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I have to measure \$R_{in}\$, the input resistance seen by the signal. It was told me, that the quickest way is measuring the voltage gain using oscilloscope, insert a resistor before \$R_B\$, measuring the new voltage gain, make the ratio and find the expression for \$r_e\$ (the input resistance seen between base and emitter looking into the emitter) and then substitute this value in the expression of \$R_{in}\$..

But in the expression of \$r_e \$ appears \$r_o||R_L\$.. and I don't know the value of \$r_o\$ (the output resistance of a BJT)... the only thing that I know is that "usually it is large"... but how much large? what about the range of its values? If \$R_L\$=1k ohm, can I neglect \$r_o\$?

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  • \$\begingroup\$ I am not sure what you want to do. In the title you speak about the output resistance - and in the text you want to "measure Rin" as "senn by the signal". Later, you again are mentioning the input resistance "looking into the emitter". \$\endgroup\$
    – LvW
    Sep 17, 2014 at 15:01
  • \$\begingroup\$ @LvW I'm looking for \$r_o\$, the output resistance of the BJT, not the output resistance of the circuit. If I find a value for \$r_o\$, I can obtain the value of \$r_e\$ and, then, the input resistance of the circuit. \$\endgroup\$
    – sunrise
    Sep 17, 2014 at 15:06
  • \$\begingroup\$ OK - if ro equals the inverse slope of the output characteristics, it is to be calculated based on the Early voltage as mentioned already. However, in parallel to an external ohmic resistor it can be neglected in most cases. \$\endgroup\$
    – LvW
    Sep 17, 2014 at 16:17
  • \$\begingroup\$ Additional remark: Don`t forget ro is also in parallel to the low-resistive input resistance at the emitter node which is [1/gm + RB/(hfe+1)]. \$\endgroup\$
    – LvW
    Sep 17, 2014 at 16:32
  • \$\begingroup\$ @LvW The first time that in the Sedra-Smith appears \$r_o\$ is when it tells about the \$i_C-v_{CE}\$ characteristics... so I think that \$r_o\$ is the inverse slope of the characteristics... do you agree? \$\endgroup\$
    – sunrise
    Sep 17, 2014 at 17:19

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The value of \$r_{o}\$ depends on the transistor's characteristics and its biasing:

$$r_{o} = \frac{V_{\text{A}} + V_{\text{CE}}}{I_{\text{C}}} \approx \frac{V_{\text{A}}}{I_{\text{C}}}$$

where \$V_{\text{A}}\$ is the transistor's Early voltage and can vary widely between transistors, but is usually on the order of \$10 - 100\text{ V}\$. Assuming a typical bias of \$I_{\text{C}} \approx 1\text{ mA}\$, \$r_{o}\$ should be on the order of \$10 - 100\text{ k}\Omega\$. If \$R_{\text{L}} = 1\text{ k}\Omega\$ and \$r_{o}\$ is only \$10\text{ k}\Omega\$ then you could have a significant error if you assume \$r_{o} \parallel R_{\text{L}} \approx R_{\text{L}}\$. But if \$V_{\text{A}}\$ is higher or \$I_{\text{C}}\$ is lower then it may be safe to ignore \$r_{o}\$, depending on how accurate you are looking to be.

Also, it's worth noting that for calculating \$R_{\text{in}}\$ the resistance \$R_{\text{E}} = r_{o}||R_{\text{L}}\$ at the emitter is magnified by \$\beta + 1\$ so an error in \$R_{\text{E}}\$ due to ignoring \$r_{o}\$ could be magnified significantly. On the other hand, for calculating \$R_{\text{out}}\$ the emitter resistance \$R_{\text{E}}\$ is divided by \$\beta + 1\$ so the error would not be as significant.

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  • \$\begingroup\$ Null, I suppose, you misunderstood the meaning of ro (see the diagram). I don´t think that this resistance is the inverse slope of the Ic=f(Vce) characteristics (as you have assumed). \$\endgroup\$
    – LvW
    Sep 17, 2014 at 15:04
  • \$\begingroup\$ \$r_{o}\$ should be connected between the collector and emitter. It is clearly connected to the emitter, and it is connected to the collector since the collector is connected to small-signal ground. So I think this is the standard meaning of \$r_{o}\$. \$\endgroup\$
    – Null
    Sep 17, 2014 at 15:10
  • \$\begingroup\$ @LvW, I'm not looking for \$R_out\$.. If I consider the \$ \pi \$-model of the BJT, there is a resistance \$r_o\$.. and \$r_o\$ there is also in the T-model.. the values of these two \$r_o\$ are the same.. isn't it? \$\endgroup\$
    – sunrise
    Sep 17, 2014 at 15:13
  • \$\begingroup\$ Null,\$ I_C= 14.6 mA \$, so \$r_o\$= goes from 685 ohm to 6.8 k ohm. \$r_o||R_L\$ goes from 406 to 871... how can I do? \$\endgroup\$
    – sunrise
    Sep 17, 2014 at 15:24
  • \$\begingroup\$ @sunrise The range of \$V_{A}\$ I gave was a general order of magnitude for the Early voltage for any transistor. The particular transistor you are using should have a constant Early voltage and you should be able to find it in the datasheet. What transistor are you using? \$\endgroup\$
    – Null
    Sep 17, 2014 at 15:28

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