to be a matching circuit (Pi network) which is meant to match the output impedance of source (oscillator i think, for resistance i can probably take R5 but not sure how to get the reactance). Pi network has a Q as well but it's different to the one from oscillator only thing it has in common is the operating frequency. I am not sure how to pick a suitable Q so can't really calculate the L C values. It's mostly meant to act as a filter and work at that operating frequency.

    C5 and C7 are meant to be trimming capacitors (not sure what they are doing in this cct for sure)

    Antenna is just a wire in this case so we are meant to pick a suitable resistance 50 ohm is appropriate.

    B is meant be about 4 V p-p and C is meant to be about 6 V p-p, not sure why it's 6 V cause i don't see any amplifiers there, meant to transfer about 5-35 is mW.

    Analogue Receiver (2):

    TP1 is meant to be about 100 mV p-p

    C1 is meant to be block any DC i think
    C2,C3 and L1 are meant to be part of a parallel tuned resonant at operating frequency (what does that mean exactly?)

    C4 no idea

    Q1 is an amplifier never seen anything like it, i have done small signal on a collector feedback circuit, someone said this is an emitter feedback?

    TP2 is meant to be 2 V p-p AC with 2 V DC (where does the DC come from?) the 2V is the amplification of 100mV AC.
    D1 is a rectifier diode with the resistor and capacitor it's a detector circuit (AM demodulation) it's half wave rectification where C is meant to boost the mean value of the output. So the wave will have a DC level and with the capacitor inside the RF bursts the square wave will be a smoothed signal?

    The first op amp is meant to be a filter not sure what else to make of it. R5 and R6 is a potential divider (why is that needed)

    C7 ensure there is no DC signal

    The opamp second one is meant to be a comparator and all that is connected to a connector. Inputting a high voltage at input to saturate the op-amp and compare the voltages at it's inputs, so when input AC is negative the output is a positive saturated square wave.

    I am not sure what R9,R10 are also potential divider (why is that needed) no clue about R7

    Both are meant to be used as non-linear op amps i think.

    TP3 is meant to be 5V manchester encoded signal that we put in from the transistor

    It's a bit to digest, my knowledge of receiver is far less then the transmitter so would like some concept cleared there and purpose of the components basically any explanations or references will be welcome. Then there is the case of analysis.

    Last edited: Aug 19, 2010
  2. gootee

    Senior Member

    Apr 24, 2007
    D1/R4/C6 form an envelope detector. The voltage at pin 3 of U1A should then be the same as the original modulating signal, without any carrier.

    U1A and R5, R6 form a simple non-inverting single-supply buffer amplifier, with R5 and R6 setting the gain.

    Depending on its value, C7 seems to be acting as a differentiator. For larger values, it behaves as a sort of edge detector and produces an exponential decaying transient for each edge in the incoming signal. For smaller values, it produces a pulse corresponding exactly with each rising and falling edge of the input.

    R7, R9, and R10 allow applying a DC offset to the + input of U1B, and set the ratio of the two inputs' voltages.

    Falling edges (or slopes) of inputs to U1A produce an upward-going rectangular pulse output from U1B, and rising input edges produce a small downward pulse at the output. If C7 is small-enough, the output pulses only occur during the rising and falling edges of a pulse input signal. Larger C7 increases the output pulses' widths without changing their leading edges' synchronization with the start of the input pulses' rising and falling edges.

    There are no negative voltages, anywhere in the circuit.

    It will help you to simulate this circuit in LTspice.

    It would have helped me if there had been component values included, and expected repetition rates, duty cycles, and frequencies etc.
  3. Efron

    Active Member

    Oct 10, 2010

    May this be usefull for anyone (only additional info has been added when required, otherwise assumptions in original message were considered as correct):

    So, R3, R1 and R2 have to be chosen in such a way that when Q2 in ON, it is effectively in saturation mode (to shunt R4)

    VDD is required to power up the circuit.

    The XTL should be chosen to operate at 12Mhz and it will stabilise the frequency. C2 and C3 will be chosen in cosequence as if the XTL was a self (knowing the frequency you can deduce the capacitance value). Be carefull, for the oscillator to work, C2 should be greater than C3, for example by a factor of 1.5 should be enough.

    Q1 is providing the feedback. The output is taken from the Emitter and the input from the Base, so C2 is acting as feedback link.

    R3 and R4 shall be chosen of high value in order not to load too much the resonant tank (XTL+ C1//C2) - example is 100K each

    There is no RFC choke

    the pi-network is effectivelly acting like a filter in order to clean up the signal coming from the oscillator phase (that includes much harmonics)

    Point A can move from 0V up to VDD (let's say 5Vcc typically). In order to have 4Vpp in B, point A should be stabilised at least at 2.5V DC, so R5 will be deduced this way if we fixe the current on it (if 1mA, R5=2500 => R3 and R4 can only now be fixed! )

    At resonant frequency, C2, C3 and L1 forms a very high impedances so the voltage in TP1 is max with respect to other frequencies (the receiver amplifier will then be effective only at the resonant frequency)

    C4 is used to isolate DC part and so allow the correct bias of Q1 (otherwise, the selft L1 will shunt to ground in DC).

    Q1 is in typical common emmitter amplifier configuration. Feedback is done through R2-R1 but this is another story.

    DC is coming from Vcc.

    Understanding this is important. At TP2, and after C5 you don't still have a square wave but a high frequency signal which envelops tends to a square wave (with distorsion). The effect of D1 together with R4 and C5 is AM demoduling.

    The next stages are used to amplify and clean up this signal to a clean square wave, where frequence will be the same as the original input.

  4. radiohead

    Senior Member

    May 28, 2009
    Ananymous11, you should consider taking an op-amp tutorial, then build some very basic circuits using op amps on a breadboard. This will help you understand why the components that satellite the chip are necessary, and what happens when you change their values.
    Neil Groves likes this.