# Non-linear amplification circuit

I've got a few questions about a task from my exam preparation paper. The sketch below shows the circuit in question. It is given that U_E varies between 0V...3V. The operating voltage U_B is 10V. The op amp and diode are ideal at the beginning (it does not say at which point in time they must be not seen as ideal anymore). The four tasks are the following:

1. Give an algebraic expression for the differential amplification $v_1=\frac{dU_A}{dU_E}$

My solution: $v_1=1+\frac{R_2}{R_1}$ (non inverting op amp)

2. What is U_E's value such that D1 is conducting? What is U_A then? Give algebraic expressions for U_E and U_A.

My solution: $U_E>U_B*\frac{R_3}{R_3+R_4}$ thus $U_A>U_B*\frac{R_3}{R_3+R_4}*(1+\frac{R2}{R1})$

3. What is the differential amplification v_2=dU_A/dU_E if D1 is conducting? Give an algebraic expression.

My solution: $\frac{dU_A}{dU_E}=\frac{R2}{R4}+\frac{R2}{R3}+1+\frac{R2}{R1}$

4. U_B is 10V. At which value U_E does U_A saturate? Give an algebraic expression for U_E. What is the differential voltage u_d as a function of U_E if saturation occurs?

I don't really know how to solve this task so I'd be glad if anyone could help me out.

Moreover I am not quite sure about my solution to question 2 and 3.

simulate this circuit – Schematic created using CircuitLab

• Please consider making your post more clear by writing the equations in MathJax. Note EE uses \$ to start and end inline Mathjax to avoid confusion when talking about dollar amounts. (We still use $\$ for display equations) Jul 25, 2017 at 16:41
• #2 is correct if your definition of an ideal diode is 0V forward voltage. You may want to clarify the forward voltage of an ideal diode with the teacher. For #3, if D1 is an ideal diode, just replace it with a short circuit. For #4 just set up an equation for U_A < 10V. Once the op amp saturates, it can no longer change the inverting input to match the non-inverting input. Jun 19, 2018 at 20:09

A few observations to simplify things:

1. The questions are in terms of $$\\frac{dU_A}{dU_E}\$$, which is the slope of the transfer characteristic, representing gain only, and ignoring any offset (y-intercept on a graph of $$\U_E\$$ vs $$\U_A\$$).

2. The amplifier is non-inverting, since the input manipulates the non-inverting input.

3. The amplifier operates in two regimes, X) where the diode is perfectly conducting, Y) where the diode is reverse biased and effectively absent.

4. When the diode is reverse biased (regime Y), its anode potential is less than the cathode. We can see that further decreases in input potential can only decrease diode anode potential (due to the resistor divider R1, R2) and the diode will remain reverse biased. This also means that the cathode potential is fixed and irrelevant to calculations.

5. When the diode is forward biased (regime X), its cathode potential becomes dynamic, and can no longer be neglected.

To find the input potential which will just place the voltage across D1 at 0V, at the threshold of transition from regime X to regime Y, it is sufficient to remove D1, and find the input potential which would produce 0V across the bridge between the two dividers R1,R2 and R3,R4. Calling the potential at the junction of R1,R2 $$\U_P\$$, and at R3,R4 $$\U_Q\$$:

\begin{aligned} U_Q &= U_B \frac{R_3}{R_3+R_4} \\ \\ &= U_B \frac{200k\Omega}{200k\Omega+800k\Omega} \\ \\ &= \frac{U_B}{5} \\ \\ \\ U_P &= U_A \frac{R_1}{R_1+R_2} \\ \\ &= U_A \frac{400k\Omega}{400k\Omega+800k\Omega} \\ \\ &= \frac{U_A}{3} \end{aligned}

$$\U_P\$$ and $$\U_Q\$$ are equal at the cross-over point between regimes:

$$U_P = U_Q = \frac{U_B}{5}$$

To find the input voltage $$\U_E\$$ that corresponds to this point, which I'll call $$\U_{EX}\$$ you have one dead simple answer, and two less trivial solutions, depending on how observant you are. Easiest first:

Use the fact that $$\u_D = 0V\$$, due to opamp action with negative feedback:

\begin{aligned} U_{EX} &= U_P = U_Q\\ \\ &= \frac{U_B}{5} \end{aligned}

Use the fact that gain is the reciprocal of feedback factor:

You might have noticed that the feedback factor β is $$\\frac{1}{3}\$$, from the relationship we derived earlier:

$$U_P = \frac{U_A}{3}$$

Gain is therefore 3, the reciprocal of $$\\frac{1}{3}\$$. Using this gain, we can say that:

\begin{aligned} \frac{U_A}{U_E} &= 3 \\ \\ U_{EX} &= \frac{U_{A}}{3} \\ \\ &= \frac{3U_P}{3} \\ \\ &= U_P \\ \\ &= U_Q \\ \\ &= \frac{U_B}{5} \end{aligned}

Use the gain equation:

If neither of the above shortcuts work for you, you can just fall back on the good old expression for gain, and follow the same procedure to derive $$\U_{EX}\$$:

\begin{aligned} \frac{U_A}{U_E} &= 1 + \frac{R_2}{R_1} \\ \\ &= 1 + \frac{800k\Omega}{400k\Omega} \\ \\ &= 3 \\ \\ \end{aligned}

So, the input voltage at the point of transition from one regime to the other is $$\U_{EX} = \frac{U_B}{5}\$$. For completeness, the output voltage at that same point, $$\U_{AX}\$$, is:

\begin{aligned} U_{AX} &= 3U_{EX} \\ \\ &= \frac{3U_B}{5} \end{aligned}

For $$\U_E \le \frac{U_B}{5}\$$ (regime Y) the diode is reverse biased, and the gain equation is simply:

$$\frac{dU_A}{dU_E} = 3$$

That was the easy part. Now onto regime X where we may not ignore D1, R3 and R4. There are a few ways of attacking the problem, such as superposition (ughh), nodal/mesh analysis (meh), and reduction by Thevenin's theorem.

To apply Thevenin's theorem, we'll assume the diode is ideal, forward biased, and therefore is effectively a wire link across the bridge between the two dividers. The resistor network starts and ends like this:

simulate this circuit – Schematic created using CircuitLab

Let me redraw the whole circuit again, this time with the Thevenin equivalent derived above:

simulate this circuit

Because of observation #1 at the beginning of this answer, I can confidently ignore $$\V_{TH2}\$$. All it does is introduce an offset into the gain equation, but I am only after the gain $$\\frac{dU_A}{dU_E}\$$. So, for $$\U_E > \frac{U_B}{5}\$$ that will be:

\begin{aligned} \frac{dU_A}{dU_E} &= 1 + \frac{R_2}{R_{TH2}} \\ \\ &= 1 + \frac{800k\Omega}{\frac{800}{7}k\Omega} \\ \\ &= 8 \end{aligned}

If it's not clear why you can ignore $$\V_{TH2}\$$, then just figure out the transfer characterstic of the above simplified amplifier:

\begin{aligned} U_A &= (U_E - V_{TH2}) (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega}) \\ \\ &= U_E\cdot \overbrace{(1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})}^{\text{gain}} \space \overbrace{ - V_{TH2} \cdot (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})}^{\text{constant offset}} \end{aligned}

The term $$\- V_{TH2} \cdot (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})\$$ is constant. If you take the derivative of $$\U_A\$$ with respect to $$\U_E\$$, that term disappears, leaving you with only the coefficient of $$\U_E\$$, which is gain.

That's my approach to this problem, which I hope was useful. I'll leave you to plug in $$\U_B = 10V\$$, and see if my answers agree with yours.

To finish with, here's a CircuitLab representation, and the output curve I get when simulating:

simulate this circuit

Okay, the first thing to realize is that this circuit is "piecewise linear"- there are two regions of operation, each of which is linear and there is a transition that takes place at the point where the diode starts to conduct.

For the op-amp balanced that point is where Ue is equal to the voltage at the junction of R3/R4 (ignoring the diode). You can calculate that from Ub and it gives you the answer to section 2 when you add in which region the diode continues to conduct.

Then you can analyze each region individually with the diode open or shorted.

Finally, the "saturated" condition occurs when the ideal op-amp output just reaches Ub- the power supply voltage- or ground. For a real op-amp it would not quite reach either one. Since they specifically mention Ub in this section, I would assume they are at least looking for the upper saturation, but also giving the lower one would be a more complete answer.

The diode's anode is the same node as the inverting input. The voltage divider at its cathode is 10x200/(200+800)=2-volts, so the maximum inverting input voltage is 2.7V. Since the feedback voltage divider provides 1/3 of the output to the inverting input, the maximum Vout will be 2.7x3=8.1 as any Vout exceeding that will forward bias the diode, so your non-inverting input is actually limited to a 2.7 volt input as well (which will output Vout = 8.1 volts.) I don't know how you fit that in an algebra equation, but that should help, it answers #2.

Your title says "non-linear", so there is likely some "non-linearities" in there because of the diode.