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In common emitter configuration, is the equation \$V_{CC}=I_{C}R_{C}+V_{CE}\$ a valid equation for all regions of operation in transistor. In other words, can I use this equation to calculate the collector current in the cutoff region?

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    \$\begingroup\$ Please don't use weird acronyms without defining them. What's a CEC? \$\endgroup\$
    – Hearth
    Jun 18, 2023 at 22:15
  • \$\begingroup\$ Sorry! I mean by CEC common emitter configuration \$\endgroup\$
    – Jack
    Jun 20, 2023 at 9:58

3 Answers 3

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Yes, assuming the emitter is at 0 VDC. If you have an emitter resistor or anything else that drops voltage in the emitter circuit then you have to take that into account as well and the formula would be $$V_{CC}=I_{C}R_{C}+V_{CE}+V_E$$

As LvW points out in the comments, if there is any other DC connection to the collector such as a DC connected load or collector feedback biasing that will affect the voltage drop across \$R_C\$. So a more accurate formula would be $$V_{CC}=I_{R_C}R_{C}+V_{CE}+V_E$$

although at this point it's not much good for finding \$I_C\$ anymore without knowing the other currents through \$R_C\$.

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  • \$\begingroup\$ Here is another requirement for the validity of the equation: No DC connection between collector and a possible load resp. 2nd transistor stage. \$\endgroup\$
    – LvW
    Jun 18, 2023 at 7:59
  • \$\begingroup\$ @LvW Yes. I'll add that to the answer. \$\endgroup\$
    – GodJihyo
    Jun 18, 2023 at 21:26
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Yes, because the resistor defines how much is VCC-Vc so the rest must be Vc-Ve. Also the resistor obeys Ohm's law so only Vcc-Vc must be IcRc.

So the equation must be true no matter what the Ic or Vce is, under normal operating conditions.

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    \$\begingroup\$ (mostly for the benefit of future readers) it may be worth adding a remark that this assumes that the equation was correct for the circuit topology (e.g. common emitter without degeneration) to begin with. \$\endgroup\$
    – nanofarad
    Jun 17, 2023 at 20:49
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Is this equation \$V_{CC}=I_CR_C+V_{CE}\$ a valid equation in all regions of operation of a transistor?

Not often does the emitter connect directly to ground so no, it isn't valid because, it doesn't account for the emitter resistor voltage \$V_{RE}\$: -

enter image description here

Image from Single Stage Common Emitter Amplifier Circuit.

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  • \$\begingroup\$ It's still a valid equation for a circuit with emitter grounded. For another circuit where emitter is not grounded it does not apply but then that can be represented with another equation. \$\endgroup\$
    – Justme
    Jun 17, 2023 at 20:53
  • \$\begingroup\$ Tell that to the OP \$\endgroup\$
    – Andy aka
    Jun 17, 2023 at 23:58

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