Please help. Having major issues with this non inverting op amp question.
A certain op-amp is used in a non-inverting amplifier having R1 = 1kohm and R2 = 3kohm (for a gain of 1+R2/R1). The source voltage is a 1 kHz sine wave with amplitude Vs. According to the manufacturer’s data sheet, the op amp has output voltage limits of +-12V and maximum output current limits of +-25mA.
a.) Determine the amplitude of the input voltage at the onset of clipping if a load resistance connected to the output is 10k ohm.
b.) Repeat if the load resistance is 100 ohm.
I am a little confused on how to setup and answer this problem with the load connected to it.
When writing the solution, could you please explain how you got to that answer. That would be a big help for my understanding of this type of amplifier. Thanks again for any help on this problem.
A certain op-amp is used in a non-inverting amplifier having R1 = 1kohm and R2 = 3kohm (for a gain of 1+R2/R1). The source voltage is a 1 kHz sine wave with amplitude Vs. According to the manufacturer’s data sheet, the op amp has output voltage limits of +-12V and maximum output current limits of +-25mA.
a.) Determine the amplitude of the input voltage at the onset of clipping if a load resistance connected to the output is 10k ohm.
b.) Repeat if the load resistance is 100 ohm.
I am a little confused on how to setup and answer this problem with the load connected to it.
When writing the solution, could you please explain how you got to that answer. That would be a big help for my understanding of this type of amplifier. Thanks again for any help on this problem.
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a.) The op amp can supply 25 mA of current so there will be no current limiting with a 1k load. Therefore, the amp will limit the output when the input exceeds/12/4=3 volts. (The gain 4.)
b.) The current limit is +/-25 mA, so this is +/-2.5 volts into 100 ohms. The amp will limit the output when the input exceed 2.5/4=.625 volts. Note that in this case I have neglected the load of the feedback network, but it is much less than the 100 ohm load.
b.) The current limit is +/-25 mA, so this is +/-2.5 volts into 100 ohms. The amp will limit the output when the input exceed 2.5/4=.625 volts. Note that in this case I have neglected the load of the feedback network, but it is much less than the 100 ohm load.