# Homework Help: 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.