7
\$\begingroup\$

I am stuck with the equation:

ICEO = (1+β) ICBO

My questions are:

  1. What is the physical interpretation of this equation?
  2. Will both the currents exist simultaneously?
  3. Are any of the reverse currents getting amplified?
\$\endgroup\$
3
  • 4
    \$\begingroup\$ Both currents cannot exist at the same time by definition. The O in I(CEO) and I(CBO) is short for 'open' and means the current from collector to emitter when base is not connected, respectively current from collector to base when emitter is not connected. Where did you read that equation? A bit more background information on what you are reading / doing might help a lot. \$\endgroup\$
    – jippie
    Commented Aug 10, 2014 at 6:58
  • \$\begingroup\$ @jippie I have been reading this book. here \$\endgroup\$
    – ibsen
    Commented Aug 10, 2014 at 7:27
  • 1
    \$\begingroup\$ Ah, let me slightly rephrase my statement then: You cannot measure the I(CBO) and I(CEO) at the same time. I thought you were looking at a datasheet. \$\endgroup\$
    – jippie
    Commented Aug 10, 2014 at 7:36

5 Answers 5

10
\$\begingroup\$

An Active Biased BJT Transistor ICBO = Reverse Leakage Current between Collector and Base while Emitter is Open. (IE=0)

ICEO = Reverse Leakage Current between Collector and Emitter while Base is Open. (IB=0)

IC=BIB+(B+1)ICBO
if(IB=0){IC=(B+1)ICBO=ICEO}

These are two different Leakage Currents for two different ways of Active Biasing of a BJT Transistor.

\$\endgroup\$
7
\$\begingroup\$

1/ ICBO is reverse leakage current going from the Collector to the Base. This current is then amplified by β to produce additional Collector current, thus the "1+β" term.

2/ Both currents exist simultaneously, but ICBO is included in ICEO.

3/ The reverse current ICBO is amplified, just like external Base current would be.

\$\endgroup\$
1
  • \$\begingroup\$ Voltage is only applied between the Collector and the Emitter, but ICEO flows from the Collector to the Base and then to the Emitter. Of course this Base current will be amplified, and the result (ICEO-ICBO) flows directly from the Collector to the Emitter. \$\endgroup\$ Commented Aug 10, 2014 at 10:28
2
\$\begingroup\$

ICEO and ICBO exist/or measured when you connect the transistor in two different configuration(common base and common emmiter respectively) so there is no question of asking whether both exist simultaneously(answer is no).Consider you have transistor say T1 to be measured.You measure both Icbo and Iceo when you connect T1 in two different config as mentioned above.So now when you want to know what the value of Icbo wrt measured Iceo Iceo=(1+beta)Icbo will be the expression for solving.In a nutshell to calculate either Icbo or Iceo you need to know to measure just either of Iceo or Icbo for which the expression you mentioned helps.Hope you get it! Cheers

\$\endgroup\$
1
\$\begingroup\$

they are leakage currents ,be so small that they are often neglected β =(IC-ICEO)/IB alpha=(IC-ICBO)IE

these two equations are used when leakage is not neglected

*but * β=IC/IB alpha= IC/IE these commonly used

\$\endgroup\$
1
\$\begingroup\$

I think the confusion with leakage currents, and other parameters, occurs because many textbooks don't explicitly state that the open terminal (b in the case of Iceo) and (e in the case of Icbo) is left 'open circuit' to allow the measurement of the parameter being measured.

This is also true for h-parameters etc. where terminals are either shorted out or left open to effectively 'freeze' parameters in order to measure another parameter under known and benign (for the measurement) conditions.

The equation relating the relationship between the parameters (Iceo and Icbo) apply when the device is in a correctly biased circuit.

It is feasible for anyone to breadboard a transistor circuit and measure these parameters but one requires accurate meters capable of measuring very small currents.

An easier way is to simulate the circuits using LTspice or similar free simulation software. Of course this last suggestion of simulation is simply to help the student understand how these parameters are measured and used.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.