Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

[xchcui]

Hello.



This is a circuit(Sziklai pair)that amplified a small current by a germanium and silicon transistor in order to turn on a led.
The germanium transistor has a collector base leakage(Icbo),which is amplified twice and cause the led to turn on even when the base of the germanium doesn't get signal.
In order to find what is the Icbo of the germanium through the datasheet,i need to know the Ucbo of the germanium transistor.
Those are the parameters in the datasheet.
Ucbo=0.5V--------->Icbo<=10uA(25°C)
Ucbo=10V---------->Icbo<=14uA(25°C)
Ucbo=32V---------->Icbo<=500uA(25°C)
So,what is the Ucbo of the germanium transistor,refer to the attached circuit,and how do you come to that value?
2)When this small leakage current flows,initially,through the circuit,before it is amplified(red arrows in the attached circuit),what should be the voltage drop on the led,the resistor and the silicon transistor emitter-base?
Thanks in advance.

 

[Harald Kapp]

Is this kind of homework or an assignment?

 

[xchcui]

I know that there is a section for homework and assignment and you may remove this question to there if you think that it is more appropriate.
But since this question is only for understanding the behavior/influence of the germanium collector-base leakage on the led/resistor voltage drop(before the amplification),without asking for any special calculations,i thought to ask it at this section.

 

[Colin Mitchell]

What is a germanium transistor??????

 

[xchcui]

What is a germanium transistor??????

 

[BobK]

What is a germanium transistor??????

I believe they are made out of flower petals:

Whoops, that is Geranium. Never mind.

Bob

 

[duke37]

Why use a leaky npn germanium transistor in the first stage when silicon npn transistors are so common?
According to your table the leakage will be about 12μA.
If you wish to keep this configuration, you could by-pass some current with a resistor from the collector to 9V

 

[xchcui]

Why use a leaky npn germanium transistor in the first stage when silicon npn transistors are so common?

This is the main idea of the question.understanding the leakage effect before calculating the by-pass resistor that you are suggested.

According to your table the leakage will be about 12μA.

I assume that you used this data:Ucbo=10V---------->Icbo<=14uA(25°C).
But,how did you come to 12uA leakage?or more accurate-how did you determine what is the Ucbo of the germanium transistor,while the source voltage is 9V?

 

[duke37]

I just guessed, I would think that there will be a wide variation between transistors.

 

[xchcui]

OK.lets say that,initially,a 12uA leakage current flows(red arrows)before it is amplified by the germanium transistor.
When that 12uA tiny current is flowing through the red led(btw,at 20mA the led voltage drop is 2V),through the 680Ω resistor and the silicon emitter-base transistor,what may be the voltage drop on the red led,the 680Ω resistor and the silicon e-b transistor?

 

[Ratch]

Hello.

View attachment 26600

This is a circuit(Sziklai pair)that amplified a small current by a germanium and silicon transistor in order to turn on a led.
The germanium transistor has a collector base leakage(Icbo),which is amplified twice and cause the led to turn on even when the base of the germanium doesn't get signal.
In order to find what is the Icbo of the germanium through the datasheet,i need to know the Ucbo of the germanium transistor.
Those are the parameters in the datasheet.
Ucbo=0.5V--------->Icbo<=10uA(25°C)
Ucbo=10V---------->Icbo<=14uA(25°C)
Ucbo=32V---------->Icbo<=500uA(25°C)
So,what is the Ucbo of the germanium transistor,refer to the attached circuit,and how do you come to that value?
2)When this small leakage current flows,initially,through the circuit,before it is amplified(red arrows in the attached circuit),what should be the voltage drop on the led,the resistor and the silicon transistor emitter-base?
Thanks in advance.

What are you trying to do with this circuit? Icbo is a internal current generator within the collector slab that is controlled by the temperature of the slab and powered by the collector-base voltage. You cannot turn it on or off. You can only design around it. Why are you using a germanium transistor, which has a much higher room temperature Icbo instead of silicon, which is much less? Who cares about the Vcbo? As an internal current generator, Icbo outputs a fairly constant current over a large range of Vcbo voltages. I would like to know more about what you are trying to accomplish before I pontificate any further.

Ratch

 

[xchcui]

What are you trying to do with this circuit? Icbo is a internal current generator within the collector slab that is controlled by the temperature of the slab and powered by the collector-base voltage. You cannot turn it on or off. You can only design around it. Why are you using a germanium transistor, which has a much higher room temperature Icbo instead of silicon, which is much less? Who cares about the Vcbo? As an internal current generator, Icbo outputs a fairly constant current over a large range of Vcbo voltages. I would like to know more about what you are trying to accomplish before I pontificate any further.

Ratch

I am trying to turn-on a led,using a very small current on the germanium base,while using minimum base-emitter voltage drop.
I aware of,that there are other possibilities for doing that,besides using the germanium transistor,but this is not what i am looking for.I would like to understand the way of calculate it.
I would like to calculate a compensating resistor for the germanium leakage as in the attached photo.

In order to do that i need to know what is the leakage current,so i checked in the data sheet and got some parameters as i mentioned before.
So,i need to know the Ucbo of the germanium transistor in order to find the Icbo.
One of the datasheet parameter says:Ucbo=10V---------->Icbo<=14uA(25°C),which is the closest parameter to the 9V power source,but on the germanium collector-base,i assume the Ucbo is smaller.So i try to understand what is the Ucbo and how do you determine it.
After that,i try to figure out what is the voltage that i should devide with the leakage current in order to calculate the compensating resistor.
For example,if i want 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.(germanium B-E voltage drop/Icbo=bypass resistor).It is simple.
But the other 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.
And should i devide that voltage with the 12uA leakage or maybe with the amplified leakage current(by the germanium transistor)?

 

[Ratch]

I am trying to turn-on a led,using a very small current on the germanium base,while using minimum base-emitter voltage drop.
I aware of,that there are other possibilities for doing that,besides using the germanium transistor,but this is not what i am looking for.I would like to understand the way of calculate it.
I would like to calculate a compensating resistor for the germanium leakage as in the attached photo.
View attachment 26651
In order to do that i need to know what is the leakage current,so i checked in the data sheet and got some parameters as i mentioned before.
So,i need to know the Ucbo of the germanium transistor in order to find the Icbo.
One of the datasheet parameter says:Ucbo=10V---------->Icbo<=14uA(25°C),which is the closest parameter to the 9V power source,but on the germanium collector-base,i assume the Ucbo is smaller.So i try to understand what is the Ucbo and how do you determine it.
After that,i try to figure out what is the voltage that i should devide with the leakage current in order to calculate the compensating resistor.
For example,if i want 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.(germanium B-E voltage drop/Icbo=bypass resistor).It is simple.
But the other 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.
And should i devide that voltage with the 12uA leakage or maybe with the amplified leakage current(by the germanium transistor)?

Right now, with that high base resistance (100 k) in Q1, almost all the Icbo will go through the base-emitter junction of Q1, and get betatized into a much larger current. You should take out Rb, and put in an emitter resistor Re. The emitter resistor will look like (beta+1) Re from the base, and the Icbo will exit harmlessly through the base instead of entering the base-emitter circuit. Then you can adjust Ie=Ic by 2/((beta+1)*Re)

Ratch

 

[Colin Mitchell]

Suppose the leakage is 20uA. The PNP will amplify this say 200 times and the emitter current will be 20 x 200 = 4000uA = 4mA and the LED will turns ON.
If you put a 1k resistor on the collector of the first transistor to the positive rail, this current will flow through the resistor and produce a voltage drop of 20mV.
If you use 10k the voltage drop will be 200mV
and 100k will be 2v.
When the voltage drop is 2v, the LED will start to turn ON if it is a red LED so you can decease the resistance slightly to prevent this.

 

[xchcui]

Suppose the leakage is 20uA. The PNP will amplify this say 200 times and the emitter current will be 20 x 200 = 4000uA = 4mA and the LED will turns ON.
If you put a 1k resistor on the collector of the first transistor to the positive rail, this current will flow through the resistor and produce a voltage drop of 20mV.
If you use 10k the voltage drop will be to 200mV
and 100k will be 2v.
When the voltage drop is 2v, the LED will start to turn ON if it is a red LED so you can decease the resistance slightly to prevent this.

In my case the leakage is 12.8uA(in case there is another compensating resistor between the germanium base-emitter transistor),so according to your explanation i need to connect a less than 166KΩ(2V/12.8uA) resistor to the germanium transistor collector.
But are you sure that,when adding the 166KΩ resistor(or 100KΩ in your case)all the 12.8uA current(in your example is 20uA) will flow only through the compensating resistor instead of through the parallel components(led,680Ω resistor and silicone E-B)?
You refer to 2v as the point that the led will turn-on,but parallel to the compensating resistor,there are,also,the 680Ω resistor and the voltage drop on the silicone E-B,sO,souldn't the voltage should be higher than 2V for the resistor calculations?

Right now, with that high base resistance (100 k) in Q1, almost all the Icbo will go through the base-emitter junction of Q1, and get betatized into a much larger current. You should take out Rb, and put in an emitter resistor Re. The emitter resistor will look like (beta+1) Re from the base, and the Icbo will exit harmlessly through the base instead of entering the base-emitter circuit. Then you can adjust Ie=Ic by 2/((beta+1)*Re)

I assume that you are talking about resistor to the germanium B-E,as can be seen at the attached circuit:

But my question is about the compensating resistor between the germanium collector and the voltage source.
The 10KΩ resistor at the germanium B-E prevent the 12.8uA to be amplified by the germanium transistor,but the silicone transistor will applified that 12.8uA leakage current if i won't add resistor between the germanium collector and the power source.
So,how do i determine the current that will flow through the compensating resistor,in order to calculate the compensating resistor value?

 

[Ratch]

In my case the leakage is 12.8uA(in case there is another compensating resistor between the germanium base-emitter transistor),so according to your explanation i need to connect a less than 166KΩ(2V/12.8uA) resistor to the germanium transistor collector.
But are you sure that,when adding the 166KΩ resistor(or 100KΩ in your case)all the 12.8uA current(in your example is 20uA) will flow only through the compensating resistor instead of through the parallel components(led,680Ω resistor and silicone E-B)?
You refer to 2v as the point that the led will turn-on,but parallel to the compensating resistor,there are,also,the 680Ω resistor and the voltage drop on the silicone E-B,sO,souldn't the voltage should be higher than 2V for the resistor calculations?

I assume that you are talking about resistor to the germanium B-E,as can be seen at the attached circuit:
View attachment 26667
But my question is about the compensating resistor between the germanium collector and the voltage source.
The 10KΩ resistor at the germanium B-E prevent the 12.8uA to be amplified by the germanium transistor,but the silicone transistor will applified that 12.8uA leakage current if i won't add resistor between the germanium collector and the power source.
So,how do i determine the current that will flow through the compensating resistor,in order to calculate the compensating resistor value?

Forget about putting any resistor across Q2. I suggested you put a resistor Re in the emitter circuit of Q1 and take out the 100k base resistor. That means putting a resistor in series with the emitter terminal of Q2, not across the base-emitter terminals like you show in latest drawing. Read post #13 again.

Ratch

 

[Colin Mitchell]

Forget about what Ratch said. He does not know what he is talking about.

 

[Ratch]

Forget about what Ratch said. He does not know what he is talking about.

See post #4 of this thread.

Ratch

 

[Colin Mitchell]

"See post #4 of this thread."

That doubly-determines Ratch doesn't know what he is talking about.

 

[Harald Kapp]

That doubly-determines Ratch doesn't know what he is talking about.

Colin, may I remind you to observe at least some netiquette? Simply stating that that someone else doesn*t know what he's talking about without prove is a sign of bad manners.
Prove your claim with some valid statements, or at least with some arguable statements. Or keep silent otherwise, please.