Another non-linear device question

1. Jun 26, 2010

jegues

1. The problem statement, all variables and given/known data
See figure.

2. Relevant equations

3. The attempt at a solution

I tried KVL (see figure), but still got stuck with only 1 equation and two unknowns. Any ideas/tips on what I should try next?

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2. Jun 26, 2010

stevenb

First, I don't like the wording of this question. It is a little confusing. I also don't like the notation. It is also a little confusing. I don't normally give full answers when no good attempt has been made, but I can see why you were confused.

My interpretation of this confusing question is that they are asking you to linearize the response for small signals. The capital V would represent a large DC signal and the small v (with a small s subscript) would be a small AC signal. The solution is to simply take the derivative of the nonlinear equation dv_Q/dv_S and this should approximately equal v_q/v_s. I get dv_q/dv_s=3v_S-3=12 when V_S=5.

Last edited: Jun 26, 2010
3. Jun 26, 2010

jegues

How did you get this?

$$\frac{3}{2}(v_{S} - 1)^{2} \Rightarrow 3v_{S} - 3 \Rightarrow 3(5 + v_{s}) -3$$

What is this equal to anyways, we don't know what vQ is?

4. Jun 27, 2010

stevenb

Sorry, I just noticed that you asked a question about my response.

First, let me say that my answer is my best guess on how to solve the problem. When I look at the wording of the problem I find it vague and misleading. It's possible that I'm missing an obvious thing, but all I can do is give you my opinion.

The way I interpret the problem is that you need to linearize the nonlinear equation and find the response of small changes in v_Q to small changes in v_S. The interpretation I make is that v_q and v_s are small signals that take place around the operating point V_S=5 V and V_Q=24 V;

Hence, v_q/v_s can be approximated by the derivative dv_Q/dv_S evaluated at the DC operating point of V_S=5 V and V_Q=24 V. Hence the following

$${{v_q}\over{v_s}}={{\Delta v_Q}\over{\Delta v_S}}\approx{{d v_Q}\over{dv_S}}\bigg|_{v_S=5V}=12$$

The fact that I arrive at one of the possible answers gives me some confidence that this is correct, but I'm not totally confident that another interpretation isn't possible.

5. Jun 27, 2010

jegues

How did you know that V_Q = 24 V?

$${{v_q}\over{v_s}}={{\Delta v_Q}\over{\Delta v_S}}\approx{{d v_Q}\over{dv_S}}\bigg|_{v_S=5V}=12$$

When you evaluate this you find that $$v_{S} = 5\cdots v_{s} = 0$$

And if $$v_{s} = 0 \cdots \frac{v_{q}}{v_{s}} = undefined$$

6. Jun 28, 2010

jegues

Bump, still looking for some help!

7. Jun 28, 2010

stevenb

Sorry, somehow I keep missing this thread.

I found V_Q simply by plugging V_S=5 in for v_S in the nonlinear equation.

I interpret v_s to mean a small signal that is not zero, but just small compared to V_S. The small response changes of v_q in response to small changes in v_s can be approximated by the derivative. Since the derivative changes at different points on a curve, if the function is nonlinear, we need to evaluate the derivative at the operating point.

This probably looks very confusing, the first time you see it, but it's a very common procedure for linearizing nonlinear circuits. If you can find a good analog circuit design book that covers transistors and diodes, you'll find this concept explained in a much better way than I've done here.

Somewhere I have a help-note I wrote on linearization. I'll try to find it and I'll post it if I find it. [EDIT: I can't seem to find it]

Last edited: Jun 28, 2010
8. Jun 28, 2010

jegues

Thanks for your responses stevenb, things have become MUCH more clear.