Determine ACL of Non-inverting amplifier.
RA = 500Ω
RB = 470Ω
RC = 20KΩ
VIN = 0.5Vp-p
The ACL (AC voltage gain) of a non-inverting amplifier is given by the following formula:
ACL = 1 + (RB/RA)
Substituting the given values, we get:
ACL = 1 + (470/500) = 1.94
Therefore, the ACL of the non-inverting amplifier is 1.94.
Note: The ACL is a dimensionless quantity and represents the ratio of output AC voltage to input AC voltage. It is also sometimes expressed in decibels (dB), which is given by 20log(ACL). In this case, the ACL in dB is 5.87dB.
I got 41 as ACL.
The calculation for the ACL using the given values for the non-inverting amplifier is as follows:
ACL = 1 + (RB/RA) = 1 + (470Ω/500Ω) = 1 + 0.94 = 1.94
Therefore, the ACL of the non-inverting amplifier is 1.94 or approximately 1.9. It is possible that you made a calculation error or used different values for RA and RB. Please double-check your calculations and ensure that you are using the correct values.
This is my answer:
acl = 1 + rf/ri
acl = 1 + 20000/500
acl = 41
Is this correct?
If you use the given values of RA = 500Ω, RB = 470Ω, and RC = 20KΩ, the calculation for the ACL is as follows:
ACL = 1 + (RB/RA)
ACL = 1 + (470Ω/500Ω)
ACL = 1 + 0.94
ACL = 1.94
Therefore, the ACL of the non-inverting amplifier is 1.94 or approximately 1.9.
Your calculation seems to have used RC instead of RB in the formula, which would result in a different ACL value. Please double-check your calculation and ensure that you are using the correct values.
Determine ACL in dB
To determine the ACL in dB, we use the following formula:
ACL (dB) = 20 log10(ACL)
Substituting the calculated value of ACL = 1.94, we get:
ACL (dB) = 20 log10(1.94)
ACL (dB) = 20 × 0.287
ACL (dB) = 5.74 dB (approx)
Therefore, the ACL of the non-inverting amplifier in decibels is approximately 5.74 dB.
If the output saturation voltage of the op-amp is ± 8V, determine Voutpp.
To determine Voutpp (the peak-to-peak output voltage), we need to consider the maximum and minimum possible output voltages of the amplifier.
Given that the output saturation voltage of the op-amp is ±8V, we know that the maximum and minimum output voltages of the amplifier are ±8V, respectively.
The peak-to-peak output voltage is then given by:
Voutpp = Vmax - Vmin
Substituting the values, we get:
Voutpp = (8V) - (-8V) = 16V
Therefore, the peak-to-peak output voltage of the amplifier would be 16V.