What is the phase difference between input and output voltages?

Question

supply voltages = ±12V
amplitude =100mV
frequency = 500Hz
vin = 100mVpk, 500 hz
r1 = 1 kΩ
r2 = 27 kΩ
vcc = 12v
vee = 12v

Av = vout/vin = ?
Av (db) = 20 log(vout/vin) = ?
𝑃ℎ𝑎𝑠𝑒 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 (𝜑) = ?

Answer 1

To find the phase difference between input and output voltages, we first need to find the gain (Av) of the circuit:

Av = -R2/R1 = -27kΩ/1kΩ = -27

Next, we can use the gain to find the output voltage:

vout = Av * vin = -27 * 100mVpk = -2.7Vpk

Now, we can find the phase difference using the formula:

Phase Difference = arctan(2πf RC)

where f is the frequency (500Hz), R is the resistance (27kΩ), and C is the capacitance (unknown).

We don't know the capacitance of the circuit, so we can't find the exact phase difference. However, we can estimate the phase difference assuming a reasonable value for C. Let's assume a capacitance of 1nF:

Phase Difference = arctan(2π(500)(27kΩ)(1nF))

Phase Difference = arctan(2π(0.0135))

Phase Difference ≈ 84.6°

So, the estimated phase difference between input and output voltages is 84.6°.