The pictures below are enhanced output impedance current mirror and a general negative feedback model. The current mirror has a negative feedback loop. However, I can't determine where is the input that is corresponding to the "input" signal in the general feedback model.

What is the input signal of the feedback network here? The output signal is Iout, and the input is not clear to me.

PS: I would like to talk more about my purpose of this post now. I want to make it clear what feedback configuration the circuit is (voltage-voltage, current-voltage, voltage-current or current-current) and then I can apply the feedback theorem to calculate output impedance of this circuit.

Rout = gm1*ro1*ro2*(1+A) where A is the loop gain and the output impedance of the network without feedback is gm1*ro1*ro2.

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closed as unclear what you're asking by Andy aka, Daniel Grillo, Voltage Spike, Dave Tweed Oct 2 '16 at 14:40

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  • \$\begingroup\$ It's like you took two partial pictures from different websites and joined them together in order to ask a question that appears irrelevant to both pictures. \$\endgroup\$ – Andy aka Sep 28 '16 at 17:50
  • \$\begingroup\$ Yes, one from my book and the other from a website. \$\endgroup\$ – anhnha Sep 28 '16 at 17:51
  • \$\begingroup\$ There is no feedback in 6.13 in the sense of the input as you've defined it. The top circuit is like two different current sources in series, with two ways to control them \$ V_{bias} \$ and \$ I_{in}\$. Even if you call the voltage controlled current source opamp circuit a constant offset there really is no feedback from input to output \$\endgroup\$ – Voltage Spike Sep 28 '16 at 18:56
  • \$\begingroup\$ So, how do you understand the feedback in this case increases output impedance by that factor (1+A)? \$\endgroup\$ – anhnha Sep 28 '16 at 19:04

The input signal is Vbias and the output signal is the voltage at the drain of Q2. The gain Aol is simply A since Q1 acts as a source follower when driven by the amplifier.

The goal is to stabilize the drain voltage of Q2 to increase the output resistance of the cascode. The technique is called "gain boosting", because the increased output resistance results in a higher gain when this circuit is used as part of an amplifier.

In order to determine the output resistance a two port model can be used as shown below, here the amplifier A is missing, but it could be taken into account by using gm A instead of simply gm. Also the current sensing path is assumed to be ideal which introduces a slight error. To fix this an additional resistor ro2 would be required in at the feedback input of the lower two-port.

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Without modification the result is: $$ R_{out} = \frac{v_x}{i_x} = ro(1 + g_m R_F) $$

  • \$\begingroup\$ Thanks, but in small signal model, Vbias is ground so how can it be input? I want to determine where is the input and output so I can apply feedback theory to calculate Rout = gm1*Rds1*Rds2*(1+A). \$\endgroup\$ – anhnha Sep 28 '16 at 17:55
  • \$\begingroup\$ In the small-signal domain the input actually is zero. And the feedback loop should keep the output at zero as well. If you want to apply feedback theory to determine Rout, then the input would be Vout and the output Iout (or maybe the other way round, whichever is easier to do). \$\endgroup\$ – Mario Sep 28 '16 at 18:08
  • \$\begingroup\$ Simple network analysis should be much easier, though. \$\endgroup\$ – Mario Sep 28 '16 at 18:11
  • \$\begingroup\$ I already derived that formula using network analysis. However, I would like to understand how feedback network can be used here to determine output impedance. As you said, Vout is input and Iout is output then you are considering the circuit as one-port network? \$\endgroup\$ – anhnha Sep 28 '16 at 18:40
  • \$\begingroup\$ The derivation using feedback theory is usually done using two-ports and a with slightly different model than the one shown in your question. \$\endgroup\$ – Mario Sep 28 '16 at 18:59

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