Differentiator

 ( Analog Circuits ) 

According to Ohm's law, the voltages at the two ends of the capacitive differentiator are related by the following transfer function (which has a zero in the origin and a pole at $s\; \{=\}\; \backslash tfrac\{\backslash text\{-\}1\}\{RC\}$):

$Y=\backslash frac\{Z\_R\}\{Z\_R+Z\_C\}X\; =\backslash frac\{R\}\{R+\backslash frac\{1\}\{sC\}\}X\; =\backslash frac\{sRC\}\{1+sRC\}X\; \backslash implies\; Y\backslash approx\; sRCX\; \backslash quad\; \backslash text\{for\}\; \backslash \; |s|\backslash ll\; \backslash frac\{1\}\{RC\}$

Similarly, the transfer function of the inductive differentiator has a zero in the origin and a pole in $s\; \{=\}\; \backslash tfrac\{\backslash text\{-\}R\}\{L\}$, corresponding to a cutoff frequency of $\backslash tfrac\{R\}\{2\backslash pi\; L\}$ in hertz or $\backslash tfrac\{R\}\{L\}$ in radians.

• $V\_\backslash text\{in\}$ is the voltage across C (from the op amp's virtual ground negative terminal).
• $V\_\backslash text\{out\}$ is the voltage across R (also from the op amp's virtual ground negative terminal).
• $I$ is the current flowing from the input through both R and C to the circuit's output.
• No current flows into the ideal op amp's inputs because they have High impedance input impedance.

According to the capacitor's current–voltage relation, this current $I$ as it flows from the input through the capacitor to the virtual ground will be proportional to the derivative of the input voltage:

$I\; =\; C\; \backslash ,\; \backslash frac\{dV\_\backslash text\{in\}\}\{dt\}\; \backslash ,\; .$

This same current $I$ is converted into a voltage when it travels from the virtual ground through the resistor to the output, according to ohm's law:

$0\; -\; V\_\backslash text\{out\}\; =\; IR\; \backslash ,\; .$

Inserting the capacitor's equation for $I$ provides the output voltage as a function of the input voltage:

$V\_\backslash text\{out\}\; =\; -RC\; \backslash frac\{dV\_\backslash text\{in\}\}\{dt\}.$

Consequently,

• The output voltage is proportional to the time derivative of the input voltage with a gain of $RC\; .$ Hence, the circuit acts as a differentiator and amplifier.
• The negative sign indicates the output has a 180° phase shift (inversion) with respect to the input.
• The equation is true for any frequency signal, assuming an ideal op amp (though a real op-amp has limited bandwidth).

The op amp's low-impedance output isolates the load of the succeeding stages, so this circuit has the same response independent of its load.

If a constant DC voltage is applied as input, the output voltage is zero. If the input voltage changes from zero to negative, the output voltage is positive. If the applied input voltage changes from zero to positive, the output voltage is negative. If a square-wave input is applied to a differentiator, then a spike waveform is obtained at the output.

• this simple differentiator circuit becomes unstable and starts to oscillate;
• the circuit becomes sensitive to high frequency noise that, when amplified, dominates the input signal.
• the limited gain–bandwidth product of real op amps will put an upper frequency limit for differentiation

Since negative feedback is present through the resistor R, we can apply the virtual ground concept, that is, the voltage at the inverting terminal is the same 0 volts at the non-inverting terminal.

Applying nodal analysis, we get

$\backslash frac\{0\; -\; V\_o\}\{R\}\; +\; \backslash frac\{0\; -\; V\_o\}\{\backslash frac\{1\}\{sC\_1\}\}\; +\; \backslash frac\{0\; -\; V\_i\}\{R\_1\; +\; \backslash frac\{1\}\{sC\}\}\; =\; 0,$

$-V\_o\; \backslash left(\backslash frac\{1\}\{R\}\; +\; sC\_1\backslash right)\; =\; \backslash frac\{V\_i\}\{R\_1\; +\; \backslash frac\{1\}\{sC\}\}.$

Therefore,

$\backslash frac\{V\_o\}\{V\_i\}\; =\; \backslash frac\{-sRC\}\{(1\; +\; sR\_1C)(1\; +\; sRC\_1)\}.$

Hence, there occurs one zero at $s\; \{=\}\; 0$ and one pole at $s\; \{=\}\; \backslash tfrac\{\backslash text\{-\}1\}\{R\_1C\}$ (corresponding to a corner frequency of $f\_1\; \{=\}\; \backslash tfrac\{1\}\{2\backslash pi\; R\_1C\}$) and another pole at $s\; \{=\}\; \backslash tfrac\{\backslash text\{-\}1\}\{RC\_1\}$ (corresponding to a corner frequency of $f\_2\; \{=\}\; \backslash tfrac\{1\}\{2\backslash pi\; RC\_1\}$).

For the above plot:

• Below $\backslash omega\_1$, the circuit attenuates, and well below $\backslash omega\_1$ acts like a differentiator.
• Between $\backslash omega\_1$ and $\backslash omega\_2$, the circuit acts as a Buffer amplifier or buffer.
• Above $\backslash omega\_2$, the circuit attenuates, and well above $\backslash omega\_2$ acts like an integrator.

Setting $RC\_1\; \{=\}\; R\_1C\; \{=\}\; RC$ will produce one zero at $s\; \{=\}\; 0$ and two poles at $s\; \{=\}\; \backslash tfrac\{\backslash text\{-\}1\}\{RC\}$ (corresponding to one corner frequency of $\backslash omega\_1\; \{=\}\; \backslash tfrac\{1\}\{RC\}$), resulting in the following frequency response (normalized using $RC\; \{=\}\; 1$):

For the above plot:

• Well below $\backslash omega\_1$, the circuit acts like a differentiator;
• Well above $\backslash omega\_1$, the circuit acts like an integrator.

A passive differentiator circuit is one of the basic electronic circuits, being widely used in circuit analysis based on the equivalent circuit method.

• Integrator
• Inverting differentiator at op amp applications