- #1

- 321

- 0

I would like to find the transconductance of such circuit. I am choosing to ignore r_0 for the time being and going with the definition

[tex]G_m = \frac {i_0}{v_i} \bigg |_{v_i = 0}[/tex],

What I did first was notice that [tex]i_0= - g_m v_\pi[/itex]

[tex]v_i[/tex] appears as the top node voltage at [itex]r_\pi[/itex] w.r.t to ground. Then I have a voltage divider and using the impedance reflection rule, I have the relationship [tex]v_\pi = \frac {r_\pi}{r_\pi + (\beta + 1)R_{EA}}v_i[/tex], then [tex]G_m = -\frac {g_m v_\pi r_\pi}{r_\pi + (\beta + 1) R_{EA} } = - \frac {g_m}{1+( \frac{\beta + 1}{r_\pi})R_{EA}}[/tex]

Then I use [tex]r_\pi = \frac {\beta}{g_m}[/tex] and [tex]\beta >> 1[/tex], so [tex]G_m = -\frac {g_m}{1 + g_mR_{EA}}[/tex]. Is this correct?

[tex]G_m = \frac {i_0}{v_i} \bigg |_{v_i = 0}[/tex],

What I did first was notice that [tex]i_0= - g_m v_\pi[/itex]

[tex]v_i[/tex] appears as the top node voltage at [itex]r_\pi[/itex] w.r.t to ground. Then I have a voltage divider and using the impedance reflection rule, I have the relationship [tex]v_\pi = \frac {r_\pi}{r_\pi + (\beta + 1)R_{EA}}v_i[/tex], then [tex]G_m = -\frac {g_m v_\pi r_\pi}{r_\pi + (\beta + 1) R_{EA} } = - \frac {g_m}{1+( \frac{\beta + 1}{r_\pi})R_{EA}}[/tex]

Then I use [tex]r_\pi = \frac {\beta}{g_m}[/tex] and [tex]\beta >> 1[/tex], so [tex]G_m = -\frac {g_m}{1 + g_mR_{EA}}[/tex]. Is this correct?

Last edited: