To determine the output voltage of the inverting Schmitt Trigger, we need to calculate the voltage at the non-inverting terminal, which is given by the formula:
Vnon-inv = VIN * (R2 / (R1 + R2))
Given VIN = +3V, R1 = 10kΩ, and R2 = 10kΩ, we can substitute the values into the formula:
Vnon-inv = (+3V) * (10kΩ / (10kΩ + 10kΩ))
= +3V * (10kΩ / 20kΩ)
= +3V * 0.5
= +1.5V
Since the supply voltage is ±12V, the output voltage of the inverting Schmitt Trigger will switch to its negative saturation level of -12V when Vnon-inv exceeds -Vsat. In this case, it does not exceed -Vsat. Therefore, the output voltage will be:
Output voltage = Vnon-inv
= +1.5V
Therefore, the correct option is c) +5V.
Inverting Schmitt Trigger, the supply voltage is ± 12V. Suppose VIN = +3 V, the output voltage will be:
R1 = 10kΩ
R2 = 10kΩ
The options are:
a) +7v
b) -10v
c) +5v
d) +10v
3 answers
are you sure?
Apologies for the mistake in my previous response. Let's recalculate the output voltage.
With VIN = +3V, R1 = 10kΩ, and R2 = 10kΩ, we can use the same formula:
Vnon-inv = VIN * (R2 / (R1 + R2))
= (+3V) * (10kΩ / (10kΩ + 10kΩ))
= +3V * (10kΩ / 20kΩ)
= +3V * 0.5
= +1.5V
However, since the supply voltage is ±12V, the output voltage of the inverting Schmitt Trigger will switch to its negative saturation level of -12V when Vnon-inv exceeds -Vsat, and will switch to its positive saturation level of +12V when Vnon-inv exceeds +Vsat.
In this case, Vnon-inv (+1.5V) is less than the positive saturation level (+Vsat) of the output, so the output will be at its positive saturation level, which is +12V.
Therefore, the correct answer is d) +10V.
With VIN = +3V, R1 = 10kΩ, and R2 = 10kΩ, we can use the same formula:
Vnon-inv = VIN * (R2 / (R1 + R2))
= (+3V) * (10kΩ / (10kΩ + 10kΩ))
= +3V * (10kΩ / 20kΩ)
= +3V * 0.5
= +1.5V
However, since the supply voltage is ±12V, the output voltage of the inverting Schmitt Trigger will switch to its negative saturation level of -12V when Vnon-inv exceeds -Vsat, and will switch to its positive saturation level of +12V when Vnon-inv exceeds +Vsat.
In this case, Vnon-inv (+1.5V) is less than the positive saturation level (+Vsat) of the output, so the output will be at its positive saturation level, which is +12V.
Therefore, the correct answer is d) +10V.