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I have the three-stage amplifier below. I am interested in finding the output resistance of it. Just wanna check if am doing it correctly? Thanks

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Analysis: \begin{equation} i_o=i_2-i_1\\i_1=i_{b3}+g_{m3}v_{\pi 3}\\i_o=i_2-i_{b3}-g_{m3}v_{\pi 3}\\i_o=\frac{v_o}{2.4k\Omega }-\frac{\left(v_{\pi 3}-v_o\right)}{r_{\pi 3}+5.6k\Omega }-g_{m3}v_{\pi 3}\\v_{\pi }=-v_o\\i_o=\frac{v_o}{2.4k\Omega \:}-\frac{\left(-v_o-v_o\right)}{r_{\pi \:3}+5.6k\Omega \:}-g_{m3}\left(-v_o\right)\\i_o=\frac{v_o}{2.4k\Omega \:}+\frac{\left(2v_o\right)}{r_{\pi \:3}+5.6k\Omega \:}+g_{m3}\left(v_o\right)\\g_{m3}=77.6m\left(\frac{A}{V}\right)\\R_o=\frac{v_o}{i_o}=12.7\Omega \end{equation}

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  • \$\begingroup\$ How are you doing it? What load current is planned? \$\endgroup\$ Aug 11 '18 at 18:04
  • \$\begingroup\$ You need to add a large capacitor in series with the V3 voltage source. \$\endgroup\$
    – G36
    Aug 11 '18 at 18:05
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    \$\begingroup\$ try 10mF at first \$\endgroup\$
    – G36
    Aug 11 '18 at 18:08
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    \$\begingroup\$ Show us your work \$\endgroup\$
    – G36
    Aug 11 '18 at 18:19
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    \$\begingroup\$ The voltage divider equation is the answer vpi3 = -Vo x rpi3/(Rc2 + rpi3) \$\endgroup\$
    – G36
    Aug 11 '18 at 19:36
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Update: if you are still having issues, remember that the DC design must be proper to get linear AC gain. The intermediate stages need proper impedance optimization for gain and DC bias and that has not been Done here. Emitter DC drops are too high and input impedance too low so extra emitter R’s are needed and lower existing values.

Emitter followers can only source current while Re sinks current. (Like LDO’s)

\$Zout = \dfrac{R_{C2}} {h_{FE3}}||R_{E3}\$ For AC coupled output to a load , looking back into Zout, a negative voltage now sources current from the external cap swing sinks current and if this exceeds I(Re3) for
and = Re for -ve voltage.

Ignoring base emitter resistance etc.

Note this design is poor due to lack of active push-pull outputs. e.g. Op Amps have these. THis makes the negative supply rather useless.

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