For this circuit, the determination is simple: the feedback from the op amp output back to its (+) input is attenuated by half (R1/(R1+R2)), while the feedback to the (-) input is reduced to 1/4 (R3/(R3/R4)) of the voltage at the output. Therefore, the positive feedback dominates. Be careful about assuming the op amp is working in its linear region: it appears to be saturated against the negative supply rail.Can someone explain to me, how to determine which feedback is active in circuit? As I can see from simulation, in this case negative feedback is active because op amp works in linear region. Is there any way to determine that with some hand-calculation?
The mere fact that the circuit contains both positive and negative feedback paths DOES NOT make it a Howland current source.I think that circuit is called a howland current source, and i'd say both positive and negative feedback are working in it.
You are right master Obi Wan, I judged too soonThe mere fact that the circuit contains both positive and negative feedback paths DOES NOT make it a Howland current source.
Ok, does it mean that when I analyze which feedback dominates (in general), first I remove input signals and then watch what is happening with (+) and (-) inputs of op amp?For this circuit, the determination is simple: the feedback from the op amp output back to its (+) input is attenuated by half (R1/(R1+R2)), while the feedback to the (-) input is reduced to 1/4 (R3/(R3/R4)) of the voltage at the output. Therefore, the positive feedback dominates. Be careful about assuming the op amp is working in its linear region: it appears to be saturated against the negative supply rail.
Disagree. The voltages all work out for battery and resistor values shown. Plus, a saturated output would be closer to the -15 V rail; 4 V headroom seems large.Be careful about assuming the op amp is working in its linear region: it appears to be saturated against the negative supply rail.
Agree. The Howland pump requires equal impedance ratios.The mere fact that the circuit contains both positive and negative feedback paths DOES NOT make it a Howland current source.
Thinking about it, I realize you're right: the op amp very likely is operating in its linear region (unless by some magic its negative saturation level were precisely -11V), because the two inputs are at equal voltages.Disagree. The voltages all work out for battery and resistor values shown. Plus, a saturated output would be closer to the -15 V rail; 4 V headroom seems large.
My simulator is giving me results like this. So V+ and V- are equal, and op amp is operating in it's linear region according with it.There is no DC reference potential for either opamp input, so the output will change due to input offset voltage error until it saturates.