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

#### Colin Mitchell

Aug 31, 2014
1,416
Here is what Ratch is saying:
"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)"

He is turning the first transistor into an emitter follower. But the input voltage is just 2v.
How is he going to get his suggestion to work ?????????????

#### Ratch

Mar 10, 2013
1,099
Here is what Ratch is saying:
"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)"

He is turning the first transistor into an emitter follower. But the input voltage is just 2v.
How is he going to get his suggestion to work ?????????????

Nonsense. Just putting in an emitter resistance does not make an emitter follower. You have to take the signal off the emitter resistor before you can say it is an emitter follower (common collector) . https://www.google.com/search?q=emi...j69i57j0l4.10524j0j8&sourceid=chrome&ie=UTF-8 The OP thinks he has a Icbo problem. Instead of compensating for it, I suggested he eliminate the amplification of it. I can explain it in more detail if necessary.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
To get the collector of the first transistor to rise high enough to prevent the second transistor turning on, the emitter resistor will have to be something like 100k.
NOW, how are you going to turn ON the first transistor ???????

#### Ratch

Mar 10, 2013
1,099
To get the collector of the first transistor to rise high enough to prevent the second transistor turning on, the emitter resistor will have to be something like 100k.
NOW, how are you going to turn ON the first transistor ???????

Why do I get the impression that you are not getting the big picture? Assuming a 1 volt drop across the collector-emitter of Q2 when it is turned on, and 0.7 volts across the LED, that makes 7.3/680 = 10 ma through the LED. The minimum beta of Q2 is 125, so a minimum 10ma/125 = 80 uamps is present in the base of Q2 and the collector of Q1. Unlike a silicon transistor, the Vbe of a germanium transistor is only about 0.4 volts. So what resistance is needed for a voltage of 2.0-0.4 = 1.6 volts to push 80 uamps? Answer is 1.6/80 uamps = 20000 ohms. Now, the minimum beta of Q1 is 50, to the emitter resistance R1 has to be 20000/50 = 400 ohms. Don't forget the resistance is betatized when looking at it from the base. So 2 volts is more than enough voltage to turn on the LED.

As for the Icbo, it is a internal current generator controlled by the temperature and powered by the collector base voltage of Q1. If it goes into the base-emitter junction, it will be betatized,and the collector current will be increased greatly. But, by making the emitter resistor look like 20000 ohms, and the base almost zero ohms, most of the Icbo exits out the base harmlessly. That is what I mean by eliminating the problem instead on compensating for it.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
When V1 is zero, the collector-base voltage of Q1 will be about 7v and about 14uA will flow and the LED will not be turned off.

#### Ratch

Mar 10, 2013
1,099
When V1 is zero, the collector-base voltage of Q1 will be about 7v and about 14uA will flow and the LED will not be turned off.

Assuming you are saying the Icbo is 14 uamps. No problem. It takes 80 uams to turn on the LED. The 14 uamps is shunted away from the base-emitter of Q1. 14 uamp*125 = 1.75 ma. Not much compared to the ON current of 10 ma, but it is the best you can do with germanium transistors, which are notorious for being leaky compared to silicon.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
The requirement is to remove the 14uA. YOU HAVE NOT DONE THAT.

BC557C is 400 to 800 not 125.

You have completely dodged the question.

You are completely wrong

I have written this up on my website under SPOT THE MISTAKE as I cannot believe what I am seeing.

#### Ratch

Mar 10, 2013
1,099
The requirement is to remove the 14uA. YOU HAVE NOT DONE THAT.

As I explained before, and which you should know by now. Icbo is an internal current generator within the collector of a BJT whose existence cannot be eliminated or turned off. It is intrinsic to semiconductors. It can only be minimized by the method I have presented. Who said the requirement was to remove Icbo? The OP wanted to compensate for it. I suggested what I think is a better method, which is minimizing its effects. You need to explain the situation correctly instead of making wild pronouncements that are irrelevant or cannot be backed up with facts.

Ratch

#### Ratch

Mar 10, 2013
1,099
The requirement is to remove the 14uA. YOU HAVE NOT DONE THAT.

BC557C is 400 to 800 not 125.

You have completely dodged the question.

You are completely wrong

I have written this up on my website under SPOT THE MISTAKE as I cannot believe what I am seeing.

There you go again, making wild statements. The datasheet disagrees with you. http://www.nxp.com/documents/data_sheet/BC556_557.pdf

Ratch

Edit: Ok I see that adding a "C" suffix increases the minimum beta to 420. That means that an it will be even easier to get the transistor to turn on the LED. So, nothing has changed as far as the circuit working and doing what it is supposed to do.

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#### Colin Mitchell

Aug 31, 2014
1,416
He is obviously trying to turn the LED off when the input to the circuit is 0v.

"That means that it will be even easier to get the transistor to turn on the LED."

We are trying to turn the LED OFF.

You have not reduced the leakage.

When the input voltage is 0v, the leakage still exists.

#### Ratch

Mar 10, 2013
1,099
He is obviously trying to turn the LED off when the input to the circuit is 0v.

"That means that it will be even easier to get the transistor to turn on the LED."

We are trying to turn the LED OFF.

We are trying to turn the LED ON and OFF. The circuit is no good if it doesn't do both.

You have not reduced the leakage.

This is about the 3rd time I said that Icbo cannot be stopped. It can only be controlled,

When the input voltage is 0v, the leakage still exists.

If he uses the BC557, then the diode current will be down to around 1.75 ma. Probably not enough for the light to be visible. Otherwise, I suggest he use a silicon transistor for Q1.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
LEAKAGE
Here's a problem with LEAKAGE.
This is when a very small current flows through a transistor (or any other component) and this current cannot be reduced or stopped.

In the following example, Q1 is a leaky transistor. It can be any type of transistor and although the leakage current flows via the collector-base junction, we can assume the transistor is exactly the same as a 470k resistor.

If we place a resistor from the supply to point A on the diagram, we will create a voltage divider and 6.4v will be at point A.

The base of Q2 will see a voltage of 6.4v but no current will flow in the base and thus the LED will not be illuminated.

The leakage comes via the 190k resistor and the LED is not turned on.

#### Colin Mitchell

Aug 31, 2014
1,416
Ratch, you have completely missed the point of the exercise.
Putting an emitter resistor on the first transistor does ABSOLUTELY NOTHING.
When input voltage is zero, the voltage between the collector and base will be about 7v and the leakage will be about 10uA.
If the gain of the BC557 is 400 to 800, the current will be 4mA to 8mA. However, when a current flows in the circuit, a voltage-drop will be produced across the 680R and this will reduce the leakage current and the actual end-result is unknown. But some LEDs illuminate on 1mA.

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#### Ratch

Mar 10, 2013
1,099
Ratch, you have completely missed the point of the exercise.
Putting an emitter resistor on the first transistor does ABSOLUTELY NOTHING.

Certainly it does. It keeps the Icbo out of the emitter circuit so it does not get betatized.

When input voltage is zero, the voltage between the collector and base will be about 7v and the leakage will be about 10uA.

If the gain of the BC557 is 400 to 800, the current will be 4mA to 8mA.

For the circuit you show above with Rb=100k and Re=0, all the Icbo will be directed through the base-emitter junction and be betatized to a collector current of 4 to 8 ma. But, if you put in a finite Re and no Rb, then all the Icbo will be directed out the base lead and not be betatized. Then the collector current will be whatever the Icbo is (10 ua).
However, when a current flows in the circuit, a voltage-drop will be produced across the 680R and this will reduce the leakage current and the actual end-result is unknown. But some LEDs illuminate on 1mA.

You can never reduce the leakage current, only its effects. The standard method of directing the Icbo away from the emitter-base junction is the time tested method used by those in the know.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
The fact is this:
Your method of putting an emitter resistor on Q1 fails to turn off the LED.
When the base voltage is zero, the transistor is turned OFF and no current flows through the emitter resistor so it does not matter what value of resistance is placed on (in) the emitter, IT WILL HAVE NO EFFECT.
If the leakage is 10uA, the current through the LED will be a few mA and it will remain illuminated.
My method of adding a diverting resistor turns the LED off.

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#### Ratch

Mar 10, 2013
1,099
The fact is this:
Your method of putting an emitter resistor on Q1 fails to turn off the LED.
When the base voltage is zero, the transistor is turned OFF and no current flows through the emitter resistor so it does not matter what value of resistance is placed on (in) the emitter, IT WILL HAVE NO EFFECT.

Sure it will. When the base voltage is zero. 10 ua of current will be present in the collector of Q1. Multiply that by 125 and Q2 will be 1.25 ma. That is sure a lot less than 10 ma it will have when Q1 is turned on. The best course of action is to use a silicon transistor for Q1, so that Icbo will be very small.

If the leakage is 10uA, the current through the LED will be a few mA and it will remain illuminated.
My method of adding a diverting resistor turns the LED off.

You can never turn off the LED with your method. You divert all the Icbo current into the base-emitter junction where it gets betatized, and makes the collector current of Q1 at least 10 ua * 50 = 0.5 ma. Then Q2 and the LED will be fully turned on. Don't you see that?

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
What absolute rubbish.
As if Q1 is going to leak 0.5mA when turned off.
If this were the case, transistors would never have been successful.

#### Ratch

Mar 10, 2013
1,099
What absolute rubbish.
As if Q1 is going to leak 0.5mA when turned off.

That's right. A germanium transistor with those specs will do so. You cannot turn off that internal Icbo generator in the collector of Q1. As long as you reverse bias the collector-base, which you do in the active region, that internal current generator will fire up and send its current into either the base lead or the base-emitter junction. That is a well known behavior, and fully documented in textbooks. It is wise to have Re/Rb >>1, so as to divert most of the Icbo away from the b-e junction.

If this were the case, transistors would never have been successful.

That is one of the reasons silicon transistors predominate. Compare the Icbo of Q1 and Q2, and notice the difference between silicon and germanium. Silicon has a much lower initial Icbo, so the problem is not so acute.

Ratch

#### Colin Mitchell

Aug 31, 2014
1,416
If Q1 is going to leak 0.5mA when turned off, the LED will never turn off.

What you are saying is absolute RUBBISH.

#### Ratch

Mar 10, 2013
1,099
If Q1 is going to leak 0.5mA when turned off, the LED will never turn off.

What you are saying is absolute RUBBISH.

It will leak 10 ua regardless. But, if you don't have any Rb, and put in some Re, then the collector current will be 10 ua when turned off. On the other hand, if we use your circuit, then all the 10 ua will pass into the b-e junction, and be betatized to to 0.5 ma Ic. I am weary of explaining that to you.

Ratch

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