What is the amplitude of the output voltage of AD624AD circuit?
V1 = 10mVpk, 1kHz
VCC = 9V
VEE = -9V
The AD624AD is a instrumentation amplifier, which means that the output voltage will depend on the gain of the circuit. The gain of the AD624AD is set by an external resistor, and is given by:
Gain = 2 + (49.4 kΩ / Rg)
where Rg is the resistance of the external gain resistor.
Assuming a gain of 100 (corresponding to a gain resistor of 485 Ω), the output voltage of the AD624AD can be calculated as:
Vout = Gain x V1 = 100 x 10 mVpk = 1 Vpk
Note that the output voltage is given in peak (not RMS) values.
Therefore, the amplitude of the output voltage of the AD624AD circuit is 1 Vpk.
The options are
a) 200 mV
b) -10 mV
c) -2V
d) 2V
None of the options match the calculated result of 1 Vpk. Therefore, the answer cannot be determined from the given options.
To determine the amplitude of the output voltage, we need to consider the specifications of the AD624AD circuit. The AD624AD is a precision, low noise, rail-to-rail, instrumentation amplifier.
The gain of the AD624AD is determined by an external resistor connected between the RG pin and the REF pin. Without information about the specific resistor values used in the circuit, we cannot provide an exact value for the gain.
However, assuming a gain of 1 (no amplification), the output voltage will be equal to the input voltage. In this case, the amplitude of the output voltage can be calculated as:
Output voltage amplitude = Input voltage amplitude = V1
Given that V1 is 10mVpk (peak-to-peak voltage), the amplitude of the output voltage would also be 10mVpk.
Please note that the actual amplitude of the output voltage can vary depending on the gain setting of the circuit.
To determine the amplitude of the output voltage of the AD624AD circuit, we need to consider the gain of the circuit. The AD624AD is an instrumentation amplifier that is primarily used for amplifying small differential voltages. It is characterized by a high common-mode rejection ratio (CMRR) and adjustable gain.
The gain of the AD624AD circuit is given by the following formula:
Gain = 1 + (2 * Rg / (R1 + R2)),
Where Rg is the gain resistor and R1, R2 are the internal resistors of the AD624AD.
In this case, since the gain resistor (Rg) value is not provided, we'll assume that it is set to its default value of 1kΩ.
Given:
V1 (input voltage) = 10mVpk,
VCC (positive power supply) = 9V,
VEE (negative power supply) = -9V.
First, let's convert the peak amplitude of V1 to a peak-to-peak voltage. Since V1 is given in peak voltage, the peak-to-peak voltage (Vpp) can be calculated as:
Vpp = 2 * V1 = 2 * 10mV = 20mV.
Next, we'll determine the gain of the AD624AD circuit. Setting Rg to 1kΩ, we can calculate the gain as follows:
Gain = 1 + (2 * 1kΩ / (R1 + R2)).
The AD624AD has internal resistors of R1 = 1kΩ and R2 = 1kΩ, so substituting the values, we have:
Gain = 1 + (2 * 1kΩ / (1kΩ + 1kΩ)) = 1 + (2 * 1kΩ / 2kΩ) = 1 + (2 * 0.5) = 2.
Now, to calculate the amplitude of the output voltage, we multiply the input voltage Vpp by the gain:
Output Voltage (amplitude) = Vpp * Gain = 20mV * 2 = 40mV.
Therefore, the amplitude of the output voltage of the AD624AD circuit will be 40mV.