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Is there a voltage divider at op-amp output?
The DAQ analog input range is from 0 to 2.4V and so I need to divide the output of the op-amp by 2. The op-amp output waveform will be similar to the one below, with the baseline being at 2.5V and the frequency in the range of 5Hz to 1kHz.
What’s the voltage across a 1 : 2 differential opamp?
I’m using above 1:2 differential opamp, when 72V applied to the 30:1 voltage divider, voltage across R1 is only 1.727V instead of 2.4V, opamp output is around 3.4V. Opamp is LMC6062IN/NOPB http://www.ti.com/lit/ds/symlink/lmc6062.pdf I think it’s because of bias current, any suggestion on solution?
What should the waveform of an op-amp be?
The op-amp output waveform will be similar to the one below, with the baseline being at 2.5V and the frequency in the range of 5Hz to 1kHz. Will a simple voltage divider at the output of the op-amp (see diagram below) divide the signal with these parameters?
How to calculate the voltage of a voltage divider?
V O U T = V I N 10 k 200 k + 100 k + 10 k . But you have a second voltage divider in parallel with R1. “R1” is now effectively 7.5 kΩ. The voltage at R1 will be V O U T = V I N 7.5 k 200 k + 100 k + 7.5 k = 72 × 2.44 % = 1.75 V and the voltage into the non-inverting input will be 2/3 of that = 1.17 V.
Why does an op amp have high gain?
If the operational amplifier were operating as an open-loop amplifier (that is, without negative feedback), a small increase in the input voltage would cause a large increase in the output voltage, because the op-amp has very high gain.
Is the op-amp the same as the voltage follower?
The Op-Amp Voltage Follower The most basic form of the voltage follower, also called a unity-gain buffer, is shown in the diagram below. As you can see, the only necessary component is the op-amp itself (however, you do need a decoupling capacitor for the IC’s power supply).