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I'm trying to calculate the input resistance of the depicted small signal equivalent circuit.

I was just wondering why I can't see the current source as an interrupt and then the input resistance will be: r_pi+R.

I recall that replacing current sources with interrupts and voltage sources with short circuits is done when calculating the resistance between two nodes in a circuit. Why isn't that allowed in the small signal equivalent?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What is the upside down "V" symbol? There is a schematic editor that you can use and you could make the math less ambiguous using latex script. \$\endgroup\$
    – Andy aka
    Jun 19 '18 at 11:05
  • \$\begingroup\$ Hint: the small-signal resistance looking into the base of a common-collector amplifier is just \$r_\pi + (1 + \beta)R_E\$ where \$R_E\$ is the resistance seen looking out of the emitter. \$\endgroup\$ Jun 19 '18 at 11:59
  • \$\begingroup\$ Thank you I've also just seen that, seems that my expression is right, however I changed my question a little bit now, I just don't undestersand why I can't replace the current source with a short circuit. \$\endgroup\$
    – Lukas
    Jun 19 '18 at 13:42
  • \$\begingroup\$ But you never can replace a current source with a short circuit. \$\endgroup\$
    – G36
    Jun 19 '18 at 13:45
  • \$\begingroup\$ And in this case, it is not allowed because gm*vbe is a dependent current source. \$\endgroup\$
    – G36
    Jun 19 '18 at 13:48
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Your final equation looks good, but you can simplify it further:

$$R_{IN} = R \left( 1 + r_\pi\left(gm + \frac{1}{R}\right)\right)$$

$$R_{IN} = R \left( 1 + r_\pi g_m + \frac{r_\pi}{R}\right)$$

$$R_{IN} = R+ r_\pi g_m R + \frac{r_\pi}{R}R $$

$$R_{IN} = R+ r_\pi g_m R + r_\pi $$

And now knowing that \$r_\pi \cdot g_m = \beta\$ because

\$ \beta= \frac{d(Ic)}{d(Ib)} \$ and \$ \textrm{d(Ib)} = \frac{d(v_{be})}{r_\pi} \$

Therefore:

\$ \frac{\beta}{r_\pi} = \frac{d(Ic)}{d(Vbe)}=g_m\$

So, we can write it like this:

$$R_{IN} = R+\beta R + r_\pi $$

$$R_{IN} = (\beta + 1) R + r_\pi $$

$$R_{IN} = r_\pi + (\beta + 1) R $$

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  • \$\begingroup\$ Thanks for the nice explanation, that makes sense now. But why aren't I allowed to replace the current source with a short circuit and just calculate the resistance by combining the remaining resistors, because this is what a friend of mine did. \$\endgroup\$
    – Lukas
    Jun 19 '18 at 13:43
  • \$\begingroup\$ Because here we are dealing with AC signals and we are using the linearized BJT small-signal model. And in this model gm*vbe is a dependent current source and its dynamic resistance is not equal to 0Ω or infinity As we have when we are dealing with ideal DC sources. For example, the ideal DC voltage source will have the dynamic resistance equal to zero. Because of a DC voltage source act just like a short for an AC signal. Change in DC voltage source current does not change the source voltage. \$\endgroup\$
    – G36
    Jun 19 '18 at 14:19

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