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I'm new to BJT and i'm trying to understand the characteristics of bjt in different modes of it. In my book they explained that in saturation mode of bjt the collector current decreases even if i increase the base current.

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but i don't understand why it should decrease. Because in the saturation mode base-collector works as a forward bias, so there should be breakdown of collector current. But instead of that collector current decreases with the increase of base current. Please help me to make my concept clear and correct.

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  • \$\begingroup\$ Your graph doesn't show any decrease in collector current does it? I'm not saying there isn't a decrease, but the graph doesn't seem relevant to your question ... \$\endgroup\$
    – brhans
    Apr 18, 2017 at 14:47
  • \$\begingroup\$ the graph simply shows the decrease of collector current in saturation mode..is it irrelevant? \$\endgroup\$
    – Leolime
    Apr 18, 2017 at 14:50
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    \$\begingroup\$ You should include the full text or reference to the textbook's text as it is unclear now what they mean. Maybe they're explaining that if you would keep Ie constant and increase Ib then Ic must decrease, which is true (since Ib + Ic = Ie) but I do not see how explaining that fact helps beginners in understanding saturation mode. \$\endgroup\$
    – Bimpelrekkie
    Apr 18, 2017 at 15:07
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    \$\begingroup\$ Try read this and you should understand what saturation in BJT is. electronics.stackexchange.com/questions/276146/… \$\endgroup\$
    – G36
    Apr 18, 2017 at 15:32
  • \$\begingroup\$ Keep increasing the base current and at some point the collector current will decrease ;-) , but I suspect they're talking about the slope decreasing. Or the book is badly written. Please post an image from your book that is causing confusion. \$\endgroup\$
    – Spehro Pefhany
    Apr 18, 2017 at 17:20

1 Answer 1

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Simply remember that Ie = Ic + Ib. So, here Ie is constant. Therefore with increase in Ib, Ic decrease and vice versa.

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  • \$\begingroup\$ Bimpelreggie already commented on the Kirchhoffs law approach. He also mentioned that the OP did not give proper context (i.e. if Ie is meant to be constant). \$\endgroup\$
    – Andreas
    Jan 23, 2018 at 20:01

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