I have been studying Studying Donald Neamen's Electronic Circuits 4th edition and Sedra Smith's Microelectronics 7th edition for quite some time now. I have been self studying. After searching extensively on Internet as well as trying to understand from the books, I have not been able to differentiate between two quantities in the context of BJT.

These two quantities are: $$I_{S}$$ $$and$$ $$I_{C B O}$$

According to Sedra Smith,

$$ I_{C B O}$$ is the collector base reverse current. Whereas $$ I_{S}=\frac{A_{E} q D_{n} n_{i}^{2}}{N_{A} W} $$

$$ i_{C}=I_{S} (e^{v_{B E} / V_{T}} -1) $$

$$I_{S}$$ is the saturation or scale current.

I have been wondering are these two quantities same or share some relation between them because I find the name for both these quantities being used inter-changeably at many places which is quite confusing and I believe they are not the same.

Also, I am confused which current is being referred to when on Internet I find the term "reverse saturation current". Searching on Internet the term "reverse saturation current" gives two kinds of sources: one referring it as $$I_{S}$$ and the other referring it as $$I_{C B O}$$ Another similar quantity is $$I_{C E O}$$.

According to another source (Balbir Kumar Electronic Devices and Circuits Second Edition), $$ I_{C}=-\alpha I_{F}+I_{C B O}\left(1-\mathrm{e}^{V_{CB} / V_{T}}\right) $$

This equation in the forward active region of npn BJT reduces to

$$ I_{C}=-\alpha I_{F}+I_{C B O} $$


[Sedra Smith Microelectronics Circuits]

[Donald Neamen Circuit Analysis and Design]

  • \$\begingroup\$ On which page did you found this equation $$ I_{C}=-\alpha I_{F}+I_{C B O}\left(1-\mathrm{e}^{V_{CB} / V_{T}}\right) $$ ? \$\endgroup\$
    – G36
    Commented Oct 8, 2021 at 18:00
  • 1
    \$\begingroup\$ In general, \$I_S\$ is reverse saturation current. And of course, the collector current isn't actually \$I_S\$ when \$V_{BE}=0V\$. But this current (\$I_S\$) has nothing to do with the leaking current \$I_{CB0}\$. \$\endgroup\$
    – G36
    Commented Oct 8, 2021 at 18:10
  • \$\begingroup\$ @G36 See the edits. Thank you for bringing that to my attention. \$\endgroup\$
    – Anubhav
    Commented Oct 9, 2021 at 3:46
  • \$\begingroup\$ Another similar thread is electronics.stackexchange.com/questions/174681/… But it does not answer my question properly. \$\endgroup\$
    – Anubhav
    Commented Oct 9, 2021 at 15:14
  • 1
    \$\begingroup\$ For the NPN we have \$I_C = I_{C_F} + I_{CBO}\$ where \$I_{C_F}\$ = "forward current" and the forward current \$I_{CF} = \alpha_F \: I_{ES} (e^\frac{V_{be}}{V_T} -1 )\$ and this term \$\alpha_F \: I_{ES}\$ equals the saturation current. And the revers current (leaking current) is \$I_{R} = \alpha_R \cdot I_{CS} (e^\frac{V_{BC}}{V_T} -1 )\$ Thus the saturation current is \$I_S =\alpha_F \: I_{ES} = \alpha_R \: I_{CS} \$ \$\endgroup\$
    – G36
    Commented Oct 9, 2021 at 16:13

1 Answer 1


They are not equal, or even particularly related.

ICBO is a characteristic of the reverse biased C-B junction. Because the base part of the depletion region of this junction does not extend to the emitter region (in most BJTs), the junction leakage characteristic is just that of a 'simple' pn junction -- a function of the area and doping (mostly that of the collector doping).

If this junction was forward biased, ICBO would be the 'IS' term in its equation.

IS (as in the Sedra/Smith book notation) is the characteristic of the B-E junction, and depends on base and emitter doping. Because this junction is more heavily doped (because this gives higher beta to the BJT), this IS term is usually smaller than the corresponding C-B term (ICBO).

The total collector current consists of both terms -- the CBO leakage and the emitter current collected by BJT action from forward bias of the BE junction.

Note that therefore at very low VBE voltages, the base current might be negative -- because the ICBO current (which flows out the base in an NPN) exceeds the base component of the IS current. This is why a BJT will leak more current C-E when the base is open rather than shorted.

With base open, the ICBO current gets multiplied by the BJT beta (which may be significantly lower at such low current levels); the total current is then some beta*ICBO.

With base shorted to emitter, VBE is 0, and the IS term is multiplied by 1, making the total current ICBO + IS

With emitter open (and measuring leakage C-B), only ICBO is observed.

  • \$\begingroup\$ When I looked in book by Neamen, I found the formula $$ i_{C}=I_{S}\left[\exp \left(\frac{v_{B E}}{V_{T}}\right)\right] \cdot\left(1+\frac{v_{C E}}{V_{A}}\right) $$ When I looked at Michael Shur titled "Physics of Semiconductor devices", I found $$ I_{\mathrm{c}}=\left(I_{\mathrm{ceo}}+\beta I_{\mathrm{b}}\right)\left(1+V_{\mathrm{ce}} / V_{\mathrm{A}}\right) $$ \$\endgroup\$
    – Anubhav
    Commented Oct 11, 2021 at 13:00
  • 1
    \$\begingroup\$ @AnubhavSingh Most of the books will ignore the I_CBO current because in silicon BJT's I_CBO is very very low ( around 1nA or less at room teperature). So we can ignore it. \$\endgroup\$
    – G36
    Commented Oct 12, 2021 at 15:15

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