In a Inverting Schmitt Trigger circuit, the supply voltage is ± 12V. The resistance is R1 = 10kΩ and R2 = 10kΩ. Suppose VIN = +7 V, the output voltage will be:
a) -10v
b) +10v
c) +7v
d) +5v
The Inverting Schmitt Trigger circuit is designed to switch the output voltage between two stable states based on the input voltage. The threshold voltages for switching states are determined by the voltage divider formed by R1 and R2.
In this circuit, the threshold voltages can be calculated using the equation:
Vthreshold = (R2 / (R1 + R2)) * Vsupply
Given that Vsupply is ±12V, R1 = 10kΩ, and R2 = 10kΩ, we can calculate the threshold voltages:
Vthreshold = (10kΩ / (10kΩ + 10kΩ)) * 12V
= (10kΩ / 20kΩ) * 12V
= 0.5 * 12V
= 6V
Therefore, the output voltage will switch when the input voltage crosses the threshold voltages of ±6V.
If VIN = +7V, the input voltage is higher than the positive threshold voltage of +6V. Therefore, the output voltage will be:
Output voltage = -Vsupply
= -(-12V)
= +12V
So the correct answer is:
b) +10V
are you sure?
Apologies for the mistake in my previous response. Let's re-evaluate the answer.
In an inverting Schmitt Trigger circuit, the output voltage switches between two stable states based on the input voltage and the thresholds set by the voltage divider formed by R1 and R2.
The threshold voltages for switching states can be calculated as follows:
Vthreshold = (R2 / (R1 + R2)) * Vsupply
Given that Vsupply is ±12V, R1 = 10kΩ, and R2 = 10kΩ, we can calculate the threshold voltages:
Vthreshold = (10kΩ / (10kΩ + 10kΩ)) * 12V
= (10kΩ / 20kΩ) * 12V
= 0.5 * 12V
= 6V
Therefore, the output voltage will switch when the input voltage crosses the threshold voltages of ±6V.
If VIN = +7V, the input voltage is higher than the positive threshold voltage of +6V. Thus, the output voltage will switch to the opposite stable state, which is the negative supply voltage.
Hence, the correct answer is:
a) -10V
what is the real answer? -10v or +10v?
Apologies for the confusion in my previous responses. Let's calculate the correct answer.
In an inverting Schmitt Trigger circuit, when the input voltage is above the positive threshold voltage, the output switches to the negative supply voltage. When the input voltage is below the negative threshold voltage, the output switches to the positive supply voltage.
Given that VIN = +7V and the positive threshold voltage is +6V, the input voltage is higher than the positive threshold voltage. Therefore, the output voltage will switch to the negative supply voltage.
Since the supply voltage is ±12V, the correct answer is:
a) -10V
To determine the output voltage of the Inverting Schmitt Trigger circuit, we need to calculate the voltage at the non-inverting input terminal (V+).
First, let's calculate the voltage divider formed by R1 and R2 to find V+. The formula for the voltage divider is:
V+ = VIN * (R2 / (R1 + R2))
Given that VIN = +7 V and R1 = R2 = 10kΩ, we can substitute the values into the formula:
V+ = 7 * (10000 / (10000 + 10000))
= 7 * (10000 / 20000)
= 7 * (0.5)
= 3.5 V
Since the Schmitt Trigger is inverted, the output voltage will be the opposite sign of the voltage at the non-inverting input terminal (V+). Therefore, the output voltage will be:
Output voltage = -3.5 V
However, since the supply voltage is ±12 V, the output voltage is limited to the available range. In this case, the output voltage will be clipped at the negative supply voltage (-12 V). Therefore, the correct answer is:
d) +5 V
To determine the output voltage in an Inverting Schmitt Trigger circuit, we need to understand its operating principle.
The Inverting Schmitt Trigger is a type of comparator circuit with hysteresis. It consists of an operational amplifier (op-amp) and two resistors: R1 and R2. The voltage at the non-inverting terminal of the op-amp is fixed to a reference voltage, while the input voltage (VIN) is applied to the inverting terminal.
In this case, the supply voltage is ± 12V, and the resistances are R1 = 10kΩ and R2 = 10kΩ. The reference voltage at the non-inverting terminal is typically set at half of the supply voltage, which is 0V in this case.
Given that VIN = +7V, it is higher than the reference voltage. In an Inverting Schmitt Trigger circuit, when the input voltage exceeds the reference voltage, the output voltage will be at its lower saturation level.
To calculate the lower saturation level, we can use the following formula:
Vout(lower saturation) = Vref × (1 + (R2 / R1))
Substituting the given values, we have:
Vout(lower saturation) = 0V × (1 + (10kΩ / 10kΩ))
= 0V × (1 + 1)
= 0V × 2
= 0V
Therefore, the output voltage will be 0V or -10V (negative saturation level) when VIN = +7V.
Therefore, the correct answer is:
a) -10V