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Why not ? A capacitor is charged by a current source, that´s all.
However, I understand the background of the question since most - if not all - books and articles on SC circuits rely on opamp circuits.
But it is easy to explain: Let`s take a simple integrator stage and let´s assume that the capacitor C1 between the OTA inverting terminal and ground was charged in phase 1 with a negative sign. During the next phase the corresponding voltage produces a positive current at the OTA output thereby charging the feedback capacitor C2 and discharging C1. This process comes to an end when C1 is completely empty and its voltage is zero. Thus, there was a charge transfer between c1 and c2 (with sign inversion).
This works with any value of transconductance gm. Only the limited output resistance can cause some errors.
I hope I could express myself clear.
LvW
I was confused whether the charge speed is decided by rout of the OTA(which is big) multiply by C or gm/C?
I think the speed is only related to rout*C in linear settling situation,will an OTA has a big output resistence also have a very slow speed when driving a cap?...and what is gm/C?
I was confused whether the charge speed is decided by rout of the OTA(which is big) multiply by C or gm/C?
I think the speed is only related to rout*C in linear settling situation,will an OTA has a big output resistence also have a very slow speed when driving a cap?...and what is gm/C?
The nature of the OTA is a current output. The ratio of the output conductance and the transconductance set the voltage gain. But that is true only without resistive load! With resistive load the voltage gain is typical to low. A second stage is needed here.
Pure capacitive load allows to put all of the bias current into the OTA cell and shorting the settling time for the same bias budget.
Or in other terms: a 2-stage opamp design have also charge the stabilization caps!
As far as I understood the original contribution from OMEsystem his question was purely if resp. why an OTA can be used in S/C circuits instead of an opamp.
In my first reply I have used a simple integrator stage as an example.
In this circuit the feedback C is charged by the output current. At the same time the input C is discharged by the same current - thereby reducing the input voltage.
For a constant input voltage the voltage across the feedback C would rise linearly, however, since the input voltage drops, there will be an increase of the output voltage following an exponential function.
The speed (that means the time constant of the exp. function) is determined neither by the "bandwidth" nor the output resistance of the OTA. The time constant related to the exp. function is C/gm (gm beeing the transconductance of the OTA).
This time constant must be small enough if compared with the time which is available for this charge transfer (sampling period).
An OTA output resistance which is not indefinit will not influence the time constant, but it will cause a remaining error, because the input C will not be completely discharged.
Hope this clarifies the situation.
LvW
I was confused whether the charge speed is decided by rout of the OTA(which is big) multiply by C or gm/C?
I think the speed is only related to rout*C in linear settling situation,will an OTA has a big output resistence also have a very slow speed when driving a cap?...and what is gm/C?
For a constant input voltage the voltage across the feedback C would rise linearly, however, since the input voltage drops, there will be an increase of the output voltage following an exponential function.
Sorry, I did not express myself clear enough. "Following an exponential function" means, of course, following the function (1-exp) . That means - similarly to charging a cap through a resistor - the voltage across the feedback capacitor (identical to the output voltage of the OTA) reaches a final value which depends only on the input voltage as well as the cap ratio.
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