c) -1.5 v/ms
Using the formula for the output voltage of an integrator circuit:
Vout = - (1 / (R1 * C1)) * ∫(Vin)dt
where Vin is the input voltage and ∫(Vin)dt is the integral of Vin with respect to time.
Since the integrator circuit has a virtual ground at the input, we can assume that Vin = 0V.
Integrating from t = 1ms to t = 2ms:
∫(Vin)dt = ∫0dt = 0
Therefore, Vout = 0V between 1<t<2ms.
In a integrator circuit, determine the output voltage between 1<t<2ms.
R1 = 2kΩ
R2 = 220kΩ
C1 = 0.36uF
VEE = -15V
VCC = 15V
a) 1.5 v/ms
b) -8.3 v/ms
c) -1.5 v/ms
d) 8.3 v/ms
1 answer