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
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.
To determine the output voltage between 1<t<2ms in an integrator circuit, we need to use the formula for the output voltage of an integrator circuit:
Vout = - (1 / R1 * C1) * ∫V(t) dt
Where V(t) is the input voltage.
In this case, the input voltage is not provided, so we cannot calculate the exact output voltage. However, we can still determine the slope of the waveform.
Given:
R1 = 2kΩ
C1 = 0.36uF
To find the output voltage slope, we need to calculate the time constant (τ) of the circuit. The time constant is given by the formula:
τ = R1 * C1
Let's calculate the time constant:
τ = 2kΩ * 0.36uF
τ = 0.72 ms
Now, we need to determine the slope of the output voltage waveform. The slope of the output voltage is given by the formula:
Slope = ΔV / Δt
Since the time interval is between 1ms and 2ms, the Δt is 1ms.
Slope = ΔV / 1ms
Since we are given the options in terms of v/ms, we need to find the change in voltage (ΔV) over the Δt interval.
To calculate the ΔV, we can use the formula:
ΔV = Vout(2ms) - Vout(1ms)
Since we do not have the exact output voltage values, we cannot calculate the exact ΔV. However, we can determine the polarity of the slope.
Based on the values given in the options, we can determine the polarity of the slope by calculating the sign of ΔV using the signs of VEE and VCC.
Given:
VEE = -15V (negative)
VCC = 15V (positive)
Let's calculate the sign of ΔV:
If VEE < VCC, then ΔV will be negative (slope will be negative).
If VEE > VCC, then ΔV will be positive (slope will be positive).
In this case, since VEE is negative and VCC is positive, ΔV will be negative.
Therefore, the correct answer is:
c) -1.5 v/ms