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I am trying to solve a problem set from MIT OpenCourseWare's 6.002 "Circuits and Electronics" course.

There is a problem about amplifiers that I would like to discuss here.

Consider the following two amplifiers

enter image description here

We are asked to show that the dependent current source sinks power for \$v_{\text{out}}>0\$ and sources power for \$v_{\text{out}}<0\$. I don't know how to do this.

There are a few questions before this one about the amplifers. In what follows, let me go through the calculations that they entailed.

Amplifier A is a single-state amplifier implemented with a voltage-dependent current source and pull-up resistor.

Let's assume that the current source parameters \$G\$ and \$V_T\$ satisfy \$G>0\$ and \$V_S>V_T>0\$. In addition, assume that \$RG<\frac{V_S}{V_S-V_T}\$.

Amplifier B is a two-stage amplifier in which each stage is identical to amplifier A.

The relationship between \$v_{\text{out}}\$ and \$v_{in}\$ for amplifier A can be obtained by applying KCL to the node with the positive terminal of \$v_{\text{out}}\$.

Assuming \$v_{\text{in}}\geq V_T\$,

$$\frac{V_S-v_{\text{out}}}{R}=G(v_{\text{in}}-V_T)^2\tag{1}$$

$$v_{\text{out}}=V_S-RG(v_{\text{in}}-V_T)^2\tag{2}$$

Graphically, this relationship looks like the light blue curve below

1707253006457.png

As \$v_{\text{in}}\$ increases, \$v_{\text{out}}\$ decreases until we have \$v_{\text{out, min}}=v_{\text{in}}-V_T\$.

$$v_{\text{out,min}}=V_S-RGv^2_{\text{out,min}}\tag{3}$$

which gives

$$v_{\text{out,min}}=\frac{-1+\sqrt{1+4RGV_S}}{2RG}\tag{4}$$

I'd like to determine \$v_{\text{out}}\$ as function of \$v_{\text{in}}\$ for amplifier B.

It seems that we have the following setup

enter image description here

Assuming \$v_{\text{in,1}}\geq V_T\$ then

$$v_{\text{in,2}}=V_S-RG(v_{\text{in,1}}-V_T)^2\tag{5}$$

and assuming \$v_{\text{in,2}}\geq V_T\$ then

$$v_{\text{out}}=V_S-RG(v_{\text{in,2}}-V_T)^2\tag{6}$$

Subbing (5) into (6) we get

$$v_{\text{out}}=V_S-RG(V_S-V_T-RG(v_{\text{in,1}}-V_T)^2)^2\tag{7}$$

$$=(V_S-RG(V_S-V_T)^2)-2(RG)^2(V_S-V_T)(v_{\text{in,1}}-V_T)^2-(RG)^3(v_{\text{in,1}}-V_T)^4\tag{8}$$

This is the relationship between \$v_{\text{out}}\$ and \$v_{\text{in}}\$ for amplifier B.

We are asked to plot this relationship. I am still working on this.

Then there is the following question.

Consider amplifier A again. Show that the dependent current source sinks power for \$v_{\text{out}}>0\$ and sources power for \$v_{\text{out}}<0\$.

If \$v_{\text{out}}\$ is negative does this mean the voltage source is such that the positive terminal is now on the bottom and \$v_{\text{in}}>V_S\$?

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  • \$\begingroup\$ Nowhere in the above do you mention this is about a mosfet. (Clearly it is an ideal mosfet, given \$V_T\$ which is also not discussed above and the shown square-law behavior expressions.) That said, you should be able to see that if vout < 0 then RG > vs/(vs-vt), which violates what you wrote in "In addition, assume that ...". I'm not reading the lessons, but I can sometimes spot logical contradictions. \$\endgroup\$ Commented Feb 7 at 2:11
  • \$\begingroup\$ I've added a link to the actual problem set. The problem set also does not mention that the voltage-controlled current source is actually a MOSFET in saturation. But yes, the context of this problem set is MOSFETs and the MOSFET amplifier. \$\endgroup\$
    – xoux
    Commented Feb 7 at 3:00
  • \$\begingroup\$ Vs is (presumably) a positive voltage source, and can therefore only supply power to the output when the output is also positive. Under those conditions the current source is sinking current away from the output in order to appropriately control Vout. If the conditions at Vin are such that the output voltage is negative, then where else, other than from the current source, could the output power come from - since Vs can only provide a positive output? In this state, the current source is (still) sinking current but its now responsible for sourcing power to the output. \$\endgroup\$
    – brhans
    Commented Feb 7 at 10:06

1 Answer 1

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First of all, the power consumed by an amplifier is not from the V_in, so you don't need to find the relationship between V_in and V_out. Normally, V_in is the small input signal with negligible power and V_S is the real power supply (S stands for source), which provides power for the whole amplifier. In the amplifier's equivalent model with a dependent current source, the two sides of the dependent current source actually could be regarded as isolated; i.e., the left side and right side has no relationship at all, like a transformer. Then you'll understand the first problem: if V_out > 0, the current source is like a resistor (voltage divider). In this case, V_out will always be smaller than Vs. If V_out < 0, the "+" line in the diagram sinks current from V_S and/or the current source, which means that the current source now acts like a voltage (power) source. For the next problem, as I mentioned, the right-hand side of the current source has no relationship with the left-hand side. They are only coupled with the equations shown in the diagram, which is only an I-V relationship. V_in could only determine the current in the current source, but can't determine the voltage. The voltage of right-hand side is determined like "locally" through KVL or KCL.

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  • \$\begingroup\$ What does it mean that the "+" line in the diagram (ie, I imagine you mean that particular node) sinks current from \$V_S\$ and/or the current source? \$\endgroup\$
    – xoux
    Commented Feb 7 at 3:03
  • \$\begingroup\$ When \$v_{out}>0\$ you say the current source is like a resistor. Is this really a good approximation? \$v_{out}\$ has a nonlinear relationship with \$v_{in}\$. \$\endgroup\$
    – xoux
    Commented Feb 7 at 3:10
  • \$\begingroup\$ 1. When V_out < 0, the "+" line becomes the common ground for V_S and the current source. Current flows from V_S, through R, to the "+" line (also possible to the current source), which is now GND. In the lower part, current flows from the current source to V_OUT side, which should be attached to a load. You'll find that both the current and voltage polarity of the current source are the same, that's the reason why it's providing power. \$\endgroup\$
    – Keyan
    Commented Feb 7 at 3:39
  • \$\begingroup\$ 2. Forget about V_in when you are talking about power. The current source acts like a voltage divider resistor for V_S, which creates a V_out that is always smaller than V_S \$\endgroup\$
    – Keyan
    Commented Feb 7 at 3:41

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