Input | Output | Supply | ||||
---|---|---|---|---|---|---|
fMin | f0dB | fMax | VoMin | VoMax | Vcc | Vee |
100 Hz | 1 kHz | 100 kHz | –2.45V | 2.45V | 2.5V | –2.5V |
The integrator circuit outputs the integral of the input signal over a frequency range based on the circuit time constant and the bandwidth of the amplifier. The input signal is applied to the inverting input so the output is inverted relative to the polarity of the input signal. The ideal integrator circuit saturates to the supply rails depending on the polarity of the input offset voltage and requires the addition of a feedback resistor, R2, to provide a stable DC operating point. The feedback resistor limits the lower frequency range over which the integration function is performed. This circuit is most commonly used as part of a larger feedback/servo loop which provides the DC feedback path, thus removing the requirement for a feedback resistor.
The ideal circuit transfer function is given below.
AC Simulation Results
Transient Simulation Results
A 1kHz sine wave input yields a 1kHz cosine output.
A 1kHz triangle wave input yields a 1kHz sine wave output.
A 1kHz square wave input yields a 1kHz triangle wave output.
Texas Instruments, Simulation for Integrator Circuit, SBOC496 tool
Texas Instruments, Instrumentation Amplifier with DC Rejection, TIPD191 reference design
TLV9002 | |
---|---|
Vcc | 1.8V to 5.5V |
VinCM | Rail-to-rail |
Vout | Rail-to-rail |
Vos | 0.4mV |
Iq | 0.06mA |
Ib | 5pA |
UGBW | 1MHz |
SR | 2V/µs |
#Channels | 1, 2, and 4 |
TLV9002 |
OPA376 | |
---|---|
Vcc | 2.2V to 5.5V |
VinCM | (Vee-0.1V) to (Vcc-1.3V) |
Vout | Rail-to-rail |
Vos | 0.005mV |
Iq | 0.76mA |
Ib | 0.2pA |
UGBW | 5.5MHz |
SR | 2V/µs |
#Channels | 1, 2, and 4 |
OPA376 |
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