How to build a sensitive crystal receiver?

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 Go to the calculator at the bottom of this page And calculate the components of your receiver for maximum sensitivity at weak signals Circuit diagram 1 Circuit diagram of the crystal receiver, which we are going to design for maximum sensitivity at weak signals. This can be a detector circuit of a 2 circuit receiver. But also a receiver with loop antenna.RP represents the losses in coil L and tuner capacitor C1 Circuit diagram 2 If an antenna is connected via a matching network (in this case C2) to the LC circuit, this will reduce the circuit Q. When optimal matched to the antenna, the Q will be halve the value of the unloaded Q (of L,C1). For a calculation of the values of C1 and C2, click here.For the calculation of maximum sensitivity: replace the LC circuit + antenna by a LC circuit with an unloaded Q equal to halve the original value. Coil value L stays the same.

You get maximum sensitivity if there is maximum power transfer from LC detector circuit to load (loudspeaker).
The LC circuit has a certain parallel resistance RP, this is not a real resistor, but a virtual resistor caused by the losses in coil and tuner capacitor.

If we know the Q-factor of the detector circuit, and the induction of the detection coil, we can calculate the value of RP as follows:

RP = 2.pi.f.L.Q (Ohm)
pi = 3.14
f = the frequency (Hertz).
L = induction of detector coil (Henry = H).
Q = the quality factor of the unloaded LC circuit.

The value of RP is depending on the frequency, use for instance the value of RP in the middle of the frequency range.
For medium wave for instance, we can use the value of RP at 1 MHz.

Maximum power transfer from LC circuit to load.
If the load resistor RL was directly connected across the LC circuit, it would be simple:
maximum power transfer would occur if RL=RP.
The loaded Q of the LC circuit is then halve the value of the unloaded Q.

But of course we also need a diode between LC circuit and load RL, to demodulate the RF signal.
In the rest of this article we assume the RF signal on the LC circuit is not modulated, so the output of de diode is a DC voltage.

Maximum sensitivity at strong signals.

With a very strong signal across the LC circuit, the diode shall work in the linear detection region.
If the input voltage is high enough, the diode will give only very few power losses, compared to the rectified power.
In the calculation (with strong signals) I assume the diode has no losses at all.
This can in practice not be reached, but it rather simplifies the calculations.
If the diode has no losses, the DC voltage across the load resistor will be equal to the peak voltage of the RF signal.
This peak voltage is 1.41 times the RMS value of the input signal.

In this situation, maximum power transfer to the load RL will occur if:
RL = 2 x R
P
The loaded Q of the LC circuit is then halve the value of the unloaded Q.

However with strong signals, maximum sensitivity is not a very important subject, output power is already high enough to be heard.
We can better design the receiver for maximum sensitivity at weak signals.

Maximum sensitivity at weak signals.

When receiving (very) weak signals, the diode shall work in the square law detection region. Circuit diagram of the diode. With the equivalent circuit of the diode at low signal levels

The input of the diode behaves like a resistor with value RD parallel to the LC circuit.
The diode output is like a DC voltage source in series with a resistor RD.
Maximum power transfer from DC voltage source to load RL occurs when:  RL = RD.

In the square law detection region, the detected DC voltage is proportional to square of the RF input voltage.
The power in the load resistor is proportional to the square of the detected DC voltage.
In other words, the power in load RL is proportional to the 4th power of the voltage across the LC circuit.
So it is important to make the voltage across the LC circuit as high as possible, this is done by making the impedance of the loaded LC circuit as high as possible.

If we make all impedances equal, so RL = RD = RP the loaded Q of the LC circuit would be equal to 0.5 times the unloaded Q.
The voltage across the LC circuit is then also 0.5 times the unloaded voltage.
The output voltage is then proportional to: 0.5^4 = 0.0625
If we however make RD and RL 3 times as high, so RL = RD = 3xRP the loaded Q of the LC circuit will be 0.75 times the unloaded Q.
The voltage across the LC circuit is then also 0.75 times the unloaded voltage.
The output power is then proportional to (0.75^4)/3 = 0.3164/3 = 0.1055.
We divide by 3 because the load resistor is now 3 times higher.
So we now have 0.1055 / 0.0625 = 1.6875 times more output power (compared to the situation: RL = RD = RP).
And also, the loaded Q is now 1.5 times higher, so a better selectivity for the receiver.

In the next Excel file: diode.xls the relative output power is calculated for many values of RD and RL, where RD=RL=3xRP is the most sensitive combination.

Conclusion: At weak signals there is maximum power transfer from LC circuit to load, so maximum sensitivity for the receiver if:

Increasing the circuit Q

Increasing the unloaded Q of the LC circuit, will make the impedance of the LC circuit higher.
Because of this the voltage across the LC circuit will increase, the diode then will work more efficient, and output power increases.
Very good crystal receivers make use of LC circuits with an unloaded Q above 1000.
A high Q also gives a selective receiver.

Increasing coil induction
The impedance of the LC circuit can also be increased by increasing the coil induction (μH value).
However making the induction higher, can also result in a lower circuit Q, compensating a part of the impedance increase.
And a lower Q of course gives worse selectivity.
Practical values for induction of medium wave receivers are
200-300 μH.

Choice of the diode
If we know the diode resistance RD, we can calculate the diode "saturation current" (Is value) with the formula:
Is = 0.000086171 x n x TK / R
D
Is = saturation current of the diode in A
n = ideality factor of the diode, if you don't know the value, then take n=1.08
TK = temperature in Kelvin (= temperature in ºC + 273)
RD = diode resistance in Ω

Now search for a diode which has a Is value close to the calculated Is value.
Or connect several diodes with a low Is parallel, to reach together the desired value.

Diode connected to a tap on the coil.
The best sensitivity is always reached if the diode is connected to top of the LC circuit, and
RL=RD= 3xRP.
But not always it will be possible to make the load resistance RL high enough.
If we don't have a load (audio transformer) with a high enough impedance, the LC circuit will be loaded to heavy, and the Q will reduce.
A method to prevent this, is to connect the diode on a tap somewhere on the coil, instead of the top of the coil.
The Q of the loaded circuit then will increase.
With the calculator at the bottom of this page (click here) the effect on output power can be calculated when using a tap on the coil.

Audio transformer.
When making a receiver with a high Q factor, the load resistance will have to be very high, for instance some Mega-Ohms.
In that case we need a transformer which transforms this high impedance down to the impedance of the used loudspeaker.
Transformers with such a high input impedance are hardly to find.
They can however be self made, look for instance at my transformer unit 2
At high input impedances, in practice both efficiency and bandwidth of a transformer will reduce.

Loudspeaker
Of course we use a loudspeaker with a sensitivity as high as possible.

Calculate optimal components for detector circuit

With calculator 1, diode and load resistor are calculated for which the sensitivity of the detector circuit at weak signals is maximum.
Fill in the yellow coloured fields, and click on "calculate calculator 1".

These values can then be taken over in calculator 2.
Then you can self change in calculator 2 all parameters of the detector circuit (all yellow fields), and calculate the effect of this on sensitivity and loaded Q.

 Frequency   f = kHz Coil value   L = μH Unloaded Q Temperature of the diode   T = ºC Ideality factor of the diode   n =
 Impedance of the unloaded LC circuit: kΩ Load resistor    RL = kΩ Diode resistance   RD = kΩ Diode "saturation current"   Is = nA  at ºC Diode "saturation current"   Is = nA at 25 ºC Loaded Q at weak signal Loaded Q at strong signal

Calculator 2: change the values yourself and compare sensitivity and Q with optimum values.

 Frequency    f = kHz Coil value   L = μH Unloaded Q factor Temperature of the diode   T = ºC Diode "saturation current"    Is = nA at ºC Ideality factor of the diode   n = Diode at tap of coil at: 100 % 95 % 90 % 85 % 80 % 75 % 70 % 65 % 60 % 55 % 50 % 45 % 40 % 35 % 30 % 25 % 20 % 15 % 10 % DC bias current through diode   Ib = nA Load resistor    RL = kΩ
 Impedance of unloaded LC circuit: kΩ Diode "saturation current"   Is = nA  at ºC Diode resistance   RD = kΩ  at ºC Output power at weak signal X = dB   with regard to calculator 1 Output power at strong signal X = dB   with regard to calculator 1 Loaded Q at weak signal = X   with regard to calculator 1 Loaded Q at strong signal = X   with regard to calculator 1

Example 1.
We have a LC circuit with a 200 μH coil, the unloaded Q is 1000 and the frequency is 1000 kHz.
As load we use a 100 kΩ audio transformer, because this is the only one we have.
What is the output power at weak signals compared to the optimal load resistor.

Fill in the following values, in calculator 1:
f = 1000 kHz
L = 200 μH
Q = 1000
Click on "calculate calculator 1" (The optimal load resistor and diode are now calculated).
Click on "take over values from calculator 1"

Change load resistor RL in calculator 2 in 100 kΩ.
Click on "calculate calculator 2"
We now see the sensitivity at weak signals is 9.97 dB lower compared to the optimal load.
It's also interesting to see, the Q at strong signals is only 38

Example 2.
We have 2 receivers, both with a 200 μH coil.
What is the difference in output power between the 2 receivers?

Fill in the following values, in calculator 1:
f = 1000 kHz
L = 200 μH
Q = 1000
Click on "calculate calculator 1" (The optimal load resistor and diode are now calculated).
Click on "take over values from calculator 1"

Reduce the Q in calculator 1 to: 100.
Click on "calculate calculator 1" (The optimal load resistor and diode are now calculated).
Click on "calculate calculator 2".

The result is, the high Q receiver gives 1000 times more output power (+30 dB) at weak signals.
At strong signals, the difference in output power is 10 times (+10 dB).

Example 3.
We have a simple crystal receiver, with a 200 μH coil, and an unloaded Q of 200.
As diode we use a germanium diode with Is = 400 nA (at 25 ºC).
The load resistor is 47 kΩ.

We are going to compare the sensitivity at 1000 kHz, with a high quality receiver, with L = 200 μH and unloaded Q = 1000.

The values of  the simple receiver can directly be entered in calculator 2.
We don't have to calculate first the optimal load and diode, because this receiver isn't using optimal load and diode.
So, in calculator 2 we fill in:
f = 1000 kHz
L = 200 μH
Q = 200
Is = 400 nA (at 25 ºC).
RL = 47 kΩ.

Fill in the values of the other receiver, in calculator 1:
f = 1000 kHz
L = 200 μH
Q = 1000
Click on "calculate calculator 1" (The optimal load resistor and diode are now calculated).
Click on "calculate calculator 2" the differences between both receiver are calculated.

Result: the simple receiver has 29 dB less output power at weak signals.
The simple receiver has a loaded Q of only 60 (at weak signals).

At strong signals, the loaded Q even reduces to 17, but with such a low sensitivity it's unlikely to get a strong signal across the LC circuit.

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