Electronic – How to electronically generate exponential and logarithmic functions

Have you considered using a look-up table of sine vs angle values and DAC (digital-to-analog converter to produce the analog output? No weinbridge or RC shift oscillators needed. This reduces your parts count and creates an elegant, simplified, precise, digitally controlled oscillator. Simply cycle thru the look-up table at the desired rate to control frequency. You may want to add a low pass filter to smooth the DAC output to reduce harmonic distortion. A lookup table can be built for 1/4 of the cycle to reduce memory needed to store values for the first quadrant (\$0-90^{\circ}\$). Then use symmetry at \$90^{\circ}\$ to build the 2nd quadrant (\$90-180^{\circ}\$). Finally, repeat the 1st and 2nd quadrant with inversion to build the 3rd and 4th quadrants (\$180-360^{\circ}\$). This completes one cycle. Higher resolution is achieved with higher bit DAC. At the max. freq of 100Hz, you should have no problem finding a DAC to do the job. Build the look-up tables values by hand calcs (error prone) or use the first few terms of the Taylor series expansion of the sine function, for example: $$\sin(x) = x-x^3/3!+x^5/5!-x^7/7!+...$$

The 1st 3 ot 4 terms should give a close approximation of the sine of the angle. \$n!\$ means n-factorial, e.g. \$5! = 5 \times 4 \times 3 \times 2 \times 1\$. So, you can actually program the equation in code and load values for each angle between 0 and 90 into memory. http://en.wikipedia.org/wiki/Sine

Let's take into account a "Basic Feedback System" (hereinafter BFS) block diagram first:

^{simulate this circuit – Schematic created using CircuitLab}

We can write:

\$ V_{OUT}=A \cdot (V_{IN}+ \beta V_{OUT}) \$

Therefore the BFS overall gain:

$$ G= \frac{V_{OUT}}{V_{IN}}=\frac{A}{1- \beta A} \> \> \> \> (=\frac{1}{\frac{1}{A}- \beta}) $$

if ( \$ 1- \beta A \$ ) → 0 , then G → \$ \infty \> \> \$ (the system becomes unstable)

so, for the stability of such a system it is required: \$ \> \> \beta A ≠ 1 \$

It shows that system stability depends on the \$ \beta \$A product - the open loop gain (see the Nyquist stability criterion for instance for more details).

(For an ideal OpAmp with A → \$ \infty \> \> \$: \$ \> \> \> G= -\frac{1}{\beta}) \$

Now let's analyze those two cases in question: (starting with case 1; an inverting amplifier)

A)

^{simulate this circuit}

\$ v_+ =0 \$

\$ V_{OUT}=A \cdot (v_+ - v_-)=-A \cdot v_- \$

=> \$ v_- = - \frac{V_{OUT}}{A} \$

\$ ( i_1 = ) \$ \$ \frac{V_{IN}-v_-}{R_2} \$ = \$ \frac{v_--V_{OUT}}{R_F} \$ \$ (=i_2) \$

then:

\$ \frac{V_{IN}}{R_2}=v_- \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

Substituting now the above expression for \$ v_- \$, we obtain:

\$ \frac{V_{IN}}{R_2}=- \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

and the overall gain is as follows:

$$ G= \frac{V_{OUT}}{V_{IN}}= \frac{(-1)}{ \frac{1}{A}(1+ \frac{R_2}{R_F})+ \frac{R_2}{R_F}} \> \> \> \> \> (1) $$

(Note that the denominator of this expression never can be 0! ; presuming A and both \$ R_2 \$ and \$ R_F \$ being positive, of course)

if A → \$ \infty \$ :

\$ G=- \frac{R_F}{R_2} \$

Comparing it now with the BFS:

\$ A'=-A \frac{R_F}{R_F+R_2} \$

\$ \beta = \frac{R_2}{R_F} \$

(here A' stands for /is analogical to/ the A in BFS)

Then:

\$ \beta A'=-A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=-A \frac{R_2}{R_F+R_2}<0 \$ always (provided A>0, of course)

=> always* stable ( \$ \beta A' \$ ≠ 1)

*For "real" OpAmps this may not apply - under certain conditions (the phase angle between \$ V_{OUT} \$ and \$ (v_+ - v_-) \$ changes with rising frequency)

Continuing with the case 3 (positive feedback):

B)

^{simulate this circuit}

\$ v_- =0 \$

\$ V_{OUT}=A \cdot (v_+ - v_-)=A \cdot v_+ \$

=> \$ v_+ = \frac{V_{OUT}}{A} \$

\$ (i_1=) \frac{V_{IN}-v_+}{R_2}= \frac{v_+-V_{OUT}}{R_F} (=i_2) \$

=> \$ \frac{V_{IN}}{R_2}=v_+ \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

Substituting now the above expression for \$ v_+ \$, we obtain:

\$ \frac{V_{IN}}{R_2}= \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$

and the overall gain is as follows:

$$ G= \frac{V_{OUT}}{V_{IN}}= \frac{1}{ \frac{1}{A}(1+ \frac{R_2}{R_F})- \frac{R_2}{R_F}} \> \> \> \> \> (2) $$

(Note that the denominator in this case can be 0!)

if A → \$ \infty \$ :

\$ G=- \frac{R_F}{R_2} \$

Now, the limit values of the overall gain G (when A is approaching \$ \infty \$ ) are the same in both the cases A) and B):

$$ G=-\frac{R_F}{R_2} $$

So it looks like it is the same at first sight...

BUT!

Comparing now the current case with the BFS:

\$ A'=A \frac{R_F}{R_F+R_2} \$

\$ \beta = \frac{R_2}{R_F} \$

(here A' again stands for /is analogical to/ the A in BFS)

\$ \beta A'=A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=A \frac{R_2}{R_F+R_2}>0 \$,

so, if \$ \frac{R_F}{R_2}=(A-1) \$ then G → \$ \infty \$ => unstable!

The exact expressions, (1) and (2), substantially differ one from another! I suppose their difference and its consequences are clearly evident from the analysis and the resulting formulas above. Due to usually very high value of A the stable case A) with negative feedback maintains, under the feedback influence, very low voltage between the Op Amp input terminal \$ v_+ \$, which is grounded, and the "live" input terminal \$ v_- \$. The latter is therefore at very low value (close to zero), that's why it is usually called virtual ground. (Maybe this "maintenance effect" is what you, sdarella, mean under the "stabilizer", am I right?) Unlike with the unstable case B), where the positive feedback leads to either oscillations or output saturation at \$ V_{OUT\_MAX} \$ or \$ V_{OUT\_MIN} \$, depending on the input conditions (see the case C) below).

C)

The case (3) with positive feedback can also be used but it works as a comparator, with input voltage comparative levels \$ V_{IN\_LH} \$ and \$ V_{IN\_HL} \$ (i.e. input voltages at which the output voltage flips rapidly from a low level (L= \$ V_{OUT\_MIN} \$) to a high level (H= \$ V_{OUT\_MAX} \$) and vice versa, resp.). However, it is usually better to use "real" comparators made/intended right for this purpose.

^{simulate this circuit}

we can write:

\$ \frac{V_{IN}-0}{R_2}= \frac{0-V_{OUT}}{R_F} \$

=> \$ V_{IN}=-\frac{R_2}{R_F}V_{OUT} \$ , (condition: \$ v_+ =0 \$ )

Provided the saturation values of \$ V_{OUT} \$ of the Op Amp are \$ V_{OUT\_MAX} \$ and \$ V_{OUT\_MIN} \$ , we obtain the following:

for \$ V_{OUT\_MIN} (<0) \$:

$$ V_{IN\_LH}=-\frac{R_2}{R_F} V_{OUT\_MIN} (>0) $$

and

for \$ V_{OUT\_MAX} (>0) \$:

$$ V_{IN\_HL}=-\frac{R_2}{R_F} V_{OUT\_MAX} (<0) $$

(it's hysteresis is then \$ V_{HYST}=V_{IN\_LH}-V_{IN\_HL}=\frac{R_2}{R_F}(V_{OUT\_MAX}-V_{OUT\_MIN}) \$)