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I have been studying transistor throughly this semester and I have always doubted this circuit schematic. First of all is this schematic correct? What about current directions are they conventional? I can't understand why collector current is heading to the transistor isn't it supposed to be output current and why they are written in small letters when the are effected by the DC sources VEE and VCC

Sorry I'm very confused!

Schematic is within the fullpage below

Addition: full page

Exam question: equestion

Detailed answers : Explanation

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    \$\begingroup\$ What makes a circuit correct or incorrect ? Any circuit doing what it is supposed to do is "correct". It might not do what you expect when you don't understand it. This is a Common Base circuit. Stop doubting it because that prevents you from understanding it. \$\endgroup\$ – Bimpelrekkie May 4 '17 at 20:39
  • \$\begingroup\$ Thank you for your advise! I'll work on it you are right. \$\endgroup\$ – Seraj May 4 '17 at 21:28
  • \$\begingroup\$ Other than the question about arrow direction, I am still not sure what the actual question is. \$\endgroup\$ – jonk May 4 '17 at 22:08
  • \$\begingroup\$ @jonk it's not actually a question just an example. I'll upload a sample question I found on a previous exam. \$\endgroup\$ – Seraj May 4 '17 at 22:10
  • \$\begingroup\$ Okay. Thanks. I couldn't figure out if I had anything to say, or not. I see a lot of photocopied details and pretty much nothing from you except "confusion" and the arrow direction (which I think you understand already from answers given, if not otherwise.) \$\endgroup\$ – jonk May 4 '17 at 22:11
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First of all is this schematic correct?

This circuit is a standard common-base amplifier using a PNP transistor.

If you wanted a D flip-flop, it's not correct.

If you wanted a PNP common-base amplifier, it is correct.

What about current directions are they conventional?

Unless you're in the US Navy or somebody specifically mentions "electron current", you can safely assume that "current" means "conventional current".

I can't understand why collector current is heading to the transistor isn't it supposed to be output current

Your text is probably using the passive sign convention. This means taking all device currents going in to the device. Using this convention simplifies calculating the power consumption of each device in the circuit.

If current is actually flowing out of the collector, you'll just find \$i_c < 0\$.

why they are written in small letters when the are effected by the DC sources VEE and VCC

What small letters and upper-case letters mean is entirely a convention decided by each author. So you'll have to read your text to find out what is intended.

It is common to designate AC currents with lower-case letters. It may be that even though DC currents are present, the author wanted to discuss AC currents, so he indicated the AC currents on the diagram.

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    \$\begingroup\$ You are right about passive sign conv. I found ic to be negative. AC-related currents of voltages are in small letters in this textbook. \$\endgroup\$ – Seraj May 4 '17 at 21:09
  • \$\begingroup\$ Please take a look at the full page, I see it as a complete mess, I hope you can clarify what is he doing for me. Thanks in advance sir! \$\endgroup\$ – Seraj May 4 '17 at 21:10
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    \$\begingroup\$ @SerajSersawi, I agree it is a mess. Look for a better book. (Maybe you made a comment about Vcc < 0, then deleted it --- Given the way Vcc is defined in the diagram with + terminal at ground, and - terminal toward the PNP collector, Vcc must be positive (making voltage at the collector negative) for this circuit to work). \$\endgroup\$ – The Photon May 5 '17 at 0:24
  • \$\begingroup\$ You may be confused by the assumption of using KCL convention for sum of all current into a node=0, but this convention is rarely used in BJT calculations as shown. We never think of collector currents as negative. That's why I considered it wrong. Just positive from + to -. So remember this in future. \$\endgroup\$ – Sunnyskyguy EE75 May 5 '17 at 2:09
  • \$\begingroup\$ yeah I was hesitant about it but it's right, I think I should see my lecturer but he barely answers else than a smile. I've searched youtube, allaboutelectronics and many other sites and I couldn't find one textbook talking about PNP BJT common base configuration. Anyways thank you all for your contributions I'll try to handle this mess. @TonyStewart.EEsince'75 , The Photon. \$\endgroup\$ – Seraj May 5 '17 at 8:03
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The arrow is wrong for \$ i_C\$ if you consider DC current flows from + to - and even ac non inverting current.

But the schematic is correct for a Common Base PNP amplifier with input via emitter (low Zin= {(Rbe/hFE) // Re} and Zout=Rc)

\$ i_C\$ flows from + to -ve same as \$i_E\$

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  • \$\begingroup\$ Thank you! Please take a look at the full page I attached. \$\endgroup\$ – Seraj May 4 '17 at 21:28
  • \$\begingroup\$ which part of your homework do you not understand? \$\endgroup\$ – Sunnyskyguy EE75 May 4 '17 at 23:39
  • \$\begingroup\$ Rc=V(Rc)/Ic (where Ic=Ie*.99) = 10V/(2mA*.99) \$\endgroup\$ – Sunnyskyguy EE75 May 4 '17 at 23:45
  • \$\begingroup\$ This is not a homework, I have a problem with sign conventions in this schematic. \$\endgroup\$ – Seraj May 4 '17 at 23:50
  • \$\begingroup\$ If Av=gain=Rc/Rs as long as Rs >rE and Re is neglected as it is for DC bias and Rs is usually selected << Re while Rc is computed above and i(Rs)-0.6mA (pk) . Dont confuse DC bias current with AC gain. \$\endgroup\$ – Sunnyskyguy EE75 May 5 '17 at 0:02

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