I´m using a transmission gate. The polarity is correct and the bulk of the nmos is connected to the gnd and the bulk of pmos is connected to the vdd. Is this the correct way? Thanks.
A "closed switch" should be "conductive" - it's its purpose!... when the switch is closed it's still conductive.
I need it's set to 0V when closed. The input signal has the range 0 - 2.56V. How can I solve this?
That's a vague term. What's the pb.? Too much charge injection during switch-off? In such case, use the standard ½W/L compensation circuit.... the result wasn't good.
That's what it is intended for, isn't it? At least when switched-on.The VSS in the switch act like a gnd and ground the capacitors.
But I'm simulating the circuit in this configuration and isn' t working. But if I replace the ground in the switchs and in the positive input for the VCM 1.65V it integrates. There is something wrong with my Folded-cascode OTA? Or this is the correct?yes, for single-ended operation.
... can I connect the positive input to the ground and the circuit still works as a integrator? (The amplifier is a folded-cascode)
I'm using a NMOS input. So, if I use a PMOS input it should work with the grounded positive input? Or should I use the VCM = 1.65V?Depends on the input circuitry of your folded-cascode amplifier. Yes, if it has PMOS inputs.
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