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Hello I am trying to understand calculating impadence of a transistor using small signal analysis. In the book I came across with following exmaple.

It tries to calculate Rout.

So I didn't understand how circuit at the left side have equivalent small signal analysis as circuit at the right side.

I mean how Vin transported to emitter as Vx?

I understand that Vin must be grounded for calculating Rout with one port input. But I can't see any relation about Vin and Vx.

enter image description here

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You use the second diagram (with vin grounded) and apply a small signal test voltage (vx) and measure the small signal test current (ix).

The important thing to realize is that you are measuring small signal parameters. Vin (large signal biasing input voltage) is assumed to be still exist, biasing the circuit in the linear region. But since you are only interested in the small signal parameters, you treat vin as grounded (measuring output resistance around the operating point of the circuit).

For this circuit (which is neglecting the c-e output resistance, ro), and summing the currents at the node (and noticing that vx = -vpi)

\$ i_x + g_m*v_{\pi} = -v_{\pi}/r_{\pi} \$

\$ i_x -g_m*v_x = v_x/r_{\pi} \$

\$ i_x = v_x/r_{\pi} + g_m*v_x \$

\$ i_x/v_x = (1/r_{\pi} + g_m) \$

\$ r_o = v_x/i_x = 1/(1/r_{\pi} + g_m) \$

Unless my algebra is incorrect somewhere. Also \$ r_{\pi} = {\beta}/g_m \$

So,

\$ r_o = 1/(1/({\beta}/g_m) + g_m) \$

\$ r_o = 1/(g_m/{\beta} + g_m) \$

\$ r_o = {\frac1{g_m(1/{\beta} + 1)}} \$

Since \$ {\beta} \$ is large, this can be easily approximated to

\$ r_o = {\frac1{g_m}} \$

Keep in mind this is ignoring both the input source resistance and the collector-emmitter resistance, so a very idealized formula.

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Vin is not equal to Vx. Vin is set to zero and a test voltage Vx is applied to the output.

In order to determine the output resistance the current Ix for this small signal voltage Vx is calculated. The resistance Rout is then found as the ratio Rout=Vx/Ix.

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