Filter circuit configuration

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If 330E is 330 ohms then an opamp cannot drive it because this very low value resistor will draw too much current from the output of the opamp.
The TL072 will not work if the DC input voltage to R5 is less than about +4V because the opamp does not have a negative supply at pin 4.
 


So I guess voltage followers will never work because their FB resistance is zero. And integrators won't work either.
 

barry said:
So I guess voltage followers will never work because their FB resistance is zero. And integrators won't work either.
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Can't you see that the negative feedback resistor value at high frequencies is only 330 ohms? Then the output at 12V peak is 36.4mA which overloads the output of most opamps. Any load on this filter increases the output current.

The TL072 and most other opamps are spec'd with a minimum load of 2000 ohms.
 


Can't you understand the workings of an opamp?

When the output is at 12 V, the input would also have to be at 12 Volts, since the gain of this sucker is approximately 1. Thus, the current through that resistor is (12-12)/330=0.

That resistor is a FEEDBACK resistor, not a LOAD.
 

Since there is no DC feedback path, the output will drift to either negative or positive saturation and stay there regardless of what the input voltages are. The rate at which it drifts will depend on the input offset voltage, input offset current and input bias current of the Op Amp.
 

It's a single pole loop filter to be used in PLL. In order to work properly R6 and R7 should have the same value. The two inputs have to be connected to "Up" and "Down" output of the Phase-Frequency detector. So they are current pulses having a duration proportional to the frequency error.
 

It would characterize the circuit as PI filter, an integrator with a zero. And the input is differential, with a slight asymmetry because R6 ≠ R7. The purpose of the asymmetry isn't obvious.
 

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