The output impedance (\$R_{out}\$) of this circuit is \$R_{S3}\$?
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If you want to find \$ R_{OUT}\$ we need to solve this circuit:
And for this circuit \$ R_{OUT} = \frac{V_X}{I_X}\$
And from KCL we have:
$$I_X =\frac{V_X}{R_S} -g_m*(-V_X)$$
$$I_X =\frac{V_X}{R_S}+g_mV_X$$
$$I_X =\frac{V_X}{R_S}+\frac{V_X}{\frac{1}{gm}} $$
So, \$R_{OUT}\$ is ??
This method can also be used to find Rth3 in your BJT circuit problem.
And another example can be found here
G36
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Do not lack r0 resistance in the model for small signals? Why did the dependent source become an independent source? Thanks for help – Carmen González May 19 '17 at 19:17
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@CarmenGonzález For simplicity I skipped ro resistance. But this is not a problem because ro is in parallel with Rs. – G36 May 19 '17 at 19:22
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We still have a dependent source, but now Vx = -Vgs and this is why Id = gm(-Vx) – G36 May 19 '17 at 19:23
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Why did \$R_g\$ disappear in the incremental model? – Carmen González Jun 04 '17 at 20:27
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@CarmenGonzález because we find Rout for Vin = 0V. So we replace the Vin with a short circuit. – G36 Jun 05 '17 at 14:46
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I think the small signal diagram is wrong. You have shorted Vout to ground (node D should be ground). That being said, looking "up" the source, the resistance should be about 1/gm in parallel with RS3, so approximately 1/gm (provided gm is large).
user1155386
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I did not realize the change you want to make in the small sign model. Could you do a drawing, please? I had already been told about this change, but without the drawing I do not understand what it means. – Carmen González May 18 '17 at 23:19
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I make a new circuit. Can you see if the circuit is well-maded? Thanks for support. – Carmen González May 19 '17 at 11:19
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