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As I work through learning the four topologies of negative feedback, I can't seem to figure out the transimpedance amplifier. enter image description here As far as I understand it, I need to consider my input as a current source. The signal input \$i_s\$ is reduced by the feedback network \$K\$ so that the current \$i_{in}\$ entering the internal amplifier is attenuated. Then the internal amplifier \$A_o\$ produces an output voltage \$v_{out}=R_oi_{in}\$. That output voltage is applied to the feedback network \$K\$, which itself produces the output current \$i_f\$ that attenuates the input because of KCL. So far, so good, I think.

What's missing for me is being able to understand how the discrete components in the circuit at right would apply to the block diagram at left. I was able to perform this type of analysis for the voltage amplifier (common-collector here) and transconductance amplifier (common emitter with degeneration here).

For this example, where would I begin if I wanted to derive the open-loop gain and closed-loop gain in terms of actual circuit values? How does \$R_C\$ in the schematic relate to \$R_o\$ in the block diagram? How does \$R_F\$ in the schematic relate to \$K\$ in the block diagram?

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  • \$\begingroup\$ The Ao on a single transistor is not very constant for large signals because it depends on the Ic. since you have no series input R the gain is limited by Rbe and Zc(f). But you ought to get near a gain about 400 or so with what you have. Rin will be very low ~20 \$\endgroup\$ – Tony Stewart EE75 Dec 23 '20 at 5:26
  • \$\begingroup\$ You need to synthesize the feedback network as a two-port network. For this topology, you need the \$y\$ parameters. Then, you will know exactly what K is and what are the load effects of the feedback on the amplifier (your BJT), which allows you to calculate the transimpedance OL gain since the feedback may be turned off. \$\endgroup\$ – edmz Dec 23 '20 at 13:36
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A) Closed-loop gain:

Here are the equations which have to be combined to find Af=vout/is:

ib=ic/hfe=is-if (hfe=beta)

if=-(vout-vbe)/Rf with vout=-icRc (simplification: rce>>Rc, Rf>>Rc)

vbe=ic/gm (gm=transconductance Ic/Vt)

This gives you the ratio ic/is which leads to vout/is with vout=-icRc.

B) Open-loop gain

Ao=vout/is=-icRc/is=-hfeRc

C) Loop gain and feedback factor

If you compare both results from A) and B) it is quite simple to derive the expressions for the loop gain and the feedback "factor" K (which is given in 1/Ohm). Just compare the closed loop gain Af with Af=Ao(1+KAo).

EDIT: Details to A):

ib=ic/hfe=is-if=is+(vout-vbe)/Rf=is-icRc/Rf-vbe/Rf=is-icRc/Rf-ic/gmRf

From this: ic/is=1/[1/hfe + (Rc+1/gm)/Rf] with 1/gm<<Rc.

ic/is=hfe/(1+hfeRc/Rf)

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  • \$\begingroup\$ Does your \$i_b\$ equal \$i_{in}\$ in my diagrams? \$\endgroup\$ – nuggethead Dec 23 '20 at 13:04
  • \$\begingroup\$ And is rce the same as what I've seen called \$r_o\$, or 100v/\$I_c\$? \$\endgroup\$ – nuggethead Dec 23 '20 at 13:14
  • \$\begingroup\$ yes, ib is the signal base current (i_in in your diagram) and rce is the dynamic output resistance of the bjt (Early voltage/Ic), which is much larger than Rc. \$\endgroup\$ – LvW Dec 23 '20 at 13:22
  • \$\begingroup\$ Ok, thanks @LvW but I still dont see how you completed part A in your answer. Can you possibly show more intermediate steps to work through those equations? \$\endgroup\$ – nuggethead Dec 23 '20 at 13:36
  • \$\begingroup\$ Please, see my EDIT. \$\endgroup\$ – LvW Dec 23 '20 at 13:56

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