Fidelity

Fidelity
The fidelity of a receiver is its ability to accurately reproduce, in
its output, the signal that appears at its input. You will usually find the broader the band passed by frequency selection circuits (due a lower Q in the inductor - or more
resistance in the tank circuit), the greater your fidelity.
You should remember that good selectivity requires that a receiver pass
a narrow frequency (or better Q/selectivity) band.
Good fidelity requires that the receiver pass a broader band to amplify
the outermost frequencies of the sidebands. Receivers you find in general use are a compromise between good selectivity and high fidelity.

FM

Frequency Modulation

The carrier frequency is constant.

The modulating frequency has the following characteristics.
1.  Amplitude - the more intense the sound, the higher the amplitude.
2. Frequency - the frequency of the modulating signal - usually 20 to 200kh for voice.

The greater the amplitude, more of the fm carrier frequency is modified.
The higher the modulating frequency, the higher the frequency imposed on the carrier wave.
For example, a modulating frequency of 10,000 Hz, will cause the carrier frequency to increase 10kHz. (The carrier will change frequency 10,000 times per second.)

Calculating the Total Current in a Mixed Circuit

First determine the resistors which are in parrallel and series.

In this case, the 10K is connected to the positive and the negative rail - making it parallel to the 10K variable Pot - which is also connected to the positive and negative rails. (the 3rd pin of the LM386 Amp is positive).
The 1K Pot is also connected parallel.

The circuit basically looks like this



As you can see, there are three parallel resistors. The total resistance is
1
---------------------------------
1/r1 + 1/r2 + 1/rn

= 833 Ohms (Assuming the variable resistors are set to maximum resistance

To calculate the current we can use ohms law

I = 9v / 833 = .01 Amp

.01 x 1,000 = 10mA

However, we still have to take into account the capacitors. How do we measure the resistance of capacitors ? Capacitors are reactive components, so their reactance or resistance must be measured when a AC signal is applied. The formula is:

Xc =          1
   _______________
          2*3.142*f*C

Where Xc is the Capacitive Reactance, f the Frequency in Hertz and C the farad.

The following capacitors are in parallel in the circuit.
100uf
.047uf
.047uf
220uf
.01uf

TOTAL:320.1uf / 1,000,000 = 
C =.00032 FARADS

Xc =              1  
_____________________
2 * 3.142 * 125,000,000 * .00032

Xc = 0.00000398 Ohms

The following are in series:
.0047uf
.1uf
Total= 220uf = .0045uf  / 1,000,000
C =.000000004.5 FARADS

Xc =                   1
______________________
2 * 3.142 * 125,000,000 * .0000000045
 
Xc = 3.97 Ohms

Now we have the series bunch of caps, and the parallel bunch of caps, in series - so calculating the capacitance and hence the impedance of the two "bunches" gives us approx:

TOTAL ~ 3 ohms

as you can see with this high frequency, the reactance of the capacitors is quite minimal

There are two or three that i dont now whether they are parallel of series.  Lets leave them out.

So lets say the total resistance in this circuit is then about 835 Ohms which gives us approximately 10mA consumption

We also have to calculate the Inductive reactance.  

The formula for Inductive Reactance is:

Xl = 2 * 3.142 * f * L

Where L is in Henries and f is in Hertz

So given our tank inductor to be approx .211 uH this equates to

.211 / 1,000,000
=.00000021 Henries

One can either calculate the individual impedances for each of the inductors (like for L1 below), or calculate the individual henries.

Xl = 2 * 3.142 * 125,000,000 * .00000021
Xl = 164 Ohms

For L2 9.3mH

Now L2 is connected from a positive to a negative, whilst L1 is connected to the positive rail so we can
say that these two inductors are connected in Parallel.
 
L1  = 1 / .211 = 4.74
L2 = 1 / 9.3 = 0.107

L = 1 / (4.74 + 0.107)
L = .206 / 1,000,000 (to convert from macroHenry to Henry)
L = .0000002

So Z (impedance for both the inductors is)
Z = 2 * 3.142 * 125,000,000 * .0000002
Z = 162 OHMS

So we now have

162 + 4 +833
= 998 Ohms

I = 9 / 998
I = 9mA drawn from the circuit not taking into account the FET (say 5mA) and IC 386 (4mA)

So total current drawn
18mA

Decibel

Decibels are measured in Logarithms which is the inverse to exponentials.  
They measure a RATIO - ValueA/Value(Reference) and they are then multiplied by LOG

Value Reference is the faintest sound a human can hear and is measure by the following:

1*10-12 W/m2    (which is defined as 0dB)

This is referred to the Threshold of Hearing.
If the sound level increases by 100 times, then we have a value of 20dB ( since 10 base to the power of 2)   -->
10 * (Log100)
Log of 100 = 2
10 * 2 = 20dB

Similary,an increase of sound of 1,000,000 from the threshold of hearing, gives us 60dB
10 * (Log1,000,000)
Log of 1,000,000 = 6
10 * 6 = 60dB


POWER = BASE10 (10(LOGX))  where logx is   Xa/Xb

Take for example, 10 exp3 (10 to the power of 3),  = 1,000

The inverse of this is the LOG - The Decibel uses base 10 logs (the factor of 10 puts the deci in the decibel)

  10Log1000
= 10*(Log1000)
= 30

E.g If speaker A produces 20 times more power than speaker B, we can measure this in decibels:

10*(Log20)
=10 * (1.3)
=13 Decibels (dB) difference between the two speakers.  

Another example
To calculate the ratio of 1 kW (one kilowatt, or 1000 watts) to 1 W in decibels

10 * (Log (1000/1))
10 * Log 1000
10 * 3
=30 dB

SOUND, AMPLITUDE, VOLTAGE AND CURRENT = BASE20 (20(LOGX))

Electric power through a resistor is defined as the square of the voltage (or current)
The square of x is written as x to the power of 2

Since the ratio between two levels are now squared, the formula becomes
20*(LogX)

Just say circuit A produces 20 Volts, and circuit B produce 10 volts
20*(Log (B/A))
20*(Log(10/20))
20*(Log(0.5))
= -6 dB
Circuit A produces 6dB less than circuit B

Calculating Inductance

To calculate the inductance of an Air, one layer coil, use the following formula:

L = 0.001 N2r2 / (228r + 254l)

where L is the inductance in henrys, r is the coil radius in metres, l is the coil length in metres (>0.8r) and N is the number of turns.
(source: http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html)

For example:
N= 7 turns
r = .004 (4mm) - radius of coil
L=.011 (1.1cm)

• .001 * 7 squared * .004 squared / (228 * .004) + (254 * .011)
  
• .000000784 / 3.706
• .00000021154
• convert to micro Henry (multiply by 1,000,000 since micro is a million)
• .21uH Inductance

Loading

The collector output may be used to drive other circuits (like the differential amplifier in the previous circuit in place of the crystal oscillator) but it may be desirable to add a few thousand ohms in series to prevent excessive loading of the oscillator.

So, loading maybe where too big or strong a signal/voltage is impressed onto a device/circuit.

http://www.techlib.com/electronics/amxmit.htm

***************

Archimedes wrote:
> When I increase the antenna length to say 30cms, from the recommeded
> 6inchs or 15cms, the circuit does not work.

The device is a super-regenerative receiver. The antenna
is part of the oscillator circuit. If you change the
length of the antenna, you change the resonant frequency
and/or possibly cause the oscillator to stop oscillating.
Either of those results will cause the receiver not to
work right.

**********************

Antenna tapped to Inductor in a Tank Circuit

The antenna acts like a capacitor.  Hence the antenna is like a capacitor connected to the inductor in series.
The capacitance of the antenna is reacting with the inductance of the coil to resonate at the frequency of the radio station.  Increasing the antenna length increases the capacitance and throws off the resonant frequency of the tank circuit.

The resonance frequency is dictated (essentially) by the contribution of the complete coil length.
The reason for the tapping is to better match the aerial impedance to the tank circuit, also (in
this case) to reduce the influence of the aerial upon the resonant frequency. You may find best
performance at an even higher tap ratio.

Best
Drew

Operation of a Monostable circuit

In its stable condition (T0 - Time 0)), Q1 is cut off and Q2 is conducting.
The input trigger (positive pulse at T1) is applied to the collector of Q1 and coupled by C1 to the base of Q2 causing Q2 to be cut off (Since these are PNP transistors, a negative signal is needed for the transistor to operate).

The collector voltage of Q2 then goes −VCC (This is because R5 produces a voltage drop and the juntion between R5 and the collector of Q2 is the source voltage -Vcc).

The more negative voltage at the collector of Q2 forward biases Q1 through R4 (this means that Q1 has a more negative signal applied to its base viaR4).

With the forward bias, Q1 conducts (from its off state), and the collector voltage of Q1 goes to about 0 volts (remember, for a transistor to operate, the base voltage is forward biased whilst the collect leg is Reverse biased - Now this means that, for a PNP transistor - Common Emitter, when Q1 is off, the Collector is -Vcc (because the resistor has the -Vcc voltage drop and this is taken with the junction of the collector leg, and the Transistor has a 0 Voltage Drop) - When Q1 turns on, the current at the collector increase, which causes a voltage drop across the transistor, which in turn reduces the current (and hence voltage) at the R3/Collector junction (from say -5v to 0v).

C1 now discharges and keeps Q2 cut off. Q2 remains cut off until C1 discharges enough to allow Q2 to conduct again (T2). When Q2 conducts again, its collector voltage goes toward 0 volts (because of the voltage drop across Q2) and Q1 is cut off. The circuit returns to its quiescent state and has completed a cycle. The circuit which in turn remains in this stable state until the next trigger arrives (T3).