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Figure

How do I calculate the equivalent Thevenin resistance of the small-signal model of a diode-connected transistor (base tied to collector)? Early-effect is neglected. The port of interest is between C and E.

In the figure, \$v_{\pi}\$ is the small-signal voltage between B and E. I did the lower two circuits, so I'm not sure if the \$v_{\text{thev}}\$ and \$i_{\text{nor}}\$ are correct. But if they are, then I get

$$Z_{\text{thev}} = \frac{v_{\text{thev}}}{i_{\text{nor}}} = -\frac{1}{g_m}$$

which isn't the same to the one I get when I apply a test voltage \$v_x\$ to the port, and set independent voltage source \$v_{\pi}\$ to zero. Then I just have a short circuit and can't calculate:

$$Z_{\text{thev}} = \frac{v_{x}}{i_{x}} $$

Null
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M_E
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  • \$v_{\pi}\$ is not independent... – Null Oct 14 '14 at 22:44
  • if B wasn't tied to C, would it still be dependent? – M_E Oct 14 '14 at 22:57
  • Yes, it depends on \$v_{\pi}\$. The angled symbol for \$g_m v_{\pi}\$ denotes a *dependent* current source, while the circular symbols in the lower diagrams denote *independent* voltage and current sources. – Null Oct 14 '14 at 23:04

1 Answers1

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Thevenin and Norton equivalents typically involve independent voltage and/or current source(s). But your only current source here is \$g_m v_{\pi}\$ and it is dependent on \$v_{\pi} = v_{be}\$ (which in this case equals \$v_{ce}\$ because of the diode connection).

To find the equivalent resistance apply a test voltage \$v_x = v_{ce}\$ across C and E, and find the current \$i_x\$ through it. The current is

$$i_x = \frac{v_x}{r_{\pi}} + g_m v_{x}$$

where the first term comes from the current through \$r_{\pi}\$ and the second from the current through the dependent source \$g_m v_{\pi}\$. Also note \$v_{\pi} = v_x\$ (again, the diode connection). Now just solve for \$v_{x}/i_{x}\$.

Null
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  • and \$v_{thev}=0\$ ? – M_E Oct 14 '14 at 23:10
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    @M_E Yes, because there are no independent sources. To find \$v_{thev}\$ you would leave the independent sources on and compute the open circuit voltage across C and E. However, the dependent voltage source is not driven by anything so \$g_m v_{\pi} = 0\$, which means no current flows through \$r_{\pi}\$, and \$v_{\pi} = v_{thev} = 0\$. – Null Oct 14 '14 at 23:14
  • Wow, if you understand doesn't that deserve an upvote? @Null (+1), I was going to guess the thermal "resistance" 25mV/Ic divided by the current gain. Solving your algebra I get rπ/(1+gm*rπ) is rπ the thermal resistance, and is gm*rπ equal to the current gain? – George Herold Oct 15 '14 at 00:10
  • @GeorgeHerold Yes, exactly. And thanks for the +1. – Null Oct 15 '14 at 00:34
  • @Null, The diode connected transistor has this "hidden" gain. (Though your answer has laid it bare.) You've got to be a bit careful 'cause the gain does change with Ic. I've been using diode connected transistors as temperature sensor's so... I'm interested. – George Herold Oct 15 '14 at 00:51
  • I can't give an upvote, when I try I get "Vote Up requires 15 reputation" – M_E Oct 15 '14 at 01:07
  • @GeorgeHerold Indeed, [diodes are often used for temperature sensing](http://en.wikipedia.org/wiki/Diode#Temperature_measurements) and [transistors can be used similarly](http://en.wikipedia.org/wiki/Silicon_bandgap_temperature_sensor) (whether diode connected or not). – Null Oct 15 '14 at 02:36
  • @M_E You can instead accept the answer by clicking the check mark icon just below the up and down vote buttons. In fact, this is encouraged and you will receive 2 reputation points for accepting an answer. – Null Oct 15 '14 at 02:38