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In output characteristics graph of common emitter transistor the current and voltage are both 0 when one of them is 0 as shown.

enter image description here

But in output characteristics graph of common base configuration it is not so.Why?

enter image description here

I just started learning transistors yesterday and I'm feeling very confused already.Please explain in as simple language as possible.Thanks!

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  • \$\begingroup\$ Where did you get those graphs from, never use that site again! Beaten to an excellent answer by Bruce. \$\endgroup\$ – Neil_UK Jan 10 '16 at 8:24
  • \$\begingroup\$ Google images :-P! @user44635 \$\endgroup\$ – user96630 Jan 10 '16 at 8:42
  • \$\begingroup\$ Google Images is an excellent source for lots of different images, but as they are not curated, you can never assume the quality of any individual image, as you have found. Keep trawling images until you have several from different sources that match, or use wikipedia, or use a dedicated electronics site. \$\endgroup\$ – Neil_UK Jan 10 '16 at 10:58
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Your top graph seems wrong. Either the horizontal axis is mislabeled (it should be Vce for Common Emitter) or this is not a Common Emitter curve trace. Also the knee voltage is very high for a bipolar transistor, and the Collector-Base leakage current is appalling.

Here's a better example of a Common Emitter curve trace:- enter image description here

The reason Collector-Emitter current goes to zero in common Emitter configuration is that the Emitter is connected to the lowest voltage in the circuit. As Collector voltage approaches this common point junction resistances start to limit current, eventually reaching zero when it cannot pull down any further.

In a common Base circuit the Base is connected to 'ground' but the Emitter is powered from a lower voltage. Therefore the Collector is able to go below zero volts as it is pulled towards the Emitter.

In both cases the transistor will saturate when Collector voltage approaches Emitter voltage and current is limited by resistance. As far as the transistor is concerned it is still acting the same, only the reference point has changed (from Emitter to Base).

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  • \$\begingroup\$ Can you explain that part "In a common Base circuit the Base is connected to 'ground' but the Emitter is powered from a lower voltage. Therefore the Collector is able to go below zero volts as it is pulled towards the Emitter." ? I did'nt get you... \$\endgroup\$ – user96630 Jan 10 '16 at 8:39
  • \$\begingroup\$ To be exact - and for a better understanding: The collector current Ic is NOT zero for Vce=0. That means: The set of curves does not touch the origin because: For Vce=0 also the B-C junction is forward biased and we have a small "negative" current Ic (oppopsite to the "normal" direction.) \$\endgroup\$ – LvW Jan 10 '16 at 9:42
  • \$\begingroup\$ @MathJack in a common Base circuit you have two power supplies of opposite polarity, one going to the Collector and the other going to the Emitter. The Base is connected to ground (zero volts). When the transistor saturates and Collector voltage approaches Emitter voltage, Collector voltage goes below ground and changes polarity. In a common Emitter circuit you only have one power supply and the Emitter is connected to ground, so Collector voltage cannot go below ground. \$\endgroup\$ – Bruce Abbott Jan 10 '16 at 10:12
  • \$\begingroup\$ I think your beta value is quite low. Realistically, in my textbook, base current values are in 100 microamps \$\endgroup\$ – Red Floyd Nov 12 '18 at 12:18
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    \$\begingroup\$ @AlphaRomeo Yes, and Icbo is also quite high - so perhaps it's an old germanium type like OC70 or just a made-up example to show what the curves should look like (I found it on Wikipedia - no idea where it originally came from). \$\endgroup\$ – Bruce Abbott Nov 13 '18 at 8:07

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