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OP- AMPAS
DIFFERETIATOR
CIRCUIT DIAGRAM:
• The input signal source of
voltage vin is connected to the
inverting input terminal through
a capacitor c.
• Non inverting input terminal is
earth connected.
• Negative feedback is given
through a resistance R
• Let vin be the signal voltage given as the input, which drives varying
current through the capacitor C.
• Due to the virtual groung at the summing point S,
Current through C =Current through R
(dQ/dT)= (0-V0)/R
= (-V/R) .......(1)
Q is the charge on the capacitor at any instant t, i.e.,Q=C.v(t),
subst. this Q value in eqn (1), and we get,
.
C.dv(t)/dt = -V0/R
There fore,
V0 = CR.dv(t)/dt
The output is proportional to the time derivative of the input signal and
the circuit function as a differentiator.
CASE-1
• This gives in the amplification of high frequency components of the
amplifier noise. This amplified noise may obscure the output
differentiated signal. To minimise the problem of noise
magnification, a small capacitor may be connected parallel to the
feed back resistor to by-pass the high frequency.
• When input signal is square
wave, it gives the output signal
of spikes.
• If the input is Triangular wave ,
we get the Rectangular output
wave.
• If the input is sine wave, the
output will be cosine wave.
CASE-2
USES:
• It can used to solve differential equation in analog computers.
• It can used due to the noise magnification problem at high
freqencies.
• Differentiating amplifiers are most commonly designed to operate on
triangular and rectangular signals.
• It also find application as wave shaping circuits, to detect high
frequency components in the input signal.
Op amp as differentiator

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Op amp as differentiator

  • 2. CIRCUIT DIAGRAM: • The input signal source of voltage vin is connected to the inverting input terminal through a capacitor c. • Non inverting input terminal is earth connected. • Negative feedback is given through a resistance R
  • 3. • Let vin be the signal voltage given as the input, which drives varying current through the capacitor C. • Due to the virtual groung at the summing point S, Current through C =Current through R (dQ/dT)= (0-V0)/R = (-V/R) .......(1) Q is the charge on the capacitor at any instant t, i.e.,Q=C.v(t), subst. this Q value in eqn (1), and we get,
  • 4. . C.dv(t)/dt = -V0/R There fore, V0 = CR.dv(t)/dt The output is proportional to the time derivative of the input signal and the circuit function as a differentiator.
  • 5. CASE-1 • This gives in the amplification of high frequency components of the amplifier noise. This amplified noise may obscure the output differentiated signal. To minimise the problem of noise magnification, a small capacitor may be connected parallel to the feed back resistor to by-pass the high frequency.
  • 6. • When input signal is square wave, it gives the output signal of spikes. • If the input is Triangular wave , we get the Rectangular output wave. • If the input is sine wave, the output will be cosine wave. CASE-2
  • 7. USES: • It can used to solve differential equation in analog computers. • It can used due to the noise magnification problem at high freqencies. • Differentiating amplifiers are most commonly designed to operate on triangular and rectangular signals. • It also find application as wave shaping circuits, to detect high frequency components in the input signal.