1. I do not know what frequency you need to fix.
For the convenience of modeling (process speed), I used a frequency of 10 megahertz.
The magnetic antenna at this frequency is an inductive impedance.
I arbitrarily pictured frame inductance and the voltage source.
Tell me what frequency you need and what kind of loop antenna you are using.
Then I may clarify an equivalent circuit.

The loop antenna should be copper wire with a square shape,it can be about 13cm length for 2.4Ghz or 32cm length for 1Ghz.I understand that the loop should be in resonance,so i don't understand why do i need to add the inductor?
I saw your third diagram and it seems that the second one(Sziklai pair with one germanium transistor and one silicon) is the most suitable.

2. Resistor 100k greatly reduces the sensitivity of the whole scheme and need for this resistor is not.

Is it safe to leave the base without any resistor that will limit the current?
Won't it damage the transistor?
Refer to adding a tank circuit for selectivity,i read about the series tank circuit and paralell one.
But i couldn't figure out,how to add it to my circuit or how to determine the value of the capacitor and the inductor.And if it won't disrupt the circuit at all.
I saw the formula:f = 1/2π√(LC),and i can find C if i know the L and opposite,but there are a lot of combinations that can give the same f,while i saw that it is necessary to add a resistor to the parallel tank circuit.
This is beyond my knowledge and i may need assistance with that.

 

Inductance I added as I simulated at a frequency of 10 MHz. In this frame rate is very high. A small frame has the inductive resistance. Impedance full frame lies in the range of 50-300 ohms. The thicker conductors, the lower the resistance of the frame (but you need to take into account the velocity factor). In your case, instead of using the inductor resistance of 100 ohms. If you need to rough up the sensitivity and allow a large input signal, then put resistor 1 kohm. It has little effect on the sensitivity, but to protect the transistor.

 

I had some difficulty understanding this post.

Inductance I added as I simulated at a frequency of 10 MHz. In this frame rate is very high. A small frame has the inductive resistance. Impedance full frame lies in the range of 50-300 ohms

When you mentioned the word"frame",does it refer to the loop,to the length of the loop in proportion to the wavelength?(while saying"full frame"was meant to a 1 wavelength loop?

So,in my case i don't need any inductor,do i?
I meant,if the loop is 13cm length,at 2400Mhz(or 30cm at 1000Mhz)it will be at resonance with pure resistance without needed any inductor component?
Also,in your circuit,if the loop was 30 meter at your 10MHz,you would not need to add the inductor,would you?

If you need to rough up the sensitivity and allow a large input signal, then put resistor 1 kohm. It has little effect on the sensitivity, but to protect the transistor.

What is the maximum voltage that can be applied on the base before it will be necessary to add a resistor?I understand the reason for omitting the resistor:

2. Resistor 100k greatly reduces the sensitivity of the whole scheme and need for this resistor is not.

But i still didn't understand why it is possible to omit the base resistor,while there isn't any resistor,at least,on the emitter?

 

You're right . Inductance is not needed. Input germanium transistor base current is approximately equal to one microampere. Applying the inclusion in base 1kOhm resistor lose 1mV. 100kOhm -100mV.

 

You're right . Inductance is not needed. Input germanium transistor base current is approximately equal to one microampere. Applying the inclusion in base 1kOhm resistor lose 1mV. 100kOhm -100mV.

Yes,i understood that.But my intention refer to omitting completely the resistor.As my question was:
"why you said before,that it is possible to omit completely the resistor from the base?"(while there isn't a load on the emitter).
And if it is possible,What is the maximum voltage that can be applied to the base,from the antenna,before it will be necessary to add a resistor?(the 1kOhm one)?

 

Maximum input voltage (amplitude) limited breakdown diode voltage and the maximum current base.
V=min(Vbr,Ibmax,Rb)

 

Bordodynov,please,take into consideration that i am not an expert and this subject is not easy for me,so its hard to step up a level before understanding the issue.I asked you 3 times the same question and i just would like to understand,why you said,that it is possible to omit the resistor,entirely,from the base and leave the base and the emitter without a resistors?
I always thought that the path along the base and the emitter must have a resistor(at the base or at the emitter),it mustn't be without aload/resistor,but now you said that it can be omitted:

2. Resistor 100k greatly reduces the sensitivity of the whole scheme and need for this resistor is not.

So,i don't understand why it can be omitted entirely?
if the voltage will be 1v from the loop,0.7v will drop on the base-emitter,what will limit the rest 0.3v(1V-0.7Vbe)?with what value should i divided the 0.3V?
Maybe you answer that question undirectly,when you mention the ac resistance of the loop has 100Ω,and this is actually acts as the limited series resistor?

 

As it turned out bad germanium diode at a frequency of 1 gigahertz. Especially for the frequency of 2.4 gigahertz. It is better to use a Schottky diode.
With this Schottky diode has a maximum signal amplitude of 15 volts. Look

 

What circuit should i go with?How should i follow your help,while you change the circuit all the time?
First you used the 100KΩ to compensate for the leak of the germanium transistor and now you change it to 47KΩ and to 4.7KΩ,why did you lower the resistor value?
2)Why did you lower the capacitor value to 1nF?
3)What is the "X2"mark refer to,at the second circuit(near the Schottky diode)?two wires?
4)And my last try:why did you say,that it is possible to omit the resistor,entirely,from the base and leave the base and the emitter without a resistors?maybe the 100Ω pure resistance of the loop is the answer,but i am not sure,so at least,may you confirm or reject my assumption?
I,really,don't understand why are you ignoring that question.

 

Resistor to compensate for leakage germanium transistors must be selected experimentally. 47kOhm better than 100kOhm. Leakage transistor depends on what is included parallel to the base-emitter junction. With Schottky diode had to put 4.7kOhm. Capacitor 1000pF enough. You can also use 0.1uF. Calculation of lower capacity requires much less time. High-frequency characteristics 1000pF better than 0.1uF. X2 - just a name. Your bad germanium diode is suitable for detector at a frequency of 1GHz and 2.4GHz. With a large signal (~ 10V), this diode will burn. I advise you to download LTspice. Advanced library (my) you can download from Ltwiki.org site. Then I can for you to upload my circuit. In can vary the quantities of resistors and capacitors. For frequencies < 30MHz your Germanium diode is good. If you do not need the ultimate sensitivity, use low cost Schottky diode bas70.

Sorry for bad english and something that is not answered all the questions. At the weekend I can only use the phone. This greatly complicates communication.

 

Your english is pretty good,no problem with that.

Leakage transistor depends on what is included parallel to the base-emitter junction. With Schottky diode had to put 4.7kOhm.

But the germanium diode or the Schottky diode is in series with the base-emitter junction ,isn't it?
It looks to me like the capacitor is in parallel to the base-emitter junction rather the diode,so i didn't understand that,may you explain?
Since i am going to solder the components on through hole pcb(rather surface mount),what is the equivalent diode that i can use instead of the Schottky HSMS-2820?

 

There is no analogue.
Use the 1N6263, BAT41 (not Diodes Corp.).
The capacitor must be immediately after the diode. Then, in the base resistor. Compensating leakage resistor must be selected. Its value should be selected so that the LED does not light without a signal.

 

At a frequency of 2400Megahertz sensitivity decreases.
You must use a high-quality a diode, with lower parasitic capacitance.
See

 

OK.i understand the diode issue and that i must to add the compensating leakage resistor.
But i didn't understand what is the connection between the value of the leakage resistor that you determined to the diode?
You said:

Leakage transistor depends on what is included parallel to the base-emitter junction. With Schottky diode had to put 4.7kOhm

So,how does the diode influence on the value of the leakage resistor determination?(4.7k with schottky diode,47k with oa91 diode).

 

See

 

It was very helpful if you might add explanation to the circuit,since i can't understand what does it mean?how did you get to this currents value? and how do the diodes influence that currents value?
In the datasheet i saw these values of the leakage currents:
Ucbo=0.5V--------->Icbo<=10uA(25°C)
Ucbo=10V---------->Icbo<=14uA(25°C)
Ucbo=32V---------->Icbo<=500uA(25°C)
Uebo=5V----------->Iebo<=550uA(75°C)
I,also,didn't entirely understood how exactly that resistor compensate the transistor leakage.
When there isn't a current on the base(germanium),the voltage after the resistor(R6/R3) is 9V,so it seems like the purpose of the resistor is to prevent from the silicon transistor to turn-on.So,i don't understand how does the resistor,exactly,make that compensation?

 

You brought the collector junction leakage. In the scheme with the common-emitter current is amplified several dozen times. When the germanium diode it is leaking a few microamperes. This current is subtracted from the leak collector junction. The result is enhanced germanium transistor is much less current. Schottky diode has low leakage and practically does not affect the amplified current. Resistor compensating leakage germanium transistor resets (catches almost all of the current) and the output of silicon transistors lacks current. In the actual circuit as a compensating resistor is necessary to use a variable resistor with nominal value of 100kohms. Before the measurement is necessary when there is no signal to set the knob of the resistor so that the LED does not light or very weak (optimal).
See

 

If i will connect a resistor between the base and the emitter of the germanium transistor,as in the attached circuit,can i give up the 47K/4.7K resistors?

 

No, the resistor should be disposed of. Even with RBE = 0 we get Ice = Icbo. For my options germanium transistor Ice = 6.5uA. The second transistor, this current will increase about 500 times (6.5uA * 500 = 3.25mA).

 

Thanks,Bordodynov,for your help.
I didn't understand how you made all the calculations and determined all of the components value or how does the leakage current flow and influence through all the circuit without the resistor and with it.
But i am going to stop here,since the thread lengthen more than i was planning.
Thanks.