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I wanted to try and build a derivative circuit using an inductor and a resistor for frequencies between 20 Hz to 20 kHz. My idea was I want a transform function of this form:

$$H(\omega)=i*\omega *const$$ $$H(\omega)=\frac{\tilde{v}_{out}}{\tilde{v}_{in}}$$

I made a circuit consisting of a voltage divider and chose the voltage at the inductor as the voltage out.

$$H(\omega)=\frac{\tilde{v}_{in}\frac{Z_L}{Z_L+Z_R}}{\tilde{v}_{in}}$$

$$H(\omega)=\frac{Z_L}{Z_L+Z_R}$$

$$H(\omega)=\frac{i\omega L}{i\omega L+R}$$

Now my idea was that if I have

$$\frac{R}{L}>>\omega$$

for the frequencies in the range then I assumed that then I would get the derivative of the input, I tried to check correctness by writing a Matlab code but everything I get has an amplitude close to zero so it seems like a straight line.

So I got stuck unsure how to check if it is built correctly especially when it is not around

$$\omega \to 0 $$ or $$\omega \to \infty $$

The numbers in the image are not the one I chose, just a schematic for the circuit I had in mind.

enter image description here

Edit:

What I tried, after reading the comments, to get constraints:

Let

$$x = \frac{j \omega L}{R}$$

Then:

$$\text{if }x \gg 1,\text{then }H(\omega) \approx 1$$

$$ \text{if }x \ll 1,\text{then } H(\omega) \approx \frac{j \omega L}{R} = \frac{j \omega}{\tau}, \tau = \frac{L}{R}. $$

Where we got Tau from:

$$ V_R + V_L = V_\text{in} \quad \Rightarrow \quad R I + L \frac{dI}{dt} = 0 $$

$$ I(t) = A e^{-t/\tau}, \quad \tau = \frac{L}{R}. $$

So the circuit behaves as a differentiator.

For x less less than 1:

$$ V_\text{out}(t) \approx \frac{L}{R} \frac{d V_\text{in}(t)}{dt}. $$

Trying to get constraints on R and L to follow the approximation:

$$ L \le \frac{0.1 R}{\omega_\text{max}} = \frac{0.1 R}{2 \pi f_\text{max}} $$

Example:

$$ R = 1\,\text{k}\Omega, \quad f_\text{max} = 20\,\text{kHz} \quad \Rightarrow \quad L \le 0.8\,\text{mH}. $$

Checking it also applies for low frequency:

$$ \omega_\text{min} L = 2 \pi \cdot 20 \cdot 0.8\,\text{mH} \approx 0.1\,\Omega \ll R $$

$$f(t)=sin(2π⋅5kHzt)+0.5sin(2π⋅15kHzt)$$

enter image description here

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  • \$\begingroup\$ You could simulate your circuit (a modern trend that ain’t going to disappear and is without doubt something invaluable to learn). \$\endgroup\$ Commented Nov 19 at 23:57
  • \$\begingroup\$ @Andyaka is not writing a code simulating the circuit , I thought what I tried is numeric simulation ? \$\endgroup\$ Commented Nov 20 at 0:00
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    \$\begingroup\$ For series RL circuit, time constant is defined as L/R. You have L= 1uH, R=10ohms, so L/R = 0.1usec. You need Freq to reach 10MHz before seeing output voltage change significantly. \$\endgroup\$ Commented Nov 20 at 4:52
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    \$\begingroup\$ Suggest xhange L to 1.59mH. It's impedance magnitude will be 10 ohms (same as the series R) at 1kHz. When you sweep F from 20Hz to 20kHz you will see Vout change with F ( won't be a straight line close to zero). \$\endgroup\$ Commented Nov 20 at 8:38
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    \$\begingroup\$ @FabioBarone i think i am starting to understand how to think about it thanks for your advice ,kindness , and patience . \$\endgroup\$ Commented Nov 20 at 9:32

3 Answers 3

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The issue here is just that you picked the wrong component parameters! The inductor you used before was too small, and the resistor too large—so the circuit’s output amplitude was super tiny at your target frequencies, which is why the simulation looked like a straight line! Just tweak the parameters—for example, swap the inductor for 800μH and the resistor for 1kΩ. This way, you’ll not only meet the requirements for the differentiator but also get a visible output amplitude. Run MATLAB again, and you’ll clearly see the differentiator effect as the frequency rises!

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  • \$\begingroup\$ it certainly works thank you , would appreciate an explanation how you found a range that work if you have time, I can’t vote up yet but will do when i can \$\endgroup\$ Commented Nov 20 at 7:13
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Inductor is not the component for differentiating a voltage, because it differentiates the current. No R/L vs. frequency selection makes your approximation good over such wide band as audible frequencies.

Use opamp based RC-differentiator. It can do the job from 0 Hz to several tens of kilohertzes without any expensive special parts. Your own approximation cannot compete.

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  • \$\begingroup\$ I was thinking that if i get (i *omega) then the transform function in that region is indicating that in the time domain the output is proportional to the derivative, i see that if i define x as $$i\omega \frac{L}{R} $$ and then if x<<1 it is proportional to a a derivative and if x>>1 then i get 1 so I am unclear why the 𝑅/L<<𝜔 is the condition I would aim for, but the real parts example was certainly helpful appreciated \$\endgroup\$ Commented Nov 20 at 1:27
  • \$\begingroup\$ @lodo Use an opamp, input serial R-C (1 kOhm-1 nF), 10 kOhm Ohm between input- and output ... input + wired to ground. It will be ok from 1 Hz to 100 kHz. \$\endgroup\$ Commented Nov 20 at 9:57
  • \$\begingroup\$ @Antonio51 thanks will model it as well , I was trying to see how to build a derivative using RL and integral us RC just to gain a different perspective and certainly comparing is the best next step \$\endgroup\$ Commented Nov 20 at 10:03
  • \$\begingroup\$ @lodo The whole answer was rubbish. It's fixed now. Sorry for wasting your time. \$\endgroup\$ Commented Nov 20 at 11:29
  • \$\begingroup\$ @oneprivate appreciate the update , as for using RC if i were to build it in reality certainly , it was more of a thought experiment to build derivative use RL and integral using RC to gain understanding of the topic \$\endgroup\$ Commented Nov 20 at 12:35
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The shown R-L circuit is nothing else than a simple 1st-order highpass with a transfer function

H(s)=s(L/R)/[1+s(L/R)

This circuit provides a leading 90deg-phase shift (identical to a differentiating process in the time domain) for frequencies far enough below the upper corner frequency wg=R/L.

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