Hi.
I am not going to get into all the details again.
But i opened the thread again,since i couldn't figure out how was the 47KΩ compensating resistor calculate?
From the data sheet i found that the leakage current is about 12.8uA and when i calculate:12.8uA*47KΩ,the result is 0.6V.
This result made me assume that you had been devided the emitter-base voltage drop of the silicon transistor(0.6V)with the leakage current(12.8uA)and determine the 47KΩ.
But the compensating resistor is parallel,not only to the silicon E-B,but also to the led and the resistor,so the voltage drop on the compensating resistor is higher than 0.6V.(maybe 2.1V?).
So,how did you calculate the compensating resistor?which parameters did you use?

 

Transistor leakage is large spread. Also, leakage is strongly dependent on temperature. It would be wise to use a variable resistor (potentiometer). I think the face value of the resistor 100 kilohms is fine.

 

Bordodynov,in practical using a variable resistor will be the best solution.
But,i just would like to know the idea,how did you calculate the compensating resistor.
I just would like to understand the way to calculate it.
If i want(for example) to determine a compensating resistor to the base-emitter of the germanium transistor,i know that i should calculate it as this:0.15V/12.8uA= ~11KΩ.
I used the 0.15V since the compensating resistor is parallel only with the voltage drop of the germanium base-emitter(0.15V).
But with the other compensating resistor(the 47KΩ),it is different.
What voltage value and current value did you use in order to get the 47KΩ value?

 

I brought a transistor leakage calculation for various displacements base. From the calculation shows that the minimum transistor leakage with my settings 12 mA (zero-resistance resistor). This current may be amplified to about 500 times (12 * 500 mA = 6000 mA). This LED glows brightly, that is not acceptable. Those. needs a resistor in the collector of the germanium transistor. To the LED light is necessary
Rc <(1.2V + 0.6V) / ic_leak.
See

 

...
Rc <(1.2V + 0.6V) / ic_leak.

The 0.6V,i assume,refer to the silicon emitter-base voltage drop.
Is the 1.2V refer to the voltage drop of the led?
The result of the calculation is 150KΩ(1.8V/12uA),
Does it mean that i need to use <150KΩ compensating resistor?

 

Yes. Close LED 1.2V refer to the voltage drop.
Not 12uA (Rbe=0) no signal.
Rbe=10k ==> Ic_leak=45ua Rc<39kOhm

See Ic_leak=F(Rbe)

 

Yes. Close LED 1.2V refer to the voltage drop.
Not 12uA (Rbe=0) no signal.
Rbe=10k ==> Ic_leak=45ua Rc<39kOhm

Ok.i understand that you calculated:
1.8V/45uA=39KΩ.So the Rc should be less than 39KΩ.
But how did you come to that 45uA current value?

The 10KΩ Rbe prevent the germanium transistor to amplify the 12uA leakage,so it seems like that only 12uA leakage current(not amplified by the germanium) flow through the Rc.
So,how did you come to 45uA?
I see in your graph that if i draws a line from 10KΩ up to the red curve,i get about 45uA on the right.But how may you come to this 45uA without the graph?
This is very important parameter to understand ,since with that parameter i should find the Rc.(1.8V/Ic leak)

 

I took a 10k resistance, as a value that is not much reduce the transmission coefficient of the diode detector. The magnitude of the current I picked up from the last schedule (graph). Analytically, it is possible to calculate, but get the system of nonlinear equations. This system of nonlinear equations is difficult to solve. One can write equations and solve them by means of mathematical programming. But it is much easier to use Spice (LTspice).
Keep in mind that you must take into account the leakage detector diode. For a germanium diode leakage is significantly greater than for the proposed Schottky diode. Schottky diode proposed higher frequency. But its arge barrier capacity is l for the frequencies you require ~ 1pF. There are diodes with much smaller parasitic capacitance, but they are for SMD (surface) mounting.

 

I had tried to use the spice(LTspice)at the first time that you recommended it,but i found it as very difficult to understand.
I saw that i need to add alot of parameters,while i don't have the knowledge to do that.I think,a kind of course should be taken in order to understand how to use it.
According to all your circuits attachment,what is,in practical,the appropriate circuit that i should use?
Circuit number 1,2 or something else?

 

If you use a germanium diode, use a circuit number 1. If you use a Schottky diode, use a circuit number 2.
At the frequencies you require a Schottky diode is more preferable. At lower frequencies (30 MHz or less) germanium diode gives the best sensitivity. The barrier capacitance 1pF ~ seriously affects the sensitivity. It is necessary to use diodes with lower parasitic capacitance. An alternative is to use the pre-amp.

 

The barrier capacitance 1pF ~ seriously affects the sensitivity

Did you mean the 1000pF capacitor?(since i didn't see any capacitor with value of 1pF at any circuit that you had been drawn along the thread).
you had recommended me before to use the 1000pF capacitor for high frequencies(+1GHz)instead of the 0.1uF capacitor,which is okay for lower frequencies.
So,why are you saying now that it is not good?(In the event that you really meant to the 1000pF,when you said 1pF).

 

1pF an exemplary diode capacitance (parasitic capacitance barrier). This capacity is not linear.
Spice model:
.model OA91 D(Is=19.43u N=2.986 Rs=65.98 Xti=3 Eg=1.11 Cjo=1.326p M=.3333 Vj=.75 Fc=.5 Isr=2.787u Nr=1.954 Bv=100 Ibv=100u Tt=5n type=germanium)
ileak =is+isr=22uA is value leak diode
Cjo=1.326p - parasitic capasitor diode Vak=0
Zc=1/(2*Pi*Freq*C)
.model HSMS-2820 d BV=15 CJO=.7p EG=.69 IBV=100u IS=22n N=1.08 RS=6 VJ=.65 XTI=2 M=.5 Vpk=15 type=Schottky mfg=Agilent
.model 1N6263 D(IS=3.8736n N=1.0099 RS=31.514 IKF=18.247 CJO=1.7738p M=.1955 VJ=.3905 ISR=6.2866n NR=3.9204 FC=0.5 TT=0 XTI=2 Iave=15m Vpk=60 mfg=STM type=Schottky
and very good
.model HSMS2865 D(Is=5e-8 Ibv=1e-5 Rs=6 N=1.05 Cjo=0.18p M=.5 Eg=.69 Xti=2 Vj=0.65 Vpk=7 mfg=Agilent type=Schottky)

 

OK.i understand.You refer to the self capacitance of the diode.
Thank you very much for your help,Bordodynov.