hyleeinhit
Member level 3
I have a question on integrator with feedback capacitor. What happens if the input current is larger than the integrator output drive current capacity? Fig 1 shows the integrator output. "net7" is the negative input node of the amplifier and "ampout" is the output node of the integrator, “rst” is the reset clock, and "vcm" is virtual ground, equal to 2.5V. Power supply is 5V. The max drive current capacity of the amplifier is a little bit less than 500uA.
I ran the simulation with Cadence when the input current Ix is 400nA, 500nA and 600nA, and got the results in Fig2. Firstly, net7 and ampout are reset to 2.5V. In the case of Ix=600nA, net7 and ampout drops to 0V rapidly. During this voltage dropping time, what happens to the feedback capacitor Cf? Is the following statement true: the discharge current on net7 is 600nA and the charge current on net "ampout" is 500nA?
I ran the simulation with Cadence when the input current Ix is 400nA, 500nA and 600nA, and got the results in Fig2. Firstly, net7 and ampout are reset to 2.5V. In the case of Ix=600nA, net7 and ampout drops to 0V rapidly. During this voltage dropping time, what happens to the feedback capacitor Cf? Is the following statement true: the discharge current on net7 is 600nA and the charge current on net "ampout" is 500nA?
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