# Why does amplifier sink power for Vout>0 and sources power for Vout<0?

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

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

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

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\$$?

• 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. Feb 7 at 2:11
• 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.
– xoux
Feb 7 at 3:00
• 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. Feb 7 at 10:06

• 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?
• 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 $v_{in}$.